The Berkeley Review MCAT Organic Chemistry Part 2

OrganicChemistry

Part IISections V-VIII

Section VCarbonyls and Alcohols

Section VICarbohydrates

Section VIINitrogen Compounds

Section VIIIOrganic Chemistry

Laboratory Techniques

BERKELEY

Specializing in MCAT Preparation

ERKELEYE • V • I • E • W

®

P.O. Box 40140, Berkeley, California 94704-0140Phone: (800) 622-8827Internet: [email protected]

(80 0) M CAT-TBRhttp://www.berkeleyreview.com

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Section V

Carbonylsand

Alcoholsby Todd Bennett

0

RX XCH3O

R CH2

O

/C\

o

O

Off/C\0

R CH2

O

O

2more times ||Off. /C\

C +1

R CH2I

R' 'CH2I ^5~" R CI3

O O

Off. /c\ + cl3R OH

O

C + CI3 ^f=^ C. 0 + HCI3R OH R O yeUowoil

Iodoform Test

Oxygen Containing Compoundsa) Alcohol Propertiesb) Alcohol Reactivityc) Alcohol Spectroscopyd) Aldehyde and Ketone Propertiese) Aldehyde and Ketone Reactivityf) Aldehyde and Ketone Spectroscopyg) Ketals and Acetals

i. Protecting Groupsh) Carboxylic Acids and Derivatives

i. Carboxylic Acidsii. Esters

iii. Lactones

iv. Acid Anhydridesv. Acid Halides

vi. Amides

Carbonyl Reactivity

a) Attack at Carbonyl Carbonb) Deprotonation of a-Protonsc) Oxidation-Reduction Reactions

Name Reactions

a) Qrignard Reactionb) Aldol Condensationc) Claisen Condensationd) Transesterificatione) Wittig Reactionf) Pinacol Rearrangementg Iodoform Reactionh) Wolff-Kishner Reaction

Synthetic Logica) Reactions of Acetoacetic Esterb) Reactions of Malonic Esterc) Decarboxylationd) Protecting Groups

Carbonyl Biochemistrya) Biological Oxidation-Reductionb) Biochemical Reagents

BERKELEYUr-E-V.KE'W®


°!>

© *

*

Carbonyls & AlcoholsSection Goals

Recognize the carbonyl functional groups and types of compounds.You must be able to recognize functional groups suchas amides, anhydrides, acidhalides, andespecially ketones and aldehydes. You must know which compounds are most reactive towardssubstitution (which isbasedon the leaving group strength), and mostelectrophilic (which isbasedon the electron withdrawing or donating capacity of the functional group).

Be able to identify infrared peaks for carbonyl compounds.Carbonyl compounds will have apeak in the infrared spectrum in the area of 1700± cm'1. This willmostlikely beuseful whencomparing twocarbonyl compounds, or identifying an unknown carbonylcompound. You may wish toknow roughly where esters, aldehydes, and ketones fall inthe IR range.

Be able to identify common name reactions involving carbonyl compounds.You must recognize common name reactions from carbonyl chemistry. Included inthis group shouldbe theAldolcondensation, Grienardreaction, Wittie reaction, and the Claisen reaction. TheAldoland Claisen reactions have biological significance, because they play a role in select biochemicalpathways such as glycolysis and beta oxidation.

Be able to recognize the acidity of alpha carbons.The carbon alpha to thecarbonyl can bedeprotonated, ifit hasa proton bonded to it. The pKa ofastandard ketone is the range 1/± 2. Once deprotonated, an enolate is formed. The enolate has anequilibrium of its own with the enol structure. The conversion of a ketone into an enol is referredto as tautomerization.

Be able to identify ketals, hemiketals, acetals and hemiacetals.Although current accepted nomenclature does notdistinguish between ketals andacetals, youshouldbe aware of the functional group. Acetals and ketalsare best described as "doubleethers." Theyplay a major role in sugar chemistry and protecting groups in carbonyl synthesis.

Understand the difference between thermodynamic and kinetic enolates.Thethermodynamic enolate is formed underconditions ofhighertemperature wherethe pathwayofgreateractivation energymaybe chosen. Thethermodynamic enolateis the moresubstitutedandthus more stable intermediate which will lead tothe more stable final product. The kinetic enolateis formed underconditions oflower temperature and greater steric hindrance where thepathwayof lower activation energy must be chosen. The kinetic enolate is the less substituted and thus lessstable intermediate which will lead to the lessstablefinalproduct.

Know the mechanisms for acidic and basic carbonyl reactions.The mechanism for transesterification ispresent inbiochemistry andorganic chemistry, soit importantthat you recognize the steps. Also recognize what catalyst is necessary to carryout the process.

©Recognize common oxidizing and reducing agents.

n a nutshell, oxidation isdefined as thegainofbondstooxygen and/or the lossofbondstohydrogen.Oxidizing agentsincludeKMn04and K2Cr207. Reduction is definedas the lossofbonds to oxygenand/or the gain of bonds to hydrogen. Reducing agents include LiAlH4 and NaBI-14.

Know common reactions by both name and reagents.The AAMC guide lists a series ofreactions that they expect you toknow byname. Iris a eood ideatonot only know the general reaction, butalso the mechanism and reaction conditions. Highlightsof this list include the Aldol reaction, the Grignard reaction, the Witting reaction, the iodoformreaction, transesterification, and the Wolff-Kishner reduction.

Organic Chemistry Carbonyls and Alcohols

Carbonyls and AlcoholsThe carbon-oxygen bond isa major part oforganic and biological chemistry. Asignificant part of organic chemistry on the MCAT involves compounds thatcontain carbon-oxygen bonds. In the case of carbonyl compounds, the carbon-oxygen 7C-bond is easily broken to form new bonds to the carbonyl carbon andsubsequently form a new compound. The carbon-oxygen a-bond found inalcohols and sugars can undergo several reactions, but it is generally not asreactive as the carbon-oxygen rc-bond. Our goalis to organize the vast multitudeof reactions involving carbonyl compounds and alcohols. Figure 5-1 showsseveral types of carbonyl compounds and carbonyl derivatives with which youshould be familiar.

Types of Carbonyl Compounds0 O

R R R HKetone Aldehyde

R'Ov OR*

\/R'Ov OH\/

R'O OR' R'O OH

\/ \/

R H R H R/CVR R/CVRAcetal Hemiacetal Ketal Hemiketal

O

IIO

IIO O

II II

R OH[ R OR' R O RCarboxylic Acid Ester Acid anhydride

it If If if.'N. R NH2 R N R'

Acid halide AmideH

Imide

Lactone

R

/NHPh

R R

Phenylhydrzine derivative

? IfR/c\

Enolate Resonance Forms

Figure 5-1

Lactam

CHn

Introduction

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OrgSlIllC ChCllllStry Carbonyls and Alcohols Oxygen Containing Compounds

Oxygen containing compounds, because of the highly electronegative nature ofoxygen, are very reactive. Much of organic chemistry revolves around alcoholsand carbonyls, so it is imperative to get a fundamental understanding of theirproperties, reactivity,and spectroscopicevidence that supports their existence.

Alcohol PropertiesBecause of their ability to form hydrogen bonds, alcohols typically have highboiling points and are generally miscible in water. Alcohols make good solventsas they are often liquids at room temperature and they have a large rangebetween their melting and boiling points. Alcohols are hydrophilic, polarmolecules that become less hydrophilic (more lipophilic) as their carbon chainlength increases. The smaller alcohols (three carbons or less) are highly watersoluble, but as the size of the alkyl group increases, their water solubilitydecreases. As with all compounds, their physical properties vary with mass andbranching, as well as the position of the hydroxyl group. As the molecular massincreases, the boiling point increases, but the effect on the melting point is lessclear. As the branching increases, the boiling point decreases. Table 5-1 showsthe physical properties of several alcohols, from which the effects of mass,branching, and positioning of thehydroxyl group on the physical properties canbe ascertained.

IsomerIUPAC Name

(Common Name)BoilingPoint

MeltingPoint

Density(g/mL)

Water

Solubility(g/lOOmL)

CH3OH Methanol 64.6°C -98°C 0.791 HighH3CCH2OH Ethanol 78.4°C -115°C 0.789 HighH3CCH2CH2OH 1-Propanol (n-Propanol) 97.2°C -127°C 0.804 High(H3Q2CHOH 2-Propanol (i-Propanol) 82.3°C -90°C 0.786 HighH3C(CH2)3OH 1-Butanol (w-Butanol) 117.3C -90°C 0.810 8.2

H3CCH(OH)CH2CH3 2-Butanol (sec-Butanol) 99.6°C -115°C 0.806 12.8

(H3C)2CHCH2OH 2-Methyl-l-propanol (f-Butanol) 107.7°C -122°C 0.802 11.3

(H3Q3COH 2-Methyl-2-propanol (f-Butanol) 82.0°C 24°C 0.789 HighH3C(CH2)4OH 1-Pentanol (n-Pentanol) 137.6°C -79°C 0.814 2.1

H3CCH(OH)CH2CH2CH3 2-Pentanol 119.3°C 0.809 5.0

(H3CCH2)2CHOH 3-Pentanol 115.9°C 0.815 5.6

H3C(CH2)4CH3 1-Hexanol (n-Hexanol) 157.5°C 0.814 0.8

C6HnOH Cyclohexanol 161.5°C 0.956 2.1

H3C(CH2)6CH2OH n-Octanol 194.7°C 0.817 1 0.3

T able 5-1

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OrgaiUC ChemiStry Carbonyls and Alcohols Oxygen Containing Compounds

Example 5.1Whattype of alcohol is the following molecule?

HiC CH„

HO-

A. Primary alcoholB. Secondary alcoholC. Tertiary alcoholD. Phenol

Solution

The compound has the alcohol functional group attached to a carbon that isbonded to two other carbons. This is definedas a secondary alcohol, so the bestanswer is choice B.

Alcohol ReactivityAlcohols are nucleophilic reagents in organic chemistry. They are not goodnucleophiles in their protonated (neutral)state, but they can be deprotonated andconverted into their anion (alkoxide) form under basic conditions. Becausealkoxides (the deprotonated form of the alcohol) are strong bases, they are notthe ideal nucleophile, but they are generally better than alcohols. Alcoholchemistry also involves oxidation into a carbonyl as we shall see later in thissection. Alcohols are commonly formed from the reduction of carbonyls, whichwe shall also postpone for the moment. The common reactions to form andconsume alcohols that do not involve carbonyl compounds center aroundnucleophilic substitution. Figure 5-2 shows nucleophilic substitution reactionsthat convert alkyl halides into alcohols. Figure 5-3 shows nucleophilicsubstitution reactions that convert alcohols into alkyl halides.

Alkyl halides to alcohols

R R

h*%/ x ~2~~^ H*yH H

R' R'

OH + X

\ 1. RCO," / „ . „^ TTR»';?-Br 2.0H-(ac,)» H°~V"R + + 2

H H

R" R"

„„t.^—X 1 • t,ii»»^— OH +HXRv / acetone Rx /R' R'

Figure 5-2

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OrCJ£UllG ChCHllStry Carbonyls and Alcohols Oxygen Containing Compounds

Alcohols to alkyl halides

O

R « R

H

,^-oh S/s^Cl ^—ClH^

H

R'

OHhr^;H

R"

PBr,

R'

Br-iH

R"

Retention

Inversion

OH ^L^.hR^K

hr^;R'

Br Racemization

Figure 5-3

Spectroscopic Evidence for AlcoholsAlcohols can be detected using either infrared or NMR spectroscopy. In IRspectra, hydroxyl groups present a distinct absorbance between 3200 and 3500cm"1 that is medium in intensity and broad due to hydrogen bonding. InXHNMR spectra, hydroxyl groups present asignal between 1and 5ppm that isbroaddue tohydrogen bonding, although the broadness varies with thesolvent.They have no definite 6-value (it varies with concentration and solvent). Thepeak slowly disappears with the addition of D2O to the NMR tube. The OHgroup does notcouple well, sowerarely consider splitting patterns foralcohols.Figure 5-4 shows the 1HNMR spectrum for 2-propanol in carbon tetrachloridesolvent. Figure 5-5 shows the IR spectrum for 2-propanol obtained neat onsaltplates.

H OH

H,C CH,

6H

1H

1H

3.0 ppm 2.0 1.0

1

Figure 5-4

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Ur£J£UllC t/llCllllStry Carbonyls and Alcohols Oxygen Containing Compounds

2984 cm

PTVl

1392 cm-'

1455 cm-1 & 1365 cm-'

Figure 5-5

Aldehyde and Ketone PropertiesBecause aldehydes and ketones do not form hydrogen bonds, they typically haveboiling points only slightly higher than alkanes of equal mass. Because of thepolarity of the carbonyl bond, they are slightly miscible in water. Aldehydes andketones are aprotic, polar molecules that become less hydrophilic as their carbonchain length increases. The smaller aldehydes and ketones (three carbons or less)are generally water soluble but as the size of the alkyl group increases, theirwater solubility decreases. Table 5-2 shows the physical properties of severalaldehydes and ketones, from which the effects of mass, branching, andpositioning of the carbonyl group on the physical properties can be ascertained.

IsomerIUPAC Name

(Common Name)BoilingPoint

MeltingPoint

Water

Solubility(g/lOOmL)

HCHO Methanal (Formaldehyde) -21°C -92°C High

H3CCHO Ethanal (Acetaldehyde) 21°C -121°C Infinite

H3CCH2CHO Propanal (Propionaldehyde) 49°C -81°C 16.3

H3C(CH2)2CHO Butanal (n-Butyraldehyde) 76°C -99°C 6.8

H3C(CH2)3CHO Pentanal 103°C -92°C 3.3

H3C(CH2)4CHO Hexanal 128°C -56°C 2.1

C6H5CHO Benzaldehyde 178°C -26°C 0.3

H3CCOCH3 Propanone (Acetone) 56°C -94°C Infinite

H3CCOCH2CH3 Butanone (Ethyl methyl ketone) 80°C -86°C 25.6

H3CCO(CH2)2CH3 2-Pentanone 102°C -78°C 5.7

(H3CCH2)2CO 3-Pentanone 1018C -41'C 4.9

H3CCO(CH2)3CH3 2-Hexanone 128°C -55°C 1.6

H3CCH2CO(CH2)2CH3 3-Hexanone 124°C 1.3

H3CCOCH2CH(CH3)2 4-Methyl-2-Pentanone 119°C -85°C 1.9

C6Hi0O Cyclohexanone 156°C °C 2.2

C6H5COCH3 Acetophenone 202°C 21°C Insoluble

Table 5-2


OrCJcllllC lÎlCllllStry Carbonyls and Alcohols Oxygen Containing Compounds

Example 5.2What is the IUPACname for the following compound?

OH

r H

O

A. l-Aldo-4-pentanolB. 4-HydroxypentanalC. 5-Oxo-2-pentanolD. 2-Hydroxypentaldehyde

Solution

The longest chain isfive carbons and the highest priority functional group is thealdehyde. The functional group with the most oxidized carbon receives thehighest priority according toIUPAC convention. For naming aldehydes, the"e"is dropped from the alkane chain of thesamelengthand an "al" suffix is added.This makes the compound pentanal, which makes choice B correct. The OH isnamed hydroxy as a substituent.

Aldehyde and Ketone ReactivityAldehydes consist ofa carbonyl with a hydrogen bonded to the carbonyl carbonalong with either an alkyl group or in the case of formaldehyde, a secondhydrogen. Ketones consist of a carbonyl group with two alkyl substituentsattached. The chemistry occurs primarily at the electrophilic carbonyl center.Aldehydes and ketones are reactive with most nucleophiles, but not by atraditional nucleophilic substitution mechanism. Once a nucleophile attacks acarbonyl carbon, it forms a four-ligand intermediate with a negative charge onoxygen known as a tetrahedral intermediate. This intermediate will be shown inthe mechanism of many carbonyl reactions in this section. The chemistry ofaldehydes is similar to the chemistry of ketones except that analdehyde can beoxidized into a carboxylic acid while ketones cannot be oxidized easily.Oxidation in carbonyl chemistry can be viewed as either the gain ofbonds tooxygen orthe loss of bonds to hydrogen. We shall thoroughly address carbonylreactions throughout this section.

Spectroscopic Evidence for Aldehydes and KetonesAldehydes have infrared absorbances inthe 1720 cm"1 to1740 cm-1 range. Theyare unique in the IR from other carbonyls due to two medium C-H stretchesaround2700 cm-1 and 2900 cm-1. Ketones haveinfrared absorbances in the1710cm-1 to 1725 cm-1 range. In *HNMR, aldehyde hydrogens are found between 9and10 ppm, which makes aldehydes easy toidentify via *HNMR. Ketones andaldehydes each have alpha protons which fall in the 2.0 to 2.5 ppm range in!HNMR. Figure 5-6 shows the *HNMR spectrum for butanone in carbontetrachloride solvent. Figure 5-7 shows the IR spectrum for butanal obtainedneat on salt plates.

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ljr£}£llllC (jtlCiniStry Carbonyls and Alcohols Oxygen Containing Compounds

O

AHgC CH^CHg

2 ppm 1 ppm Oppm

Figure 5-6

O

XH CH2CH2CH3

J2738 cm-

1726 cm"

Figure 5-7

1457 cm'1 1388 cm''and

1362 cm-1

1117 cm'1

Example 5.3Pentanal canbe distinguished from 3-pentanone by allof the following EXCEPT:

A. a signal at9-10ppm inthe aHNMR.B. fivesignals in the *HNMRrather than two signals.C. an IR absorbance at 1826 cm"1.D. an ultraviolet absorbance at 230 ran instead of 240 nm.

Solution

An aldehyde hydrogen isfound between 9ppm and 10 ppm in the 1HNMR, sochoice A is a valid way to distinguish an aldehyde from a ketone. Choice A iseliminated. Pentanal has five unique hydrogens while 3-pentanone has twounique hydrogens, so the two compounds can be distinguished by theirrespective number ofsignals in the 1HNMR. Choice Biseliminated. Aldehydesand ketones have different 7i-bonds, so they have different carbonyl absorbancesin the ultraviolet absorbance region. This eliminates choice D. Aldehydes andketones have different IR absorbances, but they are observed around 1700 cm"1,not at 1826 cm"1. This makes choiceC an invalid technique, which makes choiceC the best answer.


Ur£)21111C CllCllllStry Carbonyls and Alcohols Oxygen Containing Compounds

Ketals, Hemiketals, Acetals, and HemiacetalsKetals and hemiketals are derivatives of ketones while acetals and hemiacetalsare derivatives of aldehydes. Ketals occur when a ketone loses the carbonylgroup and gains two alkoxy functional groups (R-O). The oxidation state ofcarbon does not change, because the carbon still has two bonds to oxygen, butnow it is two sigma-bonds to two different oxygen atoms rather than a sigma-bond and pi-bond to the sameoxygen. A hemiketal occurs when the ketonehasitscarbonyl group converted into a hydroxyl group andgains one alkoxy group.Acetals are similar to ketals, except it is the aldehyde that loses its carbonylgroup to gain the two alkoxy groups. Hemiacetals are similar to hemiketals,except it again is an aldehyde, rather than a ketone, that converts its carbonylgroup into a hydroxyl group while gaining an alkoxy group. Figure 5-8 showsthe formation of the fourcompounds.

O

Ketone

xsR"OH/OH~

R"0 OH

R^R-Hemiketal

O R"0 C

II xsR"OH/H+, \/R R'

Ketone

R H

Aldehyde

O

R K

Ketal

R'O OHO R'O O

II xsR'OH/OH V/

R "H

Hemiacetal

R'O OR'

II xsTOH/H*" . V/R' H

AldehydeR H

Acetal

Figure 5-8

Acetals and ketals are useful as protecting groups in organic synthesis. Acetalsand ketals can be formed and removed only under acidic conditions, where ashemiacetals and hemiketals are formed only under basic conditions butremovedunder any conditions.


Or£)£llllC ChCmiStry Carbonyls and Alcohols Oxygen Containing Compounds

Figure 5-9 shows a general mechanism for ketal formation under acidicconditions. It is the same mechanism for the formation of an acetal from analdehyde, except that an aldehyde is the reactant, rather than a ketone.

HH+ 2 R'OH -^ ^: H.O +

R' — Q". *0—R'

R R

deprotonate

R' — OH

H

•—o+ :o—R'R1

R R

make

R' R'

make

HO R'

H

deprotonate

HO

R R

•o

R

Figure 5-9

-<—*-

R

:o

R' J R

R OH

1 break

The steps of the mechanism are labeled to emphasize the predictable nature ofacid-catalyzed mechanisms. When you draw a mechanism foran acid catalyzedreaction, the intermediates must carry positive charges and no molecule shouldever carry a negative charge. With the exception of rearrangement steps inselected cases, acid catalyzed mechanisms follow this same pattern of: 1)protonate (making the leaving group a better leaving group), 2) break (theleaving group leaves), 3) make (thenucleophile attacks the carbocation), and 4)deprotonate (returning the molecule to a neutral state). Base catalyzedmechanisms follow the exact opposite pattern of: 1) deprotonate (to make astrongnucleophile), 2)make/break (where the nucleophile attacks and dislodgesthe leaving group), and 3) protonate (returning the molecule to a neutral state).

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vlrCJ£llllC CllCIIllStry Carbonyls and Alcohols Oxygen Containing Compounds

These mechanisms should be kept as simplistic as possible. The mechanism forthe formation of a hemiacetal from an aldehydeand alcohol in the presence ofstrong base is shown in Figure 5-10.

• • • •

Q- ^~^Base H~ Q'' ;°—R'"ITS+ R'O —H

deprotonateR' H R' ~R

\ :qL R'O—h*^ ^rotonate/fH make/break

R'

• • • •

Q. «0 R'

R'p: R- -R

Figure 5-10

When you draw the mechanism for a base catalyzed reaction, be sure that theintermediates carry negative charges and no molecule ever carries a positivecharge. Ketals and acetals serve as protecting groups for carbonyl groups insynthesis involving ketones and aldehydes. Hemiacetals and hemiketals arenotuseful as protecting groups, but they are important in sugar chemistry.

Example 5.4

Addition of ethanol at apH of 4to propanal yields which organic product?A- B. C. D.

Et(V ÔEt EtO. .OEt HO. .OEt HO^ ÔH

Et H EtO H Et H Et H

Solution

Addition of an alcohol to an aldehyde under acidic conditions (pH =4is acidic)yields an acetal. The only acetal in the choices is choice A. Choice B has toomany ethoxy groups, choice Cis ahemiacetal, and choice Disageminal diol.

Example 5.5Addition of sodium methoxide inmethanol to acetone yields:A. isopropyl alcohol.B. acetaldehyde.C. aketal.

D. a hemiketal.

Solution

Addition ofan alcohol to aketone under basic conditions yields ahemiketal. Themethoxide anion attacks the carbonyl carbon of acetone to generate thetetrahedral anion intermediate, which then deprotonates the methanol to formthehemiketal and regenerate themethoxide anion. The bestansweris choice D.


OrCjaniC CJoCmiStry Carbonyls and Alcohols Oxygen Containing Compounds

Acetals and Ketals as Protecting GroupsProtecting groups are used in synthesis to prevent a reagent from reacting at asite where it is undesirable to have a reaction. Weshall only discuss protectinggroups for aldehydes and ketones at this time. Both of these carbonylcompounds employ the same reaction to add the protecting group, formingeither an acetal or ketal. As we have seen, aldehydes and ketones, in thepresence of alcohols and acid, form acetals (from aldehydes) and ketals (fromketones). Because acetals and ketals are less reactive than aldehydes andketones, they are an ideal protecting group. The alcohol that is typically used toform the protected carbonyl compound is a vicinal diol (a 1,2-diol), such asethylene glycol (HOCH2CH2OH). Figure 5-11 shows the protecting ofcyclohexanone using ethylene glycol.

HO

cone. H+

Cyclohexanone Protected as a Ketal

Figure 5-11

+ H20

The mechanism for forming the protected ketone is the same generic acid-catalyzed mechanism shown in Figure 5-9, only instead of using two moleculesof alcohol, a vicinal diol is used, so the second "make" step involves the secondhydroxyl group of the vicinal diol rather than a new alcohol.

Example 5.6The addition of ethylene glycol (HOCH2CH2OH) in the presence of acidprovidesa protecting group for ketones. Which of the following is the protectedform of 2-pentanone?

00 00 00 0.0

Solution

Theoriginal ketone (2-pentanone)has two bonds from carbon 2 to oxygen, so theproduct must also have two bonds from carbon 2 to oxygen. This eliminates allof the answer choices except choice B. The product of a diol and a ketone inanhydrous, acidic conditions is a cyclicketal.

There is not much to using an acetal or ketal protecting group. The last thing toconsider is when to use a protecting group in synthesis. As a guideline, any timethat you have a molecule with more than one reactive site, you must protect thesites at which you wish to have no reaction. The exception to this rule is whenthe site you wish to react at is significantly more reactive than any other sites onthe molecule.

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Or£}<UllC dlCmiStry Carbonyls and Alcohols Oxygen Containing Compounds

Carboxylic Acids and Their DerivativesCarboxylic acid derivatives are differentfrom aldehydes and ketones in that thecarbonyl carbon has a functional group thatpossibly can actas a leaving group.Much of the chemistry of carboxylic acids and acid derivatives centers aroundchanging the group on the carbonylcarbon. We shall address each functionalgroup starting with the carboxylic acid and look at their chemical reactions.

Carboxylic Acids

Carboxylic acids are weak acids with a pKa between 2 and 5. They are readilyconverted into esters, anhydrides, oracid halides. They carry outsimilar organicreactions as esters, but are less reactive than esters. Carboxylic acids can beformed by saponification (treating an ester with strong base in water), by treatinga methyl ketone (RCOCH3) with I2 and strong base, by oxidizing primaryalcohols and aldehydes in water, or by hydrolyzing a nitrile oran amide usingstrong acid athigh temperatures. These reactions are shown inFigure 5-12.Saponification: O

AxsH2Q/OH ^

O

A + HOR'R OR'

Ester

Ester hydrolysis: O

AR OR'

Ester

xsRgO/H*"

R O"

Carboxylate

O

AR OH

Carboxylic acid

O

+ HOR'

Iodoform reaction: O

AI2/OH

A • I3CH

Iodoform

Oxidation:

R CH3Methyl ketone

O

R O"

Carboxylate

O

XAldehyde

H H

R-ÔHPrimary alcohol

KMnQ4/OH^

AOxidation:

Amide hydrolysis: O

\R NHR'

Secondary amide

Nitrile hydrolysis

R C=N

¥L2Cr207/l?

xsÔ/H*^ A

xsHp/H*

Nitrile

Figure 5-12

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R OH

Carboxylic acid

O

AR OH

Carboxylic acid

O

AR OH

Carboxylic acid

O

+ R'NH,

A + NH,R OH

Carboxylic acid

The Berkeley Review

Or£)£llllC tUlCHliStry Carbonyls and Alcohols Oxygen Containing Compounds

Carboxylic acids can be reduced into primary alcohols or converted into othercompounds such as acid halides, acid anhydrides, or esters. Figure 5-13 showsfour reactionsof carboxylicacids with which you are expected to be familiar.

O

R Cl

Acid halide

o

AR OR'

Ester

Figure 5-13

H H

XR OH

Primary alcohol

O O

AAR O R

Acid anhydride

Example 5.7Treatment of benzoic acid with ethanol and acid yields which of the followingcompounds?

D.

OH OEt OH

EtO

OEt

Solution

Addition of an alcohol to a carboxylic acid under acidic conditions with enoughheat to overcome the activation barrier yields a new ester by way of atransesterification reaction. The final product is an ester with an ethoxy group inplace of the hydroxyl group of the carboxylic acid, alongwith a water moleculeside product. Choice B is the best answer.

Esters

Esters are carbonyl compounds with an alkyl group and an alkoxy groupattached to the carbonyl carbon. Esters have a leavinggroup (the alkoxygroup),so they undergo more reactions than ketones or aldehydes. Their reactivitycorrelates with the pKa of the conjugate acid of the leaving group. Leavinggroups that are more stable (are less basic and their conjugateacids have a lowerpKa value) are better leaving groups. Esters can easily exchange their alkoxygroup in the presence of acid and an alcohol in what is referred to as atransesterification reaction.


Or£J£UllC CoCmiStry Carbonyls and Alcohols Oxygen Containing Compounds

Example 5.8What is the majororganic product for the following reaction?

A.O O

1Et' "OCH3 EKT 'OCH, Et*" "OCH, Et' "OEt

O

H3CH2C OCH2CH3

B.

+ HoCOH/rf

C. D.HO OEt H3CO OCH3

X„ X X XSolution

At high temperature under acidic conditions, an alcohol can undergo atransesterification reaction when mixed with an ester. The final product is thenew ester with the methoxy group attached instead ofthe ethoxy group. Ethanolis the organicside product. ChoiceA is the best answer.

In Example5.8, a C-Obond and an O-H bond were both broken in the reactantsandformed in the products. This implies that the enthalpy (AH) for the reactionis close to 0. Going from an ester and non-cyclic alcohol to an ester and non-cyclic alcohol generates a change in entropy (AS) of roughly 0, because thereaction starts and finishes with roughly the same degrees of freedom. Thisimplies that the change infree energy for transesterification isaround 0 (AG =AH- TAS). Thismeans that the equilibrium constantfor a transesterification reactionis approximately 1. The reaction can be driven to products by focusing on LeChatelier's Principle. Itwill proceed inhigh yield if the products are removed ora reactant is constantly added. This also has biological significance in thattransesterification can be used for shuttles in the cell membrane. For theformation of a lactone (cyclic ester), there is a ring formed and an alcoholmolecule produced from a linear system. Depending on the reaction, there canbe either again or aloss in entropy, so lactone formation is less predictable.

Lactones

Lactones are cyclic esters, as shown in Figure 5-1. Lactones undergo the samechemistry as esters, only entropy is a factor in the reaction's favorability.Lactones can be synthesized by treating cyclic ketones with peroxyacids(RCO3H) in what is known as the Baeyer-Villiger reaction. Lactone chemistry iseasier to understand when you recognize that the compound isan ester. Figure5-14 shows the formation of a lactone from an intramolecular transesterificationreaction of a hydroxyester.

RO

H+

OHROH

Figure 5-14


Ur£J21IllC CJlClHiStry Carbonyls and Alcohols Oxygen Containing Compounds

Acid AnhydridesAcid anhydrides are formed by a condensation reaction (dehydration) of twocarboxylic acids at elevated temperatures (as shown in Figure 5-13). Therefore,when you add water to an acid anhydride, it breaks into two carboxylic acids. Itis easiest to predict the product if you focus on the inorganic side product (H2Oin this case) more than the organic product. If you circle and remove an H fromone reactant and an OH from the second reactant and then connect the two

leftover fragments, then you can derive the organic product rather effortlessly.This is shown in Figure 5-15 as a short-cut method for deducing the product ofdehydration of two carboxylic acids.

O O OO OO

X^~X ^ A A - A ACfa HOJ R' R O R* R O R'Find the water Connect the atoms Viva la anhydride

Figure 5-15

Example 5.9What is the major product from the following reaction?

O Q

\ / rûr H3COH iW- OH

-•

A. Diketone

B. Acid anhydrideC. Ester

D. Aldehyde

Solution

Addition of heat to carboxylic acids yields an acid anhydride by driving offwater. Thefinal productsare an acidanhydride and a watermolecule. Choice Bis the best answer There are three possible acid anhydrides that can form. Oneof the threepossible acid anhydrides that may form is drawn below. The othertwo acid anhydrides that can form result from the condensation of each aciduponitself. Probability saysthat theproductbelow is themost likely.

O O O O O O

CH3X,A-X ^ XX "*" XnXH„qi CfHHOj CH3HnC6^ O CH3 HnC6^ OWater is found Atoms are connected Anhydrides are good

Acid Halides

Acid halides are similar to esters, but with a halide (Cl, Br and I) in place of thealkoxy group. They are the most reactive ofall the carbonyl compounds, becausethehalide is a great leaving group. Theyundergo the samesubstitutionreactionsas other carbonyl compounds that have a leaving group, but they react faster.For some reactions, acid halides can be too reactive. They are a usefulintermediate product in many synthetic pathways, such as the conversion of acarboxylic acid into an amide for instance, where the carboxylic acid is firstconverted to the more reactive acid halide which then goes on to form the amide.


Or£}2UllC CllCllllStry Carbonyls and Alcohols Oxygen Containing Compounds

Example 5.10What are the two products formed when propanoyl chloride, H3CCH2COCI, istreated with methyl amine, H3CNH2?

A. N-methyl propanamide and hydrochloric acidB. l-amino-2-butanoneand hydrochloric acidC. Propanal and chloromethylamineD. Carbondioxide and N,N-ethylmethylamine

Solution

An amine is a good nucleophile capable of attacking the carbonyl carbon anddisplacing the chloride anion to form an amide, choice A. Amides are biologicalstructures thatyou arerequired toknow according to theMCAT Student Manual.This reaction can be used asa precursor to synthesizing an amine. Amides canbereduced to anamine using LiAlH4 inether followed by neutralization with aweak acid. There are problems with direct synthesis of primary amines usingammonia andanalkyl halide due to multiple additions. The point here is thatamides can be products themselves, orintermediates inamine synthesis.

Amides

Amides are carbonyls with an amine group bonded to the carbonyl carbon.Amides form the backbone ofproteins, andthey arefound in most ofthebases ofnucleic acids (i.e., DNA and RNA). An amide bond that links amino acidstogether is referred to as a peptide bond. Amides can be reduced into aminesusing astrong reducing agent such as lithium aluminum hydride, LiAlH4.

Example 5.11Which ofthe following compounds doesNOT contain an amidebond?A. Guanine

B. Uracil

C. Isoleucine

D. Cytochrome

Solution

Most amino acids do not contain an amide bond, although polymers of aminoacid (polypeptide chains) do. It is in proteins that amino acids form peptide(amide) bonds. The peptide bonds of proteins are broken down in acidicaqueous environments to regenerate the individual amino acids. Guanine anduracil are bases in RNA, and they contain amide bonds (as drawn below). It isthe amide functional group that forms the hydrogen bonding in base pairing.Cytochrome is an enzyme (protein), so it contains amide bonds in its peptidelinkages. The bestanswer is choice C, isoleucine, an amino acid.

9 O

£X" Otx"NÔ N^N^NHaH H

Uracil Guanine(pyrimidine base) (purine base)


Or£{£I]llC ChCmiStry Carbonyls and Alcohols Carbonyl Reactivity

Carbonyl reactions can be categorized in one of three ways. The fist type ofreaction involves a nucleophile attacking the electrophilic carbon. The secondtype of reaction involves the deprotonation of an alpha proton and thesubsequent nucleophilicity of the anion that is generated. The last type ofreaction falls into the realm of oxidation-reduction chemistry, although mostmechanisms for carbonyl redox reactions involve a nucleophile attacking thecarbonyl carbon. If you keep things simple in terms of the three types ofreactions, you should be able to summarize all of carbonyl chemistry and workthrough any questions they may present.

Attack at Carbonyl CarbonBecause carbonyl compounds contain a C=0 bond, they are good electrophiles.We shall consider ketones and aldehydes first, but other carbonyl compoundsalso act as electrophiles. The difference in reactivity between a ketone and acarboxylic acid derivative, such as an ester, centers around the presence of aleaving group on the carbonyl carbon. The C=0 bond is polar with a partialpositive charge on the carbon atom and a partial negative charge on the oxygenatom. It is the partial positive charge on the carbon that makes a carbonyl such awonderful electrophile. Figure 5-16shows a generic carbonyl reaction, where acarbonylcompound is attacked by a nucleophile to form a tetrahedral intermediate.

O Nuc O"

II Nuc:' \ /

Carbonyl Tetrahedral Intermediate

Figure 5-16

We have already seen aldehydes and ketones serving as electrophiles in theformation of acetals, hemiacetals, ketals and hemiketals. As far as electrophilicchemistry of carbonyls is concerned, there is no major variation betweenreactions. In other words, it is best to view carbonyl reactions as all basically thesame with slight variations of the nucleophile. The tetrahedral intermediate inFigure5-16 representsevery genericintermediate in carbonyl addition reactions.

Forcarbonyl compounds that have leavinggroups, the reactivity of the carbonylcompound is based on the strength of the leaving group. Stronger leavinggroups make for a more reactive (more electrophilic) carbonyl compound. Thestrength of a leaving group can be inferred from the pKa of its conjugate acid.Leaving groups are considered to be good when they form a stable compoundupon leaving. So, as the leavinggroup gets stronger, it getsmorestable, whichmakes it less basic, and thus makes its conjugate acid stronger. This means thatgoodleavinggroups generally have conjugate acids with low pKa values. Figure5-17 shows a carbonyl reactivity chart (relative reactivity of substitutedcarbonyls) for an acid halide, an acid anhydride, an ester, and an amide. Theconjugate acid of each leaving group is shown in the same diagram.


Organic Chemistry carbonyisandAicohois Carbonyl Reactivity

ThebetterXis at leavingfromH, the better it willbe at leavingfrom C,so thebestleaving group hasa conjugate acid withthe lowest pKa.

O

H^R xr iA R R cr R

pKa = -10 to -7 verygoodvery strong acid leaving group pKa - 3 to5 good leaving group

average acid strength

H OR'

pKa = 14tol7weak acid

O

R OR'

semi-poorleaving group

Relative Reactivity:

O O O

H NHR1 yC\\R NHR'

pKa = 33 to 35 very poorvery weak acid leaving group

O

O

R X R O R R' OR' R'

Figure 5-17

O

NHR'

The relative reactivity implies that anacid halide can easily beconverted into ananhydride, ester, or amide. An acid anhydride can easily be converted into anester or amide, but it is difficult to convert the anhydride into an acid halide.This technique is a good predictor of the reactivity of all carboxylic acidderivatives. The generic reaction and its tetrahedral intermediate are shown inFigure 5-18.

O^v

0R pL.G.

:Nuc

®q l.g.

^ \©R Nuc R

O

II,c.

©Nuc

©+ L.G.

Figure 5-18

Carbonylsubstitutionreactions proceedvia a tetrahedral intermediate as showninFigure 5-18. Ifthe leaving group is not agood one, then the reaction cannot gofurther than the tetrahedral intermediate and will ultimately shift back to thecarbonyl reactant. This results inanequilibrium between theoriginal carbonyland the tetrahedral intermediate. This isobserved with ketones and aldehydeswhen they are present inasolvent that has nucleophilic capability.


Or£{21111C ChCmiStry Carbonyls and Alcohols Carbonyl Reactivity

Example 5.12Whichof the following is the MOST electrophiliccarbonylcompound?

A. Amide

B. Ester

C. AldehydeD. Acid anhydride

Solution

The reactivity of a carbonyl is dictated by the leaving group on the carbon of thecarbonyl. As the acidity of the conjugate acid of the leaving group increases, sodoes the reactivity of that particular carbonyl. In the choicesabove, the conjugateacids of the leaving groups are an amine, an alcohol, H2, and a carboxylic acidrespectively. The most acidic is the carboxylic acid, so the anhydride is the mostreactive carbonyl.

Deprotonation of oc-ProtonsThe hydrogen on the alpha carbon (the carbon adjacent to the carbonyl carbon) isacidic (its pKa is approximately 19), so it can be removed using a strong base.Enolates are formed when a hydrogen on the alpha carbon is deprotonated. Theenolate can regain a proton at either the carbon or the oxygen. If it is protonatedat the oxygen,an enol is formed. There is an equilibrium between the ketone andenol. The conversion from a ketone into an enol is known as tautomerization,because a ketone and its enol are tautomers, structural isomers that vary in theposition of a 7t-bond and a hydrogen. The tautomerization of acetone is shown inFigure 5-19.

Ketone Carbanion Enolate Enol

CH,

! base

Figure 5-19

The carbanion that forms is a good nucleophile. When an alkyl halide is addedto the solution, the carbanion can attack the alkyl halide in a nucleophilicsubstitution reaction to form a new carbon-carbon bond. This results in a longerketone. The halide can be any halide, but the reaction works best with an alkyliodide compound. Alkyl bromides and chlorides yield more O-alkylation sideproducts than alkyl iodides. The generic reaction is shown in Figure 5-20.

Ketone Carbanion LongerKetone

Figure 5-20

Copyright ©byThe Berkeley Review 21 Exclusive MCAT Preparation

OrgaillC ChemiStry CarbonylsandAlcohols Carbonyl Reactivity

When the carbonyl is asymmetric, the possibility of two different enolates arises(each formed by deprotonating a different alpha hydrogen). If one of the twoalpha carbons is more substituted, then it is both more sterically hindered andmore stable. This creates a situation where the reaction can be dictated bytemperature and base size. At low temperature with a bulky base, the lesshindered (less substituted) alpha carbon gets deprotonated to form the so calledkinetic enolate (lower energy transition state). This is referred to as kinetic control.At a higher temperature with a small base, the more hindered (more substituted)alpha carbon gets deprotonated to form the so called thermodynamic enolate(leading to themore stable product). This is referred toas thermodynamic control.We shall apply this concept to aldol condensation reactions of asymmetricketones later on in this section.

Example 5.13What is the final product after acetone is treated first with NaH, followed byiodoethane, andsubsequently followed byworkup?A. 2-Methyl-2-butanolB. 3-Methyl-2-butanoneC. 2-Pentanone

D. 2-Butanone

Solution

This reaction is similar to the generic reaction in Figure 5-20. The carbon chainlength is increased by two carbons (from three to five) when the electrophile isethyl iodide. This eliminates choice D. The product is a ketone, so choice A iseliminated. The final product is 2-pentanone, as shown below, which makeschoice C the best answer.

:o: :o: :o:II __ II II

CH3 B£ CH3 H2C ^CH3H1>H^Î base ^ ^ H3CH2C

n2^ L 2-Pentanone/

H3C

Oxidation and Reduction

Oxidation and reduction are recurring inorganic chemistry, soworking from alogic-based foundation is key. If the oxophilic carbon (carbon containing abondto oxygen) hashydrogens, it canbeoxidized. Primary alcohols areoxidized intoaldehydes, which can be further oxidized into carboxylic acids. Secondaryalcohols are oxidized into ketones. Tertiary alcohols cannot be oxidized (thealcohol carbon has no hydrogen to lose). Reduction is defined as the opposite ofoxidation, so the reverse ofeach reaction just mentioned represents reduction.To make the processes more clear, we shall define oxidation and reduction interms of bonds to oxygen and bonds to hydrogen. More than just oxygencontaining compounds doredox chemistry. When two cysteine residues form acrosslink, they undergo dehydrogenation (loss of hydrogen), an oxidativeprocess. When the rc-bond in a fatty acid is hydrogenated to form an aliphaticchain, it hasundergone a reductive process.


OrgaillC ChCmistry Carbonyls and Alcohols Carbonyl Reactivity

Oxidation is defined as an increase in oxidation state,whichis caused by eitherlosing a bond to a lesselectronegative atom (inmostcases hydrogen) or gainingabond to a more electronegative atom (in most cases oxygen). Reduction isdefinedas a decreasein oxidationstate, which is causedby eithergaininga bondto a less electronegative atom (in most cases hydrogen) or losing a bond to amore electronegative atom (in most cases oxygen).

Oxidation Reduction

Gain of bonds to O Loss of bonds to O

Loss of bonds to H Gain of bonds to H

Increase in oxidation state Decrease in oxidation state

Determining oxidation states using the method learned in general chemistry is amatter of assigning oxygen a -2 (because it is more electronegative than the atomsto which it bonded, and it makes two bonds) and a +1 to hydrogen (because it isless electronegative than the atoms to which it bonded, and it makes one bond).The oxidation state of any remaining atoms is found by difference. In organicchemistry, oxidation states for specific atoms can easily be found by consideringelectron sharing in each bond. If the bond is between two atoms of unequalelectronegativity, then the more electronegative atom is assigned a -1 and the lesselectronegative atom is assigned a +1. The oxidation state of an atom is foundby summing the numbers from all of the bonds and any formal charge it mayhave. Figure 5-21shows this method as it applies to the oxidation state of carbon2 in 2-propanol and acetone.

[O] ^

lost a bond to H

gained a bond to O

Figure 5-21

Each of the four bonds to carbon is analyzed for its relative electronegativitycompared to the atoms to which it is bonded. Bonds to hydrogen give a negativeto carbon and a positive to hydrogen, because carbon is more electronegativethan hydrogen. Bonds to oxygen give a positive to carbon and a negative tooxygen, because carbon is less electronegative than oxygen. Both carbons in acarbon-carbon bond get zero, because there is no difference in electronegativity.In a secondary alcohol,the oxophiliccarbon has an oxidationstate of 0 while in aketone, the oxophilic carbon has an oxidation state of +2. This means that thecarbon was oxidized by two electrons, which is predictable, because it has lost abond to hydrogen (oxidizing it by one electron) and has gained a bond to oxygen(oxidizing it by another electron). Figure 5-22 shows that oxygen and hydrogendo not change oxidation state when going from 2-pentanol to acetone.

0C0?0CH3

O A -2

@H3C"0~0C0^CH2

Figure 5-22

Oxygen has a -2 oxidation state, as is expected. Hydrogen has a +1 oxidationstate as expected. Oxidation states should be made this simple.


OrgailiC Chemistry Carbonyls and Alcohols Carbonyl Reactivity

Oxidizing and Reducing AgentsOxidizing agents are rich in oxygen (given that they deliver oxygen to thereactant) and reducing agents are rich in hydrogen (given that they deliverhydrogen to the reactant). This perspective is applicable in both organicchemistry and biochemistry. It should be noted that NAD+ acts like Cr03 withpyridine, NADH acts like NaBH4, and FADH2 acts like H2/Pd. Ethanol can beoxidized into ethanal using either Cr03 in the presence of HC1 and pyridine(known as PCC) or by NAD+. In the reduction half-reaction, NAD+ is reduced toNADH, because for every oxidation half-reaction, there is a reduction half-reaction. Thecoupledhalf-reactions are shown in Figure5-23.

Oxidized by 2 e"

Reduced by 2 eNAD+ NADH + H+

Figure 5-23

Oxidizing agents cause oxidation (and indoing so, they get reduced). Oxidizingagents are rich in oxygen, poor in hydrogen, and have an atom in a highoxidation state with high electron affinity. Reducing agents cause reduction (andin doing so, they get oxidized). Reducing agents are poor in oxygen, rich inhydrogen, and have anatom ina low oxidation state with low ionization energy.Some common oxidizing and reducing agentsare listedbelow.Oxidizing Agents fRich inO: Poor in H) Reducing Agents TPoor in O:Rich in H)

KMn04 Cr03 O3 UAIH4 NaBH4 H2NNH2RCO3H ROOR Cl2 H2/Pd HCl/Zn HOCH2CH2SHNAD+ NADP+ FAD NADH NADPH FADH2

Typical oxidation and reduction reactions involve the conversion betweenalcohols and carbonyl compounds. Figure 5-24 is a summary of commonoxidation-reduction reactions in organic chemistry (1° alcohol -» aldehyde -»acid and2° alcohol ->ketone - 3* alcohols undergo nooxidation reactions).

Primary alcohols have twohydrogens tolose, so they are oxidizedtwice (first into an aldehyde and then into a carboxylic acid.) Thecarboxylic acid is reduced back into aprimary alcohol using LiAlH4.H

O

R J-OH oxidizing agent || oxidizing agentR- H

H

alcoholreducing;agent aldehyde

O

R- OH

carboxylic acid

reducing agent

Secondary alcohols haveonehydrogen to lose, so theyare oxidized once.

O

Copyright© by The Berkeley Review

H

»+OH oxidizing agentR- •R'

R'

2° alcoholreducing agent ketone

Figure 5-24

24 The Berkeley Review

Organic Chemistry carbonyls and Aicohois

Oxidation

From an organic chemistry perspective, oxidation is the process of gainingoxygen (and/or losing hydrogen) at the carbon which has an electronegativeatom attached. The first rule is to count the number ofcarbon-oxygen bonds tothe carbon of interest in the reactant. As the number ofcarbon-oxygen bondsincreases, theoxidation level (andoxidation state) ofthecompound increases. Interms ofoxidation levels, every carbon-oxygen bond counts asone. Ifyou lose acarbon-hydrogen bond and replace it witha carbon-oxygen bond, theoxidationlevel has increased by one while the oxidation state of the carbon has increasedby two. The oxidation of a primary alcohol into an aldehyde and then into acarboxylic acid is shown in Figure 5-25.

HO. H

/oCNH3C H

Oxidation State:

0 +1 -1 -1 = -1

Cr03/HCl

Pyridine

O

H3C/0 ^HOxidation State:0 + 1+1-1 =+1

K2Cr?07—-—^-^-»H2S04(aq)

O

+111+1

HaC^"0 ÔHOxidation State:

0 + 1 + 1 + 1=+3

Figure 5-25

Youshould be able to recognize reagents, the reaction type, and correctly predictthe product for redox reactions. As a point of interest, Cr04 with HCl inpyridine is known as pyridinium chlorochromate, or PCC for short. Mostorganic compounds lose carbon-hydrogen bonds and gain carbon-oxygen bondswhen oxidized, so the reagent (oxidizing agent) must be rich in oxygen. The twomost common oxidizing agents in organic chemistry are CrO^" and Mn04".Some common oxidation reactions of alcohols are shown in Figure 5-26.

R\

HPrimary alcohol

OH

"PCC"

CrQ3Cr

Pyridinium

N+-H

S.H*7H

Primary alcohol

DH K2Cr2°7,OH

H

^>C_ OHR

R

Secondary alcohol

Tertiary alcohol

H2S04(aq)

KMn04

KOH(aq)

[O]

Figure 5-26

O

II

R H

Aldehyde

O

II

R OH

Carboxylic acid

O

II

R R

Ketone

No Reaction

(No Hs to lose)

Carbonyl Reactivity


Organic Chemistry carbonyisandAicohois Carbonyl Reactivity

These reactions should be known well enough that you can recognize them in agroup of many reactions. For other oxidation reactions, you should simplyrecognize the oxidizing agent and know that the product is more oxidized thanthe reactant. Figure 5-27 shows some oxidation reactions of compounds otherthan alcohols. The common theme is that one of the reactants is rich in oxygen.

R H

HH R

Disubstituted alkene

(H3C)2S

HIO,

R H

\ ÔH

H R

Vicinal diol

H3COH/H20

o-C6H5C03H

Ketone

R

>- O

H

Aldehyde

R

H

Aldehyde

orLactone (Ester)

Figure 5-27

03 has excessive oxygen;alkene reactant is beingoxidized

HI04 has excessiveoxygen; diol reactantis being oxidized

C6H5C03Hhas excessiveoxygen; ketone reactantis being oxidized

The three oxidation reactions shown in Figure 5-27 are ozonolysis of an alkene,oxidative cleavage of a vicinal diol using periodic acid, and the Baeyer-Villigeroxidation of a ketone respectively. These reactions are rare enough that theydon't need to be memorized, but you should recognize that each is an oxidation.More important than memorizing organic chemistry reactions is the ability topredict results and explain observations based on the data given about thesereactions in the passagesassociatedwith them. For instance, excessiveoxygen inthe reactantshouldprovidea clue that the organicmolecule is being oxidized.

Example 5.14What is the majororganic product for the following reaction?

OH OHKMnQ4^

OH"(aq)

A- O O B' O OH c O O D* OH O

Solution

The secondary alcohol is oxidized into a ketone, which eliminates choice D. Theprimary alcohol is oxidized all the way to a carboxylicacid, which makes choiceC the best answer.

Copyright © by The BerkeleyReview 26 The Berkeley Review

OrgaillC ChemiStry Carbonyls and Alcohols Carbonyl Reactivity

Example 5.15Which of the following compounds is NOT an oxidizing agent?A. Peroxybenzoic acid (C6H5CO3H)B. Ozone (O3)C. Diisobutylaluminum hydride [((H3C)2CHCH2)2A1H]D. Chromic acid (H2Cr04)

Solution

Organic oxidizing agents generally contain oxygen atoms, like choicesA, B, andD. Diisobutylaluminum hydride (DIBAH) contains no oxygens, and in factcontains hydrogens. DIBAH is a selective reducing agent that is weaker thanlithium aluminum hydride (LLAIH4). The best answer is choice C.

Example 5.16Oxidation of an aldehyde into a carboxylic acid requires the addition of:

A. an oxidizing agent.B. a reducing agent.C. a strong base.D. a strong acid.

Solution

Oxidation requires the addition of an oxidizing agent to carry out the oxidation.Oxidation can be carried out in either acidic or basic conditions, so choices C andD are eliminated. The best answer for this question is choice A.

Example 5.17When a primary alcohol reacts with potassium permanganate, the primaryalcohol is acting as which of the following in the reaction?

A. An oxidizing agentB. A reducing agentC. A strong baseD. A strong acid

Solution

Because the alcohol is being oxidized, it is losing electrons. This means that it iscausing the reduction of Mn04" into Mn02- Reducing agents get oxidized, so theprimary alcohol is the reducing agent. The best answer is choice B.

Example 5.18Oxidation of a secondary alcohol leads to:

A. an aldehyde.B. a carboxylic acid.C. a ketone.

D. a tertiary alcohol.

Solution

As shown in Figure 5-26, a secondary alcohol is oxidized into a ketone and nofurther, because it has only one hydrogen to lose to oxidation. The best answerfor this question is choice C.


Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity

Example 5.19Which of the following compounds CANNOT be oxidized by chromic acid?

A. 2-Methyl-3-pentanolB. ButeraldehydeC. 3-Pentanol

D. Phenol

Solution

Primary and secondary alcohols can be oxidized while tertiary alcohols andphenols cannot be oxidized. Choices A and C are secondary alcohols, so they canbe oxidized into ketones. Aldehydes are oxidized into carboxyKc acids, so choiceB is eliminated. This leaves choice D, phenol, as the best answer. The carbonbonded to oxygen in phenol has no hydrogen atoms, so it cannot be oxidized.

Example 5.20To oxidize a primary alcohol into an aldehyde without further oxidation into acarboxylic acid,what reagent should you add in conjunctionwith chromic acid?

A. Pyridine and hydrochloric acidB. Toluene and hydrochloric acidC. Pyridine and sulfuric acidD. Toluene and sulfuric acid

Solution

A primary alcohol is oxidized into an aldehyde and no further when pyridiniumchlorochromate is added to the solution. This is known as PCC or Collin'sreagent. Theabsence of water fromsolution prevents the aldehyde from furtheroxidation intoa carboxylic acid. Thebestanswerfor thisquestion is choice A.

Reduction

Reduction is the process of losing oxygen (and/or gaining hydrogen) at thecarbon which has an electronegative atom attached. As with oxidation, the firststep is to count the number of carbon-oxygen bonds in the reactant to the carbonof interest. As the number of carbon-oxygen bonds decreases, the oxidation leveldecreases (as does the oxidation state). If you lose a carbon-oxygen bond andreplace it witha carbon-hydrogen bond, theoxidation level decreases by oneandthe oxidation state ofcarbon decreases by two. From a ketone or aldehyde toanalcohol, the oxidation statedecreases by twoand the leveldecreases by one. Thereduction ofan esterintoa primary alcohol is shownin Figure 5-28.

O HO H

+1"+1 1. LiAlH4(thf) +FMy0^ * xttt ^7 *• x-0^ + HOCH2CH3H^ÔCH^H, ZNH*°™ H3C^^H

Oxidation State: 0 +1 + 1-+1 = +3 Oxidation State: 0 + 1-1-1=-1

Figure 5-28

The two most common reducing agents in carbonyl chemistry are LLAIH4 andNaBH4. Aluminumis lesselectronegative than boron, so LiAlH4 is more apt todonate an H" to the carbonyl carbon, and thus is more reactive than NaBH4- Assuch, NaBH4 reduces only ketones and aldehydes while LiAlH4 reduces allcarbonylcompounds. Figures 5-29a and 5-29b show these carbonyl reductions.


Organic Chemistry carbonyls and Alcohols

R' H

Aldehyde

O

R" R'

Ketone

O

O

II

Carboxylic acid

O

II

R OR'

Ester

O

II

Aldehyde

O

II

R R'

Ketone

1. LiAlH4(thf)

2. NH4Cl(aq)

1. LiAlH4(thf)

2. NH4Cl(aq)

1.NaBH4(et2Q)^2. NH4Cl(aq)

1. NaBH4(et2Q)

2. NH4Cl(aq)

HO, H

\?

R/C^HPrimary alcohol

HO.

.C. + HOR'

R H

Primary alcohol

H

HO. H

+ H,0

R H

Primary alcohol

HO, H

R R'

Secondary alcohol

Figure 5-29a

HCl/Zn(Hg)^ Hor ^

1. HS(CH2)2SH

H

2. Raney Ni

H2NNH2^

KOH(aq) *

2 equivalentsHSCH2CH2OH

R' H

Alkane

H H

^C\ + N2 + H20W R'

Alkane

rSH

HS*

Reduced Linkage

H H

*~ R*/ Y'/H

S-CH2CH2OH

S—CH2CH2OHDisulfide linkage

R HH2 FADH2

• orPdH

H CH2COzH

Unsaturated fatty acid

H CH2C02H

Saturated fatty acid

Figure 5-29b

The aldehyde-to-alkane reactions are Clemmensen reduction and Raney nickelreduction, respectively. They reduce either an aldehyde or ketone to an alkane.

Carbonyl Reactivity


Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity

Example 5.21What is the major organic product after pentanal is treated with sodiumborohydride (NaBHi)?

A. A primary alcoholB. A secondary alcoholC. A hemiacetal

D. A carboxylic acid

Solution

Aldehydes can be oxidized into a carboxylic acid or reduced into a primaryalcohol. Sodium borohydride is a reducing agent, so an alcohol is formed. Thiseliminates choices C and D. The carbonyl in an aldehyde is on the C-l carbon, sothe hydroxyl group forms on C-l carbon, resulting in the formation of a primaryalcohol. Choice A is the best answer.

Example 5.22Acetone, when reacted with lithium aluminum hydride and quenched withweakly acidicwater, yieldswhich of the followingproducts?

A. PropaneB. 1-PropanolC. 2-PropanolD. Propene

Solution

Acetone is a ketone, thus when it is reduced, it forms a secondary alcohol. Theonly secondary alcohol listed is choice C, so choice C is the best answer.

Example 5.23To carry out the following synthetic transformation, what reagent should beadded?

Q Q O

CH2OH

tY- - &A. LiAlH4B. NaBH4C. H2N=NH2D. FADH2

Solution

In this reaction, an aldehyde has been reduced into a primary alcohol on acompound where an ester functional group is unaffected. FADH2 carries outhydrogenation of a 7c-bond, so choice D is eliminated. Hydrazine, H2NNH2reduces aldehydes and ketones to alkanes, so choice C is eliminated. BothLiAlH4 and NaBH4 reduce aldehydes to primary alcohols, but LiAlH4 wouldalso reduce the ester. This eliminates choice A and makes choice B the bestanswer. NaBH4 is selective for aldehydes and ketones.

Copyright ©byThe Berkeley Review 30 The Berkeley Review

OrganiC ChemiStry Carbonyls and Alcohols Carbonyl Reactivity

Example 5.24What is the oxidation state of aluminum in LiAlH4?A. +3

B. +1

C. -3

D. -5

Solution

The term hydride implies that hydrogen has a negativecharge associated with it.Lithium is +1, so the Al must have an oxidation stateof +3 to have the charge onthe compound be zero. The best answer is choice A.

Example 5.25What is the major organic product for the following reaction?

O O1. NaBH4(thf)

AA -•

2.NH4Cl(aq)ri

A- OH OH B' O OH C O O D- OH O

yX^ sX^J -X^KoH /k/JkoHSolution

Sodium borohydride, NaBHj, reduces aldehydes and ketones only. Both theketone and the aldehyde functional groups are reduced to alcohols (secondaryand primary respectively). This yields a 1,3-diol, so the best answer is choice A.


Organic Chemistry Carbonyls and Alcohols Name Reactions

Mh^Msi^

Name Reactions

There are name reactions such as the Grignard reaction, aldol condensation,Claisen condensation, transesterification, the Wittig reaction, pinacolrearrangement, the iodoform reaction, and the Wolff-Kishner reaction that youshould know. There are other name reactions in organic chemistry, but these arethe name reactions you must know according to the MCAT Student Manual. Wewill address each reaction according to their frequency on previous MCATexams. Where appropriate, we will emphasize the reaction mechanism or itsregioselectivity and stereoselectivity. Of most importance to the MCAT, we shalladdress any biological applications of the reactions or analogous reactions.

Grignard ReactionGrignard reactions involve the addition of a carbanion (carbanion metal halide)to an electrophilic carbon center in a carbonyl to form a new carbon-carbon bond.When alkyl magnesium bromide, the Grignard nucleophile, attacks the carbonylcarbon, a tetrahedral intermediate forms. Almost all Grignard reactions generatealcohols. No matter what the reactant, Grignard reactions generate a hydroxylgroup in the product where the carbonyl group originated. Figure 5 -30 showsthree such Grignard reactions, the third of which is a double addition to an ester(an acid anhydride or acid halide yield the same alcohol product as an ester).

Grignard Reaction

No L.G. .*. one alkyl group adds: one C —• two C (probably chiral)

5L.t H |R'MgBr

Aldehyde

BrMg+ "O R'

RX

H

work-up^

HO R'

2° Alcohol

No L.G. .-. one alkyl group adds: two C —• three C (possibly chiral)

3fO BrMg+"0 R* HO R'

X ♦ P'MgBr —»- X !5:2£!^ XR t R ] R R R R

Ketone 3° Alcohol

Has a L.G. .-. two alkyl groups add: one C —• three C (never chiral)

O BrMg+;0 OR" O

Ester

(or anhydride HQ R,or acid halide)

Copyright © by The BerkeleyReview

x-R R'

3° Alcohol

32

work-up

Figure 5-30

The Berkeley Review

Organic Chemistry carbonyls and aicohois Name Reactions

Grignard nucleophiles add once to aldehydes, which have one R-group, to formsecondary alcohols. Ketones have two R-groups, so they form tertiary alcohols.Carbonyls with a leavinggroup (esters, acid anhydrides, and acidhalides) startwith one R-group, but because they have a leaving group, they add a new R-group twice. Because the Grignard reagent adds twice to esters, acid halides andacidanhydrides, the products are achiraltertiaryalcohols. Grignardreagentsarestrongbases,so it is criticalthat no protic hydrogens are present. Thisis why thereaction uses anhydrous ether solvent. The second step in the Grignard reactionrequires a weak acid, as opposed to a strong acid, to avoid protonation of thehydroxyl group, which can lead to an Ei elimination reaction. The Grignardreaction can also take place with carbon dioxide to form a carboxylicacid.

Example 5.26When 2-butanone is added to 1-propylmagnesiumbromide in ether then workedup with ammonium chloride in water, it yields a major organic product of:

A. 2-propyl-2-butanol.B. 3-ethyl-2-pentanol.C. 3-methyl-3-hexanol.D. 3-bromo-2-butanone.

Solution

The Grignard reagent attacks the carbonyl carbon to form a new carbon-carbonbond. The reactant is a ketone, so the product is a tertiary alcohol, eliminatingchoice D. The IUPAC name in choice A is not possible, because a 2-propyl groupwould actually be part of the longest chain. The reaction is drawn below:

fO BrMgO CH2CH2CH3 _ HO 4CH2CH2CH3Q +H3CH2CH2CMgBr_^ W ^L >' ,:Xf^CH2CH3 J H3C CH2CH3 Yiff ^CH^Hgh3c4:

The longest chain in the product is six carbons, so the product is hexanol. TheIUPAC name is 3-methyl-3-hexanol, so choice C is the best answer.

Example 5.27Ethyl magnesium bromide, when added to 3-pentanone, yields which of thefollowing organic products following weak acid workup?

A. Triethyl etherB. 3-Ethyl-3-pentanoneC. 3-Ethyl-3-pentanolD. 3-Heptanol

Solution

The reactant is a ketone, so the product is a tertiary alcohol. This eliminateschoices A and B. The reaction is drawn below:

jTO BrMgO CH2CH, MJ HO 4CH2CH3G +H3CH2CMSBr_ ^ »^ >c/' ^H3CH2C { CH2CH3 J H3CH2C CH2CH3 H3CH2C CH2CH3

The longest chain in the product is five carbons, so the product is pentanol. TheIUPAC name is 3-ethyl-3-pentanol, so choice C is the best answer.

Copyright©by The Berkeley Review 33 Exclusive MCAT Preparation


Aldol Condensation

Aldol reactions start with deprotonation of an alpha hydrogen from a carbonyl(ketone or aldehyde). The pKa of most a-hydrogens that are adjacent to only onecarbonyl is between 17 and 19. This means that they are weakly acidic and canbe removed by using a strong base. The resulting anionic compound (enolate)nucleophilically attacks a neutral carbonyl compound (electrophile) to form a fi-hydroxy carbonyl compound. This species is rarely isolated, because under thereaction conditions, it undergoes elimination to yield an a,6-unsaturatedcarbonyl. Regioselectivityis involved, because the final product can have eitherE or Z structural geometry. The anionic species can also be protOnated on theoxygen, resulting in the formation of an enol, but that is in equilibrium with thecarbonyl. Conversion from a ketone (or aldehyde) into an enol is referred to astautomerization, which was shown in Figure 5-19. Figure 5-31 shows the aldolcondensation reaction of acetone.

:o:

X

Aldol Condensation

:o:

H,C

base w

X.CH2

ratio /•i H3cvsl^CH3 :o:

Lr " ^ H,C

CH3 withOH", H3Cketone-to-anion

is about 1000

A ,„,. X «& <o:CH elimination HiC

A "~H,C

H3C

a,6-Unsaturatedketone

CH3:oh

• •

6-Hydroxyketone

CH,

CH,

Figure 5-31

It is ofteneasier to predict a reaction by focusing on the side product. In an aldolcondensation that goes all the way to an a,fi-unsaturated ketone, water is the sideproduct. Finding the atoms in the reactants that make up the water makespredicting the product easy. Figure 5-32 shows how to align two acetonemolecules for the aldol condensation.

O

HO H


After aligning the molecules toorient the water, connect thecarbon atoms with a double bond

Figure 5-32

34

O

CH

HoC CH,

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The aldol condensation offers the potential for thermodynamic control versuskinetic control. When the ketone is asymmetric, then the choice of whichalphahydrogen is deprotonated is dictated by reaction conditions. The reactionconditions that are most significant are steric hindrance of the transition state andthermodynamic stability of the intermediates and product. When the reactionyields a major product that was selected to minimize steric hindrance in thetransition state, this product is said to be the kinetic product. This term refers tothe lower activation energy required to go through a less sterically hinderedtransition state. When the reaction yields a major product that was selected tomaximize stability of the intermediates or product, this product is said to be thethermodynamic product. This term refers to the greater amount of free energyreleased when forming the more stable compound. Figure 5-33shows the kineticand thermodynamic products for the aldol condensation of butanone.

Kinetic Control

less steric hindrance

lower transition state

(lower Eactivation)prefered at low temp

with strong, bulky base

-.A •CH3

A AHki Hki Hth Hth

Thermodynamic Controlmore substituted intermediate

greater energy released(more negative AG rxn)prefered at high tempwith strong, small base

OKinetic Control

O O

Name Reactions

H..AC CH.CH

•/-V2v_ii3

XC CHoCH.2v.±±3

+ X"C CH2C

3HOH|

H3CH2C CH3

t-butOK jt-butOH, -78°

H3CH2C CH3major

OThermodynamic Control

O

AH3C C

/CH3

( r + i

HO H

H3C CH2CH3

KH^50°C

A.H,C C

.xCH3

XH3C CH2CH3major

Figure 5-33

Example 5.28The following reaction is best described as:

O

A/1. Narfthf)

2. workup

O

A. a Claisen condensation via the kinetic enolate.

B. a Claisen condensation via the thermodynamic enolate.C. an aldol condensation via the kinetic enolate.

D. an aldol condensation via the thermodynamic enolate.

Copyright © by The Berkeley Review 35

H3C CH2CH3

minor

O

AH3C ^Cx-CH3

AH3CH2C CH3

mmor

Exclusive MCAT Preparation


EtO

:o:

X

Solution

The reaction involves the condensation of an asymmetric ketone upon itself,which eliminates choicesA and B,because the Claisen condensation requires thereaction of esters with esters, not aldehydes or ketones. Because the morehindered alpha carbon is involved in the addition, the reaction must have beencarried out by way of the thermodynamic enolate (more substituted enolate).This makes choice D the best answer.

When a reaction has a biological example, its likelihood of appearing on theMCAT is increased. The biological application of the aldol reaction is the fourthstep of glycolysis, which is a retro-aldol reaction. In the retro-aldol step, a six-carbon fi-hydroxyketone breaks into two three-carbonfragments. The two three-carbon species, after neutralization, are dihydroxyacetone phosphate (DHAP)and glyceraldehyde-3-phosphate (G-3-P). It is not necessary to know themechanism, but you should know that the enzyme responsible for this step isaldolase. Figure5-34 shows the fourth step of glycolysis.

HO

H

H

CH2C©

HeQ.-HOH

CH20©fi-hydroxyketone

Step IV Glycolysis

CH2C(P)

JJ HAldolase^

HO-

H-

H-

O-H

OH

CH2C©

Figure 5-34

•=>

•=>

CH20©

h°CH2OH

DHAP

HV^°

H1 OH

CH2C©G-3-P

Claisen Condensation Reaction

TheClaisen condensation reaction is essentially the aldol reaction with an ester.Esters have a leaving group, sothe product isnotana,fi-unsaturated carbonyl, asis seen with aldol condensation reactions. Figure 5-35 shows the Claisencondensation reaction. Notethat withoutacidic workup, the carbonbetween thetwo carbonyl species wouldbe deprotonatedby the ethoxidein solution.

Claisen Condensation

EtOW :Q: :Q: :°:EtOH

CH, EtO EtO

OEt

The base ischosen tomatch H3Cthe leaving group so that atranesterification reactionwon't generate a new ester.

OEt

Copyright©by The Berkeley Review 36

6-Ketoester

The 6-ketoester exists in equilibriumwith the deprotonated species untilthe solution is neutralized.

Figure 5-35

The Berkeley Review


Thefi-ketoester product has a pair of hydrogens that are now conjugated to twocarbonyl groups. As such, their acidity is enhanced, so the sodium ethoxidedeprotonates the fi-ketoester product preferentially over the ester reactant. Onceneutralized, a fi-ketoester is formed. A fi-ketoester when treated with strong acidin water can hydrolyze to form a fi-ketoacid, which when heated readilyundergoes decarboxylation into a ketone. This ultimately leads to a ketone witha longer alkyl chain than the starting ester.

Fatty acid synthesis is a biological example of a Claisen condensation reaction,where the ester is actually a thioester. Synthesis of a fatty acid involves Claisencondensation of acetyl-CoA onto another thioester. The chain grows two carbonsat a time as shown in Figure 5-36.

Fatty Acid Synthesis (Claisen Condensation with a Thiosester)

:o: :o: :o: :o:

Name Reactions

CoAS

it enzyme

^chT iCoAS

The enzyme acts as abase by deprotonatingthe thioester.

a, -CH2

H3C^1^ SCoA

Aup--

CoAS CoAS^CH2

:o:

A

H3C

H

H

:o:

A

HoC H

:o:

A

H3C

H

fi-Ketothioester

^^ reductionhydrogenation y*^ dehydration /\ NADH ,

CoAS CH2 FADH2 CoAS CH (elimination) CoAS CH2 ™*unJJ— || — -^

OH

Figure 5-36

Transesterification

Transesterification involves the exchange of one alkoxy group for another on anester. Because the equiUbrium constant is roughly one, the reaction is driven byLe Chatelier's Principle. Either a product is removed or a reactant is added. Anacid-catalyzed transesterification reaction is shown in Figure 5-37.

O

AR OR'

+ HOR" ^cat. H+

HOR'

O

OR"

Figure 5-37

Transesterification can be carried out in either acid or base. When acid is

employed, the mechanism is a typical acid-catalyzed process. In base, thealkoxide nucleophile causes the reactivity. If the base is hydroxide, a carboxylateand alcohol,are formed as the two products. When a fatty acid triglyceride istreated with strong base, glycerol and fatty acid carboxylates are formed from ahydrolysis reaction. Figure 5-38 shows the acid-catalyzed mechanism fortransesterification.



/^:o:

protonate

Transesterification mechanism

H+ ^:o:

•x+ R"OH

OR'

OR'

OR'

make

HO—R"

H

HO.^C^—R"

X OR'

deprotonate

R'OH

OR"

deprotonate

OR"

The biological applications of transesterification involves membrane transportand the conversion offatty acids into triglycerides. Fattyacidsare found in cellmembranes, where they are part of a phospholipid (comprised of glycerol, twofatty acids, and a phosphate group). When glycerol binds three fatty acids,triglycerides are formed. Fatty acids thathave been esterihed withglycerol arebroken down in basic water in what is known as saponification, the basedcatalyzed cleavage oftheester bonds ofthe triglyceride, as shown in Figure 5-39.

Saponification

Copyright© by The Berkeley Review

O

I-O^RO^RO R

38

OH-.

OH

OH

— OH

Figure 5-39

O

+ 3

'O R

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OrgaillC ChCmiStiy Carbonyls and Alcohols Name Reactions

Wittig reactionAnother useful carbonyl reaction for synthesis is the Wittig reaction. The Wittigreaction essentially undoes an ozonolysis reaction. It is yet another way to makea carbon-carbon bond. The Wittig reaction converts a carbonyl into an alkene,with stereoselectivity. You can get either the E or Z product depending whatreagents you choose. The double bond you form can be added to by electrophilicaddition reactions, with high stereoselectivity. What this means is that there arenow synthetic routes to get high yields for stereoselective synthesis. We aregoing to see it in its most simplistic form however. It goes through a four-membered intermediate, similar to the hydroborane addition reaction with analkene. Figure 5-40 shows the synthesis of a phosphonium ylide (the reactivespecies in a Wittig reaction).

Q Q Q

& & dFigure 5-40

It is the final product here that is the reactive species in a Wittig reaction. Thereis a resonance form which has a double bond between the carbon and

phosphorus. You should recall that the phosphorus is able to make more thanfour bonds due to the availability of 3d-orbitals. The overlap between a 3p-orbital (from P) and a 2p-orbital (from C) is not as strong as the overlap betweentwo 2p-orbitals however, so the rc-bond between phosphorus and carbon is notthat strong. Figure 5-41 shows the Wittig reaction and various intermediates inthe mechanism in an aprotic environment.

Wittig reaction

\__/~p=ch2 + 0=\ —• V-y-p=° + H2C=\

~d R ~d

Figure 5-41

In an aprotic environment, the transition state is a four-centered transition state.The phosphonium ylide 03P=CH2 can be synthesized from 03P: and H3CBrunder basicconditions, as shown in Figure 5-40. The Wittigreactionprefers theE-alkene product whenever there is asymmetry in the reactants.



Pinacol RearrangementThe pinacol rearrangementis a topic in the MCAT Student Manual, so we shallcover it. It is a classic example of rearrangement with some applications insynthesis. The pinacolrearrangement converts a vicinal diol into a ketone. Thereaction proceeds withan alkyl migration, which converts a tertiarycarbocationinto a resonance stabilized carbocation. Figure 5-42 shows the pinacolrearrangement of pinacol into pinacolone.

H4

_/H3C*7C V

H,C CH,

//// protonate

JCH3

K\break

Pinacol rearrangement

ho:

\

ho:

\, c

H3C-1/H3C

CH*

rOH,

ho:

\ ©c J?L.^CH3^*CH3

H,C

cone.

H2S04

rearrange

:o

^ /CH,

C

/ \"'"CHH3C

deprotonate

hô/

CH,

\CH,

^ /C C-.

/ \HoC

HON

y©c.

/H3C

i

*""/ CHo

CH3

make

,"%/

•c.

V"CH3CHo

Figure 5-42

Iodoform Reaction

The iodoform reactionfalls in a general class of reactions known as the haloformreactions, whichconverta methyl ketone into a carboxylic acid and a haloform,HCX3. The iodoform reaction is used as a chemical test for a methyl ketone,because iodoform isyellow in color and insoluble in water. Methyl ketones reactwith iodine, I2, in the presence of base to generate a carboxylic acid and adeprotonated iodoform molecule. The carboxylic acid is deprotonated to yield acarboxylate and iodoform. The iodoform reaction is shownin Figure 5-43.

Iodoform reaction

R

R

O

II

O

IIX,


CH,

+ HCl, =

o

40

R

O

II

O

IIX.

CH,

R OH

Figure 5-43

R

O

IIX,

CH,

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Organic ChCmiStiy Carbonyls and Alcohols Name Reactions

Wolff-Kishner Reduction

There are three common methods for reducing a ketone or aldehyde into analkane. There is the Clemmensen reduction, which is carried out under acidicconditions using HCl and Zn(Hg). There is the Wolff-Kishner reduction, which iscarried out under basic conditions using hydrazine (H2NNH2) and basicworkup. And last there is the treatment of the carbonyl with a 1,2 dithiol(HSCH2CH2SH) followed by treatment with Raney nickel (fine nickel in thepresence of aluminum oxide). The good thing about having three reactions isthat each is carried out at different pH values. This allows for many differentcarbonyl compounds to be reduced independent of any pH sensitive functionalgroups elsewhere on the molecule. For the MCAT, it is important to know themechanism for the Wolff-Kishner reaction. The Clemmensen reaction

mechanism is not thoroughly understood, so for the Clemmensen you really onlyneed to know that most transition metals have a positive oxidation potential andare therefore strong reducing agents. When added in their zero oxidation state,metals donate electrons to the organic molecule, causing the molecule to getreduced. Figure 5-44shows the three reactions carried out on cyclopentanone.

HS^^-X

Wolff-Kishner reduction

H2NXH H

H2Q/OH^ ( >) + N:

Raney nickel reduction

Raney Ni^ /\ +HSCH2CH2SHcat. HCl \ / EtOH

Clemmensen reduction

H H

HCl/Zn(Hg)^ X \ + H2Q + ZnCl2

Figure 5-44

You must apply the right reaction under the right conditions, which depends onthe presence of other functional groups on the molecule. The Wolff-Kishnerreaction is carried out in two steps. The first step of the Wolff-Kishner reactioninvolves the substitution of hydrazine for the carbonyloxygen, resulting in theformation of a hydrazone. Treating the hydrazone with strong base in watergenerates nitrogen gas. The first step of the reaction is reversible with theaddition of water and acid. That is to say that you can regeneratethe carbonylbyadding water and a catalytic amount of acid to the hydrazine derivative.Carbonyls can also react with secondary imines, but because there is only oneproton to lose from the amine, the reaction generatesan enamine rather than animine. We shall consider such reactions in the nitrogen compound section.Figure 5-45shows the mechanism for Wolff-Kishnerreduction.



Step 1:H2NL

•£• H H NH > '. H/Cx + :n-n: -*- c^ + h2?.

R R H H R R

Step 2:

A

N ^HIIcL R R

R/NR

t

..©:oh

H O*.

© ^-H

H H

X

t

N H ••©:n=n: + X. \

I N *-^ H N^ ®C© ^=^ | ^=^ /|

V /> R I R R | R^•H-^ H H

Figure 5-45

R/SRR | R

H


Organic Chemistry Carbonyls and Alcohols Synthetic Logic

Synthetic Logic uW*

Synthetic LogicSynthesis is a recurring topic throughout organic chemistry. The method fortestingsynthesisin most lecturecoursesis not feasible on the MCAT, becausetheMCAT is a multiple choice exam. However, synthesis appears on the MCAT.Syntheticpathways are shown in a step-wise road-map fashion similar to how apathwayappearsin biochemistry. Thepassage willpresentthesteps, which mayor may not include the reagents. Questions can focus on stereochemicalconsequences, reaction types,and the logic behinda step in the overall synthesis.The strategy behind a synthesis is based on the idea that it is easier to solve amaze backwards than forwards. The steps to analyze a synthetic pathway are:

1) Identify the new bonds (and functional groups) formedby counting theatoms and noting the changes in bonds from reactant to product.

2) Breakthe product into fragments that match the reactantskeleton.

3) Reconnect the fragments with the proper chemicalreagents.

Let'sconsider the conversion of 1-propanol into 2-pentanol. Figures5-46 through5-48 show the stepwise analysis of the synthesis and retrosynthetic analysis.

1) A five carbon secondary alcohol is formed from a three carbon primaryalcohol, so two carbons must be added to the reactant.

OH3 carbon ^ _ I 5 carbon1° alcohol ^^ OH ^ >^ 2° alcohol

Figure 5-46

2) Using retrosynthetic logic, the product is broken into two fragments.The fragment with the oxygen must have comefrom the carbonyl.

OH OH

and

retron ' retronO

Synthesisrequires: <A

nucleophile carbonyl electrophile

Figure 5-47

Thisreactioncalls for a three carbon nucleophile adding to a two carbon carbonyl(electrophile). The more oxidized fragment (oxygen containing fragment, orright fragment in this case) comes from the carbonyl. Theoxygen is bonded tothe first carbon, so it requires an aldehyde. The rest of the problem requiressynthesizing the Grignard reagent from 1-propanol and choosing the solvents.

3) This Grignard reaction requires propyl magnesium bromide and ethanal.

3. O

xl.PBr3(thf) v. ^s. HoC^^HOH 2.Mg(s)(thfT MgBr 4.Wl(«D*

Figure 5-48

OH



Protecting GroupsProtecting groups are used when a reactant has multiple reactive sites, but youonly wish to react atone of them. You must determine which functional groupscan react under the reaction conditions. In cases where more than one site canreact, the functional group you wish to not have react is converted into a lessreactive functional group. Aldehydes are converted into acetals, ketones areconverted into ketals, andalcohols areconverted into silyl ethers. You mustbeable to remove the protecting group at the conclusion of the reaction withoutinvoking extreme conditions which may cause the molecule to react. Figure 5-49shows the conversion of an alcohol into a silyl ether and then back into analcohol. The ketal protecting group has already been shown inFigure 5-11.

Protection and deprotection of an alcohol

HO H

XR R1

Alcohol

(H3C)3SiCâddition of

protecting group

(H3C)3SiO H

XR R'

Protected alcohol

(silyl ether)

NaForH,0+2—•

removal of

protecting group

HO H

XAlcohol

Figure 5-49

Reactions of Acetoacetic Ester

Acetoacetic ester has a carbon that is alpha to two carbonyl groups. As aconsequence, itsprotons are more acidic than typical alpha protons. The pKa ofthe protons alpha to both carbonyl groups inacetoacetic ester isroughly 11. As aconsequence, it canbe deprotonated in mildly basic conditions, resulting in acarbanion. The carbanion is astrong nucleophile, capable ofdisplacing a leavinggroupto form a new carbon-carbon bond. Figure 5-50 shows the addition ofanalkyl group to acetoacetic ester.

R

O O

XXOEt

1.NaOEt (HOEt)^2. R'—X *

Figure 5-50

o o

R OEt

R'

The base of choice matches the alkoxy group of the ester, so that anytransesterification reaction that takes place does notresult in any change. Thesolvent is the conjugate acid of the alkoxy group, an alcohol, because it is beingformed by the reaction anyway. Thereactantis a fi-keto ester, which is formedfrom a Claisen condensation reaction of two esters. The product can be treatedwith acidic water and heat to hydrolyze the ester into a fi-ketoacid, which canundergo decarboxylation into aketone ifthe temperature ishigh enough.



Reactions of Malonic Ester

Malonic ester, like acetoacetic ester, has a carbon that is alpha to two carbonylgroups. As a consequence, the same reactions carried out with acetoacetic estercan also be carried out with malonic ester. When malonic ester undergoes thereaction shown in Figure 5-50, the same addition takes place, although theproduct is a diester rather than a ketoester. The significant difference betweenacetoacetic acid and malonic ester is that a diester has two functional groups thatcan undergo substitution, resulting in cyclization. Figure 5-51 shows theaddition of an alkyl group to malonic ester followed by the addition of a diamineto form a six-membered ring.

O O

.XX 1. NaOEt (HOEt)

EtO ^ OEt EtO T OEt

Figure 5-51

Barbiturates form when urea is used in place of the diamine in the second step.i

DecarboxylationAs mentioned with the acetoacetic ester synthetic pathway, a fi-ketoacid canundergo decarboxylation when heated. This occurs readily, because theintramolecular hydrogen bonding of a 6-ketoacid aligns the molecule for a shiftin the rc-electrons. Figure 5-52 shows decarboxylation, the six-membered ringcreated by hydrogen bonding, and the decarboxylation products formed fromthat conformation.

Decarboxylation

O O O

o o

*• cA II

O^ ' 'R o ^ 'R H3C "RIntramolecular Carbon dioxide Enol to ketone

H-bonding in 6-ketoacid and an enol form (tautomerization)

jX« ».cX-

Figure 5-52

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OrCJcUllC ChemiStry CarbonylsandAlcohols Carbonyl Biochemistry

Carbonyl BiochemistryBiological Oxidation-ReductionThere is a great deal ofoxidation-reduction chemistry in biology. Using anorganic chemistry perspective can simplify many biological reactions. As ageneral rule, you should know that anabolism is a reductive process and theresult is the build up ofa molecule. Examples ofanaboHsm include fatty acidsynthesis and gluconeogenesis. Youshould know that catabolism is an oxidativeprocess and theresult is the breakdown ofa molecule. Examples of catabolisminclude glycolysis and fi-oxidation. Typically, catabolism releases energy whileanabolism requires, and thereby stores, energy. You should know that in us,oxidations tend to take place in the mitochondrial matrix while reductions tendto take place inthe cytoplasm, with glycolysis being a notable exception.

Example 5.29Which of the following cofactors or coenzymes is most likely required in thefollowing step ofglycolysis?

*^H °^Ô PO,2"

H- OH GAPDHÔH

H2C— O— PO,2- HoC— O— PO02A. Biotin

B. AMP

C. Zn

D. NAD+

Solution

In this reaction, carbon 1 gains a bond to oxygen, so it has been oxidized. Thismeans that it must be coupled with a reduction half-reaction. The enzymeGAPDH does not undergo reduction, so the cofactor/coenzyme must undergoreduction. Of the choices, only NAD+ (a species poor inH) isan oxidizing agentthat undergoes reduction. Choice D is the best answer.

Example 5.30Inhumans, where does fatty acid synthesis take place?A. CytoplasmB. Rough endoplasmic reticulumC. Mitochondrial matrix

D. Mitochondrial inner membrane

Solution

Fatty acid synthesis involves molecular build up, so it is a reductive process.Reductions occur mostly in thecytoplasm, so choice A is the bestanswer.

There are several examples of oxidation-reduction chemistry in biology, so weshall only focus on a few here and a few more with carbohydrate chemistry insection 6. We shall consider oxidative decarboxylation, which shows itself inaerobic and anaerobic metabolism. Figure 5-53 shows the decarboxylation ofpyruvate, a three-carbon species, into acetyl coenzyme A, where the acetylmoiety contains two carbons.


Organic Chemistry Carbonyls and Alcohols Carbonyl Biochemistry

°%/°

o=

CH3

PyruvateNAD+ NADH + H+

o=c=o

Carbon dioxide

o. . .S—CoA

ICH3

Acetyl CoA

Figure 5-53

Biochemical ReagentsFor biological redox chemistry, there are two common reversible reactions. Thefirst reaction is the inter-conversion between NAD+ and NADH, and the secondreaction is the inter-conversion between FAD and FADH2. The two moleculesand their reduction reactions are shown over the next few pages. Figure 5-54shows NAD+ (the oxidized form), which acts as an oxidizing agent inbiochemical reactions, and NADH (the reduced form), which acts as a reducingagent in biochemical reactions.

Nicotinamide

ONH2

O—P—O-CH

OHO OH

NH2

0-P— 0-CH2

1HO OH

NAD+(Nicotinamide adenine dinucleotide)

Figure 5-54

HO OH

NADH

(Nicotinamide adenine dinucleotide hydride)

NAD+serves as an oxidizing agent by picking up a hydride anion on its pyridinering to yield NADH. The species being oxidized gives off H+ as a side product.This biological redox reaction plays a major role in many cycles includingglycolysis. NADH carries out the reduction of aldehydes and ketones, much likeNaBH4 does in organic chemistry reductions. A second biological reducingagent is flavin adenine dinucleotide dihydride (FADH2), which carries outhydrogenation of rc-bonds, much like H2 and a catalyst such as palladium.Figure 5-55 shows the oxidized form, FAD, and the reduced form, FADH2.


Organic Chemistry

H3CmtH3C"^^^^N'Flavin CH

H-

H-

H-

OH

OH

OH

Carbonyls and Alcohols Carbonyl Biochemistry

CH2I

OI

-o-p=oI

O

AdenineNH2

O-p—O—CH2II liu-o^jio kj l?

H| [HHO OH J

FAD (Flavinadenine dinucleotide)

Figure 5-55

H | | HHO OH

FADH2 (Reduced form)

FAD serves as an oxidizing agent by picking up a hydrogen on the nitrogen ofthe central pyrimidine ring and a second hydrogen on the imine nitrogen of thethird ring, to yield FADH2. One hydrogen isgained asahydride, while the otheris gained as aproton, so there is no change inthe charge of the compound.

Acetyl coenzyme Ais also very common in biochemistry, so we shall considerthe acylation process ofcoenzyme A. The acetyl group is transferred from areactant toa thiol when acetyl CoA forms. The structure ofacetyl coenzyme Aisshown in Figure 5-56. The biochemical role ofcoenzyme A is in transfer ofanacetylgroup to a biological species.

H

N-

O

N-

H

Copyright ©by TheBerkeley Review

Vr

CH,

Nucleotide

NH2 *\Adenine I

H CH3 O O

II II

N'

Dc>NT NO.

CH,0 — P— O— P— O- CH,

O OH CH3

0 a Jm!

48

Acetyl Coenzyme A

Figure 5-56

HO OH

The Berkeley Review

Organic Chemistry Carbonyls and Alcohols Section Summary

Key Points for Carbonyls and Alcohols (Section 5)

Oxygen Containing Compounds

1. Alcohols (Compounds with hydroxyl groups)a) They have boiling points higher than alkanes of roughly equal mass

i. Boilingpoints of alcohols are: primary > secondary > tertiaryii. Water solubility is high for short chain alcohols

b) The hydroxyl group can be converted into a better leaving groupi. SOCI2 and PBr3 convert an alcohol into an alkyl halideii. They are mesylated or tosylated to form a good leaving group

c) They have broad IR absorbances between 3300 and 3600 cm"12. Aldehydes and Ketones (Carbonyl compounds with H or R-groups attached)

a) They have boiling points higher than alkanes of roughly equal massi. Boiling points decrease with branchingii. Water solubility is high for carbonyls up to four carbons

b) They readily form acetals and ketals with alcohol in acid and formhemiacetals and hemiketals with alcohols in base

c) Aldehydes have strong IR absorbances around 1725 cm"1 and ketoneshave strong IR absorbances around 1710 cm"1

3. Carboxylic Acids(Compounds with a COOH groupat an end of the chain)a) Theyhave meltingand boilingpointssignificantly higherthan alkanes of

roughly equal mass because they form strong H-bonds.i. They have pKa values in the 2-5 rangeii. They represent the end point of oxidation for most molecules

b) They form more reactive derivativesby converting the hydroxyl groupinto a better leaving groupi. SOCI2 converts a carboxylic acid into an acyl chloride, heat converts

two carboxylic acids into an acid anhydride, and an alcohol in acidconverts a carboxylic acid into an ester

ii. Carboxylic acids can be made by complete oxidation of a primaryalcohol or aldehyde or by hydrolysis of acid derivatives such asesters, acid anhydrides, acid halides, amides, and nitriles

4. Carboxylic Acid Derivatives (Compounds with a carbonyl group and afunctionalgroup other than OH at an end of the carbonchain)a) Esters(OHgroup of a carboxylic acid is replaced by an ORgroup)b) Lactones (Cyclic esters)c) Acid Anhydrides (forms fromthe dehydration of twocarboxylic acids)d) AcidHalides (OHgroup of a carboxylic acid is replaced by a halide)e) Amide (OHgroup of a carboxylic acid is replaced by an amine)

Carbonyl Reactivity

1. Carbonyl Compounds do Three Types of Reactionsa) They are electrophiles that can be attacked at the carbonyl carbon by a

nucleophile, forming a tetrahedral intermediateb) Strong bases deprotonate them at the oc-carbon to form a nucleophilic Cc) They undergo oxidation-reduction chemistry

i. Oxidation: gain of bonds to O and/or loss of bonds to Hii. Reduction: loss of bonds to O and/or gain of bonds to Hiii. Oxidizing agents are rich in O and reducing agents are rich in H


OrgaillC Chemistry Carbonyls and Alcohols Section Summary

Name Reactions

1. For the reactions in the MG4T Student Manual, know the basic reactiona) Grignard Reaction (addition of an alkyl magnesium halide toacarbonyl)

i. Adds one R-group toaldehydes andketones, forming alcoholsii. Adds twice tocarbonyls with a leaving group, forming a 3° alcohol

b) Aldol Condensation (deprotonation ofthe a-carbon ofanaldehyde oraketone and the subsequent addition ofthe anion to the carbonyl carbonof either a secondaldehyde or ketone)i. Forms ana,fi-unsaturated ketone orana,fi-unsaturated aldehydeii. Exhibits kinetic versus thermodynamic preference with ketonesiii. Has a biological application in thefourth stepofglycolysis

c) Claisen Condensation (deprotonation of the a-carbon of an ester and theaddition oftheanion to thecarbonyl carbon ofa second ester)i. Forms a fi-ketoester

ii. Hasa biological application in fatty acid synthesisd) Transesterification (conversion of one ester into another by exchanging

thealkoxy group, in eitheracidic or basic conditions)e) Wittig Reaction (conversion ofa ketone into analkene via 03P=CR2)f) Pinacol Rearrangement (conversion ofavicinal diol into aketone inacid)g) Iodoform Reaction (conversion of amethyl ketone into acarboxylic acid)h) Wolff-Kishner Reduction (conversion ofa ketone into analkane inbase)

Synthesis

1. Synthesis Thought: 1) changes, 2) retrosynthesis, and 3) choosing reagentsa) Retrosynthesis (product-to-reactant analysis of changes inacompound)b) Protecting groups are used on reactants with multiple functional groupsc) -OH group isprotected as-OTMS and C=0group isprotected asC(OR)2

2. Reactions ofAcetoacetic Ester (Alpha-proton has apKa around 11)a) Great nucleophile when deprotonated, soit readily attacks C=0b) When treated with acid water and heat, ithydrolyzes and decarboxylates

3. Reactions of Malonic Ester (Alpha-proton has a pKa around 11)a) Readily adds anR-group at the methylene carbonb) Can cyclize with areactant with two nucleophilic sites (i.e., barbiturates)

4. Decarboxylation (observed with fi-keto acids or 1,3 diacids)a) Proceeds through asix-membered ring held together byanH-bondb) Forms anenol which equilibrates with the carbonyl compound

Carbonyl Biochemistry

1. Oxidation andReduction in Biochemistrya) Catabolism (oxidative breakdown ofbiological molecules)

i. Occurs mostly in the mitochondrial matrix (i.e., fi-oxidation)b) Anabolism (reductive build up ofbiological molecules)

i. Occurs mostly in thecytoplasm (i.e., gluconeogenesis)2. Biochemical Reagents

a) NAD+/NADH area redox pair whereNADH is the reducedformb) FAD/FADH2 area redox pair whereFADH.2 is the reducedformc) Acetyl Coenzyme A(Picks up acetyl group toform thioester)


Carbonylsand

Alcohols

Passages14 Passages

100 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, IV, V, X, & XIGrade passages immediately after completion and log your mistakes.

II: Following Task I: Passages II, VI, IX, & XIII (26 questions in 34 minutes)Time yourself accurately, grade your answers, and review mistakes.

Ill: Review: Passages III, VII, VIII, XII, XIV, & Questions 95-100Focus on reviewing the concepts. Do not worry about timing.

R-E-V-I-E-W


I.

II.

III.

IV.

V.

VI.

VII.

VIII.

IX.

X.

XI.

XII.

XIII.

XIV.

Alcohols, Acidity, and nucleophilicity

Alcohol Reactions

Acetals and Ketals

Carbonyl Reactivity Study

Thermodynamic versus Kinetic Control

Alcohol Oxidation

Unknown OxygenContaining Compounds

Wolff-Kishner versus Clemmensen Reduction

Grignard Reaction

Aldol Condensation and Alpha Hydrogen

Claisen Condensation Reaction

Transesterification Reaction

Malonic Ester Synthesis

Carboxylic Acids

Questions not Based on a Descriptive Passage

Carbonyls and Alcohols Scoring Scale

Raw Score MCAT Score

83 - 100 13- 15

65-82 10-12

46-64 7-9

33-45 4-6

1 -32 1 -3

(1 -6)

(7 - 12)

(13- 18)

(19-25)

(26 - 32)

(33 - 39)

(40 - 46)

(47 - 53)

(54 - 60)

(61 -67)

(68 - 73)

(74 - 80)

(81 -87)

(88 - 94)

(95- 100)

Passage I (Questions 1-6)

Alcohols are a common protic solvent used in manyorganic reactions. Alcohols have organic properties (due tothe alkyl chain), and they exhibit hydrogen bonding (nottraditionally thought of as an organic property). Low-weightalcohols are the only organic solvents that can be used tosimulate aqueous conditions while dissolving organiccompounds. They are not an ideal solvent, because they canreact as a nucleophile, as an acid, or as a base. Alcohols arechosen in many reactions involving esters as a reactant,where the ester leaving group is the conjugate base of thealcohol. This is done to eliminate the chance for a

transesterification reaction.

In some reactions, alcohols serve as both the reactant andthe solvent. For instance, when an alkoxide is used as a base

(as it is in elimination reactions), it is convenient to use theconjugate acid of the alkoxide as the solvent. This is done,because the conjugate acid (an alcohol) is formed when thealkoxide is protonated. One drawback to alcohol solvents istheir relatively high boiling point, due to hydrogen bonding.When the solvent has a high boiling point, it is hard todistill the solvent away to isolate the product mixture. Oforganic compounds with equal carbons (and therefore roughlyequal molecular mass), alcohols and carboxylic acidsgenerally have the highest boiling and melting points.

1. The pKa of methanol (H3COH) is 15.5, while the pKaof phenol (H5C6OH) is 10.0. Which of the followingstatements are true?

I. Phenol is less acidic than methanol due to the

electron donation from the benzene ring throughresonance.

n.

m.

Phenol is more acidic than methanol due to

electron withdraw by the benzene ring throughresonance.

Methanol is more acidic than phenol, because ofelectron withdraw through the inductive effect.

IV. Methanol is less acidic than phenol, because ofelectron donation through the inductive effect

A. I and IV onlyB. II and m onlyC. I and m onlyD. II and IV only

2. Which of the following bases will deprotonatemethanol completely?

A. H3CC02Na

B. NH3

C. NaOH

D. H3CCH2Li

Copyright © by TheBerkeley Review® 53

Which of the alkoxides is the BEST nucleophile whenundergoing a substitution reaction with methylbromide?

A. H3CCH2ONa

B. H3CCCl2ONa

C. H5C6ONa

D. para-ClHUCgONa

Which of the following phenols is the strongest acid?

A. para-ClC6H40H

B. para-02NC6H40H

C. para-H3COC6H40H

D. C6H5OH

The pKa of H3COH is 15.5, and the pKa of CI3COHis 11.2. Which of the following values is the BESTapproximation for the pKa of F3COH?

A. 19.4

B. 13.8

C. 11.8

D. 10.4

In which of the following reactions would ethanol bethe solvent of choice?

A. H3C

H3N + Ji Br¥

H

B. O

(H3C)2CHMgBi

C.

H3C

- •.. .AH3C CH2CH3

+ Cr03/H2S04 •

O

XH3CH2CH2C H

D. O O

.M 1. KOEt

OCH2CH32. H3CCH2I

GO ON TO THE NEXT PAGE

Passage II (Questions 7-12)

Alcohols areuseful reagents in a vast number of organicreactions, most often used as a nucleophile. Their reactivityas a nucleophile correlates directly to steric hindrance. Forinstance, because of steric hindrance, a primary alcohol is abetter nucleophile than a secondary alcohol, which is in turna betternucleophile than a tertiaryalcohol.

Alcohols do not readily react as electrophiles, becausethehydroxyl group is a poor leaving group. To increase theelectrophilicity of an alcohol, the hydroxyl group is treatedwith a Br0nsted-Lowry acid or a Lewis acid to form a betterleaving group. Once a hydroxyl group is protonated toforma water molecule, it becomes a stronger leaving group thatcan be displaced by weak nucleophiles. A hydroxyl groupcan also bemade into a better leaving group by converting itinto a tosylate group according to the reaction inFigure 1.

ROH + Cl ROS02C6H4CH3

Figure 1 Tosylation of an alcohol

After a hydroxyl group isconverted into a better leavinggroup, the compound readily undergoes dehydration reactionsat elevated temperatures, which yield an alkene. Adehydration reaction is commonly known as an eliminationreaction. Elimination reactions compete with substitutionreactions in most cases where an alcohol nucleophile ispresent. The elimination reaction is most favorable withtertiary alcohols under both basic and acidic conditions,because the alkene product ishighly substituted in that case.

The nucleophilicity ofalcohols is enhanced by treatmentwith a strong base, which converts it into an alkoxide anionafter it is deprotonated. A nucleophilic substitution reactionwith an alkoxide acting as the nucleophile also hascomplications with the competing elimination reaction,which proceeds by an E2 mechanism.

7. Which ofthe following alcohols would MOST readilyundergo an E\ elimination reaction?

8.

A. B.

C.

OH

D.

or orOH R,C OHH3C OH

Which of the following alcohols will MOST readilyundergo transesterification with ethyl acetate and H+?A. H3CCH2CH2CH2CH2OHB. H3CCH2CH(CH3)CH2OHC. H3CCH2CH(OH)CH2CH3D. (H3C)2COHCH2CH3

Copyright © byTheBerkeley Review® 54

9. Which statement is NOT true about alcohols?

A. Alcohols cannot be oxidized.

B. Alcohols are among the few organic compoundscapable of forming hydrogen bonds.

C. Primary alcohols are better nucleophiles thansecondary and tertiary alcohols, because theyexhibit less steric hindrance.

D. Alcohols can be identified by their broad peak ininfraredspectroscopy between3250 and 3600cm'1.

10. Which of these alcohols can be oxidized to a ketone?

A. H3CCH2CH2OH

B. (H3C)2CHCH2OHC. H3CCH2CHOHCH3D. (H3C)2COHCH2CH3

11. What is the IUPAC name for the following compound?OH Cl

CH2CH3

A. 4-chloro-3-ethyl-2-pentanolB. 2-chloro-3-ethyl-4-pentanolC. 4-chloro-3-ethyl-sec pentanolD. 2-chloro-3-ethyl-sec-pentanol

12. Which ofthe following structures represents the produ&tafterisopropanol reacts with tosylchloride? iA.

B.

C.

D.

H3C O

„ic~irO~H3C

O

H3C O

CH3

HC-</

D—S-

IIO

•\ J -CH3

H3C ^__^

0|| /==\

/CH3H3C--s-

O

<_> O — CH\

^^•^ ^J

CH3

H3C O . .

N ii/=V 'HC-O—S—(\ />—O—CH' II \-J ^H3C » CH3

CH3


Passage III (Questions 13-18)

Acetals and ketals are formed by adding excess alcohol toeither an aldehyde (which forms an acetal) or a ketone (whichforms a ketal) in the presence of an acid. Hemiacetals andhemiketals are formed by adding an alcohol to either analdehyde or a ketone in the presence of a base. Acetals andketals do not react at high pH and cannot be formed underbasic conditions, which makes acetals and ketals idealprotecting groups for the aldehyde and ketone functionality inreactions carried out under basic conditions. Figure 1 showsa generic reaction forming a ketal protecting group.

O

A •R

ketone vicinal diolx

ketal

+ H20

water

Figure 1 Formation of a ketal from a ketone

The ketone can be regenerated by adding water to theketal, as shown by the reverse reaction in Figure 1. Theequilibrium is shifted back to the ketone by adding water.The ketal must be broken under acidic conditions. Figure 2shows the acid catalyzed conversion of a ketal into a ketone.

r^ H+O O -iL^

R^^R' R' R'OH

0+*

AH

R R*

X"

O

" R^R"+ H30+

Figure 2 Acid catalyzed conversion of a ketal into a ketone

The ketal protecting group is commonly employed inreactions such as the Grignard reaction, where the Grignardreagent could attack the carbonyl carbon. Because theGrignard reaction is carried out under basic conditions, theketal remains intact during the course of the Grignardreaction, until it is removed at the end.


13. What is the product for the following reaction?

Q

+ OH-(aq) *•CXO

A. Acetone

B. Formaldehyde

C. 1,2-Ethylenediol

D. No reaction transpires, so no product is formed


vOH °YXV1 vOH H°T^1/ÔH +O^sy^ /OH +HO"^/1

C. D.

><o >oo15. What is the product for the following reaction run under

acidic conditions?

O

OH + X —+ H3C CH3

A. 0

' O*^

c

xD

B-HQ

HO:>o

D:>o

16. Which of the following would be added to protect avicinal diol?

A. Acid and an ester

B. Base and an ester

C. Acid and a ketone

D. Base and a ketone


17. The following molecule has what functionality?

O^ .0

A. Ketal

B. Acetal

C. Hemiketal

D. Hemiacetal

QQ

18. Mixing methanol, sodium methoxide, and acetoneresults in the formation of which of the followingcompounds?

A. Ketal

B. Acetal

C. Hemiketal

D. Hemiacetal

Copyright © by The BerkeleyReview® 56

Passage IV (Questions 19-25)

The reactivity of a carbonyl compound is correlated in alinear fashion to the acidity of the conjugate acid of theleaving group. Generally speaking, the more acidic HX is,the more reactive RCOX is. This relationship allows us topredict the reactivity of a carbonyl compound using pKavalues. Figure 1 shows a generic carbonyl substitutionreactionwherethe X is a variable leaving group.

O

.ANaOCH3

HOCH3 *X R

O

X + X"

0CH3

Figure 1 Generic carbonyl substitution reaction

The leaving group can be a variety of groups fromhalides to carboxylates to alkoxy groups. The relativereactivity can be quantified by comparing reaction rates ofvarious carbonyl compounds. When comparing the relativereaction rates of compounds, a standard is selected. For thisstudy,a researcherchose the reference reaction in Figure 2.

H3C

O

AO

XNaOCHi

HOCH3*OCH2CH3 H3C OCH3

Figure 2 Reference reaction for carbonyl substitution study

A reactivity number can be calculated for a given leavinggroup by comparing the log of the reaction rate with the logof the reaction rate for the reaction with a reference leavinggroup, such as the ethoxy leaving group. The reactivitynumberis derivedby means of the followingequation:

6 = log ^^reference

Equation 1

where fi is the reactivity number, kx is the rate constantfor the reaction, and kreference is the rate constant for thereference reaction with an ethoxy leaving group. Table 1shows the pKa values along with reactivity numbers for aseries of X's (leaving groups):

X HX pKa ft

r HI -11.2 2.8

ci- HCl -7.0 2.5

h3cco2- H3CC02H 4.8 1.7

C6H5S- C6H5SH 7.8 1.1

CN" HCN 9.1 0.3

H3C0- H3COH 15.5 0.2

Table 1


19. Formic acid has a pKa of 3.78.expected for methylethanoate?

A. 2.7

B. 2.0

C. 1.8

D. 0.3

What b value is

20. In the reference reaction shown in Figure 2, the otherorganic product is:

A. -OCH2CH3

B. HOCH2CH3

C. H20

D. H3COCH2CH3

21. What b value is expected when X- is bromide?

A. 1.9

B. 2.4

C. 2.7

D. 2.9

22. For the last example in Table 1, the leaving group isH3CO". This is also the nucleophile. How could youcarry out this study?

A. By using 24Na in the NaOCH3B. By using a carbonyl with Ô labelingC. By using an R group of CD3 on the carbonyl

compound

D. By using NaOCD3/HOCD3

23. A negative 6-value:

A. occurs when the observed reaction has a greaterequilibrium constant than the reference reaction.

B. occurs when the observed reaction is slower than

the reference reaction.

C. occurs when the observed reaction is faster than thereference reaction.

D. is not possible.

Copyright © by The Berkeley Review® 57

24. Which of the following types of carbonyl compounds isthe LEAST reactive?

A. Ester

B. Anhydride

C. Amide

D. Acid halide

25. Which of the following relationships accuratelydescribes the relative reactivity of the carbonylcompounds listed?

A. Anhydride > Acetoyl iodide > Ester.

B. Acetoyl iodide > Anhydride > Ester.

C. Ester > Anhydride > Acetoyl iodide.

D. Acetoyl iodide > Ester > Anhydride.


Passage V (Questions 26 - 32)

Hydrogens on alphacarbons adjacent to a single carbonylfunctional group have pKa values between 17 and 19.Hydrogens on an alpha carbon adjacent to two carbonylfunctional group have pKa values between 11 and 13. Thismeans that alpha hydrogens can be deprotonated with theaddition of a strong base.

Upon treating a ketone with one equivalent of a strongbase, a small amount of enolate is formed. The enolateexists in the resonance shown in Figure 1 below:

:o:.. 0

:o:

R CH2 R ^CH2Figure 1 Resonance forms of deprotonated ketone

The resonance structureon the left in Figure 1 (the onecarrying a negative charge on the alpha carbon) is a strongnucleophile, capable of undergoing an Sn2 reaction with analkyl halide, resulting in the addition of the alkyl group tothe alpha carbon.

A researcher carried out a study on the effects oftemperature on the reaction of an asymmetric ketone. Theresearcher chose 2-pentanone and methylated it according tothe reactionshown in Figure 2 below:

l.LiN(CH3)2

O

XH3C^ CH2CH2CH3 2CH31B

Figure 2 Methylation of 2-pentanone

The reaction was carried out atfive different temperaturesand the percentages of the two products, A and B, wererecorded for each trial. Table 1below shows the percentagesofthe products and the corresponding reaction temperature:

Temperature Product A Product B

-78°C 93.6% 6.4%

-33°C 68.1% 31.9%

0°C 47.3% 52.7%

+25°C 27.9% 72.1%

+50°C 8.4% 91.6%

Table 1

At lower temperatures, the base deprotonates the lesshindered alpha proton, because the transition state is of lowerenergy. As the temperature of the system is increased, thereaction tends to form the more stable intermediate. Anequilibrium exists between the two intermediates. As thetemperature is increased, thepercentage of thethermodynamicintermediate increases. This shift inequilibrium explains thechange in product distribution with temperature. Themorestable intermediate leads to the more substituted enolate.


26. Which of the following CANNOT be synthesized from2-pentanone using the same reactionsequenceas shownin Figure 2 with any alkyl halide?

A. 4-Heptanone

B. 3-Ethyl-2-pentanoneC. 2-Heptanone

D. 3-Ethyl-2-hexanone

27. What is the percentage of 3-hexanone formed at 15*Cusingthe same reactantsshown in Figure 2?

A. 80%3-Hexanone

B. 65% 3-Hexanone

C. 35% 3-Hexanone

D. 20% 3-Hexanone

2 8. Enolates react in Michael addition reactions as follows

l.LiN(CH3)2

° 2 O °

-<*--• %J^R H3C^CH2 .~ kA

H3C CH3 O

3. Workup R

Which of the following products can be formed usingthe kinetic enolate of 2-butanone in a three-stepMichael addition reaction with H2C=CHC(0)CH3?

A. 6-Oxo-2-octanone

B. 3-Methyl-6-oxo-2-heptanoneC. 5-Oxo-2-heptanoneD. 6-Oxo-2-heptanone

29. The kinetic intermediate is preferable when the base isbulky. If the experiment were repeated usingLiN(CH2CH3)2 instead of LiN(CH3)2, what would beobserved?

A. Thepercentage of ProductA wouldbe greaterat alltemperatures.

B. The percentage of Product A would be less at alltemperatures.

C. The percentage of Product A would be greater attemperatures below -4°C, and it would be less attemperatures above -4°C.

D. The percentage of Product A would be less attemperaturesbelow -4°C, and it would be greater attemperatures above -4°C.


30. Product A is the result of the methyl iodide reactingwith which intermediate?

A. The transition state intermediate.

B. The kinetic intermediate.

C. The more stable intermediate.

D. The thermodynamic intermediate.

31. When treating a ketone with a small base, thethermodynamic intermediate is preferred. What is themajor organic product when 2-methylcyclohexanone isfirst treated with potassium hydride at 40°C followed bytreatment with ethyl iodide?

A. 2-Ethyl-2-methylcyclohexanone

B. 2-Ethyl-6-methylcyclohexanone

C. 3-Ethyl-2-methylcyclohexanone

D. 2-Ethyl-3-methylcyclohexanone

3 2. Which of the following reactants offers no competitionbetween the thermodynamic and kinetic enolate?

A. 2-methyl-3-heptanone

B. 2,2-dimethyl cyclopentanone

C. 2-butanone

D. 2,3-dimethyl cyclopentanone


Passage VI (Questions 33 - 39)

Alcohols can be oxidized into a carbonyl (aldehyde orketone) and/or a carboxylic acid upon the addition of anoxidizing agent. The reaction of alcohols varies with thesubstitution of the alcohol. For instance, tertiary alcohols donot oxidize. Figure 1 shows a guideline of reactivity for thealcohols.

RCH2OH1° Alcohol

R2CHOH2° Alcohol

R3COH3° Alcohol

[Ox]

[Ox]

[Ox]

O

Xo

XNo Reaction

[Ox]O

AR OH

Figure 1 Synopsis of oxidation for various alcohols

The [Ox] listed in the reactions above represents anoxidizing agent. Because an oxidizing agent gets reduced, itmust start in a high oxidation state. Oxidizing agents can berecognized by their abundance of oxygen. In the table beloware some common oxidizing agents and their yields in theoxidation of various alcohols:

Oxidizing Agent Alcohol Yield

KMnOÔH" 2-pentanol 81.7%

PCC 1-hexanol 73.9%

Cr03/H2S04 2-butanol 65.2%

Na2Cr207/H2S04 2-pentanol 87.4%

KMn04/OH- 1-butanol 92.3%

K2Cr207/H2S04 1-butanol 52.3%

Cu/CuO/H2S04 1-propanol 45.6%

H2Cr04 Cyclohexanol 97.2%

In the absence of water, primary alcohols can beconverted into aldehydes. The lower yield in the reactionsinvolving sulfuric acid is attributed to the formation of analkene side product. The choice of oxidizing reagent can bemade based on the sensitivity of the alcohol compound toeither acid or base. While the alcohol itself may not be pH-sensitive, other functionalities on the molecule may besensitive to the pH of the solution.

3 3. What is the change in oxidation state experienced by thereactive carbon in the oxidation of a secondary alcohol?

A. From 0 to +2

B. From-2 to 0

C. From 0 to-2

D . From +2 to 0


34. What is the major organic product and its approximateyield following the treatment of 2-hexanol with aqueouspermanganate solution at a pH of 10?

A. 92.3%2-hexanone

B. 81.1% 2-hexanone

C. 73.9% 2-hexanone

D. 92.3% pentanal

3 5. Which of the following alcohols will NOT undergooxidation upon treatment with chromic oxide in sulfuricacid?

A. 2-methyl-3-hexanol

B. 2-ethylcyclopentanol

C. 3,3-dimethyl-l-pentanol

D. 2-methyl-2-butanol

36. Addition of which of the following alcohols to orangeNa2Cr207 will turn the solution green?

A. 2-methylphenol

B. 1-methylcyclohexanol

C. 3-ethyl-3-heptanol

D. 2-methyl-l-octanol

3 7. Which of the following oxidation reactions is NOTpossible?

A. 1-butanol to butanone

B. 2-methyl-l-pentanol to a carboxylic acidC. 1-pentanol to an aldehydeD. Cyclohexanol to a ketone

38. How does the oxidation state of chromium change inthe treatment of chromic oxide in sulfuric acid withethanol and 2-methyl-2-butanol?

A. The oxidation state of chromium increases with theaddition of bothethanoland 2-methyl-2-butanol.

B. The oxidation state of chromium decreases with theaddition of both ethanol and 2-methyl-2-butanol.

C. The oxidation state of chromium increases with theaddition of ethanol and remains constant with theaddition of 2-methyl-2-butanol.

D. The oxidation state of chromium decreases with theaddition of ethanol and remains constant with theaddition of 2-methyl-2-butanol.


39. In the following reaction, where R = CH3, what is theoxidation and reducing agents?

OH OH O O

ÂD+TiQ2^DAJLD +Ti(OH)2R R R R

A. Oxidizing agent: C5Hi202; reducing agent: Ti02.B. Oxidizing agent: C5Hiq02; reducing agent: Ti02.C. Reducing agent: CsHi202; oxidizing agent: Ti02.D. Reducing agent: C5Hiq02; oxidizing agent: Ti02.


Passage VII (Questions 40 - 46)

A researcher wants to distinguish and identify threeunknown isomers (Compound A, Compound B, andCompound C), all of which have the molecular formula,C6Hi20. To help distinguish the structures from oneanother, the researcher performs some standard reactions usedin qualitative analysis tests for organic compounds. ToCompounds A, B, and C, she adds Jones's reagent(Cr03/H2S04). Only compound A turns green upon theaddition of Jones's reagent. Compounds B and C remainorange incolorfor an extended periodof time. The researcherconcludes thatonlyCompound A is an oxidizable compound.All three compounds maintain a brown color when treatedwith Br2 liquid in the presence of CC14 solvent, indicatingthat the brown bromine liquid did not react with any of theunknowns. The bromine liquid tests for the presence of anyunconjugated alkene rc-bonds.

Only Compound B reacts with I2 in basic medium toform a yellow oil at the bottom of the flask (the iodoformtest used to test for the 2-keto functionality). OnlyCompound A, after being treated with strong base, yields aprecipitate when added to AgN03 in water solvent. This isknown as Tollen's test, which indicates the presence of areadily oxidizable carbonyl functional group. Compound Aforms a precipitate with H3CCI in the presence of silvernitrate. The precipitate is silver chloride, formed whenchloride anion is displaced as a leaving group during anucleophilic substitution reaction.

The product of the reaction of Compound A with Jones'sreagent, was isolated, collected, and tested with blue litmuspaper. The oxidized derivative of Compound A turned theblue litmus paper red immediately. The information fromthe various testsgave clues as to the typeof compounds thateach unknown was likely to be. The researcher followingthese testswasable to conclusively identify the functionalityof each unknown. Table 1 summarizes the results of eachtest for Compounds A, B, and C.

Test -» Jones's l2/KOH(aq) Tollens RX/Ag+

Compound A Positive Negative Positive Positive

Compound B Negative Positive Negative Negative

Compound C Negative Negative Negative Negative

Table 1

40. An alcohol, when oxidized, forms a compound thatturns blue litmus paper red. In terms of substitution,the alcohol is best described as a:

A. primary alcohol.B. secondary alcohol.C. tertiary alcohol.D. primaryalcoholor secondaryalcohol.


41. Which of the following structures is NOT possible forthe formula C6Hj20?

A. Cyclic alcohol

B. Linear ketone

C. LinearaldehydeD. Cyclic ketone

42. Which of the compounds summarized in Table 1 isNOT correctly identified?

A. Compound A is a primaryor secondary alcoholB. Compound B is a methyl ketoneC. CompoundC is an aldehyde.D. None of the compounds can be a cyclic ether.

43. Which of the following compounds CANNOT beoxidized?

A. Primary alcohols

B. Aldehydes

C. Esters

D. Hemiacetals

44. All of the following are true for a compound that can beused to test for an aldehyde EXCEPT:

A. it undergoes a color change or phase change whenit is reduced.

B. it is a compound that is rich in hydrogen atoms

C. it is a transition metal in a high oxidation state.

D. it is in lower concentration than the aldehyde.

4 5. Which of the following is NOT an oxidizing agent?

A. KMn04

B. H3CCO3H

C. Mg(OH)2

D. HI04

4 6. Which of the following reagents can reduce an ester to aprimary alcohol?

A. H2/Pd

B. HCl/Zn

C. LiAlfy

D. BH3


Passage VIII (Questions 47 - 53)

Wolff-Kishner reduction converts either a ketone or an

aldehyde into an alkane. The two-step process involves theformation of a hydrazone (the hydrazine derivative) which isreduced under basic aqueous conditions. Figure 1 shows theWolff-Kishner reductionof a generic ketone.

R

R

O

X + H2N—NH2

R

NH2N

XR

R

H H

NH2N

XR

+ N2

Figure 1 Wolff-Kishner reduction of a ketone

Reduction of a carbonyl into alkane can also be carriedout under acidic conditions using Clemmensen reduction.Zinc metal provides the electrons for the reduction, as shownin Figure 2.

O

H20 + 2ZnCl2° H H

x -^ y +Figure 2 Clemmensen reduction of a ketone

4 7. The reactant in a Wolff-Kishner reduction reaction couldalso react with all of the following compoundsEXCEPT:

A. LiAlH4

B. H3CMgBr(et20)C. 03/H202

D. HCl/Zn

48. Over thecourse of a Clemmensen reduction of a ketone,how does the oxidation state of the carbonyl carbonchange?

A. It increases from -2 to +2.

B. It decreases from +2 to -2.

C. It decreases from 0 to -2.

D. It decreases from +2 to 0.

4 9. What is true following the Wolff-Kishner reduction of aketone?

A. One new chiral center is formed.

B. The hybridization ofcarbon goes from sp2 tosp3.C. Both nitrogenand carbon are reduced.D. Three net water molecules are formed asa product.


5 0. What accounts for the greater acidity associated with aproton on H2N-N=CR2 than a proton on ammonia?

A. The hydrazone has greater steric hindrance thanammonia, which increases the acidity.

B. The hydrazine molecule becomes aromatic uponlosing a proton.

C. Ammonia has less of an inductive effect thanhydrazone.

D. The lone pair formed upon deprotonation from thehydrazone is stabilizedby resonance.

51. What product is expectedwhen the following moleculeis treated with hydrazine and basic water?

O

HO

CH3

A. 3-Ethylcyclohexanol

B. 3-HydroxyethylcyclohexaneC. EthylcyclohexaneD. 3-Ethylcyclohexanone

52. Under what conditions is Wolff-Kishner reductionpreferred over Clemmensen reduction?

A. When the reactant has an acid sensitive functionalgroup.

B. When the reactant has an amine sensitivefunctional group.

C. When the reactant has a base sensitive functionalgroup.

D. When the reactant has a water sensitive functionalgroup.

53. What is the product of 2-methylcyclopentanone andphenyl hydrazine, which reacts like hydrazine?A. ph B. ph

\ \.N^ N-

n' h NX OH

H3C^/\ V^KC.

NS

Ph D.

H H

H3C H3C


Passage IX (Questions 54 - 60)

Carbonyl compounds are reactive with nucleophiles.Their electrophilicity can be attributed to the polar nature ofthe carbonyl bond which houses a partial positive charge onthe carbon in the carbonyl. Carbonyl reactivity can becorrelated to thepKaof theconjugate acidof the bestleavinggroup on the carbonyl carbon. This is to say that the betterthe leaving group the more reactive thecarbonyl compound.Table 1 shows the pKa values of conjugate acids of variousleaving groups on carbonylcompounds:

pKa Acid

-7.2 HCl

4.2 H5C6C02H

4.8 H3CC02H

9.1 HCN

10.5 H3CH2CSH

15.7 H3CH2COH

33.1 NH3

Table 1

In addition to the leaving group in a carbonyl reaction,the nucleophile is also considered. A Grignard reagent is agood nucleophile that is selective in its reactivity withcarbonyls. ThreeGrignard reactions are shown in Figure 1.

Reaction I (Grignard reaction withan aldehyde)O R' OH

1. R'MgX(et9QJ

RA

H2. NHtCKaq)

R H

ReactionII (Grignard reaction with a ketone)o r; OH

1. R'MgXfet?^2. NH^CKaq)

R R R' "R

u

XReaction HI (Grignard reaction with an ester)

Requires two equivalents of R'MgBrR' OH

1. R'MgXtebQJ \f2. NH^CKaq) ^\

O

XR X RT R'

Figure 1 Grignard reactions with three different carbonyls

Grignard reagents react most readily with acid halidesandacid anhydrides (to which twoequivalents of theGrignard canadd). Grignard reagents do not react with amides. Despitethe strength of the Grignard reagent as a nucleophile, itcannotdisplacean amine leaving group. These observationsare in accord with the acidity of the conjugate acid of theleaving groups. This information can be used to predict thedegree of product formation andthefavorability of a carbonylreaction.


54. Which of the following compounds is the LEASTreactive carbonyl compound?

A. An ester.

B. A thioester.

C. An anhydride.D. An acid chloride.

5 5. To make 2-pentanol, it would be best to use which ofthe followingreaction sequences?

A. H3CMgBr + pentanal.

B. H3CMgBr + 2-pentanone.C. H3CMgBr + butanal.

D. H3CMgBr + butanone.

56. Which of the following Grignard reactions would formstereoisomers as the product?

A. Butanone + ethyl magnesium bromide.B. 3-Pentanone + methyl magnesium bromide.C. Ethyl acetate + 2 equiv. ethyl magnesium bromide.D. Propanal + methyl magnesium bromide.

57. When treated with RMgBr, the following compoundwould yield which of the following products?

<y-A. A primary alcohol.B. A secondary alcohol.

C. A tertiary alcohol.D. A ketone.

58. A Grignard reagent adds how many times to an ester?

A. Once.

B. Twice.

C. Three times.

D. It depends on the temperature of the mixture.


5 9. Which of the following conditions is NOT true about aGrignard reaction?

A. Ether solvent is required for the Grignard reaction.

B. The Grignard reaction involves the reduction of thecarbonyl carbon.

C. The Grignard reaction can be employed to formeither primary, secondary, or tertiary alcohols.

D. For the highest yield, the solution should be acidicwhen the Grignard reagent is added.

60. If propanoic acid has a pKa value of 4.9, what can beconcluded about the reactivity of propanoic acidanhydride?

A. It is more reactive than benzoyl chloride.B. It is less reactive than N-methylbutamide.C. It is more reactive than methyl acetate.D. It is less reactive than acetic acid anhydride.


Passage X (Questions 61 - 67)

The Aldol condensation reaction is a main staple insynthetic organic chemistry. The reaction involves thedeprotonation of a hydrogen from the alpha carbon (thecarbon adjacent to the carbonyl group) and the subsequentreactions of the carbanion as a nucleophile. The carbanionform is in resonance with the enolate form. In an Aldol

condensation, the electrophile to which the carbanion isadding is a neutral carbonyl compound. The preliminaryproduct (the product of Step I in Figure 1) is referred to as afi-hydroxy carbonyl. This 6-hydroxy carbonyl can thenundergo elimination (shown as Step II in Figure 1) to form aconjugated alkene-carbonyl compound. The conjugatedalkenecarbonyl compound is referred to as an ctji-unsaturatedcarbonyl. The Aldol reaction is shown in Figure 1 below.

Step I: Formation of B-hydroxyketoneO

ARC

AH H

LDA(et2Q^ R.CH

O

KR

R. A-C

AH H

OH

H

Step II: Formation ofot,B-unstauratedketoneO O

R

R. A. R

AA

A..RCH R HCl(aqi C

A RC \ OH 'c

A H /\H H H H

Figure 1 Generic Aldol reaction

The pKa of an alpha hydrogen falls in the range of 11 to19 depending on the number of carbonyls to which it isadjacent. The hydrogens of acetone have a pKa of roughly 19while the central alpha hydrogens of 4-oxo-2-pentanone havea pKa of roughly 11. The pKa information can be used topredict the most reactive site on a molecule. It can also beused to predict the degree of reactivity for a compound.

61. Which of the labeled hydrogens on the molecule belowis the MOST acidic?

A. a

B. b

C. c

D. d

a

lH3CH2C

b o cII

\S^H


62. What is the IUPAC name for the compound that canundergo an Aldol condensation reaction with itself toform the followingcompound?

O OH

H

CH3

H3C CH3 CH3

A. ButanaldehydeB. 2-Pentanone

C. 2-MethylbutanalD. 2-Methylpropanal

63. Treatment of an aldehyde with excess alcohol in thepresence of an acid would lead to which of thefollowing products?

A. An acetal

B. A hemiacetal

C. An alkene

D. An ester

64. What is the major organic product isolated from thefollowing reaction?

O

2 II l.LiN(CH(CH3)2)2AH3C CH3 2.H30+

65. The chemical inter-conversion from a ketone to an enol

is referred to by which of the following terms?

A. Resonance

B. Conjugate pairingC. Hydroxyl exchangeD. Tautomerization


66. Which of the following bases is NOT strong enough todeprotonate the hydrogen on the alpha-carbon of aketone?

A. NaNH2

B. NaH

C. Na2C03

D. NaOCH2CH3

67. Which of the following compounds CANNOT undergoan Aldol condensation reaction?

A. A ketone

B. An aldehydeC. A carboxylic acidD. All of the above can undergo an aldol reaction


Passage XI (Questions 68 - 73)

The Claisen condensation involves the dimerization of

an ester. The first step is deprotonation of an ester to forman enolate, which is followed by the subsequent addition ofthe enolate to the carbonyl carbon of a neutral ester. Thetetrahedral intermediate that is formed loses an alkoxide

leaving group to yield a B-ketoester product.

The base chosen to deprotonate the alpha hydrogen of theester should match the alkoxide leaving group so as to avoidcomplications with transesterification. This alkoxide leavinggroup is an anion that is also a strong enough base todeprotonate the alpha hydrogen of the B-ketoester product.To isolate the B-ketoester product, the solution is neutralizedwith a weak acid. Figure 1 shows the equilibrium mixturesin a Claisen reaction.

O

A~OCH2CH3 9^

H3CH2C OCH2CH3 H3CHC'/ OCH2CH3

Q H3CH2C\/ OCH2CH3

A X-^H3CH2C CH OCH2CH3

CH3 OCH2CH3B-Ketoester ^^n. o O

^X xH3CH2C^X" OCH2CH3CH3

+ HOCH2CH3

Figure 1 Claisen reaction scheme

Without an alpha hydrogen, the last step (equilibriumstep) of the scheme above is not possible. If the B-ketoesterthat forms has no alpha hydrogen, then the yield is reducedbecause the alkoxide anion acts as a nucleophile (ratherthanas a base) and causes the retro-Claisen reaction. To increasethe yield (from as low as zero), less alkoxide base is added,and the reaction is run at a lower temperature.

68. Which proton has the lowest pKa value of thehydrogens on the following B-ketoester?

O O

H3CH2CyVH CH2CH3b c

OCH2CH3

H Ha

A. Hydrogen a.B. Hydrogen b.C. Hydrogen c.D. Hydrogen d.


69. Which of the following ester reactants would give theLOWEST yield in a Claisen reaction?

A. H3CC02CH2CH3

B. H3CCH2C02CH2CH3

C. (H3C)2CHC02CH2CH3

D. (H3C)2CHCH2C02CH2CH3

70. A Claisen condensation reaction is NOT possible withwhich of the following esters?

A. HC02CH2CH3

B. C6H5CH2C02CH2CH3

C (C6H5)2CHC02CH2CH3

D. C6H5CH2C02CH3

71. Which of the following is a product of a Claisencondensation reaction of one ester upon itself?

A. H3CCH2COCH2C02CH2CH3

B. H3CCOCH(CH3)C02CH2CH3

C. (C6H5)CH2COCH2C02CH2CH3D. (C6H5)CH2COCH(C6H5)C02CH2CH3

72. What is the major organic product for the followingreaction?

NaOEt


73. Counting stereoisomers, how many products arepossible for the following reaction?

O O

A *ÂEtO

A. 2

B. 3

C. 4

D. 6

OEt

Copyright © by TheBerkeley Review®

NaOEt

HOEt

67

Passage XII (Questions 74 - 80)

In transesterification, an alkoxy group of an ester isexchanged by the addition of an alcohol in the presence ofeither acid or base. The equilibrium constant for the reactionis approximately one, so the reaction must be driven by theaddition of excess alcohol. A researcher carries out a

transesterification reaction by adding methyl propanoate toethanol in the presence of HCl at 25°C for thirty minutes(Reaction I). The product is isolated by neutralization andthen purified using fractional distillation. The identicalreaction is then repeated under reflux conditions for thirtyminutes (Reaction II). The products from Reaction II arethen neutralized and isolated by neutralization and thenpurified by fractional distillation.

O

JL +H3CCH2OHH3CH2C OCH3

Reaction I

H+^25°C

30 min

O

11h3ch2c 0CH3

♦ H3CCH2OH -^^30 min

Reaction II

Figure 1and Figure 2 show the 'HNMR spectra of themajor components from Reactions I and II respectively.

Figure 1 Reaction I's major component 'HNMR spectra

Major component isolated from React on II

3H 3H

2H 2H

J1 J1 1u \ I—I I 1

3 2 1 0

Figure 2 Reaction II's major component *HNMR spectra


The ratio of the integrals of the peaks are 3 : 2 : 3 forthe spectrum in Figure 1 and 2 : 2 : 3 : 3 for the spectrum inFigure 2. The 'HNMR spectra were collected indeuterochloroform solvent.

74. What is the IUPAC name and formula for the

compound in the 'HNMR spectrum inFigure 1?A. Methyl propanoate (CH3CH2C02CH3)

B. Methyl ethanoate (CH3C02CH3)

C. Ethyl ethanoate (CH3C02CH2CH3)

D. Ethyl methanoate (HC02CH2CH3)

75. What can be concluded about Reaction II from the

'HNMR spectrum inFigure 2?A. Methyl propanoate, CH3CH2C02CH3, did not

react with ethanol.

B. The ethyl and methyl groups exchanged, resultingin the formation of ethyl acetate,CH3C02CH2CH3.

C. Propyl acetate, CH3C02CH2CH2CH3, formedfrom the rearrangement of a carbocation.

D. Ethyl propanoate, CH3CH2C02CH2CH3, wasgenerated in a greater than 50% yield.

7 6. An ester is more reactive as an electrophile than whichof the following compounds?

A. Acid anhydridesB. Amides

C. Acid halides

D. An ester is the most reactivecarbonyl compound.

77. Had ethyl amine (CH3CH2NH2) been used instead ofethanol is Reaction II, the final product would havebeen which of the following?

A. N-ethylpropyl amineB. 1-aminopentaneC. N-ethylpropanamideD. N-Hydroxypropanamine

78. The presence of acid in the reaction mixture serveswhich of the following purposes?

A. To protonate the ester electrophile helping todecrease reactivity.

B. To protonate the ester electrophile helping toincrease reactivity.

C. To protonate the alcohol nucleophile helping todecrease reactivity.

D. To protonate the alcohol nucleophile helping toincrease reactivity.


7 9. The presence of water in the reaction mixture wouldresult in which of the following impurities in theproduct mixture?

A. Carbonic acid (H2C03)

B. Formic acid (HC02H)

C. Acetic acid (CH3C02H)

D. Propanoic acid (CH3CH2C02H)

80. Why must the NMR be carried out in deuteratedsolvent?

A. To prevent reaction from occurring in the NMRtube.

B. To induce magnetization into solution.C. To prevent hydrogens from the solvent from being

detected by the NMR spectrophotometer.D. To decrease the pH of the solution.


Passage XIII (Questions 81 - 87)

O O

.AA1. NaOEt(HOEt).

H30+/A

O O

^, , 2. RX(et20) /*V ^*\ ' 80 to 90% yieldCH3CH20 ^ OCH2CH3 CH3CH20 y^ OCH2CH3 QR

70 to 85% yield

Urea

O

AHN NH

O

R

65 to 75% yield

Figure 1 Synthetic uses of malonic ester

Malonic ester is a common starting reagent in thesynthesis of many bio-organic compounds. Figure 1 showstwo different synthetic routes starting from the malonic ester,one leading to a carboxylic acid and the other leading to alactam(cyclic amide). The first route leads to a carboxylicacid of specific composition. The carboxylic acid can be anintermediate product that is subsequently converted into oneof several possible carbonyl compounds, such as an ester oramide. The second synthesis route leads to compounds in thebarbituric acid derivative group. These are useful in thesynthesis of barbiturates.

The first part of both synthetic pathways involves thedeprotonation of the hydrogen alpha that is adjacent to bothcarbonyl groups. The proton that is on a carbon that is alphato both carbonyls has a pKa value of roughly 11, so analkoxide is strong enough to remove that proton. Thiscreates a nucleophile that can react with alkyl halides,amongst other electrophiles. In these two examples, sodiumethoxide is chosen to prevent competition with thetransesterification reaction.

81. What is the structure of urea?

A.

C.

O O

I 1H2N^^ÔH H2N^^ NH2

1HN^^^NH2

o

B.

D.

I


82. When (CH3)2C(NH2)2 is used in lieu of urea, what isthe final product?

A.

HN

O

ANH

O O

C.

R

xHN NH

O O

H3C CH3

B.H3C CH3

yHN NH

o o

D.

R

H3C CH3

yHN NH

O O

H3C CH3

83. Which of the following molecules will NOT undergo adecarboxylation when treated with heat?

A.

H3CJU OH

C.

H3C

\MOH

D. OÔHOH T


84. Why is sodium ethoxide chosen in the synthesis?

A. Sodium ethoxide is a strong base that will notresult in transesterification of the ethyl ester.

B. Sodium ethoxide is a weak base that will not result

in transesterification of the ethyl ester.

C. Sodium ethoxide is stereo-specific.

D. Sodium ethoxide is regio-specific.

8 5. What signals would be observed in the proton NMR ofmalonic ester?

A. A 2H doublet, a 2H doublet, and a 3H tripletB. A 1H singlet, a 2H triplet, and a 3H doubletC. A 2H singlet, a 2H triplet, and a 3H quartetD. A 1H singlet a 2H quartet, and a 3H triplet

86. Which alkyl halide must be chosen to form thefollowing molecule when carrying out a synthesis frommalonic ester?

O

AHN NH

O

H3C CH2CH3

A. 2-bromopropane.B. 2-bromobutane.

C. 2-bromopentane.D. Two alkyl halides must be used; methyl bromide

and ethyl bromide.

8 7. What is the final organic product after malonic ester istreated with: 1) NaOEt, 2) CH3CH2CH2I, and 3) urea?

A. 0 B. 0

XHN NH

o^ >r "o <r y ô

CH3 CH2CH2CH3

C. 0 D. 0

AHN NH

o^ ">r ô o" yr ô

H3C CH3 H3C CH2CH2CH3

Copyright © by The BerkeleyReview®

AHN NH

AHN NH

70

Passage XIV (Questions 88 - 94)

Carboxylic acids can be produced by one of several ways.Carboxylic acids are useful in synthesis as an intermediate innumerous reaction pathways. Carboxylic acids are weakacids that when added to water will partially dissociate intotheir carboxylate conjugate base and hydronium ion.Carboxylic acids play a distinct role in biological chemistry,because of their acid-base properties and their syntheticflexibility. Figure 1 shows six of the ways in which acarboxylic acid can be synthesized:

Grignard Reduction: +RMgBr + C02 • RC02MgBr • RC02H

Oxidation of an Aldehyde:

° H2S04+ Cr03 >A

H

Hydrolysis of an Anhydride:

0 ° w> °

O

AR OH

A AR O R

AR OH

Hydrolysis of an Ester:O

+ H20 HR' 'OR'

v

XO

»• .A_•R OH

Ozonolysis of an Alkene:

R'OH

R H

Ml.Qj

H R'2.H202

R

O O

OH R OH

Hydrolysis of a Nitrile:

H+/A II _,_ KTU +RCN + H20 • JJ + NH4

Figure 1 Six reactions used to synthesize a carboxylic acid

The large number of starting reagents offers variety insynthesis based on the availability of starting compounds.The acid products can react further to form either esters(through transesterification under acidic conditions) or amides(by way of acid chloride formation using SOCl2). An amideis formed by treatment of either an acid halide or acidanhydride with an amine.

88. To synthesize butanoic acid, which of the followingsynthetic pathways will NOT work?

A. Oxidation of 1-butanol.

B. Hydrolysis of ethylbutanoate.C. Ozonolysis of 4-octene, followed by oxidative

workup with hydrogen peroxide.D. Treatment of 1-bromobutane with NaCN, followed

by strong aqueous acid treatment.


O

8 9. How does propanoic acid exist after it has been added towater given that it has a pKa value of 5.0?

A. Fully deprotonated.B. 50% deprotonated.C. 25% deprotonated.D. Less than 1% deprotonated.

90. Which of the following reactions does NOT result inthe formation of a compound that can turn blue litmuspaper red?

A. Treatment of 1-octanol with Cr03/H2S04.

B. Treatment of 3,4-dimethyl-3-hexene with ozonefollowed by reductive workup with dimethylsulfide.

C. Treatment of maleic anhydride with acidic water.D. Treatment of ethylbenzoate with acidic water.


l.PBr3(et20)H3CH2C^^.CH3 2>Mg(et2Q)

3. CCtfg) >4. H30+/A

:TOH

A. o B- , O

OH

C. o D. O

OH OH

9 2. Which of the following reactions is NOT reversible?

A. Hydrolysis of an ester.B. Grignard reagent plus carbon dioxide.C. Hydrolysis of an acid anhydride.D. Hydrolysis of a lactone.

93. Which of the following is NOT associated with acarboxylic acid?

A. An aqueous pH less than 7.0.B. A broadIR peak between 2500 and 3000cm"1.C. The loss of one peak in the JHNMR when D20 is

added to the NMR sample tube.D. A pKa value between 9.0 and 10.5.


94. What sequence of reagents will convert an aldehyde intoan amide?

A. 1. NH3 2. SOCl2

B. 1. Cr03/pyridine/HCl 2. SOCl2 3. NH3C. l.Cr03/H2S04 2. NH3 3. SOCl2D. 1. Cr03/H2S04 2. SOCl2 3. NH3


Questions 95 -100 are NOT basedon a descriptive passage

95. Which of the following structures represents theintermediate for a transesterification reaction under basic

conditions?

A. B.

R O"

R'O OR"

O

R'O — C OR"

IR

O

II+ C

IR

C. D.

O

XS,

9 6. Esters are NOT used as a reactant in which reaction?

A. Claisen condensation

B. Friedel-Craft alkylationC. Grignard reactionD. Transesterification

97. What is the majorproductof the reaction below?

O

atO

OH

O B.

0 o

OH

f

D. O 0

0 cVyÔH

M IT0 0


9 8. Which type of compound can be reduced by NaBH4?

A. Phenol

B. Tertiary alcohol

C. Ketone

D. Ether

9 9. Which of the following describes the hybridization ofthe carbon number three in 3-hexanol?

A. sp

B. sp2C. sp3D. The carbon is not hybridized.

100. The product of 2-butanol with chromium (VI) oxide insulfuric acid would have which of the following1HNMR peaks?

A. Singlet (3H), doublet (2H) and triplet (3H).B. Singlet (3H), triplet (2H) and doublet (3H).C. Singlet (3H), quartet (2H) and triplet (3H).D. Triplet (3H),doublet (2H) and triplet (3H).

"Oh, what a feeling; CHEM—1ST—RY!H"

1. D 2. D 3. A 4. B 5. D

6. D 7. D 8. A 9. A 10. C11. A 12. B 13. D 14. D 15. C16. C 17. B 18. C 19. C 20. A

21. C 22. D 23. B 24. C 25. B26. C 27. C 28. A 29. A 30. B

31. A 32. B 33. A 34. B 35. D

36. D 37. A 38. D 39. C 40. A

41. D 42. C 43. C 44. B 45. C46. C 47. C 48. B 49. B 50. D51. A 52. A 53. A 54. A 55. C56. D 57. B 58. B 59. D 60. C61. B 62. D 63. A 64. A 65. D

66. C 67. C 68. B 69. C 70. A

71. D 72. A 73. D 74. A 75. D

76. B 77. C 78. B 79. D 80. C81. B 82. B 83. C 84. A 85. D

86. B 87. B 88. D 89. D 90. B

91. D 92. B 93. D 94. D 95. A

96. B 97. A 98. C 99. C 100. C

YOU ARE SO VERY FINISHED!

Passage I (Questions 1-6) Alcohols, Acidity, and Nucleophilicity

1. Choice D is correct. According to the answer choices, either statement I or II must be true and either statementIII or IV must be true. The pKa values are given in the question, so the question centers around determining therelative acid strength of phenol and methanol. Phenol has a lower pKa, so it is a stronger acid than methanol.Methanol is therefore a weaker acid than phenol. This makes statements II and IV the true statements, whichmakes choice D the best answer. Resonance is a stronger effect than the inductive effect, so statement II is moresignificant than statement IV. Pick D, and let correctness set you free.

2. Choice D is correct. For the base to be strong enough to deprotonate methanol, its conjugate acid must be a weakeracid (have a higher pKa) than methanol. Choices A and B can immediately be eliminated, because they areboth weak bases, and therefore are not strong enough to deprotonate methanol, a very weak acid. The conjugateacid of NaOH is water, which has a pKa of 15.7 (roughly equal to methanol), so it will deprotonate about halfof the methanol molecules. The conjugate acid of H3CCH2Li is H3CCH3, which is one of the weakest acidsknown, so H3CCH2Li is definitely strong enough to deprotonate methanol. Pick D, and correct you will be. Thisquestion can be reworded to read "which of the following bases is the strongest?", because the one base strongenough to deprotonate methanol must be the strongest base of the choices.

3. Choice A is correct. Because the electrophile has only one carbon, elimination is not possible, resulting in areaction where only the substitution product is observed. The best nucleophile is also the strongest base, becausethe strongest base donates electrons most readily. Of the four answer choices, C and D can be thrown out, becausethey are phenoxides (the conjugate base of a phenol) while choices A and B are the conjugate bases of alcohols.Phenols are more acidic than standard alcohols, so phenoxides are less basic than standard alkoxides. Betweenchoice A and B, the chlorine atoms on choice B (H3CCCbONa) are electron withdrawing and thus decrease thecompound's basicity and nucleophilicity. Pick A to make your day.

4. Choice B is correct. Electron withdrawing groups increase the acidity of a compound by stabilizing the negativecharge on its conjugate base. As the conjugate base becomes more stable, the acid becomes more acidic. Thestrongest acids of the phenols listed in the question (and it could be any type of compound, not just phenols, aslong as all four choices are the same class of compound), is the one with the strongest electron withdrawinggroup. The strongest electron withdrawing group is the nitro group, which withdraws by resonance, makingchoice B the most acidic. Pick choice B and a correct choice is what you'll see.

5. Choice D is correct. Because fluorine is more electronegative than chlorine, it is more electron withdrawingthan chlorine. This means that fluorine atoms on the backbone of the molecule increase the acidity even more sothan the chlorine atoms. An increase in acidity results in a lowering of the pKa. The only value for pKa lowerthan 11.2is choice D, 10.4. You'll no doubt feel glee, if you just picked choice D.

6. Choice D is correct. The reaction in choice A is an SN2-reaction, because the electrophile is primary andammonia isa good nucleophile. Sn2 reactions require anaprotic solvent, soethanol isnota good choice. ChoiceA iseliminated. The second reaction is a Grignard reaction, which cannot be carried out in a protic solvent. In aprotic solvent, the alkyl magnesium bromide species can deprotonate the protic hydrogen, thereby destroyingthe Grignard reagent. Grignard reactions must be carried out in an aprotic solvent, so choice Bcannot use ethanolas a solvent. Choice B is eliminated. In choice C, an aldehyde is being oxidized into a carboxylic acid. Asolvent must be inert if it is to be useful. Because primary alcohols can be oxidized by Jones reagent(CrC>3/H2S04), ethanol is reactive under the reaction conditions. This makes ethanol a bad choice for thesolvent, so choice C is eliminated. The last reaction involves the deprotonation of an alpha hydrogen onacetoacetic ester. Because the ester can undergo both deprotonation and attack at the carbonyl carbon, the basethat is chosen matches the leaving group of the ester, in this case ethoxide. The ideal solvent is the conjugateacid of the alkoxide, so that it cannot generate another possible nucleophile for transesterification. Given thatethoxide is the base being used in choice D, ethanol is the ideal solvent. Choice D is your kind ofchoice.

Passage II (Questions 7 -12) Alcohol Reactions

7. Choice D is correct. The Ei elimination reaction is a two-step process that involves the formation of acarbocation intermediate. The most stable carbocation is a tertiary carbocation. The only tertiary alcohols ofthe answer choices is choice D, which makes choice D the best answer.

Copyright © by The Berkeley Review® 73 CARBONYLS & ALCOHOLS EXPLANATIONS

9.

10.

11.

12.

Choice A is correct. The best nucleophile will undergo transesterification most readily. The nucleophile thatmost readily undergoes transesterification must have the greatest partial negative charge and the least sterichindrance. Considering they are all five carbon alcohols and roughly equal in basicity, their differences innucleophilicity revolve around steric hindrance rather than partial negative charge. The least sterichindrance is associated with the primary alcohol. This eliminates choice C (a secondary alcohol) and choice D(a tertiary alcohol). Choice A is a straight chain alcohol with less steric hindrance than choice B (a primaryalcohol with a branched R group), so choice A is the best answer.

Choice A is correct. Both primary and secondary alcohols can be oxidized to carbonyl compounds, thereforestatement A is a false statement. Choice A is not true in regards to alcohols. Alcohols do form hydrogen bonds,and outside of alcohols, few other organic compounds form hydrogen bonds (amines, amides, acids are some ofthe few). Statement Bis a valid statement. Primary alcohols make the best nucleophiles, because they exhibitthe least steric hindrance in the transition state, therefore statement C is a valid statement. The infrared peakfor an alcohol isbroadened due to hydrogen bonding, and hydroxyl bonds show inthe 3400+ cm"1 range, thereforestatement D is a valid statement. Pick A to start your correct answer filled day.

Choice C is correct. Aprimary alcohol will oxidize into either an aldehyde or a carboxylic acid, depending onthe conditions, but it cannot be oxidized into a ketone. This eliminates choices Aand B. A tertiary alcohol doesnot undergo oxidation, which eliminates choice D. The oxophilic carbon (carbon bonded to the hydroxyl oxygenatom) in a secondary alcohol has two bonds to carbons, so when it is oxidized into a carbonyl, it still has twobonds to carbon. A carbonyl compound where the carbonyl carbon has two bonds to other carbons is a ketone.Choice Cispretty unbeatable when it comes to correctness on this question.

Choice Ais correct. The alcohol group takes priority according to IUPAC nomenclature rules, so the compound isautomatically something-2-pentanol. This makes only choice A possible. It is not necessary to determine theexact name in light of the answer choices. Pick A to be correct. In the interest of learning, you should note thatthe substituents are listed in alphabetical order according tonomenclature rules.

Choice Bis correct. Tosylating an alcohol, in this case isopropanol, results in the substitution of an alkoxy groupfor the chloride. The methyl group on the benzene ring should remain unaffected, so choices C and D areeliminated. Choice A is missing the oxygen atom from isopropanol, so it is eliminated. The best answer ischoice B. This could be solved by simply substituting an isopropyl group for the Rin Figure 1.

Passage III (Questions 13 -18) Acetals and Ketals

13.

14.

Choice D is correct. As stated in the passage, ketals cannot be formed under basic solutions and do not react inbasic solutions. This means that the best answer is choice D, no reaction. Under acidic conditions, both acetone(choice A) and 1,2-ethylene diol (choice C) will form.

Choice Dis correct. The reaction as drawn involves aketal and water in the presence of acid catalyst. Thiswater and acid will destroy the ketal group and consequently form a vicinal diol and ketone. The ketal carbon(carbon with two oxygen atoms bonded to it) becomes the carbonyl and the carbons with one bond to oxygenbecome the alcohols. The answer is drawn below:

Ether carbons becomethe alcohol carbons.

Ketal carbonbecomes the

ketone carbon.

The best answer is choice D.

x: > o +

HO

HO

15. Choice Cis correct. As shown in the sample reaction in Figure 1, a diol and a ketone react in the presence ofcatalytic acid to yield acyclic ketal. The difference between this question and the sample reaction in Figure 1isan extra carbon in the diol. The cyclic structure is one carbon larger, making ita six membered ring. The bestchoice is answer C.


16. Choice C is correct. According to the reaction in Figure 1, diols and ketones react in the presence of catalytic acidto yield a ketal. The ketal is a protecting group for either the diol or the ketone, depending on the objective ofthe protection reaction. To protect a ketone, you can add a diol and catalytic acid. Therefore, to protect a diol,you can add a ketone and catalytic acid. The best answer is choice C. If a vicinal diol reacts with an ester, itwill undergo transesterification. The ester formed is reactive, so it cannot serve as a protecting group. Thiseliminates choices A and B. Choice D should be eliminated, because a ketone and alcohol will only go as far asthe hemiketal when reacting under basic conditions.

17. Choice B is correct. The carbon that bridges the two rings, which has two sigma bonds to oxygen, dictates thefunctionality of the molecule. Because there is a hydrogen attached to that carbon, along with two alkoxygroups (OR groups), the compound is an acetal. The best answer is choice B. The carbon is shown in the drawingbelow.

The acetal carbon, because it is sigmabonded to two oxygens and hydrogen.

18. Choice C is correct. Sodium methoxide is a strong base, methanol is an alcohol, and acetone is a ketone. Asstated in the passage, a hemiketal is formed when mixing excess alcohol with a ketone in the presence of base.The mixture in the question results in the formation of a hemiketal. The best answer is choice C. Hemiacetalsand hemiketals are formed under basic conditions, while acetals and ketals are formed under acidic conditions.Because no acid is present in the reaction mixture, choices A and B are eliminated. No aldehyde is present, sothe product cannot be a hemiacetal, which eliminates choice D.

Passage IV (Questions 19 - 25) Carbonyl Reactivity Study

19. Choice C is correct. The formate leaving group and acetate leaving group should be fairly synonymous in thatthey are both conjugate bases of roughly comparable weak acids (there pKa values differ by about 1.0).According to Table 1 in the passage, acetic acid has a 6-value of 1.7, so formate should have a value of roughly1.7. Choice C, 1.8, is the closest to 1.7, so it is the best answer. Because a pKa of -7.0 correlates to a 6-value of2.5, to have a 6-value of 2.0, the acid would have to have a pKa value around 0.

20. Choice A is correct. The leaving group in the reference reaction shown in Figure 2 of the passage is the ethoxideanion. In the reference reaction, the ethoxide anion is displaced by a methoxide anion. Ethoxide anion is alsoformed in addition to the ester. Choose A for the result of correctitude.

21. Choice C is correct. HBr falls between HI and HCl in terms of acidity, so bromide should have a 6-valuebetween that of chlorine and iodine. Its fi-value is found between 2.5 and 2.8, making 2.7, choice C, the bestanswer.

22. Choice D is correct. This study requires that the leaving group be distinguishable from the nucleophile,otherwise we can't recognize whether a reaction has transpired. Sodium is a spectator ion, so labeling thesodium is pointless. This eliminates choice A. Because the carbonyl remains intact in this reaction, 180labeling of the carbonyl oxygen would be pointless. This eliminates choice B. Because the alkyl group of thecarbonyl moiety remains intact in this reaction, 2H labeling of the alpha hydrogens would be pointless. Thiseliminates choice C. The reaction is best monitored using an isotopically labeled nucleophile. This isNaOCD3/HOCD3, which makes choice D the best answer.

23. Choice B is correct. Taking the log of a number less than one results in a negative value. This means that if^reference were greater than kx, then the log value of their ratio would be negative and thus the 6-value wouldbe negative. A negative 6-value occurs when the observed rate constant is less than the reference rate constant.The observed rate constant is less than the reference rate constant when tiie observed rate is less than the rate ofthe reference reaction. This means that choice B is the best answer.


24. Choice Ciscorrect. The worst leaving group according to Table 1is the alkoxide group, with a fi-value of0.2, soan ester is less reactive than an anhydride and an acid halide. This eliminates choices B and D. An amide(NR2") is more basic than an alkoxide, so according to the trend between basicity and leaving group strength, itis an even worse leaving group than alkoxide. This makes the amide the least reactive carbonyl compound ofthe choices. Choice C is the best answer. The term amide has multiple meanings according to chemistrynomenclature. It is both the negatively charged conjugate base of an amine (deprotonated form) and a carbonylcompound with an amine group at the alpha-position.

Choice Bis correct. The most reactive carbonyl compound has the best leaving group attached. Using the datain Table 1, the reactivity can be inferred from the fi-value. Alarger fi-value is indicative of greater reactivity.The largest 6-value is found with iodide, thus the acetoyl iodide is the most reactive of the compounds in thequestion. This eliminates choices A and C. The anhydride is more reactive than the ester, so the best answer ischoice B.

25.

Passage V (Questions 26 - 32) Thermodynamic versus Kinetic Control

26. Choice Cis correct. To be synthesized from 2-pentanone, the final product must have the new alkyl groupbonded to either the first or third carbon of the 2-pentanone. The answer choice compounds are drawn below:

Co D. n

27.

28.

A. 0 B. 0

C.H3CI%CH2C^ ^CH2CH2CH3 H3c" ^CHCH2CH3 H3c" XCH2CH2CH2CH2CH3 ^-C\^CH2CH2CH2CH

fFrom carbon 1

CH3From carbon 3 From carbon 5

CH2CH3

From carbon 3

Choice Cis formed from the alkylation of carbon 5, which CANNOT be carried out on 2-pentanone using theenolate synthesis pathway shown in the passage. The reaction can only occur on carbon 1or carbon 3, whicheliminates choices A, B, and D and confirms that choice C is the best answer.

Choice Cis correct. The product, 3-hexanone, results from the deprotonation of carbon 1(the less hindered andthus kinetic carbon). Product Ais formed at lower temperatures, implying that Product Ais the kinetic productBy association, Product A is 3-hexanone. At 0°C, the percentage of 3-hexanone is 47.3% while at 25°C thepercentage of 3-hexanone is 27.9%. The correct choice must have a percentage that lies between these twovalues. Only 35% lies within the range, so choice C is the best answer.

Choice Ais correct. The addition of a four-carbon alpha, beta-unsaturated ketone to butanone via a Michaelreaction results in an eight-carbon product, so choices Cand D(each with only seven carbons) are eliminatedThe kinetic enolate of butanone results from the deprotonation of carbon one. For the final product to come fromthe kinetic enolate, the product must have the Rgroup added carbon one from the reactant ketone (butanone).Ihe two remaining answer choices are drawn below, with the butanone in bold script and the new Rgroup shownin standard script: or

A. 0

A

B. o o

Kinetic carbon Kinetic carbon

Only in choice A, 6-oxo-2-octanone, did the alkyl group add to the kinetic (less hindered) carbon. Choice Aisthe best answer. In choice B, the alkyl group added to the thermodynamic carbon.

29. Choice Ais correct. Because LiN(CH2CH3)2 is abulkier base than LiN(CH3)2, the kinetic product will be morefavored with LiN(CH2CH3)2 than LiN(CH3)2. This should be observed at all temperatures. Product Ais thekinetic product, so the best answer is choice A.

®Copyright © by The Berkeley Review 76 CARBONYLS & ALCOHOLS EXPLANATIONS

30. Choice B is correct. Product A is the kinetic product, so it must be formed by way of the kinetic intermediate.This can be confirmed by the data in Table 1, which shows that an increase in temperature results in an increasein Product B and a decrease in Product A. The kinetic product decreases with increasing temperature. The bestanswer is choice B. Choices C and D can both be eliminated, because they are the same answer. The termtransition state intermediate is an oxymoron at best, because there is no such thing. Pick B and move on.

31. Choice A is correct. At 40°C using the small base potassium hydride, the thermodynamic product would be themajor product of the reaction. The more substituted alpha carbon is carbon 2 of the cyclic ketone. The ethylgroup will add to this carbon, resulting in both the presence of an ethyl group and a methyl group on the C-2carbon of the ring. The final product is 2-ethyl-2-methylcyclohexanone, choice A.

32. Choice B is correct. To offer competition between kinetic and thermodynamic enolate, the molecule must havehydrogens on two alpha carbons that are unequally substituted. In choice A, the two alpha carbons are carbons 2and 4. The two carbons are unequally substituted due to the presence of the methyl substituent on carbon 2. Inchoice B, the two alpha carbons are carbons 2 and 5. The problem is that carbon 2 has no hydrogen due to thepresence of the two methyl substituents on carbon 2. There can existno competition, because only one alphacarbon can be deprotonated on the molecule. In choice C, the two alpha carbons are carbons 1 and 3. The twocarbons are unequally substituted due to the presenceof the methyl substituent on carbon 3 (the fourth carbon inthe butanone chain). In choice D, the two alpha carbons are carbons 2 and 5. The two carbons are unequallysubstituted due to the presence of the methyl substituent on carbon 2. The methyl group on carbon three plays nopart in the reaction.

Passage VI (Questions 33 - 39) Alcohol Oxidation

33. Choice A is correct. When an alcohol is oxidized into a carbonyl (ketone when the alcohol is a secondaryalcohol), the oxidation state of carbon increases (an increase in oxidation state is associated with oxidation).Because the oxidation state decreases in choices C and D, they are both eliminated. The carbon in a secondaryalcohol is sigma bonded to twocarbons, one hydrogen, and an oxygen. The active carbon neither gains from norloses electrons to carbons to which they are bonded. The active carbon gains an electron from the hydrogen towhich it is bonded and it loses an electron to the oxygen to which it is bonded. Overall, the active carbon of asecondary alcohol neither gains nor loses electrons in its bonds, thus the oxidation state ofa secondary alcohol iszero. The correct answer must be choice A. Drawn below is a short cut technique for determining the oxidationstate by considering each of the four bonds to carbon.

OH

Oxidation,

CH, CHo

Overall: 0 + 0 +1-1= 0 Overall: 0 + 0 + 1 + 1= +2

34. Choice B is correct. Basic permanganate solution will oxidize a secondary alcohol into a ketone. The ketoneformed from the oxidation of 2-hexanol has the oxygen bonded to the same carbon in the both reactant andproduct, so the organic product is 2-hexanone. Because pentanal is an aldehyde, choice D is eliminated. Thetreatment of2-pentanol (also a secondary alcohol on a straight carbon chain) with basic permanganate solutiongenerates a yield of 81.7% (as listed in Table 1.) The yield when using 2-hexanol should be about the same, so81.1% is the best choice. The best answer is therefore choice B. The additional methyl group in 2-hexanolcompared to 2-pentanol should not cause much difficulty because it is located at the end of the chain, wheresteric hindrance is minimal.

35. Choice D is correct. Primary and secondary alcohols can be oxidized, while tertiary alcohols cannot beoxidized. The compounds 2-methyl-3-hexanol and 2-ethylcyclopentanol are both secondary alcohols, so theycan both be oxidized into ketones, eliminating choices A and B. Choice C (3,3-dimethyl-l-pentanol), is aprimary alcohol so it can be oxidized into either an aldehyde (in a non-aqueous environment) or a carboxylicacid (in an aqueous environment). The only tertiary alcohol in the answer choices is choice D, 2-methyl-2-butanol.

Copyright © by The Berkeley Reviev/ 77 CARBONYLS & ALCOHOLS EXPLANATIONS

36.

37.

38.

39.

Choice Dis correct. Because the solution turned green, there is evidence that some reaction took place. The colorgreen is not important, it is the fact that a color change transpired that is important. Na2Cr207 (sodiumdichromate) is an oxidizing agent, so the alcohol that is chosen must be an alcohol that can be oxidized.Phenols cannot be oxidized, so choice A is eliminated. Both 1-methylcyclohexanol and 3-ethyl-3-heptanol aretertiary alcohols, therefore neither of those alcohols can be oxidized. This fact eliminates choices B and C.The correct compound is2-methyl-l-octanol, a primary alcohol, making choice D the best answer.

Choice A is correct. Choice A represents the conversion of a four carbon primary alcohol into a four carbonketone. This is not possible, because the carbon bonded to oxygen has changed from the first carbon to the secondcarbon. The oxygen does not migrate in oxidation reactions. Choice Bis the oxidation of a primary alcohol (2-methyl-1-pentanol) into a carboxylic acid, which is possible. Choice C is the oxidation of a primary alcohol (1-pentanol) into an aldehyde, which is possible. Choice Dis the oxidation of a secondary alcohol (cyclohexanol)intoa ketone, which is possible. Only choice Ais not possible.

Choice Dis correct. Chromium gets reduced when chromic oxide oxidizes aprimary or secondary alcohol. Whenchromium is reduced, its oxidation state is reduced. Chromic oxide will oxidize ethanol (a primary alcohol) butit will not react with 2-methyl-2-butanol, because it is a tertiary alcohol. Because of this fact, the oxidationstate of chromium decreases with the addition of ethanol but the oxidation state of chromium remains constantupon the addition of 2-methyl-2-butanol. The best answer is choice D.

Choice Ciscorrect. To be anoxidizing orreducing agent, acompound must be a reactant in the reaction. The tworeactants are 2,4-pentadiol (C5H1202) and Ti02. Because the diol gets oxidized into a diketone functionality,it must be the reducing agent. This is true, because reducing agents get oxidized. This means that Ti02 is theoxidizing agent. Choose C, for correctivity glee.

Passage VII (Questions 40 - 46) Unknown Oxygen-Containing Compounds

40. Choice A is correct. Compounds that turn litmus paper red are acidic. Primary alcohols are oxidized intocarboxylic acids while secondary alcohols are oxidized to ketones. Tertiary alcohols do not oxidize. Thismakes choice A the correct choice.

Choice Dis correct. The formula C6H120 indicates that the compound has only one degree of unsaturation. Acyclic ketone would require two degrees of unsaturation (one for the ring and one for the carbonyl ic-bond). Byhaving only one unit of unsaturation, the cyclic ketone structure is not possible. Choose D.

Choice Cis correct. Compound Ais positive with Jones test, so it can be oxidized. Compound Ais also positivefor the nucleophile test, so it is likely an alcohol, given the only heteroatom is oxygen. Primary and secondaryalcohols can be oxidized, so Compound Ais either aprimary or secondary alcohol. This makes choice Aa validcorrelation, and thereby eliminates it. The iodoform test is used to determine whether the compound is amethyl ketone. Compound Bis the only unknown that tested positive with the iodoform reagent, so Compound Bis most likely a methyl ketone. This makes choice Ba valid correlation, and thereby eliminates it. By aprocess of elimination, only choices Cand Dare possible. Because Compound Cdid not react with Jones' reagent,it cannot be oxidized. An aldehyde can be oxidized into a carboxylic acid, so Compound C cannot be analdehyde. This makes choice Can invalid correlation, so choice Cis the best answer. Acompound that isnegative with all four tests in the passage, has one unit of unsaturation, and contains one oxygen atom can beeither aketone (but not amethyl ketone) or acyclic ether. Neither acyclic ether or ketone will be oxidized byJone's reagent or Tollen's test. Neither is nucleophilic, so they will not undergo a substitution reaction with analkyl halide in the presence of silver cation. Neither is a methyl ketone, so they will yield a negativeiodoform test. As such, there is achance that Compound Cis either acyclic ether or aketone, making choice Davalid statement. Choice D is eliminated.

Choice Cis correct. This question is a test of the information that you have memorized. Both primary alcoholsand aldehydes can be oxidized into carboxylic acids, so choices A and B are eliminated. A hemiacetal willconvert back to an aldehyde in the presence of water and either an acid or a base. Given that an aldehyde canbe oxidized into a carboxylic acid under either acidic or basic conditions, choice D is also eliminated. As ageneral rule, any carbon bonded to both an oxygen and a hydrogen, can be oxidized, with the exception of anether or ester. Of the choices, only the estercannot be oxidized. Choice C is the best answer.

41.

42.

43.

Copyright © by TheBerkeley Review® 78 CARBONYLS & ALCOHOLS EXPLANATIONS

44. Choice B is correct. An aldehyde can undergo oxidation to a carboxylic acid, reduction to a primary alcohol, or asubstitution reaction to form a new 7U-bonded compound such as an imine. The most common test involves theformation of a carboxylic acid using an oxidizing agent such as OO3 in H2S04 or KMn04 in KOH. These bothinvolve the transition metal in test reagent changing oxidation state and thereby changing color. This makeschoices A and C invalid, and both are eliminated. A testing reagent should always be the limiting reagent (inlower concentration than the compound for which it tests), so that the change in color is not masked by the excessreagent that did not undergo a color change. This eliminates choice D. The best answer is choice B. Althoughan aldehyde can be reduced into a primary alcohol and reducing agents are often rich in hydrogen atoms, thecommon reducing agents (LiAlH4 and NaBH4) do not undergo any phase or color changes. This means that acompound rich in hydrogen atoms is not a good choice for the reagent used to test for the presence of an aldehydefunctional group.

45. Choice C is correct. To begin with, in organic chemistry, oxidizing agents contain oxygen and are best describedas "rich in oxygen." All four of the compounds contain oxygen, so you must recall what each reagent does.KMn04 is used to oxidize an alkene into a vicinal diol, oxidize a primary alcohol into a carboxylic acid, oxidizea secondary alcohol into a ketone, or to oxidize an aldehyde into a carboxylic acid. This eliminates choice A.H3CCO3H is a peroxy acid which can be used to oxidize a ketone into an ester (known as the Baeyer-Villigerreaction) or to oxidize an alkene into an epoxide. This eliminates choice B. HIO4 is used to oxidatively cleavevicinal diols into two carbonyl fragments. This eliminates choice D. This leaves only choice C to choose.Mg(OH)2 is used as a base, but it has no immediate use in organicchemistry as an oxidizing agent.

46. Choice C is correct. LiAlH4 reduces many things into an alcohol, such as acids, ketones, aldehydes, and oh yes...esters. PickC for that fresh, minty, "I'm alive and picking right answers" sort of feeling. The H2/Pd reagent isused to hydrogenate (reduce) alkenes. HCl/Zn is used in the reduction of a ketone into an alkane (Clemmensenreduction), and BH3 is used to add in an anti-Markovnikov fashion to an alkene.

Passage VIII (Questions 47 - 53) Wolff-Kishner Reduction versus Clemmensen Reduction

47. Choice C is correct. Wolff-Kishner reduction is carried out on both ketones and aldehydes. The question isreduced to: "Which of the following reagents can react with either an aldehyde or a ketone?" Lithiumaluminum hydride reduces both ketones ands aldehydes into alcohols, so choice A is eliminated. A Grignardreagent, such asmethyl magnesium bromide, reduces both aldehydes and ketones into alcohols ofone more alkylgroup. Choice Bis eliminated. Clemmensen reduction iscarried outon the same compounds that Wolff-Kishnerreduction is carried out on, so choice D is eliminated. Ozonolysis with oxidative workup can be used to makeketones and carboxylic acids, but it does not react with ketones (or aldehydes). This makes choice C the bestanswer.

48. Choice B is correct. Choice A can be eliminated immediately, because reduction corresponds with a decrease inoxidation state, not an increase. Tiie central carbon of a ketone has two bonds to oxygen and two to other carbons,so it starts with an oxidation state of +2. This eliminates choice C. The carbonyl carbon is converted into analkyl carbon with two bonds to carbons and two bonds to hydrogens, so it finishes with an oxidation stateof -2.This makes choice B the best answer. The conversion and oxidation determination is shown below.

Loss of 4 e

49. Choice B is correct. Following Wolff-Kishner reduction, a carbonyl carbon is converted to an alkyl carbon. Therest of the molecule remains intact, so no new chiral center is generated. This eliminates choices A. Thecarbonyl carbon starts with s/?2-hybridization (three a-bonds and one 7t-bond) and finishes with sp3-hybridization (four a-bonds), which makes choice B a true statement. The carbonyl carbon loses bonds to oxygenand gainsbonds to hydrogen in the process,so it is reduced. However, nitrogen losesbonds to hydrogen and gainsbonds to nitrogen, so it is oxidized. Each nitrogen goes from an oxidation state of -2 to 0 during the reaction, soboth nitrogen atoms are oxidized. This eliminates choice C. Figure 1 shows that one water molecule, not three,is formed overall, so choice D is invalid, and therefore eliminated.


50. Choice D is correct. The hydrazine derivative is definitely larger than ammonia, NH3, but its greater size doesnotgenerate greater acidity due tosteric hindrance. The proton isnotbeing driven off bycrowding, sochoice Aiseliminated. The hydrazine moiety does not have any cyclic array of 7t-atoms, so it does not become aromaticupon the gain or loss of a proton. This eliminates choice B. Hydrazine may in fact have a stronger inductiveeffect (from the second nitrogen) than ammonia, so choice Ccannot be eliminated. When the proton is lost fromnitrogen in the hydrazine derivative, the new lone pair is located onan atom adjacent to a 7t-bond. This allowsfor the lone pair to resonate to the carbon of the rc-bond, offering it more stability due to resonance. This makeschoice D a true statement and the best answer.

51.

52.

53.

R. ,R R. R

c)C©

H. N

0îH N

N^

Choice Ais correct. Hydrazine and basic water reduce ketones and aldehydes into alkyl groups. They leaveother functional groups intact, unless they are reactive with basic water. The hydroxyl group in unreactiveunder basic conditions, so it will remain unaffected. This means that the product must contain ahydroxyl groupand an ethyl group on the cyclohexane backbone. Choices Cand Dare eliminated, because they don't includethe hydroxyl group. The alcohol functional group gets higher priority than an ethyl group according to IUPACconvention, so 3-ethylcyclohexanol is the correct IUPAC name for the compound, making choice A the bestanswer.

Choice Ais correct. Wolff-Kishner reduction requires basic conditions while Clemmensen reduction requiresacidic conditions. Both involve an aqueous environment, so neither can be employed for water sensitivecompounds. This eliminates choice D. Wolff-Kishner reduction, by employing a hydrazine molecule, takesadvantage of acarbonyl compound's ability to react with anucleophilic nitrogen. It should not be employed fora molecule that is reactive with an amine at a site other than the carbonyl carbon. This eliminates choice B.No acid is employed in a Wolff-Kishner reduction, so it is safe for acid sensitive molecules. Clemmensenreduction uses acidic conditions, so it will react with acid sensitive groups. This means that Wolff-Kishnerreduction is preferential to Clemmensen reduction when the molecule is acid sensitive. Choice A is the bestanswer.

Choice A is correct. Water is the side product when hydrazine reacts with a ketone such as 2-methylcyclopentanone, so the product is most easily determined by removing the oxygen of the carbonyl carbonand two hydrogens from the nitrogen in phenylhydrazine that has two nitrogens and then connecting the carbonand nitrogen. This makes choice A the best answer.

N

1

OII

H3CX<-iH,C

Passage IX (Questions 54 - 60) Grignard Reaction

54. Choice A is correct. It is known that anhydrides and acid halides are highly reactive, so choices Cand Darecan be eliminated. Athio ester C—S bond is weaker than an ester C—O bond, so the least reactive carbonylcompound is the ester (choice A). The chart of pKa values can be used to verify this theory. An alcohol has ahigher pKa value than the thiol, so the alkoxy leaving group is less stable between choices Aand B, makingchoice A the best answer.

Copyright ©by The Berkeley Review® 80 CARBONYLS & ALCOHOLS EXPLANATIONS

55. Choice Cis correct. The Grignard reagent (methyl magnesium bromide) adds on one carbon (a methyl group), soin order form 2-pentanol as the product, you need to react H3CMgBr with a 4 carbon carbonyl reactant and addthe methyl group to the first carbon of the four carbon chain. Because there are six carbons in the products ofchoice A and choice B, the choices are narrowed down to either choice C or choice D based on the four-carbonelectrophile requirement. The best way to get a secondary alcohol as the product is to react H3CMgBr withbutanal, which will yield the product with the alcohol group on carbon number 2. Pick C for analtogether greatexperienceand correct answer. Choice D will form 2-methyl-2-butanol.

56. Choice D is correct. The reaction of a Grignard reagent with a carbonyl yields an alcohol. Stereoisomers areformed when the alkyl groups are all different on the alcohol carbon. For this to be true, the Grignardnucleophile must add an alkyl substituent different from the alkyl groups present on the carbonyl. Choice A iseliminated, because the ketone has an ethyl and methyl group attached. The Grignard reagent adds an ethylgroup. The alcohol that is formed is achiral (3-methyl-3-pentanol). Choice B is eliminated, because the ketoneis symmetric (has two ethyl groups), so the product must be symmetric and thus achiral. An ester when reactedwith excess Grignard reagent yields an achiral tertiary alcohol (it adds the same R-group twice), thus choice Cis eliminated. Choice D is best, because the final alcohol has an ethyl group, methyl group, and hydrogenattached. This makes the alcohol asymmetric and thus chiral. Tiie product mixture is two enantiomers.

57. Choice B is correct. Aldehydes are carbonyls with at least one hydrogen and often one carbon attached to thecarbonyl carbon. The addition of a Grignard reagent places a second carbon on the electrophilic aldehyde siteand thus reduces the aldehyde to a secondary alcohol. Pick B for the inner peace that a correct answer can bring.

HoC

RMgBiOMgBr H3C OH

H30^

cAh58. Choice B is correct. An ester has both a leaving group and a carbonyl with which to react. The first R-group to

attach converts the ester into a ketone by displacing the alkoxy leaving group. The ketone that is formed as theintermediate product can react with a second equivalent of the Grignard reagent and thus add a second R-group.The final product is a tertiary alcohol with two identical R-groups. The conclusion is that a Grignard reagentreacts twice (not just once) with an ester. Pick B to tell the world "I know Grignard!"

59. Choice D is correct. The Grignard reagent is a strong base, so an aprotic solvent such as ether is chosen. Thismakes choice A a valid statement. Because the carbonyl compound is converted into an alcohol in a Grignardreaction, a bond to oxygen has been lost. Tliis means that the carbon has been reduced, making choice Bvalid. AGrignard reagent can be used to form a secondaryor tertiary alcohol, as shown in the passage. To form a primaryalcohol, a Grignard reagent can be added to either formaldehyde or the epoxide ethylene oxide. This makeschoice C valid. Because the Grignard reagent is a strong base, it can react with the weakest of acids to form analkane. This is why the reaction must be carried under anhydrous, aprotic conditions. It is thus not true that theGrignard reagent should be added under acidic conditions to obtain the highest yield. Pick D and move on withanother correct answer under your belt.

60. Choice C is correct. Propanoic acid is the conjugateacid of propanoate, which when bonded to carbonyl groupforms an anhydride. This question is asking for the relative reactivity of an anhydride. Acid chlorides aremore reactive than an anhydride, so choice A is eliminated. Amides are less reactive than an anhydride, sochoice B is eliminated. Esters are less reactive than an anhydride, so choice C is true. The difference inreactivity between anhydrides will not be distinguishable by pKa values that only differ by 0.1. There is adifference in steric hindrance to consider. Choice D may or may not be true, but regardless, choice C is definitelytrue, thus choice D is assumed to be false within the context of this question.

Passage X (Questions 61 - 67) Aldol Condensation and Alpha Hydrogens

61. Choice B is correct. The most acidic hydrogen on the molecule is the H bonded to the alpha carbon (marked onthe molecule as carbon b). The best answer is choice B. The aldehyde proton is not as acidic as the alpha proton,because the lone pair that would form on the carbonyl carbon is not stabilized by resonance, like the alphacarbon.


62. Choice D is correct. When counting carbons starting from the carbonyl, the molecule is a 4-carbon chain with amethyl group bonded to the number 2 carbon. The compound has an aldehyde functional group, so the correctIUPAC name is 2-methylbutanal, answer choice D.

63. Choice A is correct. An aldehyde with excess alcohol in the presence of an acid will lose a water molecule, andgo on to form an acetal. It is an equilibrium reaction, but the excess alcohol will push the reaction to the acetal(product side of the reaction). A hemiacetal is formed when the reaction is carried out under basic conditions.Choice A is the best answer.

64. Choice Ais correct. The reaction isanaldol condensation. The acid workup causes elimination and thus ensuresthe formation of the a,fi-unsaturated ketone from the fi-hydroxyketone intermediate product. Choice D iseliminated, because it has seven carbons, which is not possible. Choice Cisnot ana,fi-unsaturated ketone, so itcan be eliminated. Choice B, when broken apart by aretro-aldol reaction, would yield fragments of two and fourcarbons, not three and three carbons. Choice Bis eliminated. The only cc,fi-unsaturated ketone composed of twothree carbon units is choice A.

65. Choice Dis correct. This isjust one ofthose things that you need to learn. The conversion ofa ketone into anenolor an enol into a ketone is referred to as tautomerization. Select D as your answer. Tautomerization is involvedin glycolysis in the conversion of glucose-6-phosphate into fructose-6-phosphate, which is catalyzed byisomerase.

66.

67.

Choice Cis correct. This is acase of recognizing the weakest base. Only one of the answer choices is not astrongenough base to deprotonate an alpha hydrogen, so it must be the weakest base. Carbonate, C032", is not averystrong base, because the electrons are stabilized by resonance involving the double bond (C=0 Jt-bond). It istherefore the weakest base out of the choices listed, and thus is the one not strong enough to deprotonate the Hon the alpha carbon. Choice Cis your choice for abrighter tomorrow, assuming of course that the brightness oftomorrow depends onyour answer choice for this question.

Choice Cis correct. Only a ketone and an aldehyde can undergo aldol condensation reactions. An acid wouldrather lose the proton from its hydroxyl group than lose the less acidic alpha proton to undergo the condensationreaction. Choice C, the carboxylic acid, cannot undergo aldol condensation.

Passage XI (Questions 68 - 73) Claisen Condensation Reaction

68. Choice Bis correct. The most acidic proton on afi-ketoester is the alpha hydrogen that when lost results in acarbanion conjugated to the most 7t-bonds. Choices Cand Dare eliminated, because neither is an alphahydrogen. Choice Ais only conjugated to the keto carbonyl, while choice Bis conjugated to both the ketocarbonyl and the ester carbonyl. The most acidic proton is proton b, thus you should pick choice B.

69. Choice C is correct. As stated in the passage, the lowest yield is found with the fi-ketoester anionintermediate that has no alpha hydrogen. For this to occur, the starting ester must have only one alphahydrogen. Of the choices, only choice Chas just one alpha hydrogen. You should pick C. The reaction is shownbelow.

O

H,C H

OEt'OEt.

HOEt"

HoC

Nothing protonates the alkoxide, thus theequilibrium favors the reverse reaction.

O O

H,C

H3C

OEt + "OEt

CH,

No alpha hydrogen, therefore the alkoxide leavinggroup cannot deprotonate the final fi-ketoester.

70. Choice Ais correct. AClaisen reaction is not possible with esters that have no alpha hydrogen. Of the choicesonly choice Ahas no alpha hydrogen. This makes choice Athe best answer. As apoint of interest, this moleculeisnot known toexist asa stable compound at room temperature.


71. Choice D is correct. The product of an ester reacting with itself by way ofa Claisen condensation reaction isrecognizable from the carbon chain length on thecarbonyl portion of the ester. In the answer choices, the twoalkyl groups can behighlighted to recognize the alkyl groups that must bepresent on the two esters that react,considering that only the alkoxide leaving group is lost. The alkyl groups are highlighted for choice A are:H3CCH2COCH2CO2CH2CH3, therefore choice A is formed from the condensation of two different esters:H3CCH2CO2CH2CH3 and H3CCO2CH2CH3. This same analysis of the product for choices B, C, and D willlead to the conclusion that choice D is the correct answer. In choice D, the starting ester is(C6H5)CH2C02CH2CH3 which leads to (C6H5)CH2COCH(C6H5)C02CH2CH3.

JUChoice A: O O

A ♦ AH3CH2C OEt H3C OEt

NaOEt

HOEt H3CH2C

Choice B: O O

A ♦ AH3C OEt H3CH2C OEt

NaOE£HOEt H3C

Choice C: O O

A ♦ AHgCgHjC OEt H3C OEt

NaOEt

HOEt HsCgHjC

Choice D: O O *'

OEt

AOEt HgC^C OEt HOEtNaOEt

O O

OEt

CH,

.JUOEt

OO

AH5^6H2^ H5C6H2C OEt

C6Hs

72. Choice A is correct. The starting molecule possesses two ester sites so an intramolecular Claisen reaction ispossible. An intramolecular Claisen reaction results in a cyclic fi-ketoester. This eliminates choices B and D.The chain contains six carbons, so choice C can be eliminated, because it requires seven carbons excluding thealkoxy group to form a six-carbon ring. Thebest answer is choice A. The reactionis shown below.

O O O

EtOOEt

NaOEWHOEt EtO

OEtEtO

O

73. Choice D is correct. The two esters can either react with each another or they can react with themselves. Thisresults in four structural isomers beingformed. Of the structural isomers, two have one stereocenter, thereforetwo of the four structural isomers have two possible stereoisomers (enantiomers). The total number of isomersistherefore six. Choice D is correct. The isomers are drawn below:

O

A + H3CH,C OEt

Copyright © byThe Berkeley Review®

O

A

O O

X^K.H,C '— OEt HoCHoC3 AA 3 ^ BA

O O

AÂ

NaOEtO O

AB =

CH3

O o

O O

AÂC OEt HOEt

AH H

B

H,C OEt H3CH2CBl £

CHo

O O

X^X,BB I

CH3 —3

83 CARBONYLS & ALCOHOLS EXPLANATIONS

OEt

OEt

OEt HgCHfeCT ^T OEt

CH,

Passage XII (Questions 74 - 80) Transesterification Reaction

74. Choice A is correct. The ratio of the integrals for each signal in the proton NMR of spectrum I is given as 3 : 2 : 3.This implies that the compound has either eight total hydrogens or some multiple of eight total hydrogens. Ofthe choices, choice B (methyl ethanoate) has six hydrogens and choice D (ethyl methanoate) has sixhydrogens. This narrows the question down to either choice A (methyl propanoate) or choice C (ethylethanoate). The key feature in the spectrum is the singlet far down field caused by a methyl bonded to theoxygen. This peak implies that the compound must have a methoxy group on the carbonyl carbon. Pick A foroptimal satisfaction. The structures are drawn below:

a bH3CH2C

O

OCH,

Methyl Propanoatea: Triplet (3H)b: Quartet (2H)c: Singlet (3H)

O

H3C OCHo

Methyl Ethanoatea: Singlet (3H)b: Singlet (3H)

o

H3C"b c

OCH2CH3

Ethyl Ethanoatea: Singlet (3H)b: Quartet (2H)c: Triplet (3H)

The bolded hydrogen will be found the furthest down field.

O

a/^\* b cH OCH2CH3

Ethyl Methanoatea: Singlet (1H)b: Quartet (2H)c: Triplet (3H)

75. Choice Dis correct. The ratio of the integrals for each signal in the proton NMR for spectrum inFigure 2is 2:2:3 : 3. This means that the major compound isolated from solution has ten hydrogens. Given that the reactanthas eight hydrogens, it can be concluded that some reaction took place, which eliminates choice A. Choice B(ethyl acetate) has eight hydrogens, so choice Bis eliminated. This narrows the question down to either choiceC (propyl ethanoate) or choice D (ethyl propanoate). Key features in Spectrum II include the quartet fardownfield and the absence of a singlet. Choice C(propyl ethanoate) would show a singlet for the CH3 adjacentto the carbonyl. Because Spectrum II has no singlet, choice C is eliminated. The presence of the quartetdownfield implies that the compound must have an ethyl group on the ester oxygen. Pick Dto know what rightfeels like. The structures are drawn below:

76.

77.

Methyl Propanoatea: Triplet (3H)b: Quartet (2H)c: Singlet (3H)

O

a ^^^ b cH3C OCH2CH3

Ethyl Ethanoatea: Singlet (3H)b: Quartet (2H)c: Triplet (3H)

o

a y—\ bedH3C OCH2CH2CH3

Propyl Ethanoatea: Singlet (3H)b: Triplet (2H)c: Sextet (2H)d: Triplet (3H)

a bH3CH2C

O

X c d

OCH2CH3

Ethyl Propanoatea: Triplet (3H)b: Quartet (2H)c: Quartet (2H)d: Triplet (3H)

The bolded hydrogen will be found the furthest down field.

Choice Bis correct. An ester is more reactive than all carbonyl compounds that have aleaving group worse thanthe alkoxide leaving group. Leaving groups worse than the alkoxide leaving group would be more basic thanthe alkoxide anion. Because both halides and carboxylates are good leaving groups (as well as weak bases),acid halides and acid anhydrides are both more reactive than esters. This eliminates choices A, C, and D. Theonly choice remaining is amides, choice B. The NR2" leaving group of the amide is a bad leaving group,therefore amides are not very good electrophiles. Choice Bis the best selection.

Choice C is correct. By using an amine nucleophile rather than an alcohol nucleophile, the product that isformed is an amide as opposed to an ester. The amine is a better nucleophile than the alcohol is in Reaction II,so the reaction should be more favorable than transesterification. The equilibrium constant for the reaction isgreater than one. Regardless, the question asks for the product of the reaction, which is the amide formed whenthe amine replaces the alcohol leaving group. The amine nucleophile is N-ethylamine therefore the bestanswer is N-ethylpropanamide (H3CCH2NHCOCH2CH3), choice C. This can be determined by just adding theethyl amine in same way as the ethanol is added to the ester in the sample reaction carried out at reflux.

Copyright © by The Berkeley Review( 84 CARBONYLS & ALCOHOLS EXPLANATIONS

78. Choice Bis correct. The purpose of the acid in the reaction mechanism is to protonate the carbonyl to make theester more electrophilic. In a later step, the acid protonates the alkoxy leaving group to form an alcohol, whichis abetter leaving group than alkoxide. The role of the acid in the reaction is best stated as serving to protonatethe ester electrophile and in doing so make itmore reactive. This makes choice Bcorrect and desirable to you.

79. Choice Discorrect. If water were the nucleophile in lieu of the alcohol, the product would be a carboxylic acid.When carried out under basic conditions, this reaction is known as saponification. Hydrolysis of an ester iscommon inbiological systems. The carboxylic acid formed in the reaction would have the OCH3 group replacedby an OH group. This carboxylic acid would bepropanoic acid (CH3CH2CO2H), choice D.

80. Choice C is correct. The deuterated solvent is used so that the solvent (which normally contains hydrogens)does not interfere with proton NMR spectrum of the compound. The deuterium atoms donot register onprotonNMR, so the solvent is invisible when it is deuterated. The best answer is therefore choice C. Deuteriumexhibits little to no affect on pH in organic solvents. This eliminates choice D. The solvent should have noeffect on the magnetization of the NMR sample nor should thesolvent preventany reaction as solvents are inert.Choices A and B can both be eliminated.

Passage XIII (Questions 81 - 87) Malonic Ester Synthesis

81. Choice B is correct. The structure of urea must be determined by comparing the final barbiturate structure(product of the lower pathway) with the substituted malonic ester. Urea must contain two nitrogens bonded to acentral carbonyl. The alkoxy groups of the ester are leaving groups when the urea substitutes. The structuraldeduction is drawn below. Tiie drawing represents retrosynthetic analysis of the final product. Pick choice B.

XHN NH

PO o

R

Urea

O

1H2N NH> CH3CH20

O O

OCH2CH3

R

82. Choice B is correct. Urea has a carbonyl carbon separating the two nitrogen atoms, so if the carbonyl oxygen ofurea is replaced by two methyl substituents on the carbon, then the final product will be the same as the productfrom urea, except that it will have two methyl groups at the formerly carbonyl carbon. It is not different at anysite other than the carbon between the two nitrogens (the former carbonyl carbon). The correct choice must havethe two methyl groups bonded to the central carbon in lieu of the carbonyl oxygen. The best answer is choice B.To answer this question requires knowing what urea looks like. If you don't know, then you must predict whatwould happen if the diamine were added to the diester. Tiie correct choice can be deduced if you know that theproduct of an amine and ester is an amide.

83. Choice C is correct. Decarboxylation occurs with 6-ketoacids. All four answer choices have carboxylic acidfunctionalities, but only choices A, B, and D have fi-keto groups on the backbone. The ketone functionality ofchoice C is on the gamma carbon, therefore the molecule will not undergo decarboxylation. Tiie best answer istherefore choice C.

A. O

IH3C P

O B. O O

1OH HO P

C.

OH

OH H3C £ • O

84. Choice A is correct. The strong base used to deprotonate the alpha carbon can also act as a nucleophile byattacking the carbonyl carbon. Sodium ethoxide is chosen as the base so that it matches the substituent on thecarbonyl carbon of the ester. This means that if the ethoxide undergoes transesterification, it will generate thesame ester (the leaving group of the ester is the same moiety as the nucleophile). Sodium ethoxide is a strongenough base to deprotonate the alpha hydrogen, but it will not change the ester when it undergoestransesterification. The best answer is choice A.

Copyright © byThe Berkeley Review® 85 CARBONYLS & ALCOHOLS EXPLANATIONS

85. Choice D is correct. Malonic ester has three types of protons, thus there are three signals in its proton NMRspectrum. The structure of malonic ester is shown below with each of the hydrogens labeled:

H3CH2CO'

O O

C\ /C\ b aC OCH?CHo

AHc

IIc

a: 6 hydrogens adjacent to 2H thus a 3H tripletb: 4 hydrogens adjacent to 3H thus a 2H quartetc: 2 hydrogens adjacent to OH thus a 1H singlet

Based just on the ethyl groups, resulting in a 2H a quartet and a 3H triplet, the best answer is choice D. The trueratio of hydrogens is 6 : 4 : 2, but because NMR shows only the relative quantities and not the absolutequantities, the NMR shows a ratio of 3 : 2 : 1.

86. Choice B is correct. There are four carbons added to the central carbon of malonic ester, so choices A and C canboth be eliminated. The addition of methyl bromide followed by theaddition of ethyl bromide to malonic esteronly adds three carbons total, not four. This eliminates choice D. A secondary alkyl bromide like 2-bromobutane must be added to form the tertiaryRsubstituent. Choice Bis the best choice.

87. Choice Bis correct. The product will be the exact product shown in the lower synthesis pathway in Figure 1,only the generic Rgroup is now a straightchain propyl group. This makes choice B the best answer.

Passage XIV (Questions 88 - 94) Carboxylic Acids

89.

90.

91.

Choice D is correct. To synthesize butanoic acid, one ofseveral methods can be applied. Oxidation ofa primaryalcohol will work, so choice A is feasible. Hydrolysis of an ester will form an acid, so the four carbon carbonylfragment of the ester will become a four carbon carboxylic acid. Choice Bis a viable synthetic pathway.Ozonolysis of an alkene works if the addition ofozone is followed by oxidative workup and the alkene musthave vinylic hydrogens (hydrogens on the carbons of the rc-bond). Choice C is thus valid. Treatment of an alkylbromide with the cyanide nucleophile followed by acidic workup forms a carboxylic acid. The problem withchoice D is that the carbon chain increases by one carbon with the cyanide nucleophile so the product will bepentanoic acid, not butanoic acid. The best answer is choice D.

Choice D is correct. Because the pKa for propanoic acid is 5.0, it is a weak acid, implying that it onlypartially dissociates in water. With a pKa value of 5.0, the Ka value is 1.0 x 10"5. This means that for a 1.0 Macid solution, the concentration of both H+ and conjugate base are 10"2-5, which is less than 10"2, or 1%. Thedissociation is less than 1%. The Ka value is applied as follows:

. [CH3CH2C02-][H30-] =[CH3CH2CQ2f = ^ ^ = ^[CH3CH2C02H] [CH3CH2C02H]

This implies that the relative value of [CH3CH2C02"j to [CH3CH2C02H] is VlO x 10"3 to 1which is roughly3.1 x 10"- to 1. This value is less than one percent confirming choice D.

Choice Bis correct. A carboxylic acid will turn blue litmus paper red when added to it. From the nature ofthepassage, it should be implied that the litmus paper is used to detect the acid. In choice A, a primary alcohol istreated with chromic oxide and sulfuric acid, which oxidizes the primary alcohol into a carboxylic acid.Choice A results in a product that will convert blue litmus paper to red, so choice A is valid. In choice B, thealkene is tetrasubstituted, therefore the carbonyls that are formed must beketones. They cannot be oxidized (orreduced) into a carboxylic acid. This means that the product in choice Bwill not convert blue litmus paper tored, so choice Bis the best answer. In choice C, the anhydride is readily hydrolyzed into two equivalents ofacetic acid. This eliminates choice C. In choice D, an ester is hydrolyzed into an alcohol and a carboxylic acid.This eliminates choice D. The best answer is choice B.

Choice Discorrect. The sequence of reagents used will add a carboxylic acid group to the alcohol carbon (carbon2). It required background information to know that PBr3 converts an alcohol into an alkyl bromide. If youdidn't know that before, then it's officially background knowledge now. This final product is composed of acarbon with an ethyl group, methyl group, and a carboxylic acid group attached. This makes choice D the bestanswer. Choice C should have been eliminated, because it contains six (rather than five) carbons.


92.

93.

Choice Bis correct. As a general rule, hydrolysis reactions are reversible, because each step is an equilibriumstep. The Grignard reaction is not reversible however, therefore the best answer is choice B. Choices A and Dshould have been eliminated, because a lactone is an ester (a cyclic ester), thus they are the same answer.

Choice D is correct. As a general rule, the pKa value of an acid is roughly 5.0. This makes choice D not true.Acids do result inaqueous pH values less than 7.0, because they are acidic (choice A). The broad IR peak resultsfrom the hydrogen bonding and is found between 2500 cm"1 and 3000 cm"1 (choice B), less than an alcohol stretchbecause the bond is weaker (and thus easier to stretch). Acids will exchange protons with a protic solvent, sothe acidic hydrogen can readily be deuterated. This results in the disappearance of a peak in the proton NMR(choice C).

94. Choice D is correct. As listed in the passage, the addition of thionyl chloride (SOCl2) followed by an amine toa carboxylic acid will result in the formation of an amide. This eliminates choices A and C. An aldehyde isoxidized into a carboxylic acid by the addition of H2S04 and Cr03 as shown in the passage. The best answer istherefore choice D.

Not Based on a Passage (Questions 95 -100) Carbonyls and Alcohols

95. Choice A is correct. Under basic conditions, the alkoxy nucleophile group attacks the s/?2-hybridized carbonylcarbon of the ester (see below) to form an s/?3-hybridized tetrahedral intermediate with the negative chargeshifted to the carbonyl oxygen. The best answer is choice A.

aR'O

R"0"

R OR"

Intermediate

0

R OR'

96. Choice B is correct. The Claisen condensation reaction involves the deprotonation of an alpha hydrogen on anester, resulting in a highly nucleophilic carbanion, which attacks another ester in solution. The ultimateproduct of a Claisen condensation reaction is a 6-ketoester. The Claisen condensation reaction definitely uses anester reactant, so choice A is eliminated. A Grignard reaction involves the addition of a nucleophilic alkylmagnesium bromide species to a carbonyl compound. An ester certainly qualifies as a carbonyl compound. TiieGrignard reagent can add twice to an ester, resulting in the synthesis of a tertiary alcohol with at least twoidentical alkyl groups. The Grignard reaction can use an ester reactant, so choice C is eliminated.Transesterification, as the name implies, involves the changing of an ester by changing the alkoxy group. Thisis done by substituting one alkoxy group for another on the carbonyl carbon. A transesterification reactiondefinitely uses an ester reactant, so choice D is eliminated. Friedel-Kraft alkylation is not a reaction you arerequired to know, but that is irrelevant in this question. The term alkylation implies that an alkyl group isbeing added, and this does not lend itself to the use of an ester. An ester is likely to add an acyl group when itreacts. This means that a Friedel-Kraft alkylation does not require an ester reactant, so choice B is the bestanswer.

97. Choice A is correct. Treatment of a fi-diacid with heat will drive off water (dehydrate the 6-diacid) to formthe corresponding acid anhydride. Choice A is an acid anhydride, so it is the best answer. The reaction isreversible, so acids can be formed from anhydrides upon the addition of water.

98. Choice C is correct. In order to be reduced, the compound must have a carbon with at least two bonds to oxygen(both a CT-bond and a 7t-bond from carbon to oxygen). Only choice C (a ketone) has two bonds to oxygen, so it mustbe the best answer. Phenols, tertiary alcohols, and ethers all involve carbons that only have one bond to oxygen,so they are all eliminated.

99. Choice C is correct. Carbon number three, the third carbon, of the alcohol is involved in no 7t-bonds and it isbonded to four other atoms, so it must have sp3-hybridization. Pick C for optimal results.


100. Choice C is correct. Secondaryalcohols, when oxidized, form ketones. The product following the oxidation of2-butanolusing a chromium VI species is 2-butanone. Theproduct is drawn below along with its proton NMRpeaks. Pick choice C.

O|| a: 3 hydrogens next to a carbonwith no hydrogens:Singlet (3H)C b ^*CH3 b: 2hydrogens next toa carbon with3hydrogens: Quartet (2H)

H-C CH2 c:3 hydrogens next to a carbonwith 2 hydrogens:Triplet (3H)

C4J%0 f 1a

/ C

b

i I1\ \ 1 1

I2 ppm

11 ppm

1Oppm

:y,\-/r}-4??'K'^l ^?^W'^>]Ay --"• iT> '%'>'* >

"Better living through chemistry!"

Copyright © byTheBerkeley Review® 88 CARBONYLS & ALCOHOLS EXPLANATIONS

Section VI

Carbohydratesby Todd Bennett

D-Glucose Hj>0

OÔH

H-

HO-

H-

H«

OH

H

OH

OH

CBjOH

H

4

HO

OH HOH2C

} 3 2H 3 OH3

H

2 H

OH

fi-D-glucopyranose

HOH2C HOH2C

OH

4-0-(6-D-glucop3n:anosyl)-6-D-glucopyranoside

Monosaccharides

a) Straight Chain Formi. Aldoses and Ketoses

ii. Pentoses and Hexoses

iii. Common D-Sugarsb) Stereoisomerism

Epimersii. Anomers

iii.Haworth Projectionsiv. Mutarotation

v. Isomerization

Oligo- and Polysaccharides

a) Linkages (Acetal Connectivity)b) Disaccharides

c) Trisaccharides

d) PolysaccharidesLinkages

ii. Cellulose and Starches

iii. Branching

Chemical Reactions and Tests

a) nitric Acid Sugar Oxidationb) Osazone Test

c) Benedict's Test

d) Tollen's Test

e) Periodate Oxidation

f) Kiliani-Fisher Synthesisg) Ruff Degradationh) Wohl Degradation

Biological Applications

a) Blood Typesb) Glycolysis

BerkeleyJLJr-E-V-KE'W®


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CarbohydratesSection Goals

Be able to recognize common monosaccharides.

There are certain monosaccharides that recur in both organic chemistry and bicKhemistrythat youshould recognize andbeable todraw. These include glucose, ribose, fructose, mannose, andgalactose.It may be easiest to recall how the other sugars relate to glucose. For instance, mannose is the C-2epimer of glucose. If you know the structure of glucose, then you know the structure of mannose.

Be able to inter-convert between Fischer and Haworth Projections.Some passages and questions will assume that you can translate between structures. For instance,the Fischer projection may be given in the passage, but the question may center around the structureina Haworth projection. In translating structures, there arethree separate points to observe. Thechirality ofthe anomeric carbon isdependent onthe direction ofthe attack. The chirality ofthepenultimate carbon (carbon five in aldohexoses) isconstant according to theDorLstatus, andthebackbone hydroxyl groups on the left in the Fischer projection are up in the Haworth projection.

Be able to solve the chirality of an unknown sugar from reaction data.Fromorganicchemistryclass,you may recall solvingthe identityofa sugar by evaluatingthe reactiondata presented. This makes for an ideal passage, although the MCAT writers rarely use thisinformation in a passage, justsingle questions. \ou mayrecall evaluating whether a sugarwouldoxidize into an optically active or inactive diacid.

Be able to recognize the chirality of cyclic and linear sugars.It is rather simple to solve for the chirality of selected carbons in both the Fischer and Haworthprojections, after you have done this once. Therules are the same, so determine a quick way toidentifythe R and Sconfiguration ofeachhydroxyl group of tine sugar.

Be able to identify common disaccharides.As with the monosaccharides, there are certain disaccharides thatrecur in organic chemistry andbiochemistry. Thedisaccharides you must recognize includesucrose, lactose, and maltose. Knowthe twomonosaccharides that comprise thedisaccharide and theglycosidic linkage bindingthem.

Be able to recognize the linkage associated with a given disaccharide.The linkageof a disaccharide is definedby the chiralityof the anomericcarbon and the carbonofthe sugar containing oxygen ofthe linkage. The disaccharide that presents thegreatest difficultyis sucrose, which is composed of two anomeric carbons (onethat is a-1 and the other that is fi-2).

Be able to distinguish epimers and anomers.

You mustbe able to identify thedifference in chirality between diastereomers, whichin the sugarquestions are commonly presented as eitheralpha or beta in regards to an anomer or epimers.

Know the common reactions involving sugars and sugar derivatives.This booklet contains a largenumberofsugarreactions, perhapsmorethanyouneed. Keyreactionsinclude theKiliani-Fischer synthesis, Ruff degradation, osazone formation, andnitric acid oxidation.The passage usually provides plenty of reactions, so don't memorize them, understand them.

Know the common polysaccharides and their biological significance.You must recognize thedifference between glycogen and cellulose. Thestructuraldifference simplyinvolves the linkage,but the difference in reactivity is significant.

Organic Chemistry Carbohydrates

CarbohydratesCarbohydrates are organic compounds that contain carbon, oxygen, andhydrogen in a 1 : 1 : 2 ratio. The term is derived from carbon that has beenhydrated in a ratio of one carbon to one water,which explains the C : H : O ratioof 1 : 2 : 1. When we consider carbohydrates, we typically think of them assugars. For this section, we will focus exclusively on sugars and their organicchemistry and biochemistry applications and examples. While we take adecidedly organic chemistry perspective on this material, you should pay extraattention to any subjects that bridge organic chemistryand biochemistry. Someof this material will overlap with topics in the metabolic components andmetabolic pathways sections of the biology books, but it will be presented in amanner that aims at chemistry concepts and short cuts.

We shall consider rules and definitions for sugars first. There is genericnomenclature that describe sugars by their functionality and carbon count. Inaddition to knowing generic nomenclature, you must be able to identify specificexamples of aldohexoses, aldopentoses, and ketohexoses. Common sugars thatyou should recognize in all traditional structural representations (Fischer,Haworth, and chair) are glucose, ribose, fructose, galactose, and mannose. Wewill present mnemonic devices to help with the recall of these structures.Stereochemistry is a significant part of sugar nomenclature, so it is necessary toknow the stereochemistry associated with the prefixes of D and L.

Monosaccharides of five carbons or more are typically found in cyclicstructuresrather than straight chain structures. To form the cyclic structure, a hydroxylgroup attacks a carbonyl carbon to form either a hemiacetal (if the sugar is analdose) or a hemiketal (if the sugar is a ketose). Converting between linear andcyclic representations involves knowing how to draw the hydroxyl groups ateach carbon. We shall also consider the relationship of sugars. In the cyclic form,you must be able to determine what anomer is represented. In both the straightchain and cyclic forms, epimers are possible. We shall emphasizestereochemistry terminology.

Considering much of sugar chemistry involves multiple sugars, we will addressglycosidic linkages and how to recognize the type of linkage. We will considermany examples from disaccharides, to trisaccharides, to oligosaccharides, andfinally polysaccharides. Alpha and beta linkages impact the reactivity andstructural nature. To cleave linkages, specific enzymes are required. Of mostnotoriety is the enzyme to cleave alpha linkages in glycogen, amylose, andamylopectin, which allows us to break down starches. We lack the enzyme tocleave beta-linkages, so we are unable to breakdown cellulose.

We will also consider chemical reactions of sugars. There are three types ofchemical reactions we shall address. One type is used to identify the chirality atspecific stereocenters, such as treatment with HNO3 and osazone formation.Another type increases the length of the carbon chain of the sugar. The last typebreaks down the sugar, one carbon at a time. The last feature we will consider isthe biological reactivity. We will look at glycolysis from a practical perspective,rather than the detail seen in biochemistry. We will also consider blood typesand the impact of stereochemistry on the recognition of the glycoproteins thatdetermine the blood type.

Introduction


Organic Chemistry Carbohydrates Monosaccharides

Monosaccharides

Straight Chain MonosaccharidesMonosaccharides have the basic formula CnH2nOn, giving them one oxygenattached to every carbon and one unit of unsaturation. The one unit ofunsaturation is present as either a carbonyl group or a ring. To form the cyclicstructure from the carbonyl structure, a hydroxyl group must attack the carbonylto form a hemiacetal or hemiketal, depending on whether the carbonyl form is analdehyde or a ketone. Monosaccharides have generic nomenclature for theirstructures that depends on their carbon chain length and the type carbonylfunctional group. Aldehyde sugars are aldoses and ketone sugars are ketoses. Thelength of the sugar is also considered in the name. Four carbon sugars aretetroses, five carbon sugars are pentoses, and six carbon sugars are hexoses.Combining length and functional group yields sugar names such as:

Aldohexose: Aldo - hex - ose = aldehyde - 6 carbon - sugar.Ketotetrose: Keto - tetr - ose = ketone - 4 carbon - sugar.

Within each chain length, there are a set number of stereoisomers, each with adifferent common name. The number of stereoisomers can be derived from 2n

(where n = the number of chiral centers in the molecule). For example, analdotetroses has two chiral centers (carbon 2 and carbon 3), so there are fourpossible aldotetroses. The Fischer projection for each of the four aldotetroses areshown in Figure 6-1.

H^°

H-

H-

OH

OH

CH2OH

D-(-)-Erythrose

H^°

H-

HO-

OH

H

CH2OH

L-(+)-Threose

H^°

HO-

H-

H

OH

CH2OH

D-(-)-Threose

H^°

HO-

HO-

H

H

CH2OH

L-(+)-Erythrose

Figure 6-1

The last chiral center (on the penultimate carbon) in the D-isomers of the sugarshas R-stereochemistry, while the last chiral center in the L-isomers of the sugarshas S-stereochemistry. There is no correlation between D and L with (+) and (-)optical rotation. As seen when comparing L-threose to D-threose, D- and L-sugars with the same name are enantiomers of one another. That is why one isdesignated as (+) and the other as (-), indicating they have opposite directionrotations of plane polarized light.

Sugars are routinely represented by Fischer projections, which are short hand forthe top view (top perspective) of a compound in its fully eclipsed conformation.The Fisher projection is derived from the straight chain form which is not themost stable form, but is the easiest form to draw. Any substituent on the right orleft side of the main backbone is projecting out at you. In bold wedge drawing,Fischer projections look like bow ties. When the fourth priority substituent iscoming out at you, you may recall that the chiral center is the opposite of whatyou see. This means that in Fisher projections, if the fourth priority substituent ison the side of the chain, then you must take the opposite rotation of the arc that isviewed for the chiral center in two dimensions to ascertain its chirality. Thetranslation into a Fisher projection from a three dimensional structure is shownin Figure 6-2.

Copyright © by Tiie Berkeley Review 92 The Berkeley Review


OH OHRotate the sugar tothe all eclipsed form

A ViewingPerspective

CH2OH

OH OH H

Figure 6-2

Aldoses and Ketoses

Aldoses are sugarswith an aldehydefunctional group. Ketoses aresugarswith aketone functional group. Figure 6-3 shows two aldotetroses and twoketotetroses. A hydroxyl group in parenthesis indicates that the chirality isunknown. The only ketotetrose known is erythulose.

Hv^°

H-

H-

"(OH)

OH

CHjOH

D-Aldotetrose

CH,OH

H—f— OHCHjOH

D-Ketotetrose

H^°

H-

HO-

-(OH)

H

CHjOH

L-Aldotetrose

HO-

CH2OH

CB,OH

L-Ketotetrose

Figure 6-3

Pentoses and Hexoses

Pentoses are five carbon sugars, whether it be a ketose or aldose. Hexoses are sixcarbon sugars. Most sugars encountered in biology are pentoses and hexoses.Figure 6-4shows a pair of D-pentoses and a pair of D-hexoses.

HV^°

H-

H-

H-

(OH)

(OH)

OH

CH2OH

D-Aldopentose

*N^°H-

H-

H-

H-

-(OH)

-(OH)

"(OH)

OH

CH2OH

D-Aldohexose

H-

H-

CH2OH

= o

-(OH)

-OH

CH2OH

D-Ketopentose

Figure 6-4

H-

H-

H-

CH2OH

= o

"(OH)

"(OH)

OH

CH2OH

D-Ketohexose

Monosaccharides



Example 6.1How many aldohexoses are possible?

A. 4

B. 8

C. 16

D. 32

Solution

An aldohexose is a six carbon aldehyde sugar. Carbon one is an aldehyde, andcarbon six is a primary alcohol, so neither carbon 1 nor carbon 6 is chiral.Carbons 2, 3,4, and5 are all chiral, so there are16 (24) possible combinations ofchiral centers (i.e., 2R, 3R, 4R, 5R etc.), which makes choice C the best answer.There are sixteen aldohexoses of which 8 are D-stereoisomers (carbon 5 has Rstereochemistry) and 8 are L-stereoisomers (carbon5 has S stereochemistry).

Example 6.2How many 2-ketohexoses are possible?

A. 4

B. 8

C. 16

D. 32

Solution

A 2-ketohexose is a six carbon ketone sugar. Carbons 1 and 6 are primaryalcohols, and carbon 2 is a ketone, so neither carbon 1, carbon 2, nor carbon 6 ischiral. Carbons 3,4, and5 arechiral, so there are8 (23) possible combinations ofchiral centers (i.e., 3S, 4S, 5S, etc.) which makes choice B the best answer. Thereare eight 2-ketohexoses of which 4 are D-stereoisomers and 4 are L-stereoisomers.

Example 6.3How many D-aldopentoses are there?

A. 2

B. 4

C. 8

D. 16

Solution

A D-aldopentose is a five carbon aldehyde sugar in which carbon 4 (thepenultimatecarbon) has R-stereochemistry. Carbon1 is an aldehyde and carbon5 is a primary alcohol, so neither carbon 1 nor carbon 5 is chiral. Carbons 2, 3,and 4 are chiral, so there are8 (23) possible combinations ofchiral centers (i.e.,2R, 3S, 4S; 2R, 3R, 4S; 2R, 3S, 4R, 2R, 3R, 4R; 2S, 3S, 4S; 2S, 3R, 4S; 2S, 3S, 4R; and2S, 3R,4R). There are a total of eight aldopentoses of which four are D-isomersand 4 are L-isomers, which makes choice B the best answer. If you are wise, pickchoice B.



Figure 6-5 shows the eight aldopentoses along with their common names. Again,the D-isomer and L-isomer of aparticular sugar are enantiomers (mirror images)of one another and have opposite optical rotations from one another (the +and -in parenthesis indicates the direction of the rotation of plane polarized light.)

Hvf0HÔ HVÔ Hvô

H— -OH H— — OH HO— — H HO— — H

H— -OH HO— — H H— — OH HO^ — H

HO— — H HO— — H HO— — H HO— — H

CH2OH CH2OH cnpn CH2OHL-(+)-Lyxose L-(+)-Arabinose L-(-)-Xylose L-(+)-Ribose

*Nf0 Hv.f° *Nf°Hvô

H— -OH H— — OH HO— — H HO— — H

H— -OH HO— — H H— -OH HO— L-HH— UoH H— -OH H— — OH H— — OH

CH2OH CH2OH CH2OH CH2OH

D-(-)-Ribose D-(+)-•Xylose D-(-)-Arabinose D-(-)-Lyxose

Figure 6-5

The above sugars represent the complete group of aldopentoses (there are eighttotal). As pointed out in Example 6.3, there are equalamounts of D-sugars andL-sugarsin each group of equivalent carbon sugars.

Example 6.4Which of the following terms best describes the relationship between D-(-)-Ribose and L-(+)-Ribose?

A. Anomers

B. EpimersC. Enantiomers

D. Diastereomers

Solution

Because D-(-)-ribose and L-(+)-ribose vary at all of their chiral centers (comparethe two structures in Figure 6-5), they are enantiomers. When all of the chiralcenters vary, the two structures are mirror images, which makes themenantiomers. This makes choice C correct. Choice A is eliminated, because to beanomers, the structures would have to be cyclic. If they were cyclic structures,the nomenclature would have included either alpha or beta at the start and thename would have been ribofuranose, which implies that it is a five-membered ringsugar. Choice B is eliminated, because the sugars vary at more than one carbon(they vary at carbons 2, 3, and 4). Diastereomers refer to stereoisomers in whichnot all of the chiral centers change (diastereomers are not mirror images). Thiseliminates choiceD. Pick C for optimal results.

Monosaccharides



Example 6.5Which of the following terms best describes L-(+)-Lyxose and L-(-)-Xylose?

A. Anomers

B. C-2 EpimersC. C-3 EpimersD. Enantiomers

Solution

Because L-(+)-Lyxose and L-(-)-Xylose vary at only the second carbon (comparethe two structures in Figure 6-5), they are referred to as epimers. If all of thechiral centers vary, the two structures are mirror images, which makes themenantiomers. Not all of the chiral centers differ, so choice D is incorrect. ChoiceA is eliminated, because to be anomers, the structures would have to be cyclic,which they are not, as implied by the names. Choice C is eliminated, because thesugars vary at carbon 2 and not carbon 3. This narrows it down to choice B. Thetwo structures vary at carbon 2, so they are C-2 epimers.

To complete our coverage of monosaccharides, let us consider aldohexoses.Figure 6-6shows the eight D-aldohexoses. Do enjoy looking at them.

Hô

H-

H-

HO-

H-

OH

OH

H

OH

CH2OH

D-(-)-Gulose

Hô

H-

H-

H-

H-

OH

OH

OH

OH

CH2OH

D-(+)-Allose

HNÔ

H-

HO-

HO-

H-

OH

H

H

OH

CH2OH

D-(+)-Galactose

HNÔ

H-

HO-

H-

H-

OH

H

OH

OH

CH2OH

D-(+)-Glucose

Figure 6-6

Hô

HO-

H-

HO-

H-

•H

OH

H

OH

CH2OH

D-(-)-Idose

H^,o

HO-

H-

H-

H-

H

OH

OH

OH

CH2OH

D-(+)-Altrose

Example 6.6D-mannose and L-gulosevary in chirality at which carbon(s)?A. 2 onlyB. 3 and 4 onlyC. 5 onlyD. 2 and 5 only

HvÔ

HO-

HO-

HO-

H-

H

H

H

OH

CH2OH

D-(+)-Talose

Hô

HO-

HO-

H-

H-

H

H

OH

OH

CH2OH

D-(+)-Mannose

Copyright ©by The BerkeleyReview 96 The Berkeley Review

OrgaillC Chemistry Carbohydrates Monosaccharides

Solution

Because D-sugars differ from L-sugars at the penultimate carbon, the twoaldohexoses must differ in chirality at carbon 5. Thiseliminateschoices A and B.The question comes down towhether D-mannose andL-gulose vary at carbon 2or not. According to Figure 6-6, D-mannose and D-gulose differat carbon2, soL-gulose and D-mannose must have the same chirality at carbon 2, because D-gulose and L-gulose differ at all chiral centers. This makes choice C the bestanswer. The two structures are shown below.

D-(+)-Mannose H^

f0L-(+)-Gulose HvÔ

HO— — H HO — — H

HO— — H HO— — H

H— — OH H— — OH

H— — OH HO— — H

CH2OH CH2OH

Example 6.7Which of the following sugars contains the fewest carbons?

A. D-Threose

B. D-Galactose

C. L-Fructose

D. L-Idose

Solution

Galactose and idose are listed in Figure 6-6, so both are aldohexoses. They eachhave the same number of carbons, so choices B and D are eliminated. Fructose isa ketohexose, which means it too has six carbons, like the aldohexoses. Thiseliminates choice C. Threose is an aldotetrose. Becausealdotetroses contain onlyfour carbons, choice A is the best choice.

Example 6.8D-Allose is all of the following EXCEPT the:

A. C-2 epimer of D-altrose.B. C-3 epimer of D-glucose.C. C-4 epimer of D-galactose.D. C-5 epimer of L-talose.

Solution

According to Figure 6-6, D-allose has all of its hydroxyl groups on the right in itsFischer projection. D-Altrose has all of its hydroxyl groups on the right exceptfor hydroxyl 2, so it is in fact the C-2 epimer of D-allose. This eliminates choiceA. D-Glucose has all of its hydroxyl groups on the right except for hydroxyl 3,so it is the C-3 epimer of D-allose. This eliminates choice B. D-Talose has all ofits hydroxyl groups on the left except for hydroxyl 5, so L-talose must have all ofits hydroxyl groups on the right except for hydroxyl 5. L-Talose and D-allose areC-5epimers, which eliminates choice D. Galactose has two hydroxyl groups onthe left and two on the right, so it differs from D-allose at two chiral centers. Assuch, it is not the C-4 epimer of D-allose, making choice C the best answer.



Common Monosaccharides

Memorizing the names and structures for all sugars is no doubt a little pointless,but knowing the common monosaccharides is not a bad idea. Commonmonosaccharides in biology include ribose, glucose, mannose, fructose andgalactose. To help recall the common sugars, we shall present a mnemonic for afew (glucose, mannose, and ribose) and then know how the others relate to them.Tohelp recallglucose, we have a not so nice way to remember the position of thehydroxyl groups. If you flip the page off with the fingernail of your middlefinger pointing towards the page, your finger tips are positioned just as thehydroxyl groups of glucose are positioned in the Fischerprojection. This helpswith enantiomers too. D-glucose is found using your right hand while L-glucoseis found using your left hand. Given that your left and right hands areenantiomers (mirror images), D-glucose and L-glucose are also enantiomers.Mannose can be recognized using a similar finger trick. By making a gunfigurine usingyour middlefinger and indexfingeras the barrel,your finger tipsagain mimic the positions of the hydroxyl groups in the Fischer projection formannose. You can say "man uses guns, so the gun hand position representsmannose." Ribose has all the hydroxyl groups on the right side of the Fischerprojection, so "ribose is all right." Thisprovides three solid mnemonic devicesforrecalling three common sugars. Other monosaccharides can be learned relativeto those three. ThestraightchainFischer projection of some of the more famousD-monosaccharides (famous implies that they are found in biological systems)and a suggested memory aid foreachoneare shownin Figure6-7.

Hvf HNf° HVf HY° CH2OHH- -OH H- -OH ho- -H H- -OH =o

H- -OH HO- -H HO- -H HO- -H HO- -H

H- -OH H- -OH H- -OH HO- -H H- -OH

CH2OH H- -OH H- -OH H- -OH H- -OH

D-Ri

'Ribose is

bose

all right!"(

D-Gl

:h2ohucose

(

D-Ma

:h2ohmnose

CH2OHD-Galactose

CH2OH

D-Fructose

"F*** glucose!" "Man uses his gun!" C-4 epimer ofglucose ketose ofglucose

Figure 6-7

The terms below the name ofeach sugar refer to a way in whicheachstructurecanbe recalled. "F***" ofcourse standsfor"flip off, right?

Stereoisomerism in SugarsThe majority of sugar chemistry centers around the stereochemistry of thehydroxyl groups. As such, we shall focus on stereochemistry of sugars morethan the chemistry of alcohols, carbonyls, hemiacetals, or acetals. In sugarchemistry, it is criticalthat you be able to determine how stereoisomers relate toone another, whether they are enantiomers, epimers, anomers, or structuralisomers. The term epimers should not be mistaken with anomers, which refers totwo sugars in their ring forms that vary at the hemiacetal (in the case of aldoses)or hemiketal (in the case of ketoses) carbon in the ring. It is easiest to say thatanomers vary in chirality at the most oxidized carbon while epimers vary inchirality at any one of the less oxidized carbons.

Copyright ©by The Berkeley Review

Anomers vary in chirality at the most oxidizedcarbonwhile epimersvary in chirality at any one of the less oxidized carbons.



EpimersEpimers aresugar diastereomers that vary at one stereocenter (chiral center) inthe carbon backbone. D-Threose and D-erythrose are C-2 epimers, because theyvary atcarbon 2. D-Glucose and D-Galactose are C-4 epimers, because they varyatcarbon 4. In the straight chain form ofa monosaccharide, the carbonyl carbonis most oxidized, but because ithas s^-hybridization, ithas no chirality. The lastcarbon hasnochirality, because the carbon isprimary with two hydrogens. Thismeans that two sugars cannot be epimers at the last carbon or the carbonylcarbon in the straight chain form. Thestraight chainformis not the most stableform, but it is an intermediate when one anomer is converted into the otheranomer.

Anomers

The most stable form of a sugar is the cyclic form, which is formed when ahydroxyl group attacks the carbonyl carbonyielding a hemiacetal. A new chiralcenteris formed when the hydroxyl attacks the carbonyl. Because the attackcanoccur from either top or bottom, there are two diastereomers formed. The twodiastereomers that result when the ring structure is formed are referred to asanomers and the new chiral carbon is the anomeric carbon. The two anomers arereferred to as the alpha (axial for D-pyranoses) and the beta (equatorial for D-pyranoses) anomers. The beta anomer is typically more stable than the alphaanomer. It is only in the cyclic form (furanose formfor five-membered rings andpyranose form for six-membered rings) that anomers are possible. Figure 6-8shows the anomers of glucose, oc-D-glucopyranose and 6-D-glucopyranose.

HOH2C

OH

C-D-glucopyranoseAll substituents have

equatorial orientation

OH

HOH2C

OHot-D-glucopyranose

All substituents have equatorialorientation, except the anomeric OH

Figure 6-8

Six membered rings are referred to as pyrans while five membered rings arereferred to as furans. The name fi-D-glucopyranose comes from the anomerichydroxyl group being cis to carbon 6 (6), D because the last chiral center (carbon5) has R-stereochemistry, gluco from the sugar glucose, and pyranose because itis a six-membered ring sugar. All cyclic sugars are named in a similar fashion.Monosaccharides come as either five-membered or six-membered rings, so youshould be familiar with the term furanose as well as pyranose. Ribose in itscyclic form is a furanose as you are no doubt familiar with frombiology.

In cyclic sugar structures, the term alpha technically means that theanomeric oxygen is trans with the last carbon and the term betatechnically means that the anomeric oxygen is cis with the lastcarbon.

Monosaccharides



HOH2C,

Mutarotation

The two anomers of a sugar exist in equiUbrium in vivo and can interchange via astraight chain intermediate. Conversion from one anomer into the other isknown as mutarotation. Figure 6-9 shows the key structures in mutarotation.

2 HOH2C

OH •*

OH

C-D-glucopyranose(64%)

OHa-D-glucopyranose

(36%)

Straight chain(<0.1%)

Figure 6-9

Example 6.9Mutarotation involves which of the following?

A. Sugar decomposition.B. Sugar dehydration.C. Sugar oxidation.D. Stereoisomerization.

Solution

As shown in Figure 6-9,mutarotation involves the conversion between anomers.Sugar decomposition occurs with degradation or digestion, which does notdescribe mutarotation, so choice A is eliminated. Sugardehydrationinvolves theloss of water by the sugar, which does not occur with mutarotation, so choice Biseliminated. Dehydration is accomplished chemically by a highly hygroscopicsubstance suchas concentrated sulfuric acid. Sugaroxidation would requireanoxidizing agent, which is not present in mutarotation. Choice C is eliminated.The process of mutarotation involves the conversion from one stereoisomer intoanother, which is described by the term stereoisomerization. Choice D is oneterrific answer for a question like this.

Example 6.10In vivo,D-glucose is MOST stable in its:

A. alpha anomer of the pyranose form.B. beta anomer of the pyranose form.C. straight chain form.D. furanose form.

Solution

Thepyranose(six-membered) ring formis more favorable than the straight chainand the furanose (fivemembered) ring form. This eliminates choices C and D.The beta anomer,with the anomerichydroxyl group in equatorial orientation, ismore favorable than the alpha anomer. This eliminates choice A and makeschoice B the best answer.

Copyright © by The Berkeley Review 100 Tiie Berkeley Review


Haworth ProjectionsThe Haworth projection is a simplified representation of cyclic sugars. Theyhave no physical relevance, but are easy to draw (just like Fisher projections).Theydo not show the three dimensional spacing that chair conformations show,but theyemphasize positionsaboveand belowthe ring,whichis the majorfocuswhenconsideringsugars. Convertingfrom the Fischer projection to the Haworthprojectionis easy once you get the hang of it. To go from the straight chain to thering form, recall the phrase "downright uplefting," which applies to carbons 2,3,and 4 in an aldohexose, but not the penultimate carbon (which has its oxygeninthe ring) or the anomeric carbon. Any hydroxyl groups on the right side in theFischer projection end up below the ring in its cyclicstructure. Any hydroxylgroups on the left in the Fischer projection end up above the ring in its cyclicstructure. Numbering the carbons makes it is easier to convert to the Haworthprojection, because it helps to keep track of carbons 2, 3, and 4. The anomericcarbon is given the far right position in the Haworth projection. Figure 6-10shows the generation of the Haworth projection of D-galactopyranose.

on left H

••• «P HO

on rightOH .-. down HO

HO

; •=>

CHjOH

Figure 6-10

The anomeric carbon is shown at the far right in furanose rings just as it is inpyranose rings. Fructose and ribose are classic examples of naturally occurringfuranose rings. The carbonyl group of fructose is on carbon 2, so when oxygen ofhydroxyl 5 attacks the carbonyl, the ring is five atoms in size. Figure 6-11 showsthe Haworth projections of a-D-mannopyranose and fi-D-fructofuranose.

6„ 6

HOH2CH

5

1

HO

*CH2OH

OH

H Ha-D-mannopyranose

Figure 6-11

OH

CH2OH

HO H

p-D-fructofuranose

Monosaccharides

OH



Example 6.11The following structure is the Haworth projection of what sugar?

CH2OH

OH

HO

A. fi-D-glucopyranoseB. fi-D-mannopyranoseC. cc-D-glucofuranoseD. a-D-mannofuranose

Solution

Choices C and D are eliminated immediately, because the ring in the question issix-membered, not five-membered. Translating from the Haworth projection tothe Fischer projection might make it easier to determine the identity of the sugar.Hydroxyl 2 is up, hydroxyl 3 is up, and hydroxyl 4 is down in the Haworthprojection, so hydroxyl 2 is left, hydroxyl 3 is left, and hydroxyl 4 is right in thestraight chain form. Thiscorresponds to the Fischerprojection of mannose. Thebest answer is choice B. It might be easier to decide whether the compound isglucose (choice A) than it is to deduce the sugar's identity. fi-D-glucose has thehydroxyls alternatingbetween up and down as you move from carbon to carbon.The structure in this question doesn't show this, so it can't be glucose. The onlychoice left is mannose, the C-2epimer of glucose. You may want to verify thatonly the second carbon is different from C-D-glucose. The structures of thevarious sugars are drawn below:

OH 6CH2OH QH

H

^Ô

HO

H OH

6-D-glucopyranose

HOH^°

H- OH

HO- H

OH

H- OH

CH2OHD-glucose

CH2OH

HO

OH

H OH

oc-D-glucofuranose

H H

G-D-mannopyranose

HO

'CH2OH

HI I OH

H H

a-D-mannofuranose

HO- H

HO- H

H- OH

H- OH

CH2OHD-mannose

The important skill to develop here is the ability to quickly convert from the ringstructure to the straight chain structure, as these are the common ways in whichyou will see the sugars.



Isomerization

In addition to mutarotation, monosaccharides can convert from an aldose into aketose via an enediol intermediate. This conversion is known as ketose-aldoseisomerization In glycolysis, the enzyme isomerase convertsglucose-6-phosphateinto fructose-6-phosphate via isomerization. Also in glycolysis, an isomerizationprocess similar to tautomerization converts dihydroxyacetone phosphate intoglyceraldehyde 3-phosphate. Figure 6-12 shows the step-by-step conversion ofan aldose into a ketose in both acid and base.

Ketose-Aldose Isomerization (base and acid catalyzed)

Monosaccharides

V*O

H-

HO-

H-

H-

H

deprotonateOH Hon C-2

I— nHGSfH

I /-vrr deprotonateOH OH on C-2

protonate

OH O-on C-lH w HO

OH "^ H

OH H

HO H

H

OH

OH

CH2OH

in

OH'w HO-

^ H-

H-

H

OH

OH

CH2OH

^ HO-H-

H-

protonateO on C-l

'V- H HO

CH2OH

H

OH H-H+0

H-

HO-

H-

H-

OH

•H

•OH

•OH

CH2OH

deprotonateH on C-2

•H

•OH

•OH

CH2OH

protonateC-l

H-

HO-

H-

H-

= 0+H deprotonateOHonC-2H0.

HO

H

H

•OH

•OH

CH2OH

Figure 6-12

Thebolded hydrogen atoms are the active hydrogens that are being either gainedor lost in the reaction mechanism. The requirement for the conversion is theformation of either the enol (under acidic conditions) where the carbonyl oxygenis protonated and the alpha hydrogen is deprotonated, or the formation of theenolate (under basic conditions) where the alpha hydrogen is deprotonated.When isomerization is catalyzed by an enzyme, it often proceeds through anenamine intermediate. As Figure 6-13 shows, conversion between C-2 epimers(glucose into mannose) also proceeds through an enediol intermediate.

Isomerization of glucose into mannose (acid catalyzed)

H-

H-

O

H-

HO-

H-

H-

^V- H

OH

H +RI

OH

H+0

H-

HO-

H-

H-

<^ H

OH

H -H+w HOOH

OH

CHjOH

HO^H H+0'-OH HO

H +H+W HOOH^ H

*^ H

H

H -H-1

OH^"

OH

H-

H- OH H-

CHpH

D-Glucose

CH2OH

Enediol intermediate

Figure 6-13

CHjOH

•H

•OH

•OH

CH2OH

protonateC-l

H

-OH

F=0

•H

OH

OH

CH2OH

O<V H

HO

HO

H

H

H

•H

OH

•OH

CHgOH

D-Mannose


OrgailiC Chemistry Carbohydrates Monosaccharides

Example 6.12In the conversion from glucose into mannose, one of the intermediates is:

A. ribose.

B. a six carbon enediol.

C. galactose.D. a five-carbon ketose.

Solution

Under basic conditions, the conversion involves the deprotonation of an alphahydrogen, which forms an enolate intermediate. Protonation of carbon 2regenerates the monosaccharide as one of two epimers, mannose or glucose.Under acidic conditions, the conversion involves the protonation of the carbonyloxygen and the deprotonation of an alpha hydrogen, which forms an enediolintermediate. Protonation of carbon 2 regenerates the monosaccharide in one oftwo stereoisomer forms, which leads to either mannose or glucose. The bestanswer is therefore an enediol, choice B. If enolate and enediol were bothchoices, then the reaction conditions would have been required. Becauseenolateis not a choice, it is safe to assume that the conditions are acidic. Figure 6-13shows the conversion. No carbon is gained or lost during the conversion process,so choices A and D are eliminated.

Example 6.13What is the name of the process that converts 3-phosphoglyceraldehyde into 1-phospho-l,3-dihydroxyacetone?A. AldehydosisB. Enediolosis

C. Ketose-aldose isomerization

D. Transphosphorylation

Solution

The numeral precedingthe phosphate group changes from 3 to 1, which wouldlead one to believe that the phosphate group changed positions. However, thechange in numeral is a result of a change in the functional group prioritynumbers,and the phosphate group in fact never changes its position. As such,transphosphorylation is an incorrect answer choice designed to trip up peoplewho didn't draw thestructures and assumed thenumberingchangedbecause thegroup moved. Choice D is eliminated. An aldehyde is converted into a ketone,so enediolosis is not an apt description. This eliminates choice B. The overallconversion is from an aldose into a ketose, so the best term to describe theprocess is ketose-aldose isomerization. This eliminates choice A and makeschoice C the best answer.

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Organic Chemistry Carbohydrates Oligo- and Polysaccharides

Oligosaccharides and PolysaccharidesThis section addresses disaccharides, trisaccharides, oUgosaccharides, andpolysaccharides. Disaccharides result from combining two monosaccharides,trisaccharides result from combining three monosaccharides, oligosaccharidesresult from the combination of two to eight monosaccharides, andpolysaccharides result from combining several monosaccharides.

Linkages (Acetal Connectivity)Sugars combine when a hydroxyl group of one sugar attacks the hemiacetalcarbon of a second sugar, displacing the hydroxyl group from the hemiacetalcarbon to form an acetal. This forms a glycosidic linkage, resulting from adehydration reaction of the two hydroxyls involved. The terms used to describethe orientation of the anomeric hydroxyl group in a cyclicmonosaccharide (alphaand beta) are also used to describe the orientation of the oxygen in the glycosidiclinkage. When the anomeric oxygen is cis with respect to the last carbon (carbonnumber 6 in an aldohexapyranose), the linkage is said to be beta.

Disaccharides

The names for disaccharides are derived from the two component sugars wherethe left sugar (using hydroxyl 1 of an aldopyranose) is considered a substituenton the right sugar (using either hydroxyl 4 or 6). The left sugar gets an "osyl"suffix and the right sugar gets an "oside" suffix. For instance, sucroseconsistsofglucopyranosyl and fructofuranoside. In sucrose, the linkage is 1,2, whichis rare.In most cases, the linkage is 1,4, as in the disaccharides maltose, cellobiose(formed from the hydrolysis of cellulose), and lactose. These commondisaccharidesall contain at least one glucose residue in one of its anomeric forms.Linkages are named for the carbons involved as well as the actual structural(stereochemical orientation) of the anomeric carbon. A picture says so muchmore than words (at last check the picture was valued at 1000 words), so for thatreason, Figure6-14 is a picture of four commondisaccharides.

CH2OH

HO

H OH « BSucrose

(a-D-glucopyranosyl)-G-D-fructofuranoside)(oc-l-fi-2-linked disaccharide)

OH H

CH2OH

HOI CH2OHWA-^Q, HOHzC

HO^-^0^^°vOH HO-\^-V^|6-glycosidic OH |

linkage OH

Lactose

(4-0-(fi-D-galactopyranosyl)-o>D-glucopyranoside)(fi-1,4-linked disaccharide)

HOH2CHO—^-^-°x HOHzCHoX^ÔÂ^°v

OH HO-^^ÂÔH

HOHzCHO-"-^"--^— °»a-glycosidic

ho\—^Â linkase

fi-glycosidic OHlinkage HO-\^^ÂCellobiose

ohIMaltose OH

(4-0-(fi-D-glucopyranosyl)-B-D-glucopyranoside)(C-l, 4-linked disaccharide)

(4-O-(a-D-glucopyranosyl)-0t-D-glucopyranoside)(a-1,4-linked disaccharide)

Figure 6-14



Trisaccharides

Trisaccharides are an oligosaccharide comprised of three monosaccharides.Trisaccharides contain two glycosidic linkages. One of the most commontrisaccharides is raffinose, which entails a cc-D-galactopyranose forming a linkagewith its anomeric hydroxyl group with hydroxyl6 of sucrose. Figure 6-15 showsraffinose in the Haworth projection and three dimensionally more valid chairform.

CHjOH

HO

a N?H v^ ^V? wHo\—If I^Q^ll fCH2>H OH OB QH jl

Raffinose

Trisacdiarideof a-D-galactose, a-D-glucose, and fi-D-fructose(a-l,6-and o-l-fi-2-linked trisaccharide)

OH

ohVo 1 ] Jf CH2OH

OH H

Figure 6-15

PolysaccharidesPolysaccharides are exactly what the name implies: multiple sugars. We shalltreat polysaccharidesas polymers made of four or more monosaccharides. Mostare linked from the anomeric carbon of the sugar on the left (carbon 1) to thefourth carbon on the sugar on the right (carbon 4), a 1,4 glycosidic linkage. Longchains of sugars (starches like amylose and amylopectin) are common inbiological systems. We store our excess sugar inlong polymers such as glycogen(poly a-D-glucopyranose), which has 1,4-alpha linkages between adjacentglucopyranoses and 1,6-branching about one out of every ten glucopyranoseresidues in the polymer. The mostabundant of all polysaccharides is cellulose(poly fi-D-glucopyranose) which is used for structural purposes in plants. Weare unable to digest cellulose, because we lack theenzyme neededtobreak6-1,4-glycosidic linkages. We produce alpha glucosidase (which cleaves only alpha-linkages of poly glucopyranoses) but not beta glucosidase.- Figure 6-16 showscellulose, amylose,and glycogen.

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HOH2C

HOH2C OH HOH2CCellulose

(B-1,4-linked polysaccharide)

CH2OH

OH

CH2OH CH2OH CH2OH CH2OH

HJr— °vH hJ—"<\H hJ—Oh hJ—<\hVOH HV VOH HV VOH HV VOH H>

O-1 •-0-1

H OH

HO

H OH

Amylose(a-1,4-linked polysaccharide~no branching)

H OH

Glycogen(a-1,4-linked polysaccharide-1,6-branching)

Figure 6-16

There are no 1,6-linkages found in cellulose, only 1,4-linkages. This means thatcellulose is a linear polymer with no branching. Strands of cellulose are heldtogether by hydrogen bonding between the ring oxygen and the hydrogen onhydroxyl 3, making the structure uniform on the microscopic level. Eachglucoseresidue is approximately 5.1 angstroms from carbon 4 to carbon 1. Theglucopyranose monomers alternate orientation in cellulose to allow hydrogenbonding, as shown in Figure 6-16.

Example 6.14Branching in amylopectin is attributed to what type of linkage?

A. 4,4B. 4,1

C. 1,6D. 6,4

Solution

Branching is attributed to 1,6-linkages, although in dextran the monomers arelinked by 1,6-glycosidic linkages. This is memorization for the most part.Amylopectin and glycogen have branching, while cellulose and amylose do nothave branching. The best choice is C.

H OH

1


Organic Chemistry Carbohydrates Chemical Reactions and Tests

Nitric Acid Sugar OxidationNitric acidis richin oxygen, making it an oxidizing agent. When an aldohexoseis treated with nitric acid, both terminal carbons areoxidized intocarboxylic acidgroups. If the product, analdaric acid, is meso, then the compound is opticallyinactive. Using retro-analysis, thechirality ofthehydroxyl groups in theoriginalmonosaccharide can be determined. Figure 6-17 shows the aldaric acidformation reactions of four aldohexoses.

Nitric acid sugar oxidation

O H<vO

H-

HO-

H-

H-

<VOH

•OH

•H

OH

OH

O

HO-

HO-

H-

H-

^ H

•H

H

Ô

HO-

HO-

H-

H-

OH

•H

•H

OH

OH

H

HO

H

H

•OH

•H

OH

OH

CH2OH

Glucose

HN03

H20,A

OÔHGlucaric acid

(optically active)

HNO3

OH H20,A

OH

CH2OH

MannoseO^ÔH

Mannaric acid

(optically active)O H*V

O

H-

HO-

HO-

H-

V- OH

•OH

•H

•H

OH

O

H

H

H

H

V- H

OH

OH HNO,

O

H-

H-

H-

H-

'V- OH

•OH

OH

OH

OH

H

HO

HO

H

•OH

•H

•H

OH

CH2OH

Galactose

HNO3

H20,A

O^ÔHGalactaric acid

(optically inactive)

OH H20,A

OH

CH2OH

Allose

Figure 6-17

O^ÔHAllaric acid

(optically inactive)

Example 6.15The addition ofnitric acid togalactose results in:A. an optically active compound.B. an optically inactive compound.C. the loss of a carbon.

D. one formaldehyde and five formic acids.

Solution

Nitric acid oxidizes an aldose into a diacid (aldaric acid) product. Galactosebecomes optically inactive, because the diacid form is meso (it has amirror planethrough the bond between carbon 3and carbon 4). This can be seen in Figure 6-17. Choice B is the best answer.



Osazone Test

The osazone test has appeared frequently on the MCAT, despite its obscurity intheminds ofmostorganic chemists. Therepeated useofthe reaction in passagesis a testament to the test writer's goal to present topicsthat are not commonplacefor the sake of asking questions that at initial glance look unfamiliar. Theosazone test should be more aptly named the epimer test. When treated withthree equivalents of phenyl hydrazine, an aldose or 2-ketose undergoes asubstitution reaction to form an imine, followed by the oxidation of an alcoholinto a carbonyl, and lastly a second substitution reactionforming a second iminegroup. Imine moieties form at carbons 1 and 2, so any chiralityat either carbon islost. As a consequence,C-2 epimers yield the same osazone. Figure 6-18 showsthat D-mannose, D-fructose, and D-glucose, all yield the same osazone.

Osazone (C-2 epimer) test

HNô HvÔ CIÔH

-H

-OH

-OH

0 H\ /

3eq N—N/ \

_H H,HOAc

HN^N—NH3

HO-

HO-

H-

H-

H

H

OH

OH

or

H-

HO-

H-

H-

OH

H

OH

OH

HO-

or h-

H-

HO-

H-

H-

t=N—NH3

-H

-OH

-OH

CH2OH

D-Mannose

CrfyDH

D-Glucose

CHjOH

D-FructoseCH2OH

D-Glucose

phenylosazone

Figure 6-18

The osazone test works in conjunction with other information to zero in on theidentity of an unknown sugar. For instance, if an unknown D-aldohexose yieldsthe same osazone as D-galactose, but does not have the same optical rotation asD-galactose,then it must be either the C-2epimer of D-galactose(D-talose) or theketose of D-galactose (D-tagatose.)

Benedict's Test (Fehling's solution)Benedict's test oxidizes the sugar as copper dication gets reduced. In gettingreduced, the copper solution goes from blue to red for a positive test. The aldosebecomes an aldonic acid, which under basic conditions is deprotonated to forman aldonate. Figure 6-19shows Benedict's test.

Benedict's test (Fehling's solution)

O*V H

H— — OH

HO— — H

H— — OH

H— — OH

CH2OH

Gliucose

2Cu

5 OH"

2+

Figure 6-19

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O*vO"

H— — OH

HO— — H

H— — OH

H— — OH

CH2OH

Ghaconate

109 Exclusive MCAT Preparation


Tollen's Test

The Tollen's test involves the reduction ofsilver ionby an oxidizable sugar. Assilver ion is reduced to silvermetal, if the solution is spun within the container,then silver precipitates on the inside walls of the container, resulting in a socalled silver mirror. As silver is reduced, the sugar is oxidized. Thealdehydefunctional group oxidizes quickly, resultingin a rapid reaction with an aldose.Primary and secondary alcohols canalso be oxidized, so eventually a ketose willreact too. The Tollen's test issaid tobepositive ifthesilver mirror forms rapidly,indicatingthe presence ofan aldose. Figure6-20 shows the Tollen's test.

Tollens test

% rH°^0"NH4+ °-^H O. O-NH/

H— — OH H— — OH HO — H HO— — H

HO— — H 2 Ag(NH3)2-•• HO— — H BO— -H 2Ag(NH3)2^ HO- — H

H— — OH 2 OH" H— — OH H— — OH 2 OH" H— — OH

H— — OH H— — OH H— — OH H— — OH

CHjOH

Glucose

CH2OH

Ammonium gluconateCHpH CH2OH

Mannose Ammonium mannonate

Figure 6-20

Example 6.16The Tollen's test tests for:

A. a reducing sugar.B. an oxidizingsugar.C. an acidic sugar.D. a ketohexose.

Solution

The Tollen's test results in the reduction of silver, so the sugar must be areducingagent. This makes the Tollen's test a test for a reducing sugar (hemiacetal oraldose). Choice A is the best answer.

Periodate Oxidation

Astrong oxidizing agent such as periodate, IO4-, oxidatively cleaves the carbon-carbon bonds ofa sugar. After complete oxidation, the terminal carbons increaseby one bond to oxygen and internal carbons increase by two bonds to oxygen,forming several one carbon compounds as products. The complete oxidation oferythrose, analdotetrose, isshown inFigure 6-21.

HN^°H-

H-

OH

OH

CH2OH

Erythrose

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Periodate oxidation

IO, 3HC02H + H2CO + 3IO3-

Figure 6-21


OrganiC ChCmiStry Carbohydrates Chemical Reactions and Tests

Example 6.17The addition of excess periodic acid (HIO4) to fructose yields:A. one formaldehyde and five formic acids.B. one formaldehyde, four formic acids, and carbon dioxide.C. one formaldehyde and four formic acids.D. two formaldehydes, three formic acids, and carbon dioxide.

Solution

Fructose is a 2-ketohexose. With excess periodic acid, carbon-carbon bonds areoxidatively cleaved. The terminal carbons (carbons1 and 6) are oxidized up onelevel from primary alcohols to formaldehydes (from one C-O bond to two C-Obonds). Because two formaldehyde molecules are formed, the best answer ischoice D. The carbonyl carbon (carbon 2) is oxidized up two levels from a ketoneto carbon dioxide (from two C-O bonds to four C-O bonds). The secondaryalcohol carbons (carbons 3,4, and 5) are oxidized up two levels from alcohols toformic acids (from one C-O bond to three C-O bonds), resulting in the formationof three formic acid molecules. The result is two formaldehydes, one carbondioxide, and three formic acids, choice D. An reaction summary is shown below.

HO-CHjOH = H

HO

CH2OH L=oh°

H

AA/* JJQ = o=C=0

= 0 QH H

«aa/{aa/» HO"

H OH HO

OH H

H—h-OH

OH HO

OH H

»AA/

HO

OHH

«/W AA/»

H" OH

«AA/AA/»

'+H " )=°OH HO

OH H

[—hOH = Y=0

H—L-OH = V=0OH HOCH2OH H

HO- CH2OH = >°H

Kiliani-Fischer SynthesisThe Kiliani-Fischer synthesis increases the carbon chain of an aldose by one. Thenew carbon is added as a cyano group to the carbonyl carbon, so the chainincreases by one carbon at the aldehyde end. After reduction to an imine andhydrolysis, the cyano group is converted into an aldehyde. The result is theformation of two aldose epimers with an additional carbon. For instance, whenusing D-arabinose in Kiliani-Fischer synthesis, both D-mannose and D-glucose(C-2epimers) are formed. Glucose is 72% of the product mixture, because of thechiral influence of carbon 2 of the aldopentose (D-arabinose). Figure 6-22 showsthe Kiliani-Fischer synthesis of two D-aldohexoses from D-arabinose. An OHgroup in parenthesis represents the possibility of both chiral centers.



Kiliani-Fischer synthesis

NIII

°yH °yn°yH c

H— — OH HO— — H

HO— —H KCN^ H~" — (OH) 1. H2/Pd-BaS04HO— — H HO— — H

H— -OH H2° H°- —H ZHjO/H1"H—

+

— OH H— — OH

H— -OH H"~ — OHH— — OH H— — OH

CH2OH H"~ — OHCH2OH CH2OH

D-Arabinose CH2OH D-Glucose D-Mannose

(72%) (28%)

Figure 6-22

Example 6.18The Kiliani-Fischer synthesis ona D-aldopentose produces:A. a fifty-fifty mixture of enantiomeric D-aldohexoses.B. a major-minor mixture of diastereomeric D-aldohexoses.C. a major-minor mixture of diastereomeric D-ketohexoses.D. a fifty-fifty mixture of diastereomeric D-aldotetroses.

Solution

Fischer-Kiliani synthesis adds a carbon to an aldose, so with a five-carbonreactant, a six-carbon product isformed. This eliminates choice D. The productis an aldose, because the carbonyl group must form at a terminal carbon, sochoice Ciseliminated. The products are diastereomers (C-2 epimers), so choiceA is eliminated. Because attack at the carbonyl carbon is influenced by theasymmetry ofthe rest ofthe molecule, the mixture isnotfifty-fifty. Choice Bisthe best answer.

Example 6.19The addition of cyanide to a sugar is involved in which of the followingprocesses?

A. Nitric acid oxidationB. Nguyen degradationC. Kiliani-Fischer synthesisD. Ruff degradation

Solution

As seen in the first step of the Kiliani-Fischer synthesis inFigure 6-22, the cyanideanion adds to the carbonyl carbon ofan aldose toyield a cyano extension of theoriginal aldose. The best answer is choice C. Because of the influence of theother chiral centers, the cyano group does not add ina fifty-fifty manner to thetop and bottom of the carbonyl. This results in a diastereomeric mixture ofproducts. Nitric acid oxidation involves nitric acid, so choice A should havebeen eliminated. Degradation ofa sugarresults in the loss ofa carbon, not thegain, so thereagents should notcontain carbon. This eliminates choices Band D,although choice Bisnotanactual sugar reaction, so you may have eliminated itfor that reason instead.



Ruff DegradationThe word degradation implies that something is breaking down. In Ruffdegradation, an aldose loses its first carbon, resulting in an aldose of one lesscarbon. As we saw in the carbonylsectionand in severalbiochemistry examples,losing one carbon is readily accomplished by decarboxylation. In order todecarboxylate the terminal carbon, it must be oxidized first to a carboxylic acidand then to carbon dioxide. Bromine in water effectively converts the aldehydeinto a carboxylic acid (or in the ring form, a hemiacetal into a lactone). Thecarboxylic acid is oxidized into carbon dioxide by peroxide in the presence of aferric cation catalyst. CO2 leaves and decreases the chain length by one carbon.Figure 6-23 shows the Ruff degradation of D-glucose. It is rarely used withaldopentoses, because when the product is an aldotetrose, the yield is low, likelydo to the difficulty encountered with a smaller ring in the cyclic form.

Ruff degradation

°1H — — OH

°1H — — OH °^H

HO—

H —

—H

— OH

Br2

H20^Ca(OH)2 HO_

H20 * H_—H

— OH

H2°2_Fe3+*"

HO—

H —

—H

— OH

H — — OH H — — OH H — — OH

CH2OH CH2OH CH2OH

D-Glucose D-Gluconate

Figure 6-23

D-Arabinose

Example 6.20Ruff degradation results in:

A. the gain of a carbon onto carbon 1 of the chain.B. the loss of carbon 1 from the chain.

C. the gain of a carbon onto the last carbon of the chain.D. the loss of the last carbon from the chain.

Solution

Degradation impliesbreakdown, so Ruffdegradation is associated with a loss ofcarbon, not a gain. This eliminates choices A and C. As shown in Figure 6-23, itis the first carbon that is lost, so choice B is the best answer.

Example 6.21The addition of bromine in water to a sugar is involved in which of the followingprocesses?

A. Nitric acid oxidation

B. Wohl degradationC. Kiliani-Fischer synthesisD. Ruff degradation

Solution

As seen in Figure 6-23, the combination of bromine and water will oxidize amonosaccharide into an aldonic acid (monoacid) in the first step of the Ruffdegradation. The best answer is choice D.



°^H

Wohl DegradationWohl degradation is essentially a Kiliani-Fischer synthesis in reverse. An aldoseis first converted into its oxime usinghydroxylamine to which acetic anhydrideand acetate are added to acylate the hydroxyl groups and dehydrate the oximeinto a cyano group. The cyano group is removed under basic conditions, whichalso hydrolyze the esterson the sugar. The result is an aldose of one less carbon.Figure 6-24shows the Wohl degradation of D-ribose.

Wohl degradation

HON^^H N °^HH-

H-

H-

OH

OH H2NOH

OH

H-

H-

H-

OH

OH

O O

X0K H-

H-

H*

OAc NaOCH3H-

H-

OH

OH

CH2OH

D-Ribose

H20

Copyright ©byTheBerkeley Review

>OH Na02CCH3

CH2OH

Figure 6-24

114

•OAc

OAc

CH2OAc

CH2OH

D-Erythrose

The Berkeley Review


Blood TypesAs a general rule, natural sugars have D-chirality. This is because our enzymesrecognize D-sugars, and therefore select for D-sugars. One notable exception tothe "D only" rule includes the presence of L-fucose, a reduced form of L-galactose, in the glycoproteins of blood type markers. An oligosaccharide(defined as a polysaccharide with anywhere from two to eight monosaccharides)is bound to the amino terminal of a protein on the surface of a red blood cell tofunction as the antigenic determinant. Figure 6-25shows the terminal sugars ofthe three common antigenic markers. You should be able to ascertain from thestructures why type O is the universal donor.

Type O (Glycoprotein)

HO cupn cupu

I HO

pit protem

HO

Types A and B (Glycoprotein)

HO

H3Cvô|= OH I

for Type B) ' OyF ; NH/

• CHnOHH° H3C

CH2OHCHpH

protem

Figure 6-25

L-Fucose is the sugar with a CH3 group, rather than a CH2OH group, at thecarbon 6 position. It is bonded to the central galactose residue via an alpha-1,2-glycosidic linkage. Where type O differs from types A and B is that is does nothave a fourth monosaccharide linked to hydroxyl 3 of the central galactoseresidue. Types A and B differ from one another at the carbon 2 position of thefourth sugar. Type Bhas an ordinary galactoseresidue as the fourth sugar, andthus has a hydroxyl group on carbon 2, while type A has the carbon 2 hydroxylgroup replaced by an N-acyl group.



Organic Chemistry Carbohydrates Biological Applications

CHjOH

H OH

Glucose

O H

xHO—r—H

Glycolysis HighlightsGlycolysis isa ten-step metabolic process thatconverts glucose into two pyruvatemolecules. During theprocess, two net equivalents of adenosine triphosphate(ATP) areformed from adenosine diphosphate (ADP) and inorganic phosphate.Glycolysis is oxidative in nature (given that glucose is beingbroken down), sotwo equivalents of the oxidizingagent NAD+ are required. However, becauseglycolysis takes place in the cytoplasm, the NAD+ levels are kept low, soit getsusedup quickly. Figure 6-26 shows the overall glycolysis reaction.

CH2OH

H OH

Glucose

2 ATP 2 ADP

2NAD+

2H20 <V-a

4 ADP 4 ATP CH3

Pyruvate2 NADH

AG =

Figure 6-26

Glycolysis ispresented indetail in the Biology IIbook. The goal inpresentingaspects ofglycolysis inthe organic chemistry book isprovide biological examplesof some common organic reactions. The MCAT is a thinking test, so we willfocus on the big picture and the logic behind selected steps. The energy valuesfor each step are derived from in vivo conditions, not standard conditions.Glycolysis can be segmented into three stages, the first of which converts glucoseinto two equivalents ofglyceraldehyde-3-phosphate (Steps I - V), thesecond ofwhich converts glyceraldehyde-3-phosphate into 3-phosphoglycerate (Steps VIand VII), and the last of which converts 3-phosphoglycerate into pyruvate (StepsVm - X). Figure 6-27 shows Stage I,the first five steps, ofglycolysis.

Step I:-8.0 kcal/mole

CHjOPOÔ.

CHjOH

HO

Step II:AG = -0.6 kcal/mole

Phosphoglucoseisomerase^

OH

Hexokinase

ATP ADPH OH

Glucose-6-phosphate

OH

OH H

Fructose-6-phosphate

"ATP

°^^H

)—I—HHO

Step HI:AG = -5.3 kcal/mole

Phosphofructokinase

CHpPOg2"Glyceraldehyde 3-phosphate Aldolase^

Triose phosphate isomerase CH2OP03?~ " **

~ f=°CHjOH

CH2OP03iO-

fÂDPCHpPOg

OH

2-CHpPOgGlyceraldehyde-3-phosphate

Step V:AG = 0.6 kcal/mole

Step IV:AG = -0.3 kcal/mole

OH H

Fructose-l,6-bisphosphateDihydroxyacetone phosphate

Figure 6-27

Copyright ©byTheBerkeley Review 116 The Berkeley Review

OrganiC GhemiStry Carbohydrates Biological Applications

StepI (First committedstep in glycolysis): Conversion of glucose into glucose-6-phosphate releases a great deal of energy, so it is thefirst committed step inglycolysis. A pathway's first committed step is the first essentially irreversiblereaction in the pathway. This is true in many mechanisms. Once a highlyfavorable step transpires, the rest of the steps in the mechanism rapidly follow.In the case of Step I, in addition to the energetics not allowing for the backreaction, once glucose is phosphorylated, it carriesa negativecharge,so it cannotdiffuse through the plasmamembrane. Although glucose is toopolar to diffusethrough the membrane on its own, it crosses with the aid of proteins calledglucose transporters. Glucose transporters act like enzymes in that theiractivityis reversible (if glucose levels get too high, then they can pump it out) andspecific (they recognize glucose, not glucose-6-phosphate). Step I is a regulatorystep, meaning that the activity of the enzyme is controlled (allosterically) by thelevels of ATP and other metabolites. Regulation always occurs at irreversiblesteps (rather than reversible steps), so as a helpful hint, you could look for stepscoupled with the conversion of ATP into ADP, because that generally makes thestep exergonic (AG < 0). Steps I and III are points of regulation.

Steps IV (Reverse aldol reaction): Following isomerization of an aldose into aketose and the addition of another phosphate group in Steps II and III, themolecule is poised to undergo a reverse aldol reaction. In Step IV, a six-carbonmolecule is cleaved into two three-carbon molecules by a reverse Aldol reactionthat breaks the bond between carbon 3 and carbon 4 of fructose-l,6-diphosphateto form dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate(G-3-P). Fructose-l,6-bisphosphate is a 6-hydroxyketone, which undergoes thereverse aldol reaction. In Step V, DHAP is converted into G-3-P by way of anisomerization reaction. By having so many steps, some of the intermediateproducts in glycolysis can be used in other pathways. One example is theconversion of G-3-P into glycerol and then into fatty acid triglycerides.

As of Step V, no oxidation has occurred. In Stage II, the carbon-basedcompounds are oxidized, releasing energy. The energy is "stored" in a phosphatebond. It is at this point that NAD+ is required. Figure 6-28 shows Stage II ofglycolysis, in which G-3-P is converted into 1,3-bisphopshoglycerate.

H. -O PiVGlyceraldehyde-3-phosphate q^ OPO,2" . . . , O- .O"^f^ V dehydrogenase ^*^ Phosphoglycerate '^s^

OH f "S* H

kinase

H OH * ^ ^ H OHT^T S~*XCH2OP032" NAD+ NADH +H+ CH2OP032_ ADP ATP CH2OP032_

Glyceraldehyde-3-phosphate Step VI: 1,3-Bisphosphoglycerate Step VII: 3-PhosphoglycerateAG = -0.4 kcal/mole AG = 0.3 kcal/mole

Figure 6-28

Step VI (Oxidation of an aldehyde into a carboxylic acid derivative): Oxidativephosphorylation of G-3-P into 1,3-bisphosphoglycerate is the only oxidation stepin glycolysis. It is the only step where the number of bonds from carbon tooxygen increases. Oxidation releases energy, which is "stored" by adding aphosphate group. Oxidation occurs at carbon 1 as it gains a bond to oxygenwhen converted from an aldehyde into an acid carboxylate. The term P,indicates that there is an inorganic phosphate gained by the reactant. At pH =7.2, inorganic phosphate exists mostly in its dianion form. Dephosphorylation of1,3-bisphosphoglycerate into 3-phosphoglycerate in Step VII, releases aphosphate, which serves to regenerate ATP that was invested in earlier steps.


Organic Chemistry Carbohydrates Biological Applications

0<v°"

H- OH

In the last stage of glycolysis, the compound loses a phosphate to generate moreATP and then undergoes isomerization and elimination to generate the finalproduct, pyruvate. These three steps are shown in figure 6-29.

Step VIII:AG = 0.2 kcal/mole

Phosphoglyceromutase

°^°"

H- OPO,

Step IX:AG = -0.8 kcal/mole

O. .O"

OPO,

CBjOPO/"

3-Phosphoglycerate

CH2OH

2-Phosphoglycerate

Enolasew

HoO

ICHj

Phosphoenolpyruvate

L w^ÂDP +Ho

CH3

Pyruvate

f Step X:AfP AG = -4.0 kcal/mole

Figure 6-29

Steps IX (Enolate formation): Conversion of 3-phosphoglycerate into 2-phosphoglycerate positions thephosphate forenolformation in Step IX. Becausethese two structural isomers have the same bonds, this reaction results in aminimal change in energy. Conversion of2-phosphoglycerate intophosphoenolpyruvate involves dehydration. The double bond formed from elimination has aphosphorylated hydroxyl group attached, making the compound an enol.Conversion ofphosphoenolpyruvate intopyruvate completes glycolysis.

You should not memorize glycolysis, as the amount of time invested to do sowouldnot havea good returnin terms of increasing your MCAT performance.However, you should have a basic idea of what happens overall. It is moreimportant thatyourecognize the typeofreaction than it is to knowdetails ofanyparticular step. A passage maypresentone or more steps of glycolysis and askyou questions from both an organic chemistry and biochemistry perspective.You should know that glucose is converted into pyruvate and that there is a netgeneration of 2 ATP and a release of energy. It is an oxidative process, so amolecule is brokendown. You canapply the generalprinciplesdiscussedhere toother pathways,so learnthe concepts and insights.

1) Regulatorysteps usually involve ATP.

2) Multipstep pathways produce intermediate products that can feedother pathways.

3) Oxidationbreaksdown molecules and requires an oxidizing agent (aspecies poor in H, such as NAD+). Oxidation releases energy.



Oxidation and Reduction of PyruvateOnceformed, pyruvate has a few potential fates. It can undergo either aerobic oranaerobic breakdown. Pyruvate can undergo fermentation, where carbon 2 isreduced from a ketone into an alcohol, resulting in the formation of lactate,which regenerates NAD+. Wecannot reduce lactate any further (only yeast andbacteria can convert pyruvate into ethanol), so it is excreted as waste. Pyruvatecan also undergo oxidative decarboxylation. Converting a carboxylic acid groupinto CO2 results in the gain of a bond to oxygen, so it is oxidation and musttherefore be coupled with reduction of NAD+ into NADH. To make thesereactions easier to grasp, you should focus on the oxidative or reductive nature ofthe pathway. Figure 6-30shows glycolysis, followed by three potential pathwaysthat pyruvate can take.

HOH2C

OHI

GlucoseOH

Glycolysis

^^ Lactate

H- OH

CH,

C02 + CH3CH2OH

C02 + Acetyl CoA

Figure 6-30

Catabolism is the biological process of breaking down a sugar. Given that thegeneration of ATP is considered critical to energy storage, let us finish by brieflylooking at an ATP molecule. It is made of a nucleoside (comprised of adenineand ribose) and a triphosphate group linked at hydroxyl 5. Adenosinetriphosphate is shown in Figure 6-31.

ro

II

Triphosphatex

o o^

Nucleoside (Sugar + Base)A*.

O — P-rO—P—O—P —O—CB

♦ IO"O" O"

Hydrolysis of ATP cleaves hereand releases -7.3 kcal/mole

Figure 6-31

OH HO

Ribose

A nucleoside is a monosaccharide (usually ribose) bonded at carbon 1 to one ofthe bases in either DNA or RNA. Linkages between phosphate groups generate7.3kcal/mole when they are hydrolyzed, meaning that when ATP gains H2O, itcleaves to form ADP, Pi (inorganic phosphate), and H+ with a AG = -7.3kcal/mole. This is a valuable tool for storing energy in small increments.



OrganiC ChemiStry Carbohydrates Section Summary

Key Points for Carbohydrates (Section 6)

Monosaccharides

1. Sugars are named for carbon count and carbonyl typea) Tetrose = 4 C, Pentose = 5 C, and Hexose = 6 Cb) Aldose = aldehyde sugar; Ketose= ketone sugarc) D-sugar has last chiral center = R;L-sugar has last chiral center = Sd) Commonmonosaccharides from biology should be memorized

i. Ribose:"Riboseis all right!"ii. Glucose:"F*** glucose!", and flip it off.iii. Mannose: "Man goes with gun."iv. Galactose: C-4epimer of glucosev. Fructose: ketose of glucose

2. Cyclic sugars are either furanoses or pyranosesa) Pyranosesare six-membered ring sugars

i. Beta anomerhas last carboncis to anomerichydroxyl groupii. Anomeric carbon is the most oxidized carbon (has two bonds to O)iii. 6-Glucopyranose hasallsubstituents in the equatorial position

b) Furanosesare five-membered ringsi. Ribose and fructose are the two most common examples

Oligosaccharides and Polysaccharides

1. Oligosaccharidescontain 2-8 monosaccharides, polysaccharides contain ?a) Glycosidic linkages bind them (wateris releasedwhen linkageforms)

i. Disaccharides include sucrose, lactose, and maltoseii. Polysaccharides include cellulose, amylose, and glycogeniii. Linkages are a or C, named for the componenthydroxyl groupsiv. Enzymes that cleave glycosidic linkages are specific

b) Branching is typically doneusinghydroxyl 6 ofa sugar in the polymeri. Cellulose (6-linked) and amylose (oc-linked) have no branchingii. Amylopectin (a-linked) has branching about 1 out of 30 monomersiii. Glycogen(a-linked) has branching about 1 out of 10 monomers

Chemical Reactions and Tests

1. Chemical reactions and testsdetermine chirality,add Cs, or remove Cs.a) Nitricacid (HNO3) oxidizes terminalCs to acids: opticalactivitytestb) Phenylhydrazine (0HNNH2) forms an osazone: C-2epimer testc) Tollen's test (Ag+)oxidizes C-l of aldoses: aldose testd) Periodate test (IO4") oxidizessugar by cleaving all C-Cbondse) Kiliani-Fischer synthesis increases an aldoseby oneC;forms epimersf) Ruff and Wohl degradation decrease an aldoseby one C

Biological Examples

1. Blood types have glycoproteins with different stereochemistrya) Type O contains3 sugars (Glc-Gal-Fuc) and Types A and Bcontain 4b) L-Fucose is the third sugar, deviating from the natural sugars are D rule

2. Glycolysis breaks down glucoseinto two pyruvate moleculesand energya) Forms a net of 2 ATP and 2 pyruvates for further reactionb) Has one oxidation step;mostly isomerization and phosphate exchange


Carbohydrates

Passages13 Passages

100 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, III, V, VI, & XGrade passages immediately after completion and log your mistakes.

following Task I: Passages II, IV, VII, & XI (28 questions in 36 minutes)Time yourself accurately, grade your answers, and review mistakes.

Review: Passages VIII, IX, XII, XIII, & Questions 92 - 100focus on reviewing the concepts. Do not worry about timing.

II

m

R.E-V-I.E-W .Specializing in MCAT Preparation

I. Glucose and Glucopyranose

II. Fischer and Haworth Projections

III. Sugar Conventions

IV. Monosaccharides versus Disaccharides

V. Amylose, Amylopectin, and Cellulose

VI. Unknown L-Aldopentose Elucidation

VII. Unknown D-Aldohexose Elucidation

VIII. Kiliani-Fischer Synthesis

IX. Osazone Derivative Test

X. Glycolysis Reactions

XI. Blood Types

XII. Biochemistry of Sugars

XIII. Combustion of Sugars


Carbohydrates Scoring Scale


85 - 100 13- 15

65-84 10-12

47 -64 7-9

33-46 4-6

1 -32 1 -3

(1 -7)

(8 - 14)

(15 - 22)

(23 - 29)

(30 - 36)

(37 - 43)

(44 - 50)

(51 -57)

(58 - 64)

(65-71)

(72 - 78)

(79 - 85)

(86-91)

(92 - 100)

Passage I (Questions 1 - 7)

The most stable form of glucose is a hemiacetal ringstructure. The ring is formed when the hydroxyl group oncarbon 5 attacks the aldehyde group to form a hemiacetal.Thehydroxyl can attack thecarbonyl carbon (sp2-center) fromeither the top or bottom of the rc-bond. As a result of thisattack, there are two possible diastereomers that can form.The aldehyde carbon forms the new stereocenter. These twopossible diastereomers are referred to as the alpha and betaanomers of the glucopyranose. The beta isomer (anomer) isdefined as having the anomeric hydroxyl (on carbon one) ciswithrespect to carbon number six in the ring form. Becausemost natural sugarshaveD-configuration, it is perhaps easierto think of the beta anomer as the one with the hydroxylgroup up (equatorial) in the ring form. Both anomers of D-glucopyranose are drawn in Figure 1 below.

OH

6-D-glucopyranose a-D-glucopyranose

Figure 1 Anomeric forms of D-glucopyranose

The 6-anomer is favored by a ratio of 64% to 36% overthe oc-anomer. Hydroxyl groups on adjacent carbons in thering can form hydrogen bonds from all orientations exceptdiaxial. When the hydroxyl groups are equatorial theyexperience less steric repulsion than when they have axialorientation, but hydrogen bonding is slightly reduced.Because the 6-anomer is more abundant, the reduced sterichindrance of the equatorial orientation must be of greaterimportance than the slight loss in hydrogen bonding.Hydrogen bonding is possible from the gauche orientationassociated with diequatorial hydroxylgroups.

1. Whichis the more stable anomer of D-glucose?

A. The alpha-anomer, becausethe hydroxyl group hasequatorial orientation.

B. The beta-anomer, because the hydroxyl group hasequatorial orientation.

C. The alpha-anomer, becausethe hydroxyl grouphasaxial orientation.

D. The beta-anomer, because the hydroxyl group hasaxial orientation.


2. D- Glucose and D-galactose are best described as whichof the following?

A. Anomers

B. C-2 epimersC. C-4 epimersD. Enantiomers

3. D-Glucoseand L-glucose are:

A. anomers.

B. C-5 epimers.C. diastereomers.

D. enantiomers.

4. D-Glucose has which of the following stereochemicalarrangements?

A. 2R, 3R, 4R, 5R

B. 2R, 3S, 4R, 5R

C. 2S, 3R, 4R, 5R

D. 2S, 3S, 4R, 5R

5. How many stereoisomers are possible for a linearaldohexose?

A. 4

B. 8

C. 16

D. 32

6. In the beta-anomer of D-glucopyranose, the first carbonhas the hydroxyl with what orientation?

A. Cis

B. Planar

C. Axial

D. Equatorial

7. The oxygen present in the 1,4-linkage in thedisaccharide below is part of what functionality?

HOH2C

HO^\^Q HOH2C*oA^\Aô-^X*^\HO

HO^X--ÂOHI

A. Ketal

B. Acetal

C. Hemiketal

D. Hemiacetal

OH

OH



The most stable form of most monosaccharides is the

cyclic form, rather than the straight chain form. This isparticularly true of the aldopentoses and aldohexoses.Although it is common to represent aldohexoses in theirstraight chain form, for most sugars, less than one percentexist naturally in the straight chain form. The straight chainform of a monosaccharide is represented by the Fischerprojection, while the cyclic form is often represented by theHaworth projection. Cyclic structures come in five and six-membered rings, referred to as furanose and pyranoserespectively. The Fischer projection of D-mannose and theHaworth projection of B-D-mannopyranose are shown inFigure 1.

H^°HO

BO-

H

H

H

OH

H OH

CH2OH

D-mannose

Figure 1 Fischer and Haworth projections of mannose

Polysaccharides, such as cellulose, are formed from adehydration reaction of cyclic monosaccharides. One water islost for every glycosidic linkage that is formed. Theglycosidic linkage between the two monosaccharides in adisaccharide involves the anomeric hydroxyl of one sugar(referred to as theglycosyl group) and a hydroxyl group of asecond monosaccharide (referred to as the glycoside). Forinstance, 4-0-(B-D-galactopyranosyl)-a-D-glucopyranoside,lactose, results from a B-1,4-linkage between hydroxy1-4 ofa-D-glucopyranose and carbon-1 of B-D-galactopyranose. Aglycosidic linkage is named for the two hydroxyl groups thatundergodehydration to form the linkage.

8. Which of the following Haworth projections representsthe C-3 epimerof B-D-mannopyranose?

A- CH2OH B- CH2OH

C.

OH H/l IV H HOxHOM l/OH HO\ l/OH

H OH

CH2OH

H H

B-D-mannopyranose

OH H

D.CH2OH CH2OH

\oh Hyi \H HO/1oV /h ho\[ i/hH OH OH H


9. A polysaccharide consisting of only B-anomers with1,4-glycosidic linkages is commonly known as:

A. glycogen.B. cellulose.

C. sucrose.

D. lactose.

10. All of the following are true of aldopentoses EXCEPT:

A. they are likely to form furanose rings.B. carbon 4 determines the designation as D or L.C. carbon 2 is the anomeric carbon in the cyclic form.D. they have four chiral centers in their cyclic form.

11. What is the molecular weight for the trisaccharide madefrom two glucoses and a mannose with 1,4-B-linkages?

A. 474grams/moleB. 486 grams/moleC. 504 grams/moleD. 540 grams/mole

12. The glycosidic linkage in most polysaccharides is whichof the following?

A. 1,1

B. 4,4

C. 1,4

D. 2,5

13. The cyclic form of an aldopentose is a:

A. furanose ring with an acetal functional group.B. furanose ring with an ketal functional group.C. pyranosering with an acetal functional group.D. pyranose ring with an ketal functional group.

14. The following structurehas what glycosidicbond?

CH2OH

I\OH HO/1HO\j l/OCH,

H H

A. An O-glycosidic bond between the carbon ofmethanol and the anomeric oxygen of a-mannose.

B. An O-glycosidic bond between the carbon ofmethanol and the anomeric oxygen of B-mannose.

C. An O-glycosidic bond between the oxygen ofmethanol and the anomeric carbon of a-mannose.

D. An O-glycosidic bond between the oxygen ofmethanol and the anomeric carbon of B-mannose.



Sugars are organic compounds made exclusively ofcarbon,hydrogen,and oxygen. In a monosaccharide, the ratioof carbon-to-hydrogen-to-oxygen is 1 : 2 : 1. As the namecarbohydrate implies, for every carbon atom, there is a watermolecule in the formula. By convention, sugars are namedfor the number of carbons and given the suffix "ose" toindicate thattheyare a sugar. The highest priority functionalgroupof the sugar is indicated by the prefix. For instance,analdohexose is a six carbon sugar with an aldehydefunctionalgroup. In an aldohexose, the first carbon has an aldehydefunctional group and the remaining five carbons havehydroxyl groups. The stereocenters have variety, but theconnectivity is the same. Figure 1 shows the eight possibleD-aldohexoses.

HNô

H-

HO-

HO-

H-

OH

H

H

OH

CH2OH

D-(+)-Galactose

H^°

H-

H-

HO-

H-

OH

OH

H

OH

CH2OH

D-(-)-Gulose

HxÔ

H-

HO-

H-

H-

OH

H

OH

OH

CH2OH

D-(+)-Glucose

HV^°

HO-

H-

HO-

H-

H

OH

H

OH

CH2OH

D-(->Idose

HVÔ

HO-

HO-

H-

H-

H

H

OH

OH

CH2OH

D-(+)-Mannose

Hvô

HO-

H-

H-

H-

H

OH

OH

OH

CH2OH

D-(+>Altrose

Hvô HvÔ

HO-

HO-

HO-

H-

H

H

H

OH

H-

H-

H-

H-

OH

OH

OH

OH

CH2OH CH2OH

D-(+)-Talose D-(+)-Allose

Figure 1 Eight D-aldohexoses

Fructose is also a six-carbon sugar, but it has a ketonefunctionality on carbon2, rather than an aldehyde functionalgroup. It is referred to as a ketohexose.


15. All of the following statements about a D-aldohexoseare true EXCEPT:

A. It contains fivesecondary alcohol groups.B. Its straight chain form has four chiral centers.C. Its most oxidized carbon is carbon 1.

D. Its fifth carbon has R-chirality.

16. Which of the following sugars can be classified as aketohexose?

A. D-Mannose

B. D-Fructose

C. D-Glucose

D. D-Galactose

17. What is the C-5 epimer of L-glucose?

A. D-Idose

B. L-Mannose

C. D-Allose

D. D-Glucose

18. The C-2 epimer of D-Gulose is also the:

A. C-3 epimer of D-GlucoseB. C-3 epimer of D-IdoseC. C-4 epimer of D-AlloseD. C-4 epimer of D-Altrose

19. The Tollen's test involves an oxidizable sugar reducingsilver cation to silver metal, which can then be

identified by its precipitate. Which of the followingsugars yields no silver precipitate?

A. Mannose

B. Sucrose

C. Glucose

D. Galactose

20. Sugars naturally exist in cyclic conformations in theirmost stable form. The ring is formed when an OHgroup attacks the carbonyl carbon. The most stableform of fructose would be which of the following?

A. Hemiacetal

B. Acetal

C. Hemiketal

D. Ketal


21. Which of the following aldohexoses is NOT an epimerofD-talose?

A. D-Galactose.

B. D-Idose.

C. D-Mannose.

D. D-Glucose.

22. Treatment of which of the following sugars withNaBH4 will yield an optically inactive product?

A. Glucose

B. Mannose

C. Ribose

D. Fructose


Passage IV (Questions 23 - 29)

Monosaccharides typically have a formula of CnH2nOn,where n is an integer. Depending on the number of carbonsin the monosaccharide, it may have a preference for a cyclicstructure (a hemiacetal or hemiketal) over the straight chainform. To cyclize, the hydroxyl group on the penultimatecarbon attacks the carbonyl carbon, which turns the carbonylcarbon into the anomericcarbon. Generally, aldopentoses andketohexoses form furanose rings, while aldohexoses formpyranose rings, as shown in Figure 1.

CH2OH

(HO)/| <\(OH)

,(OH)(HOV

H H

D-aldohexapyranose

CH2OH (OH)

H H

D-aldopentafuranose

CH2OH (OH)

CH2OH

H H

CH2OH D-ketohexafuranose

D-ketohexose

Figure 1 Generic cyclic monosaccharides

When two cyclic monosaccahides combine, they lose onewater molecule as a side product. The resulting disaccharideis described by the common oxygen between the rings. Thisglycosidic linkage is described by its respective position oneach ring. In almost all disaccharides, one sugar uses itanomeric hydroxyl group to form the glycosidic linkage. Adisaccharide is typically drawn with the sugar using itsanomeric oxygen in the linkage on the left and the othersugar on the right. When drawn in this standard method, thesugar on the left is given an "osyl" suffix while the sugar onthe right is given an "ide" suffix.


23. The C-2 epimer of galactose is the same as which of thefollowing sugars?

A. The C-2 epimer of glucose.B. The C-2 epimer of mannose.C. The C-4 epimer of mannose.D. The C-4 epimer of glucose.

2 4. The linkage in sucrose is best described as:

CH2OH

Q H

A. al-a2

B. al-B2

C. Bl-a2

D. B2-a2

CH2OH

25. The glycosidic linkage of maltose is what type offunctional group?

HOH2C

HO--^V^--'0vho-A.^----v-A

UH0H2C*

ohIOH

A. Hemiacetal

B. Hemiketal

C. Acetal

D. Ketal

26. The following monosaccharide relates in what mannerto B-D-glucopyranose?

HOH2C

OHO

OH

OHHO

A. It is an anomer of B-D-glucopyranose.

B. It is an conformer of B-D-glucopyranose.C. It is an enantiomer of B-D-glucopyranose.

D. It is an epimer of B-D-glucopyranose.


27. A trisaccharide madefrom threeunique aldohexoses has:

A. two acetal linkages.

B. three acetal linkages.

C. two hemiacetal linkages.D. three hemiacetal linkages.

2 8. Exhaustive methylation of a disaccharide converts all ofthe hydroxyl groups into methoxy groups. Uponhydrolysis under acidic conditions, what is observed forthe fully methylated disaccharide?

A. All of the OCH3 groups return to being OHgroups.

B. Only the glycoside loses OCH3 groups, not theglycosyl group.

C. The disaccharideracemizesat every chiralcenter.D. The two carbons involved in the linkage will have

hydroxyl groups and no methoxy groups.

29. In the following disaccharide, the glycoside is:

CH2OH CH2OH

A. D-galactose.

B. L-galactose.

C. D-mannose.

D. L-mannose.

1—0—»



Human beings possess the enzyme for cleaving thealpha-1,4-glycosidic linkage of starches. Wecan break downpolysaccharides suchas amylose, amylopectin, andglycogen.However, we are unable to digest cellulose, because itcontains B-D-glucopyranose residues connected by B-l-4-glycosidic linkages. Foods rich in cellulose pass through usundigested. Bothamylose and amylopectin are madefrom a-D-glucopyranose, drawn in Figure 1.

HOH2CO

Figure 1 a-D-Glucopyranose

Amylose is a polysaccharide made of a-D-glucopyranosewith exclusively a-1,4-glycosidic linkages. Amylopectin islike amylose, except that about one out of every thirty of theglucopyranose monomers has- an a-l,6-glycosidic linkage toa branched strand of a-glucopyranoses. Glycogen also hasbranching, but significantly more than amylopectin. Allthree polysaccharides are ultimately broken down into smallerfragments, such as the disaccharide maltose. We absorb onlymonosaccharides into our bloodstream, so starches must beenzymatically cleaved in the mouth and small intestine.Even disaccharides are broken down into monosaccharides to

be absorbed from the intestine into the blood.

3 0. Which of the following disaccharides is maltose?

A. HOH2C

HO--^-\--—°v HOH2CHO-\^-•-Y-\---o---T^ ~~\OH HO-X^--\--A-OH

OHB. HOH2C

HO--V-\-"°\HO-X--tAohIhoh2co—V-a—--0

HOA—-l^-\-OHOH

C. HOH2CHO^V^X-— 0 HOH2C

HOA---OH HO-\-

OHlOH

D. HOH2CHO~~^-\-"\

hoA^-tAoh1hoh2c•\

Ho\îohI

OH


31. Why is amylase released in both saliva and by thepancreas into the small intestine?

A. Salivary amylase is destroyed by the gastric fluids.B. Pancreatic amylase is weaker than salivary amylase.

C. Salivary amylase is weaker than pancreatic amylase.

D. Pancreatic amylase cleaves amylose with B-linkageswhile salivary amylase cleaves a-linkages.

3 2. Which sugar below CANNOT be absorbed into blood?

A. Fructose

B. Galactose

C. Glucose

D. Lactose

3 3. The difference between amylose and amylopectin is that:

A. amylose has more branching.B. amylose has B-l,6-branching, while amylopectin

has a-l,6-branching.

C. amylose has no branching, while amylopectin hasa-l,6-branching.

D. amylopectin has no branching, while amylose hasa-l,6-branching.

3 4. What is NOT found in glycogen?

A. Monomers of a-D-glucopyranose.

B. a-l,3-linkages

C. Acetal functional groups

D. a-1,6-linkages

35. Which of the labeled hydroxyl groups are involved inlinkages in amylopectin?

CH2OH IVml

HO/II/Hk OH

-°VH

H/l

H

•Jf oh^H1

A. I andIIonly

B. I and m only

C. I, II, and m only

D. I, m, and IV only

3 6. Amylose is eventually broken down into

A. cellulose.

B. lactose.

C. pyruvate.

D. sucrose.



Compound A, an unknownL-aldopentose, is treatedwithnitric acid to yield Compound B, a diacid referred to as analdaric acid, as shown in Reaction 1 below. Compound Bshows no specific rotation ([ap] of 0°). This is to say thatthe diacid does not rotate plane polarized light when anaqueoussolution of the diacid is analyzed using a polarimeter.

HNQ3 Compound B([aD]ofO')

Reaction 1

Compound A

Compound A, when treated with a sequence of reagentsthat is a slight modification of the Kiliani-Fischer synthesis(which increasesthe length of an aldopentose by one carbon),yields two diastereomers that vary in their chirality at onlyone carbon. The two epimers are referred to as Compound Cand Compound D. Carbon 1 of Compound A is labeled with,3C. The I3CNMR peak that starts at 202 ppm inCompound A but is found at 78 ppm in both Compound Cand Compound D. This implies that the ,3C label is not onthe aldehyde carbon in Compounds C and D. Reaction 2 isshown below.

Compound A

l.HCN

2.H2/Pd(BaS04)

3. H30+/A '

Reaction 2

Compound C+

Compound D

Compound C when treated with nitric acid yields a diacidproduct (Compound E) that shows a specific rotation ([ap])of -26°. Compound D when treated with nitric acid yields adiacid product (Compound F) that shows a specific rotation([&d]) of 0*. This is to say that compound E is opticallyactive, while compound F is optically inactive. Thereactions are shown below as Reaction 3 and Reaction 4:

Compound C HN03

Reaction 3

Compound D HNO3

Reaction 4

Compound E([aD]of-26°)

Compound F([aD]ofO')

Using the optical rotation data presented, it is possible todeducethe identityof the unknown L-aldopentose(CompoundA). The chirality of each hydroxyl group can be determinedone at a time from the optical activity data.

37. How many stereoisomers are possible foraldopentose?

A. 2

B. 4

C. 8

D. 16

Copyright © by The BerkeleyReview®

an

129

3 8. Compounds C and D are BEST described as which ofthe following?

A. Anomers

B. C-2 epimersC. C-3 epimersD. Enantiomers

3 9. Which of the following sugars is NOT possible for theidentity of Compound A?

A. Ribose

B. Xylose

C. Glucose

D. Lyxose

40. Compound B has which of the following molecularmasses?

A. 132 grams/moleB. 148 grams/moleC. 164 grams/moleD. 180 grams/mole

41. The melting point of Compound B:

A. is greater than the melting point of Compound A.B. is equal to the melting point of Compound A.C. is less than the melting point of Compound A.D. is less than the boiling point of Compound A.

42. Aldohexoses in their most stable form would NOT

show which of the following IR peaks?

A. 1140 cm"1

B. 1715 cm-1

C. 2980 cm"1

D. 3480cm-1

43. The second carbon (C-2) for Compound A:

A. must have R stereochemistry.

B. must have S stereochemistry.C. can have either R or S stereochemistry.D. is achiral.



Compound X, an unknown D-aldohexose, undergoes asequence of chemical reactions. Compound X is first treatedwith nitric acid and the purified product (Compound Y) isplaced into a polarimeter to analyze its optical activity.Compound Y shows an optical rotation, [arj> = 26°].Compound X is then treated with the following sequence ofreagents: first bromine in water, second ferric carbonate inwater, and finally hydrogen peroxide. The product(Compound Z) has a molecular mass of 150 grams/mole.Compound Z is then treated with nitric acid to yield a newproduct (Compound W), which when placed into apolarimeter shows no optical activity.

Compound X is then treated with one equivalent ofperiodic acid, which breaks the bond between carbons 3 and 4,oxidizing carbons 3 and 4 up one level to aldehydes. Thissix-carbon hemiacetal fragment (Compound Q) is thenhydrolyzed out of the hemiacetal form and converted into twosmaller fragments, both three carbons long (Compound R andCompound S). From the optical rotation data, the researcherconcludes that carbons 3 and 4 have opposite chirality. Thereactions are summarized in Figure 1 below:

VH(OH)

(OH)

(OH)

OH

CH2OH

Compound X

HN03

l.Br2(aq)

Compound Yoptically active

Compound Z(MW=150g/mol)Fe2(C03)3(aq)

H202(aq)

IHNO3

HIO4 Compound Woptically inactive

Compound Q

H*<aq)o o o

AA h-Vch2°h \A A Com]Compound R

&

Compound SH OH

Compound R

H OH

Compound S

Figure 1 Reaction summary of Compound X experiment

The reaction involving the Br2 liquid in water is knownas Ruff degradation, which removes carbon 1 and oxidizescarbon 2 into an aldehyde, while leaving the sugar chainintact from carbon 3. The stereochemistry of the backbonedoes not change during this reaction. The structure ofCompound X can be deduced from this information.

44. Treatment of Compound X with sodium borohydride(NaBH4) would yield which of the following?

A. An optically active diacid.B. An optically inactive diacid.C. An optically active hexa-ol.D. An optically inactive hexa-ol.


4 5. The stereochemistry of carbon 3 in the Compound X is:

A. definitelyR.B. definitely S.C. variable (either R or S).

D. achiral (no chiral center).

46. Compound X, when treated with hydrogen cyanidefollowed by hydrogen gas over palladium metal onbarium sulfate and then treated with acidic water would

yield which of the following?

A. AD-aldopentose.B. An L-aldopentose.C. A D-aldoheptose.D. An L-aldoheptose.

4 7. A reverse aldol reaction involving Compound X yieldsan aldotetrose (formed by breaking the C-2:C-3 bond).The aldotetrose when treated with sodium borohydrideyields an optically active tetra-ol. This confirms whichof the following conclusions about the chiral centers ofCompound X?

A. Compound X has chirality of 3R and 4R.B. Compound X has chirality of 3S and 4R.C. Compound X has chirality of 4R and 5R.D. Compound X has chirality of 4S and 5R.

4 8. How many stereoisomers are possible for the straightchain form of a D-aldohexose?

A. 4

B. 8

C. 16

D. 32

4 9. Which of the following reagents, when added to analdohexose, would yield the same conclusion about itschirality as adding HNO3?

A. NaBH4

B. H2S04

C. Br2(aq)

D. KMn04(aq)

5 0. Treatment of an aldohexapyranose with one equivalentof HIO4 breaks the sugar between:

A. carbons 1 and 2.

B. carbons 5 and 6.

C. carbons 2 and 3.

D. carbons 3 and 4.



The length of a sugar can be increased by one carbonusing the Kiliani-Fischer synthesis. Kiliani-Fischersynthesis employs a cyano nucleophile in the first step. Thecyano group attacks the carbonyl carbon in an additionreaction. The cyano group is then reduced into an iminewhich is subsequently hydrolyzed into an aldehyde. Anexample using ribose is shown in Figure 1.

Hô HvÔv°

OH

OH ^[5]

HO- H OH

OH

OH

OH

l.HCN/KCN<jN h OH + H

OH 3. H30+/A

CH2OHD-(+)-Ribose

H-

H-

OH

OH

CH2OH CH2OHD-(+)-Altrose D-(+)-Allose

Figure 1 Kiliani-Fischer synthesis starting with ribose

A researchercarried out the Kiliani-Fischersynthesis onfour different D-aldopentoses to determine the effects of sterichindrance on stereoselectivity. The four D-aldopentoses areshown in Figure 2.

v° HvÔ

H OH HO-

H OH H OH HO

H OH H—I—OH H-CH2OH CH2OH

D-Ribose D-Arabinose

H

Hvô HvÔ

H- OH HO-

•H HO-

OH H-

CH2OH

H

H

OH

CH2OH

D-Xylose D-Lyxose

Figure 2 Four D-aldopentoses

Table 1 lists the product distribution for each synthesis.

Aldopentose Major Product Major Product

D-Ribose 81%D-Altrose 19% D-Allose

D-Arabinose 72% D-Glucose 28% D-Mannose

D-Xylose 71% D-Idose 29% D-Gulose

D-Lyxose 76% D-Galactose 24% D-Talose

Table 1

51. To synthesize the following D-aldohexose, whataldopentose should be used?

Q OH OH H OH

>H H H OH H

A. D-Arabinose

B. D-Lyxose

C. D-Ribose

D. D-Xylose

Copyright © by The Berkeley Review®

CH2OH

131

52. Which of thefollowing pairof sugars areC-2 epimers?

A. D-Ribose and D-Altrose

B. D-Glucose and D-Talose

C. D-Mannose and D-Allose

D. D-Idose and D-Gulose

53. In which of the steps in Kiliani-Fischersynthesis is thesugar reduced?

A. Steps I only

B. Steps II only

C. Steps I and II onlyD. Steps I, II, and IE only.

54. Which of the following is an intermediate in theKiliani-Fischer synthesis?

A. Aldehyde

B. Amine

C. Ester

D. Imine

55. The product mixture following Kiliani- Fischersynthesis is best described as:

A. two structural isomers in equal concentration.

B. two structural isomers in unequal concentration.

C. two diastereomers in unequal concentration.

D. two enantiomers in equal concentration.

5 6. Which structure represents B-D-talopyranose?

A- CH2OH B- CH2OH

OH HO A °\ OHOH H

H HXI IXH

HO XI \ OH

OH HOHXI IXH

OH HOHO XI lyoH

5 7. Kiliani-Fischer synthesis converts an:

A. aldohexose into an aldopentose.

B. aldohexose into a ketohexose.

C. aldopentose into an aldofuranose.

D. aldopentose into an aldohexose.



An unknown sugar is extracted from the fruit collectedfrom the Racaniqui Tree native to Willoughby, Montana.The sugar is isolated by column chromatography in 99.9%purity as shown by ^HNMR. Themolecular mass was foundto be approximately 342 ±2 grams per mole. To have thatmolecular mass, the unknown sugar must be a twelve-carbondisaccharide. The sugar shows a specific rotation of+126.2*.

The disaccharide linkage is cleaved rather easily withmild acid, and two six-carbon monosaccharides, Compound Iand Compound II, are isolated. When treated with threeequivalents of phenylhydrazine, both monosaccharides formosazones. Compound I yields an osazone with a specificrotation of 442.6° while Compound II yields an osazone witha specific rotation of +37.2°. Figure 1 shows the reaction ofglucose with three equivalents of phenylhydrazine.

*v°H

HO

H

H

OH

H

OH

OH

CH2OH

HN^N—NH0

F=N—NH0

H

OH

OH

CH2OH

+ NH3 + C6H5NH2

Figure 1 Osazone formation starting from glucose

HO-

H-

H-

30HN—NH2 HHOAc *~

The specific rotation of the osazone of D-glucose is+54.6°. Both Compound I and Compound II are naturallyoccurring sugars, so they are assumed to be D-sugars. Whenboth sugars are treated with nitric acid, each one is oxidizedinto an aldaric acid (a diacid with carboxylic acid groups atboth carbon one and carbon six). Both of the aldaric acidsformed are optically active. This implies that neitheraldaricacid is meso and that both sugars are aldohexoses, rather thanketohexoses. A 2-ketohexose can also react with threeequivalents of phenylhydrazine to form an osazone derivative.

Theinformation from thederivatives narrows theidentityof each aldohexose to one of four choices. To identify theoriginal sugar, the physical properties such as melting pointandspecific rotation can be compared to the values of the fourpossible D-aldohexoses. To determine the linkage of thedisaccharide, the hydroxyl groups can be glycosidicallylabeled. After hydrolyzing the labeled disaccharide into itstwo monosaccharides, the sugar with hydroxyl four unlabeledis the glycoside (sugar on the right in standard notation).Thisallows for thedisaccharide to be identified precisely.

58. How many stereocenters are present in the originaldisaccharide?

A. 7

B. 8

C. 10

D. 11


59. Which of the following pairs of structures couldrepresent Compound I and Compound n?

HN^°H-

HO-

H-

H-

OH

H

OH

OH

CH2OH

B.

Hs^°HO-

H-

H-

H-

H

OH

OH

OH

CH2OH

C.HN^

o

HO-

HO-

H-

H-

H

H

OH

OH

CH2OH

D.

Hs^°H-

H-

HO-

H-

OH

OH

H

OH

CH2OH

H-

HO-

HO-

H-

OH

H

H

OH

CH2OH

"x^0HO-

HO-

HO-

H-

H

H

H

OH

CH2OH

%*°HO-

H-

HO-

H-

H

OH

H

OH

CH2OH

*\f°HO-

H-

HO-

H-

H

OH

H

OH

CH2OH

60. Which of the following aldopentoses will yield anoptically active aldaric acid?

A.

HN*°H-

H-

H-

OH

OH

OH

CH2OH

C.

HN^°HO-

H-

HO-

H

OH

H

CH2OH

B.

%*°H-

HO-

H-

OH

H

OH

CH2OH

D.

H^°HO-

H-

H-

H

OH

OH

CH2OH


61. Which of thefollowing sugars CANNOT beCompoundI or Compound n?

I. Mannose(the C-2 epimer of glucose)

II. Talose (the C-4 epimer of mannose)

IE. Galactose (the C-4 epimer of glucose)

A. I onlyB. HonlyC. I and HonlyD. I and m only

62. Which of the following choices shows a sugar pairedwith the incorrect reason for eliminating it as a possiblestructurefor either CompoundI or Compoundn?

A.

Hv^°H-

H-

H-

H-

OH

OH

•OH

OH

CH2OHWill yield an opticallyinactive aldaric acid

C.

Hv^°HO-

H-

HO-

H-

H

OH

H

OH

CH2OHWill yield an opticallyinactive aldaric acid

B.

Hv^0HO-

HO-

H-

H-

H

H

OH

OH

CH2OHWill yield the sameosazone as glucose

D.

HO-

H-

H-

CH2OH

= 0

H

OH

OH

CH2OHWill yield the sameosazone as glucose

63. Which of the following choices contains two sugarsthat will form the same osazone?

A. D-ribose and D-xylose (C-3 epimers).B. D-idose and D-altrose (C-4 epimers).C. D-talose and L-talose.

D. D-mannose (C-2 epimer of D-glucose) and D-fructose (ketose of D-glucose).

64. What happens to the three phenylhydrazine molecules?

A. One adds to the sugar and two are oxidized.B. Two add to the sugar and one is oxidized.C. One adds to the sugar and two are reduced.D. Two add to the sugar and one is reduced.



The early part of glycolysis involves the conversion ofD-glucose into glyceraldehyde-3-phosphate. Glycolysis iscarried out in ten enzymatically catalyzed steps. The firstfour steps ofglycolysis, forming glyceraldehyde-3-phosphate,are shown below:

H^° H^°H-

HO-

H-

H-

OH

H

OH

OH

CH2OH

H-

HO-

H-

H-

Hexokinase.

~0H f \ H~OH . __ T h

ATP ADP

OH

H

OH

OH

CH2OP032"

STEP I: Glucose to Glucose-6-phosphateAG" = - 4.0 kcal/mole

HY°H— — OH

HO — —H

H— — OH

H— — OH

IsomeraseHO-

H-

H-

CH2OH

=o

•H

OH

OH

CH2OP03:CH2OP03^

STEP II: Glucose-6-phosphate to Fructose-6-phosphateAG" = + 0.4 kcal/mole

CH2OH

= 0

H Phospofructokinase_ HO-

OH

OH

CH2OP03^

CH2OP032'= 0

H

OH

OH

CH2OP032"

HO-

H-

H-

H Fhosporructokinase HO -

~0H f >v * «•OH T H-

2 ATP ADP

STEPIII: Fructose-6-phosphate to 1,6-BisfructophosphateAG° = - 3.4 kcal/mole

HO-

H-

H-

2-CH2OP03

=0

•H Aldolase

OH ^~

OH

CH2OP032' H^^°I I^r f=0 + h—j— OHCH2OH 2-CH2OP03

CH2OP032'STEPIV: 1,6-Bisfructophosphate to Dihydroxyacetone-

phosphate and Glyceraldehyde-3-phosphateAG8 = + 5.7 kcal/mole

Dihydroxyacetonephosphate undergoes isomerization toform glyceraldehyde-3-phosphate. Steps I-IV correspond tocommon reactions in organic chemistry, so the mechanismfor each step follows standard organic chemistry processes.


The addition of the phosphate group in Step I is similar toesterification of a carboxylic acid with an alcohol. The roleof the enzyme is to align the molecules so that the hydroxylgroup on carbon 6 reacts to gain the phosphate, rather thananother hydroxyl group. The conversion of an aldehyde intoa ketone in Step II goes through a process known astautomerization. Step III is similar to Step I. The breakingof the carbon-carbon bond in Step IV is an example of theretroaldol reaction. The formation of a B-hydroxyketone froma carbonyl is a common reaction in synthetic chemistry.

65. G-3-P, glyceraldehyde-3-phosphate, and DHAP,dihydroacetonephopshate are best described as:

A. configurational isomers.

B. diastereomers.

C. epimers

D. structural isomers.

66. If D-galactose (the C-4 epimerof glucose) undergoes thesame four reactions shown in the passage, how will theproducts differ from the products of glucose?

A. G-3-P will have a different chiral center.

B. DHAP will have a different chiral center.

C. Both G-3-P and DHAP will have different chiralcenters.

D. The products will be identical to the productsformed by the glucose reaction.

67. If carbon 5 of a single D-glucose is enriched with 13C,where will the enriched carbon (label) be found in theproducts?

A. Carbon 1 of DHAP.

B. Carbon 3 of DHAP.

C. Carbon2ofG-3-P.

D. Carbon3ofG-3-P.

68. What is the overall free energy change from Step I toStep IV of glycolysis?

A. +2.1 kcals per mole.B. + 1.3 kcals per mole.C. - 1.3 kcals per mole.D. - 2.1 kcals per mole.

69. How many chiral centers are present in the two finalproducts (G-3-P and DHAP)?

A. 0

B. 1

C. 2

D. 3


7 0. Which of the following sugars is fructose-6-phosphate?A. B.

CH2OP032" CH2OH CH2OH CH2OP032

D.

CH2OP03z" CH2OH CH2OH CH20P03O

71. Glycolysis converts glucose into:

A. two pyruvates and generates a net two ATPs.

B. two pyruvates and generates a net four ATPs.

C. three pyruvates and generates a net two ATPs.

D. three pyruvates and generates a net four ATPs.

2-



Blood types in human beings are determined by thepolysaccharide derivative attached to the amino terminal of aprotein in the wall of the red blood cell. The success of ablood transfusiondepends on the compatibility of antibodiesthatrecognize the markers on red blood cellsas equivalent orforeign. If the antibodies do not recognize the foreign redblood cells as invaders, then the transfusion is possible.Blood type O, the universal donor, contains a trisaccharide,whiletypesA and B contain tetrasaccharides. Figure 1showsthe polysaccharide associated with Type O blood, Figure 2shows the polysaccharide associatedwith Type A blood,andFigure 3 shows the polysaccharide associated with Type Bblood.

H3Cy^r^/-oH[ho

HO

Figure 1 The polysaccharide associated with Type O blood

HO

HaCvôl

NHiO

CH2OH

CH2OH

protein

CH2OH

HO-«X^ÂNH|

S ?\ proteinCH3

Figure 2 The polysaccharideassociated with Type A blood

HLS2\ _s*H0H0J»Kô^\HO-7^^-7 o nh|

Hoîc-ppÔH kr\HO

HO

Figure 3 The polysaccharide associated with Type B blood

In the three blood types, the first sugar attached to theprotein is N-acetylglucosamine. Attached to the first sugarvia a 1,4-glycosidic linkage is galactose.


In all threeblood types, the galactose forms a bridge viahydroxyl two to a fucose residue. In blood type A, thegalactosehas a N-acetylglucosamine linked to hydroxyl threewhile in blood type B, the galactose has a second galactoseresidue linked to hydroxyl three. These polysaccharides arereferred to as the antigenicdeterminants of the red blood cells.These, along with the Rh factor, determine the feasibility ofa transfusion.

72. What linkage is present between N-acetylglucosamineand galactose in blood type O?

A. cc-l,2-linkageB. 6-1,2-linkageC. a-1,4-linkageD. 6-1,4-linkage

73. How many acetal and ketal functionalities are present inthe type O blood antigenic determinant?

A. 0

B. 1

C. 2

D. 3

74. If the CH3 group of fucose were replaced by CH2OH,how would the molecule differ from glucose?

A. It would be the C-2 epimer of D-glucose.B. It would be the C-2 epimer of L-glucose.C. It would be the C-4 epimer of D-glucose.D. It would be the C-4 epimer of L-glucose.

7 5. Which of the following molecules is B-D-2-acetamino-2-deoxyglucose?

A* CH2OH

HO-^V^"°\ho-\^^*\-AnhI

/ OH0=C

\CH3

C* CH2OHHO-"-V^°\ho^^-Â

/ OHo=c

NH2

B.

HOHO

D.

HOHO

CH2OHO

OH

NH

0=C\

CH3

CH2OHO

OH

/O

o=c

NH2


7 6. The formation of L-fucose from an L-aldohexose would

require what type of reaction?

A. Reduction specific for carbon 6.

B. Oxidation specific for carbon 6.C. Reduction specific for carbon 4.D. Oxidation specific for carbon 4.

7 7. In the type B antigenic determinant, what has been addedto hydroxyl three of the central galactose?

A. cc-galactofuranose

B. B-galactofuranoseC a-galactopyranoseD. B-galactopyranose

78. How many chiral centers are present in the type Aantigenic determinant?

A. 12

B. 16

C 20

D. 24



Glycolysis is an example of a biological process thatgenerates energy that is ultimately stored in the form ofadenosinetriphosphate, ATP. ATP is able to store energybecause its phosphodiester bonds are very weak and result inthe release of energy when they under go hydrolysis. Thehydrolysis of ATP into ADP releases 7.3 kcal/mole ofenergy. Table 1 shows the energy of hydrolysis for selectedphophorylated compounds that are commonin biochemistry.

Compound AG0' (kcal/mole)

Phosphoenolpyruvate -14.8

Acetyl phosphate -10.3

Pyrophosphate -7.9

Adenosine triphosphate (ATP) -7.3

Glucose 1-phosphate -5.0

Glucose 6-phosphate -3.2

Glycerol 3-phosphate -2.2

Table 1

The AG" termrepresents the standardfree energychangefor the reaction at a pH of 7.

The conversion from one structure into another structure

in Table 1 involves a phosphate group transfer. Thefavorability of the transfer reaction can be ascertained fromthe energies of hydrolysis for the two reactions. The bestphosphate group transferagent is phosphoenolpyruvate. Theloss of the phosphate group by phosphoenolpyruvate andsubsequentconversion from pyruvate enol into pyruvate isshown in Figure 1.

°^-0'

H-

k—OP032

-H

°^°"

-• I—OH

H—U—H

°^°"

O

CH3

Figure 1 Conversion of phosphoenolpyruvate intopyruvate enol and pyruvate

The substantially large energy of hydration associatedwith phosphoenol pyruvate is attributed in part to thefavorable conversion from an enol into a ketone. The ketoneis the more stable of the two tautomers, so the equilibriumconstant is greater than 1 for the conversion of an enol into aketone.

7 9. What is the approximate change in free energy for theconversion of pyruvate enol into pyruvate?

A. +16.4 kcal/mole

B. +6.7 kcal/mole

C. -6.7 kcal/mole

D. -16.4 kcal/mole


8 0. Which of the following compounds can couple with theconversion of ADP into ATP in an overall favorable

fashion?

A. Acetyl phosphate

B. Glucose 1-phosphate

C. Glucose 6-phosphate

D. Glycerol 3-phosphate

81. The high energy associated with the hydrolysis of acompound with a phosphate group can be attributed to:

A. the strong phosphodiester bond, where the strengthcan be attributed to its aromaticity.

B. the strong phosphodiester bond, where the strengthcan be attributed to its strong resonance.

C. the weak phosphodiester bond, where the weaknesscan be attributed to electrostatic repulsion.

D. the weak phosphodiester bond, where the weaknesscan be attributed to the inductive effect.

8 2. Which enzyme catalyzes the conversion of phosphoenolpyruvate, PEP, into pyruvate?

AG = -15 kcal/mole

ATP ADP

A. PEP dehydrogenase

B. PEP reductase

C. Pyruvate kinase

D. PEP aldolase

°^°-

= 0

CH3

83. When glyceraldehyde-3-phosphate, G-3-P, converts to1,3-bisphosphoglycerate, what is NOT true?

%^H Q*^

OP03^

OH

2-CH2OP03

A. Carbon 1 is oxidized.

B. NAD+is required.C. Carbon 2 undergoes no redox chemistry

D. Biotin is required.


OH

CÔPO*2

137

84. Hydrolysis of which of the following compoundsreleases approximately 5 kcal/moles of free energy?

C.CH2OH

Q OP032"

D.

HO

CH2OP032"

H OH

85. How is the structure for fructose-1,6-bisphosphate bestdescribed?

CH2OP032' CH2OP032-

OH H

A. It has a furanose ring with alpha orientation.

B. It has a furanose ring with beta orientation.

C. It has a pyranose ring with alpha orientation.D. It has a pyranose ring with beta orientation.


Passage XIII (Questions 86-91)

Sugars, when metabolized by animals, generate asubstantial amount of energy. Many naturally occurringmonosaccharides and disaccharides can be extracted from the

fruits of plants. Among the most common naturallyoccurring sugars are fructose, sucrose,and glucose. Figure 1showssix common biologicalmonosaccharides.

HNf HNf H^°H— — OH H— — OH HO —H

HO — —H HO —H BO— —H

H— — OH BO— —H H— — OH

H— — OH H— — OH H— — OH

CH2OH CH2OH CH2OHD-Glucose D-Galactose D-Mannose

HY° HY° CH2OH

= 0H— — OH H— — OH

HO — —HHO — —-H H— — OH

H — — OHH — — OH H— — OH

H— — OHCH2OH CH2OH

D-Xylose D-Ribose Cri2UHD-Fructose

Figure 1 Six common monosaccharides

The more heat released per gram of a sugar, the moreeffective the sugar is as a fuel. Most thermodynamic data,such as theheat of combustion (AH) and free energy changes(AG) under standard conditions, are listed in terms ofkcals/mole. Table 1 shows the thermodynamic valuesassociated with the sugars in Figure 1.

Sugar AHcombustion(kcal/mole)

AGcombustjon(kcal/mole)

Glucose -2538 -2827

Mannose -2512 -2801

Fructose -2482 -2771

Galactose -2470 -2759

Xylose -2102 -2341

Ribose -2076 -2315

Table 1

86. What is the stereo-configuration for D-xylose?A. 2R, 3R, 4R

B. 2R, 3S, 4R

C. 2S, 3R, 4R

D. 2S, 3S, 4R


87. How much C02(g) is producedfrom the metabolism of10.0grams of D-glucose, if all of the glucose is usedup in the reaction?

A. 10 x-L- x 44 grams C02(g)180

B. 10 x-J—x ^-x 44 grams C07(g)180 1

C. 10 x1£Q- x 6 grams C02(g)44

D. 10 x 180 x1 x -X. grams C02(g)6 44

88. The BEST explanation for the differences in heats ofcombustion for the aldohexoses is which of thefollowing?

A. Thebonds aredifferent in thedifferent sugars.B. Thearrangement of the hydroxyl groupsis different

between sugars, causing differences in hydrogenbonding.

C. Only someof the sugars haveresonance stability.D. Only some of the sugars are aromatic.

89. If the heat capacity of water is 4.18 J/g K, how muchwill the temperature of 1000 g of water insulating abomb calorimeter rise if it absorbs all of the heatgenerated by burning 15.0 grams of ribose?

A. 4.97 eC

B. 41.4 °C

C. 49.7 °C

D. 207 °C

90. The amount of C02(g) produced by combustion wouldbe greatest from 1.0 gram of which of the followingsugars?

A. Glucose

B. Sucrose

C. Ribose

D. All monosaccharides produce the same amount ofC02(g) per gram.

91. The amount of energy produced by combustion wouldbe greatest using 1.0 gram of which of the followingsugars?

A. Glucose

B. Xylose

C. Ribose

D. All monosaccharides produce the same amount ofenergy per gram.


Questions 92 - 100 are NOT basedon a descriptive passage

92. Decomposition of glucose with excess HIO4 wouldyield which of following products?

A. 4 formic acids and 2 formaldehydes.B. 5 formic acids and carbon dioxide.

C. 4 formic acids, carbon dioxide and 1 formaldehyde.D. 5 formic acids and 1 formaldehyde.

9 3. What is the mass percent of carbon in an aldopentose?

A. 33% C by mass

B. 37% C by mass

C. 40% C by mass

D. 44% C by mass

94. After treatment with nitric acid, a sugar is convertedinto a diacid (aldaric acid). Which of the followingsugars would yield an optically active diacid upontreatment with nitric acid?

A. D-Mannose

B. D-Galactose

C. D-Ribose

D. D-Allose

95. In order for a D-aldopentose to become opticallyinactive upon treatment with nitric acid, it must havewhat stereochemistry?

A. 2R, 3S

B. 2S, 3R

C. 2S, 3S

D. All of the choices above will lead to an opticallyinactive diacid

9 6. Ribose differs from deoxyribose in all of the followingways EXCEPT:

A. Ribose has more stereoisomers.

B. Ribose has one more chiral carbon.

C. Ribose is the C-2 epimer of deoxyribose.

D. Ribose is more oxidized than deoxyribose.

9 7. When two monosaccharides form a disaccharide, what

does NOT occur during the reaction?

A. The loss of a water molecule

B. The formation of an acetal from a hemiacetal

C. The formation of a glycosidic linkage

D. The oxidation of the anomeric carbon


9 8. The Fehling's test is positive when copper goes from anoxidation state of +2 to +1. What change in the sugarcorresponds to a change is in color for the solution inFehling's test?

A. An aldose becomes a carboxylate

B. An aldose becomes a hexa alcohol

C. An aldose becomes an acetal

D. A ketose becomes a carboxylate

9 9. What is the molecular formula of a trisaccharide formed

from two aldohexoses and one aldopentose?

A. C17H30O15

B. C,7H34Oi7

C. C18H32O16

D. Ci8H36Oi8

100. An aldopentose has how many units of unsaturation?

A. 0

B. 1

C. 2

D. 3

"Just study it!"

1. B 2. C 3. D 4. B 5. C

6. D 7. B 8. D 9. B 10. C

11. C 12. C 13. A 14. C 15. A

16. B 17. A 18. D 19. B 20. C

21. D 22. C 23. C 24. B 25. C

26. D 27. A 28. D 29. C 30. D

31. A 32. D 33. C 34. B 35. D

36. C 37. C 38. B 39. C 40. D

41. A 42. B 43. B 44. C 45. A

46. C 47. D 48. B 49. A 50. D

51. D 52. D 53. C 54. D 55. C

56. C 57. D 58. C 59. B 60. D

61. D 62. C 63. D 64. D 65. D

66. D 67. C 68. C 69. B 70. A

71. A 72. D 73. D 74. D 75. B

76. A 77. C 78. C 79. C 80. A

81. C 82. C 83. D 84. A 85. A

86. B 87. B 88. B 89. C 90. B

91. A 92. D 93. C 94. A 95. A

96. C 97. D 98. A 99. A 100. B

NO MORE CARBOHYDRATES!

Passage I (Questions 1-7) Glucose and Glucopyranose

1.

2.

3.

4.

5.

6.

7.

Choice B is correct. According to the rules of sterichindrance, the equatorial orientation is more stable than theaxial orientation for the anomeric hydroxyl group on carbon 1. As stated in the passage, the fi-anomer of D-glucopyranosehas its anomeric hydroxyl group in the equatorialposition. Because there is 64% beta anomer andonly 36% alpha anomer formed upon cyclization, the beta anomer must be the more stable of the two anomers.From the drawing in Figure 1and the information presented in the passage, this question should have been easy.You really should choose B.

Choice C is correct. Anomers are diastereomers of the same sugar that vary in chirality at the anomeric carbon.To be the same sugar, the sugar must have the same name. This eliminates choice A. Enantiomers vary at all ofthe chiral centers. In the case of enantiomeric sugars, they are named the same, except for the D or L prefix.This eliminates choice D. Glucose and Galactose are C-4 epimers of one another. Epimers are sugardiastereomers that differ in chirality at only one of the carbons in backbone of the straight chain form of thesugar. To get this correct requires knowing some sugar facts. Welcome to MCAT preparation and have anice day,after you choose C.

Choice D is correct. By definition, the Dand Lforms ofthe same sugar aremirror images ofone another, whichmeans (by definition) that the two compounds are enantiomers of one another. The Dand Lforms of a sugar willhave opposite values (differing only in sign) for their optical rotations of plane-polarized light. This meansthat one rotates plane-polarized light in a clockwise fashion while its enantiomer rotates plane-polarizedlight in the counterclockwise direction. In this question, as would be the case when comparing the Dand Lisomersof any other sugar, D-glucose and L-glucose are enantiomers of one another. ChooseD for best results.

Choice Bis correct. Hopefully the structure of glucose is permanent in your memory (using the "right-handmethod" for D-glucose), glucose in its straight chain Fischer projection is 2 right (R), 3 left (S), 4 right (R) and 5right (R). It is important that you remember that hydroxyls on the left are Sand hydroxyls on the right are Rinthe Fischer projection of standard aldohexoses. This makes choice B the best answer.

Choice C is correct. An aldohexose has chiral centers present at carbons 2, 3, 4, and 5 for a total of four chiralcenters in the molecule. The maximum number of stereoisomers is found using the formula 2n where n is thenumber of chiral centers present in the molecule. In this question, plugging in four for the value of n yields 2n =24, which equals 16, choice C.

Choice D is correct. Looking at the structures in Figure 1, it can be seen that the anomeric carbon (carbon 1) hasthe hydroxyl group in the equatorial position in the 12-anomer. Choose Dfor optimal satisfaction. Cis and transare poor answers, because to be cis or trans, it must be relative to something. In actuality, the fi-orien ration isdefined as cis with respect to carbon six in an aldohexapyranose, but that was not implied in this question.

Choice Bis correct. The 1,4-glycosidic linkage between any two monosaccharide units involves the hemiacetalOH of carbon-1 of one sugar with ahydroxyl of carbon 4of the other sugar. These combine to form an acetal whencarbon 1is anomeric. Pick Bfor correctitude sensations. Had the anomeric carbon been carbon-2 (as in the case ofa cyclic ketose), the glycosidic linkage would be 2,4 and the functionality would be a ketal.

HOH,C

HO

HO

Acetals are formed from an aldehydegroup, so the monosaccharide is an aldose

Anomeric C is bonded to:

OR, OR', C, and H

OH

Copyright © by The Berkeley Review® 140 CARBOHYDRATES EXPLANATIONS!

Passage II (Questions 8 -14) Fischer and Haworth Projections

8. Choice D is correct. When the straight chain form of a sugar is converted into its Haworth projection, thehydroxyl groups originally on the left side of the sugar are found above the ring and hydroxyl groups originallyon the right side of the sugar are found below the ring (recall the mnemonic "downright uplefting"). Mannosehas the hydroxyls on carbons 2 and 3 on the left. The C-epimer of D-mannose therefore has a straight chainwith only the hydroxyl on carbon 2 on the left. In the Haworth projection of the C-3 epimer of 6-D-mannopyranose, the hydroxyl groups on carbons 1 and 2 are above the ring and the hydroxyl groups on carbons 3and 4 are below the ring. This makes choice D the best answer. Choices A and B should have been eliminatedearly on, because they have a-orientation at carbon 1. As a point of interest, choice C is fi-D-glucopyranose.

9. Choice B is correct. This question is best answered from straight memorization, although a good guess can bederived from the passage. Lactose is a disaccharide made from glucose and galactose, not a polysaccharide, sochoice D is eliminated. Sucrose is a disaccharide made from glucose and fructose and not a polysaccharide, sochoice C is eliminated. Glycogen has alpha linkages between saccharide monomers, while cellulose (thepolysaccharide we can't digest) has beta linkages between the saccharide monomers. We lack the enzyme tocleave the 6-1,4-glycosidic linkage. Choice B is the best answer.

10. Choice C is correct. When you consider the most common aldopyranose, ribose, it is reasonable to assume that afuranose ring is the most stable cyclic form for aldopentoses in general. Choice A can be eliminated. Thepenultimate carbon determines the designation of D or L for a sugar. In a five carbon sugar, carbon 4 is thepenultimate carbon, so choice B can be eliminated. In the cyclic form, furanose ring, all of the carbons exceptcarbon 5 have chiral centers, so it is true that a cyclic aldopentose has four chiral centers. Choice D iseliminated. In an aldopentose, carbon 1 has two bonds to oxygen, so in its cyclicform, carbon 1 still has two bondsto oxygen, making it the anomeric carbon. Carbon 2 is not the anomeric carbon, so choice C is correct.

11. Choice C is correct. For any trisaccharide, the total weight is the weight of the three monosaccharides minusthe weight of the two waters lost in forming the linkages. When three 6-carbon monosaccharides are combined,there are two bridging linkages. Calculating this value gives 3(180) - 2(18) = 540 - 36 = 504 grams/mole.Choose C to show off those math skills of yours. The fact that the linkages were 1,4-linkages is a moot pointwhen considering the mass of the compound. Either way you look at it, the polysaccharide loses one water perlinkage, regardless of the exact linkage.

12. Choice C is correct. This question tests straight memorization, so remember it correctly, then choose C. Shouldyou not recall the fact that polysaccharides form linkages from carbon one to four (when dealing withpyranoses), then the passage also gives you a subtle hint, by giving only one example of a linkage, and theexample is a 1,4-linkage.

13. Choice A is correct. An aldopentose is most likely to form a furanose ring when the hydroxyl on the penultimatecarbon (carbon4) attacks the aldehyde carbon. This eliminateschoices C and D. Because the original sugar is analdehyde, the structure is a hemiacetal and not a hemiketal. The best answer is choice A.

14. Choice C is correct. Tiie glycosidic bond forms when a nucleophile displaces the anomeric hydroxyl group of thesugar. The nucleophile is methanol, which attacks using the lone pairs on its oxygen atom. This eliminateschoices A and B. This question now focuses on whetherit is the a-anomer or the 6-anomer. The methoxy group istrans to carbon 6, so it is the a-anomer, making choice C the best answer.

Passage III (Questions 15 - 22) Sugar Conventions

15. Choice A is correct. An aldohexose has one aldehyde functional group, one primary alcohol functional group,and four secondary alcohol functional groups. This makes choice A the best answer. An aldohexose has chiralcenters at carbons 2, 3, 4, and 5, so choice B is eliminated. Tiie aldehyde carbon has the most bonds to oxygen ofany carbon, so it is most oxidized. This eliminates choice C. For a D-sugar, the penultimate carbon has R-chirality, so for a D-aldohexose, carbon 5 has R-chirality. Choice D is eliminated.

16. Choice B is correct. A ketohexose is a six-carbon sugar with a ketone functionality (likely on carbon 2) in itslinear structure. If it were in the cyclic form, the term becomes pyranose or furanose depending on the ring size.Since fructose is the only ketose listed in the choices given, it is best for you to choose B.


17. Choice A is correct. When the hydroxyl group oncarbon 5 ofan aldohexose is inverted, thesugarbecomes theopposite type ofsugar. This means that theC-5 epimer ofL-glucose mustbe a D-sugar, eliminating choice B. Inorder to form D-glucose from L-glucose, all of the chiral centers must differ, because the two structures areenantiomers. Only one chiral center differs, so choice D is eliminated. The C-5 epimer of L-glucose hashydroxyl groups on the right oncarbons 3 and 5, and hydroxyl groups 2 and 4 on the left, which according toFigure 1 is D-idose. The best answer is choice A.

18. Choice Dis correct. D-Gulose has hydroxyl groups onthe right for all carbons except carbon 4. The C-2 epimer ofD-gulose has hydroxyl groups on left for carbons 2 and 4. The C-3 epimer of glucose has all of the hydroxylgroups on the right, so choice A is eliminated. The C-3 epimer ofD-idose has thehydroxyl groupon the leftforcarbons 2,3, and 4, so choice Bis eliminated. The C-4 epimer ofD-allose has thehydroxyl group on the left oncarbon 4 only, sochoice C is eliminated. The C-4 epimer ofD-altrose hashydroxyl groups on the leftforcarbons2 and 4, so choice D is the best answer.

19. Choice Bis correct. Ketals cannotbeoxidized byTollen's reagent, because they arestable underbasic conditionsand they cannot be converted into aldehydes (which can be oxidized). The key feature in a ketal preventingoxidation is the absence of a C-H bond on the ketal carbon (which is the feature thatprevents a ketone frombeing oxidized) . The C-H bond is necessary for oxidation to occur. Oxidation from the organic chemistryperspective involves the loss ofhydrogen aswell as the gain ofoxygen. For this question, choose Bfor greatestsatisfaction.

20. Choice C is correct. Fructose is a ketose, so its cyclic structure cannot be an acetal or hemiacetal. This eliminateschoices Aand B. Ketones, when added to alcohols in basic medium, go on to form hemiketals. The hemiketalwill be in equilibrium with ketone. The question does not state whether the medium isbasic oracidic, so youmust answer this from experience and memory. Monosaccharides for hemiketals (orhemiacetals if the structurestarts as an aldose). This means C is the best choice.

21. Choice D is correct. Epimers differ by one chiral center in their backbones. Talose differs from galactose atcarbon 2, differs from idose at carbon 3, and differs from mannose at carbon 4. This means talose is an epimer ofgalactose, idose, and mannose, eliminating choices A, B, and C. It differs from glucose at carbons 2 and 4,meaning it is not an epimer of glucose. Thismakeschoice D the best answer.

22.

*s^°

HO-

HO-

HO-

H-

•H

•H

H

OH

CHjOH

D-(+)-Talose

*s^°

H-

HO-

HO-

H-

OH

•H

•H

OH

CBjOH

D-(+)-Galactose

H^*°

HO-

H-

HO-

H-

•H

OH

•H

OH

CH2OH

D-(-)-Idose

Hx^°

HO-

HO-

H-

H-

•H

•H

OH

OH

CH2OH

D-(+)-Mannose

H^°

H-

HO'

H-

H-

OH

•H

OH

OH

CH2OH

D-(+)-Glucose

Choice Cis correct. Treatment of an aldehyde with areducing agent such as NaBHj reduces the carbonyl group(C=0) into a primary alcohol. Following that same reactivity, treatment of an aldohexose (or ketohexose) willconvert the sugar into a hexa-ol (a six carbon chain containing six hydroxyls, one on each carbon). Treatment ofan aldopentose will convert it into a penta-ol. The results from an analytical reasoning perspective are similarto treatment of the sugar with nitric acid, because the product now has the possibility ofbeing meso (indicatedby a lack of optical activity). Fructose is eliminated, because the new hydroxyl would result in the formation ofa new chiral center, thus there would be the formation of two diastereomers. At least one of the twodiastereomers has to be optically active. Glucose and mannose are eliminated, because they have no mirrorplane between carbons three and four. This makes ribose the best choice. Ribose will generate a penta-ol with amirror plane slicing through carbon three. Pick choice C for the tingly sensation of correctivity. Galactosewould also yield an optically inactive product although it isnot given asananswer selection.

Copyright © by The BerkeleyReview® 142 CARBOHYDRATES EXPLANATIONS!

Passage IV (Questions 23 - 29) Monosaccharides versus Disaccharides

23. Choice C is correct. Galactose is the C-4 epimer of glucose. The C-2 epimer of galactose varies from glucose atboth carbon 2 and carbon 4, so choices A and D are eliminated. The C-2 epimer of mannose is glucose, so choice Bis also eliminated. The only remaining choice, and the correct answer, is choice C. Talose (which is the C-2epimer of galactose as shown in Figure 1) is the C-4 epimer of mannose. Pick choice C.

H^°

OH

H C-4

H-

HO-

H-

H-

QH epimers

OH

CH2OH

D-Glucose

H-^°

H-

HO-

HO-

H-

OH

H

rj epimers

OH

CH2OH

D-Galactose

C-2

H\^°

HO-

HO"

HO-

H-

H

H

„ epimers

OH

CH2OH

D-Talose

C-4

H-^°

HO-

HO-

H-

H-

H

H

OH

OH

CH2OH

D-Mannose

24. Choice B is correct. The glycosidic linkage involves carbon 1 of the glucose on the left and carbon 2 of the fructoseon the right. This eliminates choice D. The glucopyranose ring has a-orientation, so choice C is eliminated. Inthe fructose structure, the anomeric oxygen (involved in the linkage) is cis with carbon 6, so it has 6-orientation.This makes choice B the best answer.

25. Choice C is correct. Glycosidic linkages involve an anomeric carbon with two OR groups, so the functionalitycannot contain the prefix "hemi". This eliminates choices A and B. Because maltose is formed from glucose, analdose, the functional group is an acetal. This makes choice C the best answer. Aldoses go on to formhemiacetals as monosaccharides and acetals as polysaccharides. Ketoses go on to form hemiketals asmonosaccharides and ketals as polysaccharides.

26. Choice D is correct. In fi-D-glucopyranose, all of the substituents on the pyranose ring have equatorialorientation. This is a piece of information you should have committed to memory. The structure in questiondiffers from 6-D-glucopyranose at carbon 3, where the hydroxyl group has axial orientation. This makes thestructure a C-3 epimer of 6-D-glucopyranose, making choice D the best answer. An anomer would vary inchirality at the anomeric carbon. This eliminates choice A. A conformer is the identical molecule rotated orcontorted. If only one substituent changes from equatorial to axial while keeping the ring in the sameorientation, then the structures are not conformers. This eliminates choice B. Enantiomers differ at every chiralcenter, so all of the centers would need to be axial for it to be an enantiomer. This eliminates choice C.

27. Choice A is correct. A trisaccharide has two glycosidic linkages. This eliminates choices B and D. When thetrisaccharide is made from three unique aldohexoses, then the linkages will be acetal linkages, rather thanketal linkages. The linkages involve carbons with two OR groups, so they are acetal linkages, rather thanhemiacetal groups. This makes choice A the best answer.

28. Choice D is correct. Hydrolysis under acidic conditions hydrolyzes acetals and hemiacetals, but does not affectethers. When a disaccharide is exhaustively methylated, all of the hydroxyl groups become methoxy groups.All of the methoxy groups are ethers except for the anomeric carbon of the glycoside (sugar on the right), whichgoes from a hemiacetal to an acetal. This means that the anomeric carbon of the glycoside and the anomericcarbon of the glycosyl group making the linkage are both acetals. When hydrolyzed, they will generatehydroxyl groups. All of the other oxygen atoms will be part of methoxy groups. Not all of the OCH3 groupsreturn to being hydroxyl groups, so choice A is eliminated. The glycoside loses the methoxy group on its anomericcarbon, but not any others. The glycosyl group does in fact not lose any methoxy groups. Choice B is not a goodanswer, but it cannot be eliminated. The disaccharide only racemizes at the anomeric carbons, so choice C iseliminated. Following hydrolysis, the two carbons involved in the glycosidic linkage gain hydroxyl groupswhen water is added. The best answer is choice D.


29. Choice C is correct. The glycoside is typically the sugar on the right, recognized because it does not use itsanomeric carbon for the linkage. fi-D-Glucopyranose has hydroxyl groups that alternate up-to-down-to-up-to-down in the Haworth project. The sugar on the right follows the same pattern as fi-D-glucopyranose, except forcarbon 2, which has its hydroxyl group up instead of down. This means that the glycoside is the C-2 epimer of6-D-glucopyranose, making choice C, D-mannose, the best answer.

Passage V (Questions 30 - 36) Amylose, Amylopectin, and Cellulose

30. Choice Dis correct. Maltose is adisaccharide made from two a-D-glucopyranose structures. The linkage must bealpha, so choices Aand Care eliminated. Choice Bis eliminated, because the hydroxyl group on the glycoside(sugar on the right) has beta orientation. The best answer is choice D.

Choice Ais correct. Amylase breaks down the a-glycosidic linkage. However, it is not stable under highlyacidic conditions, where it is readily hydrolyzed. Gastric fluids are highly acidic, so amylase is destroyed inthe stomach. It must be released again by the pancreas. This makes choice Athe best answer. Both pancreaticand salivary amylase cleave a-glycosidic linkages, so choice Dis absolutely wrong. The two forms of amylaseare identical enzymes, so theyare equivalent in strength. This eliminates choices Band C.

Choice D is correct. The blood can only absorb monosaccharides. Fructose, galactose, and glucose aremonosaccharides, so choices A, B, and Care eliminated. Lactose is a disaccharide of galactose and glucose, so itcannot be absorbed into the blood. Choice D is the best answer. In lactose intolerance, we don't have thedisaccharase lactase, therefore you cannot break lactose into galactose and glucose. Since we cannot absorb thedisaccharide, the lactose does not get absorbed and instead goes to feed our bacteria (which metabolize it to gasand toxic metabolites). This results in pain and an increase in the number of molecules in the gut and lumen,which draws water in and results in diarrhea.

Choice C is correct. It is stated in the passage that amylose has no branching, so choices A, B, and D areeliminated. It is also stated in the passage that amylopectin has a-l,6-branching at about one out of everythirty glucose residues. This confirms that choice C is the best answer.

Choice Bis correct. Glycogen is astarch made from exclusively a-D-glucopyranose with a significant amount of1,6-branching. This eliminates choice A. The standard glycosidic linkage in the polysaccharide is a-1,4, soglycogen contains both a-l,4-linkages and a-l,6-linkages. This eliminates choice Dand makes choice Bthe bestanswer. Because glucose is an aldehyde, all of the linkages involve anomeric carbons that are part of an acetalfunctional group. This eliminates choice C.

Choice Dis correct. Amylopectin is a polysaccharide of a-D-glucopyranose that is held together by a-1,4-glycosidic linkages with an occasional 1,6-glycosidic linkage for branching. This means that the oxygen atomson carbons 1, 4, and 6are involved in glycosidic linkages. Hydroxyl I is on carbon 1, so the correct choice mustcontain I. This does not help, because every answer selection contains I. Hydroxyl II is on carbon 2, so the correctchoice must not contain II. This eliminates choices Aand C. Hydroxyl III is on carbon 4, so the correct choicemust contain III. This does not help, because the remaining answer selections (choices Band D) both contains III.Hydroxyl IV is on carbon 6, so the correct choice must contain IV. Only choice Dcontains IV, so it is correct.

Choice Cis correct. Cellulose is another polymer of glucose, similar to amylose, but with 6-1,4-glycosidicImkages. This eliminates choice A. Lactose is adisaccharide made from glucose and galactose, so lactose cannotbe formed from the break down of amylose. Choice Bis eliminated. Sucrose is adisaccharide made from glucoseand fructose, so sucrose cannot be formed from the break down of amylose. Choice Dis eliminated. Amylose isbroken down into glucose, which is absorbed by the blood and transported. Some of the glucose undergoesglycolysis (that which isn't stored), resulting in the formation of pyruvate. Choice Cis the best answer.

31.

32.

33.

34.

35.

36.

Passage VI (Questions 37 - 43) Unknown L-Aldopentose Elucidation37. Choice Ciscorrect. An aldopentose has three stereocenters, located atcarbons 2,3, and 4. The maximum number

ofstereoisomers possible is found by2n where n is the number ofstereocenters. The reason the word "maximum"is chosen is that if one of the stereoisomers happens to be ameso compound, then the total number of possiblestereoisomers will decrease by one. This won't be a problem with the sugars, because sugars are not meso.Because there are three stereocenters, there will be eight possible stereoisomers. Choose C.

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38. Choice B is correct. Because the cyano nucleophile can attack the carbonyl carbon from either side (notnecessarily equally from either side however), carbon one in the reactant monosaccharide (which becomescarbon 2 in the product) generates a mixture of two chiral centers on carbon two of the product. No other chiralcenters will change, so only the chiral center at carbon two differs between the two product molecules. Thismakes the two compounds diastereomers (non-superimposable and not mirror images). More specific thandiastereomers when dealing with sugars is the term epimer, given to sugar diastereomers that vary at only onecarbon in the backbone. This makes choice B the best answer.

39. Choice C is correct. Compound A is said to be an L-aldopentose. You should know from your information basethat glucoseis a sixcarbon sugar (aldohexose). Thismeans that the unknown sugar cannot be glucose. The correctanswer is choice C. Ribose you should know is an aldopentose, so choice A definitely should have beeneliminated. Without knowing exactly what xylose and lyxose are, they cannot be eliminated. Xylose is the C-3epimer of ribose and lyxose is the C-2 epimer of xylose.

40. Choice D is correct. Compound A is an aldopentose with a molecularmass of 150 grams per mole. When oxidizedat both terminal carbons, the mass increases. This eliminates choices A and B. The formula for the aldaric acidformed is C5H8O7 so the molecular mass is equal to 60 + 8 + 112 = 180 grams per mole. Choice D is therefore thecorrect answer. You could also have solved this knowing that when it is oxidized, it gains two oxygen atoms andloses two hydrogen atoms, resulting in a net gain in massof 30grams/mole.

41. Choice Ais correct. Compound Bhas increased hydrogen bonding compared to Compound A,because carboxylicacids form stronger hydrogen bonds than alcohols. The increase in hydrogen bonding will manifest itself as anincrease in melting point. The best answer is therefore choice A. Choices C and D should exclude one another,because if choice C were true, then choice D would have to be true.

42. Choice B is correct. Aldohexoses in their most stable form are actually aldohexapyranoses, the six-memberedring form ofa monosaccharide. The ring forms when the hydroxyl group oncarbon 5 attacks the carbonyl carbon.This means that there is no carbonyl absorbance observed in the IRspectrumof the most stable form of the sugar.The carbonyl peak in the IR isjust above 1700 cm-1, so no peak is observed there. Choice Bis the best answer.

43. Choice B is correct. Because Compound Bis optically inactive, it mustbe a meso compound. In order for thecompound to be meso, there must be a mirror plane through the molecule. The molecule contains five carbons sothe mirror plane must slice through carbon three reflecting carbon 2onto carbon 4. Compound Bis drawn withcarbon 2 having accurate stereochemistry as determined by the optical inactivity of the diacid.

O

On the left because

it is an L-suga

H^°

H-

H-

HO-

"(OH)

-(OH)

H

5 CH-.OH

HO,S^

HO

HNQ

HO

On the left because

of the mirror plane

Compound A Compound BUnknown L-aldopentose optically inactive diacid

If the hydroxyl group on carbon 2is on the left in the Fischer projection of compound B, then itmust be on the leftin the Fischer projection of compound A (the original sugar). Hydroxyl groups on the left in the Fischerprojection are assigned S stereochemistry. Pick B.

Passage VII (Questions 44 - 50) Unknown D-Aldohexose Elucidation

44. Choice C is correct. Treatment of Compound X, an aldohexose, with NaBH4 will reduce the aldehydefunctionality of carbon 1 to a primary alcohol. Because carbon 6 is also a primary alcohol, the two groups areidentical. Generating matching groups at the terminal ends ofthe sugar issimilar to what occurs when a sugar isoxidized with nitric acids (in which case both terminal carbons become carboxylic acid groups). Because thediacid of Compound X is optically active (it has an optical rotation associated with it), the reduced form (ahexa-alcohol) is also optically active. The best answer is choice C.

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45.

46.

Choice A is correct. Because Compound W (the oxidized aldaric acid derivative of Compound Z, thealdopentose formed from the Ruff degradation ofCompound X) shows no optical activity, it must be a mesodiacid. This means that the second and fourth carbons ofCompound Whave opposite stereochemistry.This means that the two hydroxyl groups in question are on the same side of the backbone in the Fischerprojection. Carbons 2 and 4 in Compound Ware originally carbons 3 and 5 from Compound X(the originalaldohexose). Because the original aldohexose is a D-sugar, the fifth carbon has R chirality. This makes thethird carbon R, because the hydroxyl group is on the same side as hydroxyl five in the Fischer projection. Allsugars drawn on the rightin a Fischer projection have Rstereochemistry. Pick A.

Choice C is correct. Treatment of an aldohexose with the reagents for the Kiliani-Fischer synthesis generates asugar with one more carbon in the backbone (an aldoheptose). Because the chain increases by one carbon, analdopentose is not possible, so choices A and B are eliminated. The addition of CsN" occurs at carbon 1 of theoriginal aldohexose. The chirality of the penultimate carbon remains the same, because the penultimate carbonis still the same carbon. This makes choice C the best answer.

CH2OH CH2OH CHpH CH2OHThe penultimate carbon (next to the last carbon) has Dchirality in the final sugar (aldoheptose).

47. Choice Dis correct. Areverse aldol reaction cleaves between the alpha and beta carbons of a beta hydroxycarbonyl compound. This means that Compound X, an aldohexose, is cleaved into a two carbon fragment (fromcarbons 1and 2) and afour carbon fragment (from carbons 3through 6). The four carbon fragment as stated in thequestion is an aldotetrose.

After the aldotetrose is reduced with sodium borohydride, the four carbons (3 through 6from the originalaldohexose) have hydroxyl groups. Because the tetra-ol of the aldotetrose is optically active, its two chiralcenters have like stereochemistry (If they were Rand S, then the compound would be meso and opticallyinactive). This means that the two hydroxyl groups are on opposite sides of the sugar backbone in the Fischerprojection. Carbon 3of the aldotetrose (carbon 5from the original aldohexose) must be Rgiven that the originalaldohexose is aD-sugar. This means that carbon 4from the original D-aldohexose has Schirality (which putsiton the opposite side from the fifth hydroxyl in the Fischer projection). Pick Dfor best results. The drawingbelow summarizes the reactions.

:0

— (OH)

Lqh Reverse aldol3 •

— (OH)

TOH

H

H-

H-

(OH)

(OH)

(OH) Hn

(OH) Pd (BaSQ)

OH

r3—(OH) NaBF|

OH

H-

H-

H"

JCH2OH6CH2OH

aldotetrose

HVNH

H-

H-

H-

H-

H-

(OH)

(OH)

(OH) rf

-(OH) H20

OH

CH2OH

H-

H-

(OH)

rOH

CH2OH

HO-

H-

Hv^°H-

H-

H-

H-

H-

CH2OH

rOHCH2OH

"(OH)

(OH)

(OH)

(OH)

OH

tetra-ol optically active tetra-ol

48. Choice Bis correct The number ofpossible stereoisomers is2n where n is the number ofchiral centers. The valueof n in the case of a straight chain aldohexose is four. 24 =16, so there are sixteen possible stereoisomers for analdohexose. Of these 16 possible stereoisomers for an aldohexose straight chain, eight are D-sugars and eightare L-sugars. This make 8the best answer. Pick B, and do what is best... in terms of answering that is.

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49. Choice A is correct. Theaddition ofHNO3 results in a compound with identical groups on the terminal carbons(with HNO3 they are both carboxylic acid groups). The addition of NaBH4 to the sugar will reduce thealdehyde to a primary alcohol making both terminal carbons identical without affecting the other carbons.This is analogous to the results when using nitric acid. Nitric acid is cheaper than sodium borohydride, but as ageneral rule, the MCAT has stayed away from questions about cost effectiveness of chemical reactions. Pick Aand show off your correct answer selection skills. Choice B(H2SO4) dehydrates a sugar to carbon, while choiceC (Br2(aq)) will oxidize only carbon 1 into a carboxylic acid, not both terminal carbons. Choice D (KMnOâq))will oxidize everything on the sugar, so choice Dis a bad answer. It is bad to pickbad answers; don't be bad!

50. Choice D is correct. The moststable form ofan aldohexose is thepyranose ringstructure. This ties up theOHofcarbon 5. Because periodic acid (HIO4) cleaves 1,2-diols, the sites are limited to neighboring diols at Ci and C2,C2 and C3, and C3 and C4. The least sterically hindered is the C3 and C4 pair of alcohols. Pick D. Thisinformation could be extracted directly from the passage, which is common on the MCAT.

Passage VIII (Questions 51 - 57) Kiliani-Fischer Synthesis

51.

52.

53.

Choice D is correct. Using the Kiliani-Fischer method, a D-aldopentose is used to synthesize a D-aldohexose.The addition occurs at carbon 1 of the D-aldopentose, turning carbon 1 into carbon 2 in the D-aldohexose products.This means that the chirality of carbons 2, 3, and 4 in the D-aldopentose match that of carbons 3, 4, and 5 in theD-aldohexose. To determine the D-aldopentose needed to synthesize the D-aldohexose in the question, usingretrosynthetic analysis, carbon 1 must be removed and carbon 2 must be turned into a aldehyde group.

H^°H-

H-

HO-

H-

OH

OH

H

OH

CH2OH

P

Hv^°H-

HO-

H-

OH

H

OH

CH2OH

D-Xylose

Upon matching structures with the four D-aldopentoses shown in Figure 2, the best answer is D-xylose, choice D.

Choice D is correct. To be C-2 epimers, the sugars must be diastereomers that vary in their chirality at carbon 2.Ribose is an aldopentose and altrose is an aldohexose, so they are not even isomers, let alone epimers. Choice Ais eliminated. The only way to answer this question, without actually knowing what the sugars are, is toemploy the data in Table 1. The products of Kiliani-Fischer synthesis are C-2 epimers. Glucose and talose arenot C-2 epimers and neither are mannose and allose. This eliminates choices B and C. Idose and gulose areformed as products when xylose undergoes Kiliani-Fischer synthesis, so idose and gulose are C-2 epimers,making choice D the best answer.

Choice C is correct. Reduction in organic chemistry is most simply viewed as either the gain of bonds tohydrogen or the loss of bonds to oxygen. In step 1, the carbonyl carbon loses one bond to oxygen (in forming a newbond to carbon), so carbon 1 is reduced in Step I. This eliminates choice B. In Step II, reduction occurs, because thecyano group gains a hydrogen on carbon and a hydrogen on nitrogen. The symbol [H] refers to reduction, so Step IIis definitely reduction. This eliminates choice A. Step III involves the hydrolysis of an imine into analdehyde. This is where the general definition needs to be a bit more specific. Both oxygen and nitrogen aremore electronegative than carbon, so when carbon goes from being bonded to nitrogen to being bonded to oxygen, ithas not been oxidized or reduced. This means that Step III is not oxidation or reduction, so choice C is the bestanswer. This is a case where oversimplification can hurt. Technically speaking then, oxidation occurs when thenumber of bonds to more electronegative atoms increases and /or the number of bonds to less electronegative atomsdecreases. So, for those of you who considered Step III to be oxidation, you still got lucky in that you didn'tfigure it to be reduction and choice C was still your choice. But for edification, know the more correct definitionof oxidation and reduction for organic chemistry.

54. Choice D is correct. In the Kiliani-Fischer synthesis as shown in Figure 1, and aldehyde is first converted into ahydroxy nitrile compound. The nitrile is reduced into an amine. The amine is hydrolyzed into the aldehyde.Of the choices, only choice D, an imine, is an intermediate in the synthesis. Choice D is the best answer.

Copyright © by The BerkeleyReview® 147 CARBOHYDRATES EXPLANATIONS!

55. Choice C is correct. The two products formed from Kiliani-Fischer synthesis, according Table 1,are C-2 epimersof one another. Epimers are diastereomers, which eliminates choices A, B, and D. The two diastereomers areformed in unequal concentration, because the presence of a chiral center in the reactant makes one side of thealdehyde easier to attack than the other. Choice C is the best answer.

56.

57.

Choice C is correct. Talose differs inchirality from glucose at carbons 2and 4. In the Haworth projection offi-D-glucopyranose, the hydroxyl groups are up (above the pyranose ring) on carbons 1 and 3, and down in carbons 2and 4. This means that talose has the hydroxyl groups on carbons 2 and 4, up, soall of the hydroxyl groups areup in 13-D-talopyranose. For talopyranose, the hydroxyl groups oncarbons 2, 3, and 4 must be up, sochoice C isthe best answer.

Choice D is correct. Kiliani-Fischer synthesis adds one carbon to the sugar structure. Choice A is eliminated,because a carbon is lost, not gained. Choice Bis isomerization, where the number of carbons does not change.This eliminates choice B. In choice C, the number of carbons in the product is not listed, so the choice can neitherbeconfirmed nor refuted. Inchoice D, the number ofcarbons has increased byone. This makes choice D the bestanswer.

Passage IX (Questions 58 - 64) Osazone Derivative Test

58.

59.

60.

61.

62.

Choice Ciscorrect. The unknown disaccharide inthe passage is comprised oftwo cyclic monosaccharides formedfrom aldohexoses (aldopyranose structures). Acyclic aldohexose molecule has five stereocenters (one at everycarbon except the terminal carbon, carbon 6). Given that each monosaccharide has five chiral carbons, adisaccharide must have ten stereocenters total. Pick choice C.

Choice Bis correct. Choice Ais eliminated, because the structure on the left is glucose. Neither sugar can beglucose, because the specific rotations of the osazones of both sugars do not match the specific rotation of theosazone of glucose. Choice C is eliminated, because the structure on the left is mannose, the C-2 epimer ofglucose. Neither sugar can be mannose, because the specific rotations of the osazones of both sugars do not matchthe specific rotation of the osazone of glucose (which would also be the osazone of mannose). Choice D iseliminated, because the two structures are C-2 epimers of one another. Because the two sugars form osazoneswith different specific rotations, the two sugars cannot be C-2 epimers ofone another. Inchoice B, neither ofthesugars will lead to an osazone that matches either glucose or mannose, neither sugar leads to an opticallyinactive aldaric acid, and the two sugars arenot epimers ofoneanother.

Choice D is correct. When analdopentose isoxidized with nitric acid to form analdaric acid, the two carbons toobserve for optical activity are carbons 2and 4. If the hydroxyl groups on carbons 2and 4 are on the same side ofthe chain in the Fischer projection, then the compound is meso and thus optically inactive. It is only in choice Dthat the two hydroxyls oncarbons 2and 4are on opposite sides. Choice Dis the best answer.

Choice Dis correct. The osazone of Compound Ihas specific rotation +42.6°, the osazone of Compound II hasspecific rotation +37.2°, and glucose yields anosazone with a specific rotation of +54.6°. The osazone of mannosewill also show an optical rotation of +54.6°, because mannose is the C-2 epimer of glucose. This means that theunknown sugars cannot be either glucose or mannose (selection I). It is possible for the unknown sugars to be eithertalose or galactose according to the osazone test. Because the aldaric acids of both Compound Iand Compound IIare optically active, they are not meso. Galactose leads to an optically inactive (meso) aldaric acid. Thismeans that the unknown sugars cannot be galactose (selection III). Compounds I and II cannot be mannose orgalactose, making choice D the best answer. For those of you who chose B, because you forgot it was a"CANNOT" question, try to limit your scream to a few seconds... "AAAAHHHHHHHHHHHHHHHHH!!!"

Choice C is correct. Choice A will yield an optically inactive aldaric acid when treated with nitric acid,because it will have a mirror plane slicing the carbon-carbon bond between carbons three and four. NeitherCompound Inor Compound II yields a meso aldaric acid, so choice Acannot be Compound I or Compound II.Choice Bis mannose, the C-2 epimer of glucose. Choice Dis fructose, the ketose isomer of glucose. Both choices Band Dare eliminated, because they will both yield the same osazone as glucose and Compounds Iand II generatea different osazone than glucose. Choice Cwill not lead to an optically inactive aldaric acid, because it will notbemeso after oxidation. The best option for you Is choice C.

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63. Choice D is correct. Identical osazones will form from sugars with matching chiral centers from carbon threedown. This means that C-3 and C-4 epimers will form different osazones, so choices A and B are eliminated.Choice C is a pair of enantiomers, which would generated enantiomeric osazones, so choice C eliminated.Fructose is the ketose formed when mannose isomerizes from an aldose into a ketose. Mannose and fructose areidentical from carbon 3 on, so they form identical osazones. The best choice is thus answer D.

H^°HO-

HO"

H-

H-

H 0

•H 3eq N~ N

H h H

OH

OH

CH2OH

D-mannose

H^ .N—NH3

CH2OHOsazone product

CH2OHD-fructose

64. Choice D is correct. Because the final osazone product after treatment of the sugar with three equivalents ofphenyl hydrazine contains two phenyl hydrazine moieties, it is safe to assume that two of the threeequivalents are substituting onto the sugar. The third phenyl hydrazine gains two hydrogens to form ammonia(NH3) and aniline (C6H5NH2). The gain of hydrogen is reduction, thus one of the three phenylhydrazinemolecules is reduced. Tiie best answer is thus choice D.

Passage X (Questions 65 - 71) Glycolysis Reactions

65. Choice D is correct. Glyceraldehyde-3-phosphate and dihydroxyacetonephosphate are linear sugar structureswith the same formula, so they are isomers of some sort. Dihydroxyacetonephosphate has no chiral center,while glyceraldehyde-3-phosphate has one chiral center, so the two structures cannot be configuration isomers(isomers that have the same bonds, but different spatial arrangement). Epimers and diastereomers areconfigurational isomers, so if the two structures are not configurational isomers, they cannot be diastereomers orepimers. This eliminates choices A, B, and C. An aldose and ketose of the same carbon length are structuralisomers. In this example, they each have one phosphate group, so choice D is the best answer.

ketose with no

chiral centers

CHnOPOo

CH7OH

C3H506P2-(same formula but

different bonds)

H^°H- OH alcbse with one

chiral center2-CH20P03

Dihydroxyacetonephosphate Glyceraldehyde-3-phosphate

66. Choice D is correct. Carbon 4 of glucose becomes the aldehyde carbon in glyceraldehyde-3-phosphate. It losesits original chirality when it takes on sp2-hybridization (associated with the aldehyde carbon), thus it has nochirality in the end. Galactose (the C-4 epimer of glucose) should therefore yield the exact same products asglucose if the enzymes were able to recognize galactose. The remaining chiral centers are identical between D-glucose and D-galactose, so the chirality of the products is the same. The best answer is choice D.

67. Choice C is correct. The lower three carbons (carbons 4, 5, and 6) of D-glucose form the glyceraldehyde-3-phosphate molecule, so carbon 5 finishes as the middle carbon (C-2) of glyceraldehyde-3-phosphate. Choice Cis the best answer. Because of isomerization, the label is also found at C-2 of DHAP, but that is not a choice.

HN^°H-

HO-

H-

OH

-H

OH

Step I

HV^°H-

HO-

H-

OH

-H

OH

H-*C-OH1

SteP g HO-H-

CH2OH

O

H Step HI HQ.

H-OH

OH

CH2OP032"

_H Step IV

OH

2-CH20P03

(=°CH2OH

H-*C-OH1

H-*C-OH1

CH2OH CH2OP032-

H-*C1

2-CH2OP03

H-*C-OH1

CH2OP032-


2-CH2OP03

CARBOHYDRATES EXPLANATIONS!

68. Choice C is correct. This is a question of Hess's Law as applied in thermodynamics. The overall free energychange is the sum of the individual free energy changes for Step I through Step IV. The math to determine thesum of the four steps is carried out as follows: -4.0 + (0.4) + (-3.4) + (5.7) = -7.4 + 6.1 = -1.3, which is choice C.

69. Choice B is correct. The two terminal carbons in dihydroxyacetone phosphate (DHAP) have two equivalenthydrogens and the middle carbon has sp2-hybridization, so none of the carbons are chiral. The first carbon inglyceraldehyde-3-phosphate (G-3-P) has s/?2-hybridization and the third carbon has two equivalenthydrogens. Only the second carbon of G-3-P has four different substituents attached to it. The total number ofchiral centers between the two compounds is therefore just one. The best answer is choice B.

70. Choice A is correct. Fructose is a ketohexose with a

furanose ring structure. The phosphate group isFructose, a-D-fructofuranose, fructose-6-phosphate,

CH2OH

ketone on carbon 2, so carbon 2 is the anomeric carbon in thelocated on the last hydroxyl group (carbon 6) of fructose,and a-D-fructofuranose-6-phosphate are shown below.

CH2OH

= o

— H

O

H

CH2OH CH2OH

HO-

H-

H-

OH

OH

6CH2OH

Fructose

POH

OH

a-DFructofuranse

Choice C should be eliminated, because carbon 3because the phosphate is on the wrong carbon.

HO-

H-

H-

OH

OH

2-6CH2OP03^

Fructose-6-phosphate a-D-Fructofuranse-6-phosphab

has no hydroxyl group. Choices B and D are eliminated,

^>

6CH2OP032"•O

OH

CH2OH

OH

71. Choice A is correct. This question tests your knowledge of glycolysis. Glucose is a six-carbon sugar whilepyruvate is a three-carbon structure, so twopyruvates form from oneglucose molecule. Thiseliminates choices Cand D. Early in glycolysis, two ATPs are invested, but two ATPs are formed per pyruvate, so the net result isthat two ATPs are formed in glycolysis. This makes choice A the best answer.

Passage XI (Questions 72 - 78) Blood Types

72. Choice D is correct. The linkage can be determined directly by looking at the two sugars of interest out of thethree sugars in the type-O blood antigen. N-acetylglucosamine is the first sugar linked to the protein (the N-acetyl should give this away), thus the galactose must be the second sugar from the protein in the antigenicdeterminant trisaccharide. The linkage is from carbon 1of the galactose to carbon 4of N-acetylglucosamine byway of the 6-anomer ofgalactose. This means that the linkage is a 6-1,4- glycosidic linkage, so choice D is best.The structure is drawn below.

HO 6I CH2OH

hoVvt %°^TîO 1 3 NF

IHOHO

oW 9protein

CH,

73. Choice D is correct. All three sugars of the type Oblood antigenic determinant are formed from the cyclizationof aldoses, therefore there are three acetal functionalities (one for each sugar) rather than ketalfunctionalities. All three sugars are linked to something, so they are all acetals rather than hemiacetals. Thebest answer is therefore choice D.

Copyright © hy The Berkeley Review® 150 CARBOHYDRATES EXPLANATIONS!

74. Choice Discorrect. Fucose, differs from anormal sugar inthat carbon 6has been reduced from aprimary alcoholinto a methyl group. If the methyl group is replaced with a primary alcohol group, then it becomes analdohexapyranose. This question centers on the relationship of the aldohexapyranose derivative of fucose andglucose. 6-D-Glucopyranose hasallofitssubstituents inequatorial position. However, because carbon 5 offucosein blood type O has S-chirality, it is an L-sugar rather than a D-sugar. 6-L-Glucopyranose also has all of itssubstituents in equatorial orientation. The fucose derivative with the hydroxyl group on carbon 6 has all of itssubstituents in equatorial orientation except the hydroxyl groups on carbons 1and 4. The group on carbon 1 isirrelevant in thisquestion. The bestanswer ischoice D, theC-4 epimer of glucose. This means that whencarbon6 of galactose is reduced from a primary alcohol into an alkyl group, fucose is formed. L-Fucopyranose, L-galactopyranose (L-fucopyranose with a hydroxyl group added tocarbon 6), and L-glucopyranose are all drawnbelow.

HO

HO

cc-L-Fucopyranose

OH

HOP^C

HO

HO

a-L-Galactopyranose(oc-L-Fucopyranose w/ 6-OH)

OH

HOP^CHO

OH

HO

HO

cc-L-Glucopyranose

The second structure (L-fucopyranose with a hydroxyl group added to carbon 6) is similar to that of L-glucopyranose, with the difference coming at the chiral center of carbon 4. It is thus concluded that the secondstructure (thesugarformed whenthe CH3 group offucose is replaced by CH2OH) is the C-4 epimer ofL-glucose.The correct answer is choice D.

75. Choice B is correct. The2-deoxyglucose portionof thenamehints that the secondcarbonhas no oxygen bonded toit, so choices C and D can be eliminated. Onlychoices A and Bare deoxygenated at carbon2 of the pyranose ring.ChoiceA is an alpha sugar while choice B is a beta sugar. ChoiceB is therefore the best answer.

76. Choice A is correct. A normal L-aldohexose has a hydroxyl group present on carbon 6, while fucose lacks ahydroxyl group on carbon 6. This means that fucose is more reduced than a normal aldohexose, so fucose resultsfrom the reduction (loss of oxygen) of carbon 6. The best answer is choice A.

77. Choice C is correct. The sugar is a six-membered ring (a pyranose), rather than a five-membered ring (afuranose), so choices A and B are eliminated. All of the choices are cyclic galactoses, so identifying the sugar isnot necessary. As drawn in Figure 3, the galactose that has been added to hydroxyl group of carbon 3 of the thirdsugar, is upside down compared to the standard chair projection. The hydroxyl linkage is still from the axialorientation (trans to carbon 6), so it is the alpha anomer of galactose. The sugar is an a-galactopyranose, so thebest answer is choice C.

78. Choice C is correct. The type A antigenic determinant has four sugar rings, each of which has chiral centers atcarbons 1 through 5. This means that there are five chiral centers per sugar ring, and four rings total. Fivecenters times four sugars results in a total of twenty chiral centers. The correct answer is choice C.

Passage XII (Questions 79 - 85) Biochemistry of Sugars

79. Choice C is correct. According to Table 1, conversion of phosphoenolpyruvate into pyruvate releases 14.8kcal/mole. Most of the other phosphate compounds in Table 1 release between 5 and 10 kcals per mole whenhydrolyzed. It can be assumed that hydrolysis of phosphoenolpyruvate into pyruvate enol releases between 5 to10 kcal/mole. This means that the conversion from pyruvate enol into pyruvate releases between 5 and 10kcals/mole, given that the overall conversion from phosphoenolpyruvate into pyruvate releases about 15kcals/mole. The release of energy makes AG a negative value, so choices A and B are eliminated. The bestanswer is choice C, given that it is the only choice that fits within the range.

80. Choice A is correct. The conversion of ADP into ATPinvolves the gain of a phosphate, which according to Table1 requires 7.3 kcals/mole. Only acetyl phosphate releases more than 7.3 kcals/mole when it is hydrolyzed, sochoice A is the best answer. Choices B, C, and D all release less than 7.3 kcals/mole.


81. Choice C is correct. When bonds arebroken, energy must beadded to thesystem. When bonds are formed, energyis released from the system. An energy releasing reaction such as hydrolysis of a phosphodiester thus resultsfrom the breaking of weaker bonds and formation of stronger bonds. This means that the phosphodiester bondmust be weak, eliminating choices A and B. The weakness can be attributed to repulsion from the negativecharges located on the oxygen atoms in the phosphate groups. Repulsion causes the bond to elongate, wherelonger bonds are weaker bonds. Choice C is the best answer.

82. Choice C is correct. Hydrogens are not gained or lost when phosphoenol pyruvate, PEP, loses its phosphategroup to ADP, so the enzyme is not PEP dehydrogenase. ChoiceA is eliminated. Phosphoenol pyruvate, PEP, isnot oxidized or reduced in the conversion, so the enzyme is not PEP reductase. Choice B is eliminated.Phosphoenol pyruvate, PEP, does not make or break any carbon-carbon bonds in the conversion, so the enzyme isnot PEP aldolase. Choice D is eliminated. A phosphate group is transferred, so the enzyme is a kinase. The bestanswer is pyruvate kinase, making choice C the best answer.

83.

84.

Choice D is correct. When glyceraldehyde-3-phosphate, G-3-P, converts to 1,3-bisphosphoglycerate, thenumber ofbonds to oxygen on carbon 1 increases. This means that carbon 1 has beenoxidized, eliminating choiceA. The oxidizing agent used to oxidize the aldehyde into a carboxylic acid is NAD+ (which is poor in H, aproperty of an oxidizing agent), so choice B can also be eliminated. Carbon 2 starts and finishes with the exactsame bonds, so it is neither oxidized nor reduced. This eliminateschoice C. By process of elimination, choice Dis the best answer. The passage provided no information to know this fact, so you either had to know frommemory that biotin is not required or you had to use your chemistry logic to eliminate incorrect answer choices.

Choice A is correct. According to Table 1, 5 kcals/mole is released when glucose-1-phosphate is hydrolyzed.This means that you need to determine which structure represents glucose-1-phosphate. Choices B and D areeliminated, because the phosphate group is on carbon 6. In fi-D-glucopyranose, the hydroxyl groups alternatefrom up-to-down-to-up-to-down when viewing from carbon 1 through carbon 4. This is only observed in choice A,so choice A is the best answer.

85. Choice A is correct. The ring is a five-membered ring, so choices C and D are eliminated. Tiieanomeric carbon iscarbon 2, on the right half of the sugar as drawn. The hydroxyl group oncarbon 2 is down, while carbon 6 is up.This means that the two groups are trans to one another, making the sugar an alpha sugar. The best answer ischoice A.

Passage XIII (Questions 86 - 91) Combustion of Sugars

86. Choice B is correct. A shortcut for evaluating the chirality ofsugars in the Fischer projection is to realize thatall hydroxyl groups on the right side of the backbone are located on an Rstereocenter while all hydroxyl groupson the left side of tiie backbone are located on anSstereocenter for standard sugars drawn according to conventionwith the carbonyl carbon at or near the top. The procedure for obtaining the stereochemistry of one of thestereocenters is drawn below:

Method 1

H-

HO-

H-

S for hydroxyls R for hydroxylsonthe left. CH2OH ontrie right.

D-Xylose

For sugar is in the standard Fisherprojection, hydroxyls on the righthave Rchiralityand hydroxyls onthe left have S chirality

OH

•H

OH

or Method 2

HÔ

H

HO

HiOH

H

OH

Solving the firstchiral center yields:

CH2OHD-Xylose

Draw the Fisher projection with3-dimensional accuracy. The Hsare in front in Fisher projections.

HO

CH2OHD-Xylose

Counterclockwise normally indicatesan S stereocenter, but because thehydrogen is coming out of the plane,the stereocenter is reversed to R.

The compound has chirality of 2R,3S,4R,so choice B is the best answer.


87. Choice Bis correct. From the balanced equation, there are six C02(g) formed for every one glucose thatreacts. Tosolve this question, you must convert the glucose from grams tomoles, convert moles ofglucose tomoles ofC02(g),and thenconvert moles ofC02(g) intograms ofC02(g). This process requires dividing by the molecular mass ofglucose (180), multiplying by the ratiofrom thebalanced equation for carbon dioxide and glucose (6:1), and thenmultiplyingby the molecularmass of carbondioxide (44). The long-hand math goes as follows:

10.00gC6H12O6 x Imole x 6C°2 x**gE™ =lOx^x* x44gC02180 grams 1C6H1206 lmole 180 1

The set up is what is requested in the question, so the best answer is choice B.

88. Choice B is correct. The best explanation for the difference in heats of combustion between the aldohexoses(glucose, galactose, and mannose) is structural in nature. They all have the exact same bonds (thus they arestereoisomers) so according to the simplistic application of Hess's Law, they should yield the same amount ofheat. Because the heats are so different, there must be a difference in stability. The more stable the molecule,the less heat that will be given off. Because all of the aldohexoses have the same bonds, choice A iseliminated. Sugars are neither aromatic, nor do they have any stabilizing resonance forms. This eliminateschoices C and D. The only choice left is answer B, stating that the sugars have different orientationis spaceandthus different hydrogen bonding. The sugars also exhibit differences in their steric interactions. The best answerof the choices is choice B.

89. Choice C is correct. The given in this question is 15 grams ribose which, when divided by the molecular weightof ribose (150 g/mole), gives 0.10 moles of ribose. The AHcombustion for ribose is listed in Table 1 as -2076kj/mole. This means that the 0.10 moles of ribose generates 207.6 kj (207,600 J) of heat when it is burned

(oxidized). Plugging into the equation E = mCAT yields - K, which is approximately1000 x 4.18 1000 x 4

(roughly50), making choiceC the best choice. The mathematical layout is drawn for you below:

15gC5H10O5 x lmole x 2Q76kT = 2076kJ &E=mCAT /. AT =-E-= 207'600 =2HIA = 200 s 5(rc150 grams lmole mC 1000x4.18 4.18 4

It is doubtful that you will see a calculation this lengthy on your MCAT, but you should still understand the setup and the theory behind the math.

90. Choice B is correct. The most C02(g) results from the combustion of the compound with the largest mass percentof carbon. This is because for every carbon within a sugar, one C02(g) molecule is formed. Therefore, the morecarbon by mass in the compound there is, the more C02(g) that is formed upon complete oxidation. The masspercent of carbon in any monosaccharide is 40%. The mass percent of carbon in sucrose (C12H22O11) is greaterthan the mass percent of monosaccharide, because the difference in the ratio of C:H:0 between sucrose and themonosaccharides is due to a loss in H and O from the water. This increases the relative abundance of carbon in

sucrose compared to a monosaccharide giving sucrose a greater mass percent of carbon than a monosaccharide.

The mass percent ofcarbon insucrose is: 1^4 _ 144 _ _7_2_ an(j _72_ > _72_ =0.40 .-. -Z2_ > 40%. Because the144 + 22 + 176 342 171 171 180 171

mass percent of carbon in sucrose is greater than 40%, more C02(g) will be produced from the combustion of onegram of sucrose than from the combustion of one gram of any monosaccharide. Choose B and be happy.

91. Choice A is correct. For this, you must use the AH in terms of kj per gram, where the kj/g value is found bydividing the kj/moles by the molecular mass of the sugar. Monosaccharides have masses that are multiples of30 g/mole, so the numerators should be simple to work with. The following calculations show the energy pergram values for each of the sugars given in the answer choices. The correct answer is the sugar with the highestnumerical value.

Glucose:2538=423 Xylose :21Q2= 4204 Ribose :2QZfc=il5£180 30 150 30 150 30

Ribose gets eliminated immediately, because it is equal in mass to xylose but produces less energy. The possibleanswers are narrowed down to either glucose or xylose. Instead of solving the math, it is easier to reduce the twovalues to something over a common denominator (in this case 30). Glucose results in the greatest numerator,making choice A correct.

Copyright © by TheBerkeley Review® 153 CARBOHYDRATES EXPLANATIONS!

Not Based on a Passage (Questions 92 -100) Carbohydrates

92. Choice D is correct. Addition of an excess of periodic acid to a monosaccharide results in the oxidative cleavageof all of the carbon-carbon bonds in the sugar. Aldehydes are oxidized into formic acid , primary alcohols areoxidized to formaldehyde and the secondary alcohols associated with internal carbons are oxidized by losingboth carbon-carbon bonds to form formic acid. Tiie product distribution for the decomposition of glucose is drawnbelow:

H^°H-

HO"

H-

H-

OH

H

OH

OH

CH2OH

D-glucose

O

>• 5H" OH

O

M H

In the end, the full oxidation of glucose yields 5 formic acids and 1 formaldehyde. Choose D for correctness.

93. Choice C is correct. The molecular formula for xylose is C5H10O5. The molecular weight of xylose (or any otheraldopentose or ketopentose) is 60 + 10 + 80 = 150 grams/mole. The mass of the five carbons is 60 grams. Thismeans that the mass percent of carbon in xylose is 60/150 x 100% = 40%. As a point of trivial interest, the masspercent of carbon in any t7ionosaccharide is always 40%. Choose C and feel happily correct.

94. Choice A is correct. In order to be optically inactive, the compound must be either meso or achiral. In the case ofsugar derivatives, the diacid (aldaric acid) formed after oxidation of the terminal carbons must be meso to beoptically inactive. D-mannose has stereochemistry of 2S, 3S, 4R, 5R, which does not yield a meso aldaric acidwhen oxidized, so the aldaric acid derivative of mannose is optically active. D-galactose has stereochemistryof 2R, 3S, 4S, 5R, which yields a meso aldaric acid which is optically inactive. D-ribose has stereochemistry of2R, 3R, 4Rand is fivecarbons long (which you should know from memory), so the third carbon would no longerbea chiral center in the aldaric acid form. The aldaric acid of D-ribose is meso and therefore optically inactive.D-allose has stereochemistry of 2R, 3R, 4R, 5R, which would form a meso aldaric acid which is opticallyinactive. The aldaric acid of D-allose is meso and therefore optically inactive. The only sugar to form anoptically active aldaric acid derivative of the given choices is D-mannose. The best answer is choice A.

HO. .O

HO-

HO-

H-

H-

•H

•H

OH

OH

HOÔAladirc acid derivative

of D-Mannose

(not meso .\ active)

HO. .O

H-

HO-

HO-

H-

OH

•H

•H

OH


of D-Galactose

(meso .•. inactive)


HO. .O

H-

—H-

H-

•OH

OH—

OH


of D-Ribose

(meso /. inactive)

HO. .O

H-

H-

H-

H-

OH

OH

OH

OH


of D-Allose

(meso /. inactive)

CARBOHYDRATES EXPLANATIONS!

95. Choice A is correct. When treated with nitric acid, the two terminal carbons (carbon 1 and carbon 5) of analdopentose get oxidized into carboxylic acid functionalities. This results in the formation ofa 1,5-dicarboxylicacid (aldaric acid). After the oxidation reaction, only carbons 2 and 4 can be chiral centers, because the thirdcarbon now has twoidentical substituents attached to it (making carbon 3 achiral.) In a D-aldopentose, carbon 4has Rstereochemistry. This would mean that carbon 2 must also have Rstereochemistry in the original sugar,although that changes in the aldaric acid, because the priorities are different. Only choice A has this feature.The stereochemistry of carbon 3 is irrelevant.

HO^.0 HOÔHV° HO..O

H-

H-

H-

(OH) H-

H-

(OH)

i-ony

H-

H-

OH

•(•Grr>

OH

H- OH

-(OH)•(OH)

OH

CH2OH

D-Aldopentose(opticallyactive)

HNQ

OH

HO O HOÔAldaric acid derivative Inactive, so it must be meso

of D-Aldopentose OH-4on right .*. OH-2 on right(opticallyinactive) OH-3 is unknown b/c C-3 is achiral

P»H- OH

CHjOH

2R, 3 unknown

96. Choice C is correct. Ribose differs from deoxyribose in that carbon 2 has no hydroxylgroup in deoxyribose. Thismeans that carbon 2 does not have four unique substituents, so it is not a chiral center. This makes choice B avalid statement, which eliminates it. By having one more chiralcenter, ribose has more possible stereoisomers,making choice A a valid statement, and thus it is eliminated. Ribose, by having a hydroxyl group rather than ahydrogen on carbon 2, is more oxidized than deoxyribose. This makes choice D a valid statement, whicheliminates it. Ribose and deoxyribose do not have the same molecular formula, so they cannot be stereoisomers,let alone epimers. This makes choice C a false statement and therefore the best answer.

97. Choice D is correct. When two cyclic monosaccharides form a disaccharide, a glycosidic linkage is formed.Linkage formation occurs when a hydroxyl on one sugar attacks the anomeric carbon on the other sugar. A watermolecule is released as the leaving group, as the hemiacetal with the reactive anomeric carbon gets convertedinto an acetal. This makes choices A, B, and C all true. The anomeric carbon starts and finishes with two bondsto oxygen, one bond to carbon, and one bond to hydrogen. This means it is neither oxidized or reduced, so choice Dis not observed, making it the best answer.

HOP^C

HO

OH + HO

HOH,C H2? HOH>C Glycosidic linkage

h\ ori\ OH

Acetal because

anomeric C is bonded to:

OR, OR', C, and H

Hemiacetal because

anomeric C is bonded to:

OR, OH, C, and H

OH

98. Choice A is correct. When copper goes from an oxidation state of +2 to +1, it has been reduced. The color of thesolution changes when copper changes oxidation state, so a color change corresponds to an oxidation-reductionreaction involving copper ion and the sugar. Because copper is reduced, the sugar must be oxidized. If an aldoseforms a carboxylate (deprotonated form of the carboxylic acid), it has been oxidized. This makes choice A acorrect answer. If a sugar forms a compound with only hydroxyl functional groups and no carbonyl group, it hasbeen reduced. This eliminates choice B. If an aldose becomes an acetal, the number of bonds to oxygen andhydrogen has not changed, so it has neither been oxidized or reduced. This eliminates choice C. Ketones do notoxidize into carboxylic acids, so choice D is eliminated.


99. Choice A is correct. A trisaccharide has two glycosidic linkages. For every glycosidic linkage, one watermolecule is lost. An aldohexose has a formula of C6H12O6 while a aldopentose has a formula C5H10O5. Atrisaccharide from two aldohexoses and one aldopentose has seventeen carbons, so choices C and D areeliminated. Choice B represents the sum of the formulae of all three monosaccharides, but with the loss of twowater molecules, the formula is C17H30O15, making choice A the best answer.

100. Choice B is correct. An aldopentose has one 7t-bond and no rings. It can form a cyclic structure, but that has no it-bonds and one ring. Either way, the compound has only one unit of unsaturation. The best answer is choice B.

"How sweet it is to finish sugars!"

Copyright© by The Berkeley Review® 156 CARBOHYDRATES EXPLANATIONS!

Section VII

NitrogenCompounds

by Todd Bennett

Histidine

PKa(R) =607 (Basic Amino Acid)

Nitrogen Compounds (Non-Biological)a) Amines

i. Basicity and Acidityii. Huelcophilicityiii. Formation of Amines

iv. Reactions of Amines

v. Hofmann Elimination

b) Imine Chemistry

i. Formation of Imines

ii. Imine-Enamine Isomerization

c) Amides

i. Structure

ii. Formation of Amides

iii. Reactions of Amides

d) Amino Acid Synthesis

i. Hell-Volhard-Zelinskii Synthesisii. Strecker Synthesisiii. Reductive Amination of a-Keto Acids

Nitrogen Compounds (Biological)

a) Amino Acids

i. Strcutures and Classification

ii. Isoelectric Points

iii. Easy Calculation of pi

b) Proteins

i. Structure Features

ii. Structural Levels

c) Biological Protein Processes

d) Biochemistry Lab Techniques

i. Gel Electrophoresisii. Affinity Chromatographyiii. Sequencing (Primary Structure)iv. Cutting/Fragmentationv. Edman's Reagent

BERKELEYJjR-E.y«KE-W*


*

Nitrogen-Containing CompoundsSection Goals

Know the amino acids that affect the tertiary structure of proteins.You must know the structuresforcysteine and proline. You should know that cysteineformsdisulfidebridges thatresult incross-linldng within proteins. Proline, because ofitscyclic structure, will causestructural abnormalities likes bends, kinks, and turns. You should have an idea of what structuralfeatures are most affected.

Be able to recognize and classify amino acids according to their side chain.Amino acids are classified in many ways, including hydrophobic, hydrophilic, acidic, basic, polar,and aromatic. Each of the classificationsgives you information about the reactivity of the aminoacid side chain. With proteins, the side chain is most important, because the amino and carboxylterminals are involved in the peptide linkage.

Be able to determine the isoelectric point for amino acids and proteins.Theisoelectric pH is the pH at which the compoundcarriesno net charge. For all amino acidsexceptthose witha basic sidechain, it canbe determined by averaging thevalues forpKai and pK^. Forproteins and the basic amino acids (histidine,lysine,and argirune),you must average the two pKavalues that involve the zwitterion (neutral molecule). This is most easily found by first determiningthe charge of the protein (or amino acid) when it is fully protonated.

Have an understanding of lab techniques such as gel electrophoresis.The basics of gel electrophoresis and isoelectricfocusing involve placing the protein or amino acidbetween thecharged plates ofa capacitor, andallowing thecompound tomigrate through a viscousgel that offers resistance. Thenature of the chargeon the compound is determined by the migrationfeatures.

Know the structural definitions and structural features of proteins.It is important to know whatis meant byprimary, secondary, tertiary, and quaternary structure,although some definitions overlap. You should recognize 6-pleated sheets and how they are arranged(parallel or anti-parallel). Have a basic idea of a-turns and oc-helices.

Be able to determine the sequence of a peptide from chemical information.Sequencing a protein involves treating theprotein withdenaturing reagents, and thensequentiallydeterminingthe componentaminoadds or peptides that fragmentfrom trieprotein when it is treatedwith a sequencing enzyme (digestive enzyme). Youdon't necessarily need to memorize the reagents,but you must be able to sequence a protein when provided with the reagent (or enzyme) and itsfunction. You must know Edman's reagent.

Know the basicity of amines and the effects of alkyl substituents.Perhaps the most common weak base in organic chemistry is the amine. The pKa of a standardprotonated alkyl amine is in the range of 9 to 11. This makes the amine a weak base, and theammonium cation a weak acid.

Know that amine compounds can exist in either the free base or acid salt form.

There areseveral common amine compounds usedformedicinal purposes asantibiotics, antipyretics,and analgesics. TheMCATlikesto present organicmoleculesthat have a definitebiologicalapplication.

Know how amines react to form amides.There are several instances in biochemistry where an amine will react with a carbonyl compoundto form an amide bond. The most obvious case is the formation of proteins from amino acids. Forin vitro reactions, you must be aware of the solvent, because of the acid/base properties of the amines.

Organic Chemistry Nitrogen Compounds

Nitrogen CompoundsCompounds containing nitrogen make up a surprising large percentage of theknown organic molecules. Manyoxygen-containing compounds suchas alcoholsand carbonyls have equivalent compounds where the oxygen is replaced by anitrogen. For instance, alcohols and primary amines are similar in that sense.Because nitrogen has two hydrogen atoms instead of one, like an alcohol, itsreactivity is not identical to an alcohol, but it is nonetheless similar. Ketones andaldehydes share a strong structural similarity to imines, and their relativereactivity is close to one another. The significant difference in an imine is that itstautomer, an enamine,has different reactivity than an enol,because nitrogen andoxygen are different in electronegativity. We shall draw as many analogies tooxygen-containing compounds as we can.

Wewill start by considering amines of all types. Unlike the drastic differenceinreactivity between an ether and an alcohol, aminesdo not change their reactivitysignificantly as they become more substituted. Steric hindrance and electrondensity on the nitrogen are impactedby substitution, but not to the degree thattertiary amines are significantly more or less reactive than primary amines. Ofsignificance when considering amines is theirbasicity and theirnucleophilicity.This makes up the bulk of reactions carried out by amines. Amine basicity andnucleophilicity isnot as clear cut and easyto understand asonemight hope. Theimpact of steric hindrance, hydrogen bonding in solution, and electronics is notas predictable as it is with most other compounds.

After considering amines and imines, we will address some of the less commonnitrogen-containing compounds such as hydrazines, oximes, and azides. Thereare not a great deal of examplesof their reactivity, and it should be expectedthatif they appearon the MCAT, then the test writers will provide a substantial poolof information with them. Weshall consider amides last, as they willserveas atransition from non-biological examples to biological examples. Amide bonds,referred to as peptide bonds when shared between two amino acids, makeup thefoundation of the primary structure of a protein. Amides are formed andhydrolyzed readily under biological conditions, so they are an ideal buildingblock for much of biological chemistry. We will address in vitro synthesis ofamino acidsas our seguefromnon-biological examples into biological examplesof nitrogen-containing compounds.

We shall spend a significant amount of time covering biological examples ofnitrogen-containing compounds. The majority of the text will be devoted to theamino acids and proteins, but we shall also address briefly other compoundssuch as the common bases in DNA and RNA as well as some common classes ofcompounds found in neurobiology. We will focus on the structure and functionof the compounds, leaving the biological aspects to the biology books. Ofgreatestinterestwill be isoelectric points and an easyway to determinethem.

Lastly, we will address biochemistry laboratory techniques that pertain toproteins and amino acids. The MCAT has placed an emphasis on laboratoryexperiments in their previous exams, so we will address the logic and theorybehind electrophoresis, affinity chromatography, and sequencing. Althoughthere are many other techniques in biochemistry, these threeare chosen becauseof their frequency on the MCAT and their applicability to proteins and aminoacids. All otherbiochemistry laboratory techniques covered byourcourse canbefound in the biology books.

Introduction


Organic Chemistry INitrogen Compounds Non-Biological

Non-Biological Nitrogen CompoundsAmines

Amines are compounds with a central nitrogen that has three sigma bonds toeither carbon or hydrogen. The degree of substitution of the amine is describedas primary, secondary, or tertiary, where a nitrogen with all three sigma bonds tocarbon is known as a tertiary amine. To complete the octet about the nitrogen,there is also a lone pair of electrons on the nitrogen. The lone pair is said to be inan sp3-hybrid orbital. If the nitrogen has a double bond (both a pi and a sigmabond) to carbon, the compound is known as an imine. If a hydrogen on the aminenitrogen is replaced by a hydroxyl group, the compound is known as ahydroxyamine. Nitrogen when bonded to nitrogen by a sigma bond is a hydrazine.The structures of some nitrogen-containing compounds are shown in Figure 7-1.

H \ R \ R'

Ammonia 1° Amine 2° Amine

H HH/\"NH2 R CR

Hydroxylamine Hydrazine Imine

Figure 7-1

Example 7.1Which of the following compounds is an amine?

A.

H3C

\ /C=

/

H

H3C

C.

NH-

N

INH

3° Amine

HO'

,N^

CR-

Oxime

D.

N"

I-I O

Solution

Compound A has a double bond between carbon and nitrogen making it animine. Imines can be reduced to form amines, but they are not aminesthemselves. Choice A is thus eliminated. Choice D has the nitrogen conjugatedto a carbonyl group. This functionality is known as an amide. Choice D is thuseliminated. Choice C is a hydrazine, because of the nitrogen-nitrogen singlebond. Choice C is thus eliminated. The only choice left is choice B. Thecompound is a primary amine, because only one carbon is bonded to nitrogen.

Amines are named in a similar fashion as ethers. The alkyl groups attached tonitrogen are named first, followed by the word amine. They can also be namedas amino alkanes. Because of the multiple name option, the likelihood of aminenomenclature questions is not too high. We'll do one example just to cover allbases.



Example 7.2What is theever-so-wonderful IUPAC namefor thefollowing compound?

CH,

A. N-memylphenylamineB. N-benzylmethanamineC. Methyl anilineD. Methyl amino benzene

Solution

Thecompoundhas a benzenering (phenylgroup)and a methylgroup bonded toan sp3-hybridized nitrogen, making the compound an amine. The compound isthus named N-methylphenylamine according to the wild and adventurousIUPAC naming gang. The best answer is choice A.

Amines are both weak bases and good nucleophiles. They are found in manyneurobiology compounds, because many amines have stimulatingeffects on thebrain. Some of the morebiologically famous amines areshownin Figure 7-2.

HO

CH,

HO

(-)-Morphine

H2N

O-S"*Amphetamine

CYTn

J H CH,

Nicotine

OH

hoyV-iEpinephrine (adrenaline)

Figure 7-2

OCHa

Cocaine

J9ôch3Mescaline

The above compounds are found in various medical and household products,some of which you may recognize. Because of their medicinal and industrialuses,syntheticpathways to each are of some interest. Natural product synthesisis a significant part of organic chemistry. What is meant by the term "naturalproduct" is a compound that is synthesized in a plant or animal. Thesecompounds are targets of a great deal of synthetic labor. In some caseshowever,it is far cheaper to extract the amines from plants (i.e., nicotine) usingboth acid-baseextractiontechniques,as well as organicsolvent extraction. For instance, itis through extraction that caffeine is removed from the coffee bean. Extractioncan be far easier than synthesis. For this reason, it is more useful to consider thenatural synthetic pathway rather than an invitro synthesis.

Non-Biological

NHr


OrgailiC ChCmiStiy Nitrogen Compounds Non-Biological

Because the nitrogen of the amine has a non-bonding pair of electrons that it isfree to donate, nitrogen atoms can serve as a nucleophilic center for the molecule.The lone pair of electrons can be donated to a proton making an aminecompound a Bronsted-Lowry base. The basicity of amines is well-documented.The protonated form of the amine is often isolated from aqueous solutions as theammonium salt. For instance, ethyl amine (CH3CH2NH2) can also be found asthe ammonium chloride salt (CH3CH2NH3+C1") after the free base is treated withhydrochloric acid. The free-base form is defined as the deprotonated form of theamine. It is with the basicity of amines that we shall start their study.

Amine BasicityA common use for amines is as weak bases. Amines have pKb values in the 4 to5 range, which means that the Kb associated with the protonation of an amine isbetween 1(Hand10*5. The varying basicity ofamines can beattributed tofactorssuch as the inductive effect, steric hindrance, hydrogen bonding, and solvation.The seemingly chaotic order of relative amine basicity demonstrates the presenceof both the inductive effect and steric factors. Steric factors would favor a smaller

base, because of its ability to bind a proton with little to no hindrance. Theinductive effect predicts that the more substituted amine is the most basic. Therelative basicity of the methyl amines is: (H3Q2NH (pKb = 3.27) > H3CNH2 (pKb= 3.35)> (H3Q3N (pKb= 4.21)> NH3 (pKb = 4.74).

A lower pKb value corresponds to a stronger base, so according to the pKb data,a secondary amine is the most basic of all amines in water (with the same alkylgroup on all of the amines). In organic chemistry, pK data is applied todetermine which way an acid-base equiUbrium will lie (as we considered insection I, pages 37-40 of Organic Chemistry book I). Reaction 7.1 shows afavorable proton transfer reaction involving amines of different substitution.

H3CNH2 + NH4+ ••* ^ H3CNH3+ + NH3

Reaction 7.1

Because the methyl group is electron donating, methyl amine is more basic thanammonia. This means that the ammonium cation is more acidic than the methylammonium cation. The stronger base and the stronger acid lie on the reactantside, therefore the product side of the reaction is more stable than the reactantside of the reaction. This means that the reaction will proceed in the forwarddirection as written when starting from standard conditions. At equilibrium, theproducts are more abundant than the reactants, so theequilibrium constant (Keq)is greater than 1. The equilibrium constant for a proton-transfer reaction can bedetermined using Equation 7.1(listed also as Equation 1.13in section I).

Kea =10Pâ(Pr°duct acJd) "pKa(reactant acid) (7.1)

If you don't recall the common acid-base equations, you should consult page 40of Organic Chemistry book I. In organic chemistry, we are concerned with thepredominant species in solution, so it is a good idea that you fully understandthe Henderson-Hasselbalch equation for buffers, shown in a modified form asEquation 7.2.

tt „ , [Deprotonated species] t„n.pH = pKa (Protonated species) + log——f- - r (7.2)

[Protonated species]

Acid-base chemistry has been thoroughly covered in section I and in generalchemistry, so we shall assume you have reviewed it thoroughly.


Organic ChCmiStiy Nitrogen Compounds Non-Biological

Example 7.3Given that the pKbof NH3is 4.74, what is the approximate pKbof Et3N?

A. -0.32

B. 3.25

C. 5.31

D. 9.26

Solution

Alkyl groups are electron donating, so triethyl amine (Et3N) is more basic thanammonia. As a consequence, the pKb for triethyl amine must be less than 4.74.This eliminates choices C and D. A negative pKb corresponds to a strong baseand amines are weak bases, so choice A is too low. Choice B is the best answer.

Example 7.4Diethyl amine is morebasic than ethylamine in an aqueous solution. If the pKbfor ethyl amine is 3.29, then the pKa for diethyl ammoniumchlorideis:

A. 3.05.

B. 3.89.

C 10.11.

D. 10.95.

Solution

The question is asking for the pKa of a protonated amine, which lies between 9and 11. This eliminates choices A and B. Because diethyl amine is more basicthanethyl amine, thepKb fordiethyl amine is lower than thepKb for ethyl amine(which is 3.29). The sum of pKa for diethyl ammonium cation and pKb fordiethyl amine is 14. For the math to work out, the pKa for diethyl ammoniumcation must be greater than 10.71. Thismakes choice D the best of all possibleanswers in this the bestofallpossible acid-base questions involving amines.

Example 7.5Whatis the relativebase strength of the following threecompounds?

O

H3CCompound I Compound II Compound III

A. Compound II > Compound I > Compound IIIB. Compound I > Compound II > Compound inC. Compound III> Compound I > Compound IID. Compound III > Compound II > Compound I

Solution

All three compounds are amines. The question comes down to the electrondonating and withdrawing nature of the other functional group. Becausecarbonyls are withdrawing by resonance, Compound II is less basic thanCompound I. Because alkoxy groupsare donating by resonance, Compound HIis more basic than Compound I. The relative order for basicity is thusCompound in > CompoundI > Compound II, making choice C correct.


OrgaillC CllCmiStry Nitrogen Compounds Non-Biological

Example 7.6Whatis thepH of0.1 Maniline, which has a pKb of4.2?A. 2.6

B. 8.8

C 11.4

D. 13.0

Solution

Aniline is a base, so the pH is greater than 7.0. This eliminates choice A. Anilineis not a strong base, so a 0.10 M solution must have a pH less than 13. Thiseliminateschoice D. Fora pure weak acid or pure weak base, it is best to use thesimplified equation learned in the general chemistry book. For a weak base, weuse:pOH = V2 pKb - V2 log [Base]. Plugging the values into the equation weget: pOH= 1/2 (4.2) - V2 (-1) = 2.1 + 0.5 = 2.6. If the pOHis 2.6, then the pH is11.4,making choice C the best answer.

NucleophilicityWeak bases, such as amines, make good nucleophiles and the nucleophilicity ofamines parallels their basicity for the most part. Deviations from this trend aredue to sterichindrance. The majority of amine reactions in organic chemistryinvolve the amine behaving as a nucleophile by attacking an electrophile andeither replacing a leaving group orbreaking a Tt-bond. Amines do nucleophilicsubstitution reactions without competition from elimination reactions. Oneproblem with amines is that they are increasingly nucleophilic as more alkylgroups are added. Thismakes the synthesis of primary amines rather difficult,because once formed, they are more reactive than ammonia. Typical aminereactivity questions involve comparing the relative basicities andnucleophilicitiesof a series of amines, looking at the effects of substituents boththrough resonance and the inductive effect on reactivity, and amine structurequestions.

Example 7.7What is the major organicproduct when ammonia, NH3, is treated excessethyliodide, H3CCH2I, following workup?

A. H2C=CH2B. H3CCH2NH2C. (H3CCH2)2NHD. (H3CCH2)3N

Solution

Becausesecondary amines are better nucleophiles than primary amines which inturn are better nucleophiles than ammonia, the reaction proceeds until the alkylhalide is exhausted. With excessalkyl iodide, the reaction is free to add as manyethyl groups to the amine as possible. The best choice shows three alkyl groupsadded to the amine. This makes the correct answer choice answer D. Choice Awould not form without a strong base, so it should have been eliminatedimmediately. It is possible that (^CC^^N"1", a quaternary amine, would form,but it is not listed as an answer choice.


OrgaiHC Chemistry Nitrogen Compounds Non-Biological

Example 7.8An amine can react with all of the following EXCEPT:A. an anhydride.B. benzoic acid.

C. an alkyl halide.D. an ether.

Solution

Anamine can reactas a nucleophile with a highlyreactive electrophile such as ananhydride. This eliminates choice A. An amine can react as a base, so whenadded to a carboxylic acid such as benzoicacid, it carries out a proton transferreaction. This eliminates choice B. An amine can react as a nucleophile with analkyl halide in a nucleophilic substitution reaction. This eliminates choice C.Ether is usually a solvent for many of these reactions, because an amine does notreact with an ether. The best answer is choice D. The three reactions for choicesA, B, and C are shown below:

X x +nh3 — X + XR O R R OH H2N R

fl +NH3 • Jf +NH4+H5C<r OH *#+ &

Jf +NH3 • If +HXR X R NH,

Example 7.9What is the major organic product for the following reaction?

O

NH3

A- NH B- O c O D'

r-^v/^W UxT-^X/^H,N ^ OH H,N ^ NH,NH

Solution

Ammonia is a good nucleophile that can break open the strained four-memberedring by attacking the carbonyl carbon. The final product is not going to have afour-membered ring, so choices A and D are eliminated. Amines react withesters to yield amides, so choice C is eliminated and the best answer is choice B.


Organic ChCmistiy Nitrogen Compounds Non-Biological

Example 7.10What is the hybridization of nitrogen in a tertiary amine?

A. spB. sp2C. sp3D. d2sp3

Solution

In a tertiary amine, nitrogen makes three bonds and has a lone pair of electrons.This means that there are four electron pairs surrounding nitrogen, so thehybridization must involve four atomic orbitals. The hybridization of nitrogen inany amine, whether it is primary, secondary, or tertiary, is sp3. This makeschoice C the best answer.

Example 7.11How can it be explained that Compound AWL8386 is more basic thanCompound KMF8889?

CH3 CH,I I

O OCompound AWL8386 Compound KMF8889

A. Aromaticity reduces the availability of the electron pair for donation inCompound AWL8386.

B. The inductive effectis greater in Compound KMF8889.C. Sterichindrance is greater in Compound AWL8386.D. The nitrogen in Compound KMF8889 lackshybridization.

Solution

Thelonepair on nitrogen in Compound KMF8889 is tied up in the aromatic ringof the pyrrole. This makes it less available for sharingwith a protichydrogen,and thus makes the compound less basic. This makes choice A the best answer.



Formation of Amines

Primary amines may be synthesized in a variety of ways. The synthesis ofprimary amines is not as simple as it may initially appear. You cannot just addammonia to an alkyl halide, because the product (an alkyl amine) is a betternucleophile than ammonia and thus it can react a second and third time withanother alkyl halide. This results in the formation of multi-substituted amineproducts (2° and 3° amines) with random distribution in the product mixture.Figure 7-3 shows six synthetic methods for forming primary amines. Some ofthe reactions can also be used to form secondary and tertiary amines.

Gabriel phthalimide synthesis (for 1°amines only)

O O

N-HKOH

H20+ RX

\ 1 +H3N+R^^X. OH^^>^

Azide reduction (for 1° amines only)

NaN3 - ♦ H2/ MetalR—X 2-^. R —N— NSN —2 »• R-NH2 + N2(g)

Nitrile reduction (for 1° amines only)

Nar=N Ho / MetalR_X Nal-N» R_(;=N _jx ^ R—CH2-NH2

Hofmann rearrangement (for 1° amines only)

O

X2 / KOH(aq)RNH2 + C02 + KX + HX

R NH^

Imine reduction (for 1° or 2° amines)

O N

|| J^ llR R' R R'

1. NaBH4(thf)

2. NH4Cl(aq)

Amide reduction (for 1°, 2°, or 3° amines)

O H H

II \ ^J! i.LiAiH4 y H1L N^ 2.NH4Cl(aq) R^ ^N^

H H

Figure 7-3

H,

R

N

I

H

H

R'

Non-Biological


Organic Chemistry Nitrogen Compounds Non-Biological

Reduction of an azide or nitrile, the Gabriel phthalimide synthesis, and Hofmannrearrangement all produce a primary amine as their major product after workup.Imine reduction can form either a primary or secondary amine, depending on theimine, and amide reduction can form either a primary, secondary, or tertiaryamine. Figure 7-4shows the formation of secondary and tertiary amines.

R'

RImine reduction „ ,

/R ^N^N 1o

II

R"

RNH2_A *~

|| l.NaBH4(thf) ln^C^ 2.NH4Cl(aq)> r-Îî

R R" R2° amine

Amide reduction

O

II

N1

H H

l.LiAlH4 > V p.R' 2.NH4Cl(aqf r^ ^N^

11

H1

H

2° amine

o

II

R N1

H H

l.LiAlH4 ^ V p.,2.NH4Cl(aqf r^ XN^

1

R1

R'3° amine

R"

Figure 7-4

Example 7.12Which of the followingcompounds CANNOT be reduced to form an amine?A. An amide

B. An imine

C. A nitro compoundD. An alkyl hydrazine

Solution

According to Figure 7-3, when an amide is treated with lithium aluminumhydride, it is reduced into an amine. When an imine (R2C=NH) is treated withsodium borohydride, it is reduced into an amine. This eliminates choices A andB immediately. Figure 7-3 shows neither a nitro group nor hydrazine beingreduced into an amine, so memorization won't finish this question. A nitrocompound contains onenitrogen, while hydrazine contains two nitrogen atoms.Considering an amine only contains onenitrogen, it is more likely that a nitrogroup is reduced intoan amine (primary amine to bespecific) than hydrazine. Anitro group can be reduced to form an amineby adding Clemmensen reagents(HCl(aq) and a metal catalyst), which eliminates choice C. An alkyl hydrazine(RHN-NH2) cannot be reduced to form an amine, because reduction will notcleave the nitrogen-nitrogen bond. This makes choice D the best answer. Thisalsoemphasizes a pointaboutyour review. Despite our best efforts to trackthe

Copyright © by The BerkeleyReview 168 Tiie Berkeley Review

OrgaillC Chemistry Nitrogen Compounds Non-Biological

MCAT and incorporate information they release about the exam, there will besome topics we miss. If you plan on memorizing all the information in ourmaterials, you will be fine for the most part. But there will some subjects thatthey introduce that may surprise you. To prepare for this exam, you mustemphasize general concepts and logic more than memorization.

Reactions of Amines

The lone pair of electrons on nitrogen is readily shared, making an amine a goodnucleophile. The crux of amine reactivity involves the lone pair on nitrogenattacking an electrophilic carbon. As a general rule, either a leaving group isdisplaced or water is formed as a side product. This should help to predict theproducts of amine reactions. We will consider the addition of amines to alkylhalides, diazonium formation and the Hofmann elimination first and imineformation and hydrazine reactions later. Figure 7-5shows a few amine reactions.

Addition of an amine to an alkyl halide

H3N: + R—X • H3N+RX", H2N+R2X", HN+R3X",This reaction results in multiple additions, because the moresubstituted an amine, the better it is as a nucleophile.

Reactions of diazonium salts

NaNO,R—NH2 HQ *• R—N+=N:C1" + NaCl + 2H20N2is about as good a leavinggroup as there is,so diazonium salts are excellent electrophiles.

NaNOo ._ CuX _Ar-NH2 -hcT^ Ar-N+=N:Cr (x =cl-Br, c5N:)» Ar-X + N2(g)

Hofmann Elimination

CH2^ NH2 3CH3I CH2^ ^+N(CH3)3rR- CUT ^*" CH2Vg20/A

RHC=CH2 + (H3C)3N + Agl

Least substituted alkene

Figure 7-5

In the first reaction, addition to an alkyl halide, there is a leaving group. Hence,the nitrogen displaces the leaving group by way of a nucleophilic substitutionreaction to form an amine. As noted in Figure 7-5, there is the good chance thatmultiple alkyl groups will add tothe amine. Inthe second reaction, formation ofa diazoniumsalt, there is no leavinggroup. Hence, we need to lookfor the waterthat forms. One nitrogen has the hydrogen atomsand the other has the oxygenatoms, so upon losing the hydrogens and oxygens, nitrogen atoms in need ofbonds remain, so this reaction must form nitrogen-nitrogen bonds. In the lastreaction, Hofmann elimination, there is a leaving group, so the nitrogen displacestheleaving group to form an amine. With excess methyl iodide, it exhaustivelymethylates to form a quaternary amine. The quaternary aminecarriesa positivecharge, making it an excellent leaving group. The long alkyl chain attached tothe nitrogen can undergo E2 elimination, kicking out the leaving group andforming a leastsubstitutedalkene. WeshalladdresstheHofmann elimination inmore detail, as it has appeared on previous versions of the MCAT.


OrgaiUC ChCmiStiy Nitrogen Compounds Non-Biological

Hofmann Elimination

The mechanism for the Hofmann elimination reaction is a straightforward E2mechanism. The Hofmann elimination reaction involves forming a cationicquaternary amine (nitrogen with four alkyl groups attached). When thequaternary amine isantitoa fi-hydrogen (hydrogen on thecarbon adjacent to thecarbon bonded to the quaternary amine), it can beeliminated with strong base.Ag20, when hydrated by water, yields two hydroxide anions (OH-) and twosilvercations (Ag+), so it is a strongbase. Thesilvercationcan bind the iodideanion andprecipitate outofsolution while thehydroxide anion deprotonates the6-hydrogen resulting in elimination. The E2 reaction does not haverearrangement associated with it, therefore the product is the least substitutedalkene. This is one of few ways to synthesize a terminal alkene. Figure 7-6showsthe orientation ofa generic molecule in a Hofmannelimination reaction.

^N+(CH3)3r N(CH3)3The beta hydrogen and the H' /v'-^H "'-q-**quaternary amine are anti to R . ii^R' ^" **one another for E2 reaction. p? J r ^"C^ Rl

H "OH H-0N

Figure 7-6

H

It is important that the compound have fi-hydrogens relative to the quaternaryamine, otherwise there can be no elimination product. Be particularly aware ofthis when dealing with cyclic amines. Figure 7-7 shows Hofmann elimination ofa cyclic amine.

H H3C CH3I 3\ /+ 3NN 2CH3I^ .N Ag20

OÔ(H3C)3N + Agl + <A/^

Figure 7-7

Copyright©by The Berkeley Review 170 Tne Berkeley Review


Imines

When amines are added to a carbonyl, a nucleophilic attack at the carbonylcarbon takes place. If there is no leaving group on the carbonyl, then the reactionforms an imine and water and is driven by its equilibrium. If water or imine isremoved (usually by distilling away the product with the lower boiling point),then the reaction is pushed forward. This is one of several reactions that can berecognized by the water side product more easily than the organic product. Theformation of an imine, oxime, and hydrazone are shown in Figure 7-8, to presentthe ease of predicting the product after identifying the water that is formed.

1° Amine

K

H H

R H

Aldehyde

Hydroxyamine

OH

H H

XAldehyde

Hydrazine

H2N

H H

XAldehyde

R

N'

1° Amine

R'

H H

R R

Ketone

Hydroxyamine

OH

.OH I

H H

R H

OximeR R

Ketone

Hydrazine

H2N

H,N„N

H H

R'

N'

II " R^H 1 cat.HCl*_xVyS*^ Imine x^s.

R R

Imine

.OH

N'

A.. "j? "cat.HC, rArOxime

H,N,N

«XX

Ketone

cat. HClR R

HydrazoneH

Hydrazone

Figure 7-8

These reactions proceed considerably better with an aldehyde than a ketone.This is attributed to the greater electrophilicity of an aldehyde than a ketone.When a primaryamine is used, the imine that is formed is referred to as a Shiffbase. Hydrazones are used in the identification of carbonyl compounds,particularly ketones and aldehydes. Othercarbonyl compounds have derivativesas well, but the majority of derivatives are formed by substitution reactions withnitrogen compounds. Thebasic idea here is that by saturating an aldehyde orketone with a compound that contains at least two hydrogen atoms on thenitrogen (ammonia, primary amines, hydroxyamine, or hydrazine), a C=0 isconverted into C=N. With excess water, the reverse reaction is observed.

Non-Biological



O

Example 7.13What is the major organic product of the reaction shown below?

O

,XA.

Ph HN /

+ N-N —r-/ \ A

R H H

D.

R

O B' NH

X XNH R RNH

IPh

c NPh

XR'

RA,

NHPh

Solution

Find the water and all will be fine. The oxygen comes from the ketone, so thetwo hydrogens must come from therutrogen-containing species. Only oneofthetwo nitrogen atoms has two hydrogen atoms, so thatis thenitrogen thatreplacesthe oxygen inthe ketone. The result after removing the water and connecting theatoms is choice D. Connect-the-dot skills you developed in kindergarten mayprove more useful ontheMCAT thanthevolume ofinformation youmemorizedduringcollege. Probably not, but it felt good tosay.

Imine-Enamine Tautomerization

Just as ketones andaldehydes undergo tautomerization in thepresence ofmildacid or mild base to form an enol, imines also undergo tautomerization to forman enamine. Figure 7-9 shows the equilibrium between an imine and anenamine.

Figure 7-9

An imine can form only if the nitrogen compound has two hydrogen atoms onthe nitrogen. When a secondary amine reacts with an aldehyde or ketone, theproduct is an enamine, because an imine cannot form. The formation of anenamine from a secondary amine and ketone isshown in Figure 7-10.

R' ..^R"N

R' .. .R"N

HR

H

I

R' R"

2° AmineHO / \

R Elimination^

H H

Aldehyde

Copyright ©by TheBerkeley Review 172

HOH H

Carbinolamine

Figure 7-10

HR

H

Enamine

The Berkeley Review

OrgaillC ChCmiStiy Nitrogen Compounds Non-Biological

Amides

Amides are carbonyl compounds with a nitrogen bonded to the carbonyl carbon.They are formed when an amine or ammonia is added to a carbonyl compoundwith a leaving group. Figure 7-11shows the three types of amides.

O O O

II II II

R^N^ R^N"*" R^N^'I I I

H H R"

1° Amide 2° Amide 3" Amide

Figure 7-11

IUPAC nomenclature rules call for naming the alkyl groups on the nitrogen witha prefix of "N" first, followed by the carbonyl chain. The degree of substitutionimpacts the physical properties. Acetamide, H3CCONH2, has a melting point of82°C and a boiling point of 221°C. N-Methylacetamide, H3CCONHCH3, has amelting point of 28°C and a boiling point of 204oC. N,N-Dimethylacetamide,H3CCON(CH3)2, has a melting point of -20°C and a boiling point of 165°C.Amides form H-bonds between oxygen's lone pair and the hydrogen on nitrogen,which explains the drop in boiling and melting points as substitution increases.

Formation of Amides

Amides are typically formed from the addition of an amine to an acidanhydride,acid halide, or an ester, as covered in section V of Organic Chemistry book II. Aprimary amidecanbe formed by partialhydrolysis ofa nitrile (R—ON)and canalso be formed from an oxime by way of the Beckmann rearrangement. TheBeckmann rearrangement results in the formation of an amide from an oximethroughthe addition ofeither a strong acidor PCI3O/PCI5. An oxime is formedby the addition of hydroxyamine to either an aldehyde or a ketone, as shown inFigure 7-8. Beckmannrearrangement is shown in Figure 7-12.

H

1 H°-

Beckmann Rearrangement:o:

*63 .7 tf" X».«H H~ U^ r n

^ / R " «N *•

AR R* „ / H

9b H

h—.0.. T

H/

H :P.' H C^=N/ \

R RX VH ^ XR— C=N+— R' ^ ^ R R

Figure 7-12

<0^H H



Reactions of Amides

We have seen that amides can be reduced into amines using lithium aluminumhydride. In addition to reduction, an amide can be hydrolyzed into a carboxylicacid and amine. Primaryamidescanundergo Hofmannrearrangement to form aprimary amine of one less carbon and carbon dioxide. The reaction does notwork with substituted amides; it worksonly with primary amides. Figure7-13shows the step-by-step mechanism for Hofmann rearrangement.

Hofmann Rearrangement

n- KJ?.-H!?:" # h

P :o—h

\§- o

H

o=c=o + r—n:• • • • \

H

h—o:

Isocyanate

H— o:y R

Figure 7-13

In the Curtius Rearrangement an acid chloride is treated with sodium azideNaN3 to form an isocyanate. The conversion goes through an acyl azide thatwhen heated decomposes to yield an isocyanate and nitrogen gas (a great leavinggroup). Just as seen inthe Hofmann rearrangement, anisocyanate can undergodecarboxylation in basic conditions to generate a primary amine. Figure 7-14shows thekey intermediates oftheCurtius rearrangement.


OrgailiC ChCmiStiy Nitrogen Compounds Non-Biological

R^Cl<In.-

Curtius RearrangementO H

/

0=C=0 + R—N

H

HO"

O N=N = N 0=C=N

RT N=N = N R^ ^N—N= N

Figure 7-14

Amino Acid SynthesisAn amino acid, as the name implies, is a molecule with an amine group and acarboxylic acid group. In an aqueous environment, the carboxyl terminal isdeprotonated and the aminoterminal is protonated, so it hasat leasttwochargedsitesdespite having no net charge. However, in organic solvents they existas anuncharged molecule. There are three synthetic pathways for making aminoacids: Strecker synthesis, Hell-Volhard-Zelinskii synthesis, and reductiveamination of a-keto acids. The three syntheses are shown in Figure 7-15.

Hell-Volhard-Zelinskii synthesisO O O

R u i_Dr2lr R ii NH3 R

R

R

, [^ l.Br»P RvIV*^ OH 2.H20 ^Y 'OH y OH

Br NH2

Strecker synthesis

O N H2N C=N H2N C02HH NH3 I] KCN W H3Q+ y

R^H R^H1** R^H A R^HReductive amination of a-keto acids

O N H2N HANH3 |j NaBH^ NH4C1 \/

K C02H R^C02H Et>° H*° R^CCtfJFigure 7-15


OrgaillC ChCmiStiy Nitrogen Compounds Biological

Amino Acids

Natural amino acids (S-2-amino carboxylic acids) are a critical partofbiology asthebuilding blocks ofproteins. There area totalof twenty standard amino acidsthat humans code for. Although there are over 500 amino acids in nature,humans code for just twenty. InmRNA, amino acids are coded for bya sequenceof3bases (codons) using adenine (A), cytosine (C), guanine (G), and uracil (U).For instance, histidine is coded as CAC or CAU. With 64 (43) combinations ofcodon triplets, there are multiple codes for some amino acids. Figure 7-16 showsthe generic structure ofan S-amino acids inboth aqueous environment and lipidenvironment.

Q o

H3NV ^Cs. H2NC O- ^ ÔH

A AH R H R

Zwitterion form ofamino acid Uncharged form ofamino acid(aqueous environment) (lipid environment)

Figure 7-16

Because amino acids are compounds which contain an amino group on thesecond carbon and acarboxylic acid, they have acid-base properties that you areexpected toknow. Amino acids are essentially justpolyprotic acids where the allofthe protons are weakly acidic. The amino terminal when protonated forms acation with apKa of roughly 9.7 ±0.9. The carboxylic acid terminal has a pKa ofroughly 2.2 ±0.4. In long chains, they form proteins, which you are also expectedtoknow. This section covers functional features oftheamino acids, theacid-baseproperties of the amino acids, protein structure, the structural effects of hydrogenbonding, cysteine, and proline, protein sequencing, and some enzymaticchemistry. We shall start with terminology and classification. Two terms thatwill be used repeatedly to describe the amino acids are hydrophilic andhydrophobic (based on their side chains). Within the hydrophilic group, someamino acids are acidic, some are basic, and some the others are polar.

Structures and Classification

There is a chiral center at the a-carbon in all amino acids except glycine.Naturally occurring amino acids have S-stereochemistry atcarbon 2, although inbiochemistry we refer to that as an L-amino acid. The exception to the Lcorresponding to S rule is cysteine, where L-cysteine is the natural form, but ithas R-stereochemistry at the a-carbon. For the MCAT, you should recognize thetwenty amino acids for which we code, along with general pKa data for theamino terminal, the carboxyl terminal, and theside chain ofprotic amino acids.Amino acids are categorized according to behavior, and although it may varyfrom textbook to textbook, itis generally based on water solubility and degree ofprotic activity. We shall classify amino acids into one of five categories:hydrophobic, semi-hydrophobic, hydrophilic, acidic, andbasic. There areaminoacids that seemingly should fit into multiple categories, like aspartic acid, whichishydrophilic and acidic, but we shall have each amino acid appear only once.



Hydrophobic amino acidsHydrophobic amino acids are "water-fearing", which means that they are notwater soluble and they prefer to be situated in an organic (lipid) environmentrather than an aqueous environment. In aqueous environments, hydrophobicamino acids are found in the core of a folded protein. Figure 7-17 shows thehydrophobic amino acids glycine, alanine, valine, leucine,and isoleucine as theyexist in an aqueous environment at pH = 7.

Hydrophobic Amino Acids

O OPKa2=9.87 ||

c o-

pKa2=9.78

H3N*,

pKa2=9.72

HgN-,

Biological

o

OpKal=2.33

H3N+,

J\ pKai=2.29H CH(CH3)2

Valine

(Val,V)

H H

Glycine(Gly,G) pKa2=9.74

H3N+.

O

y^ pKal=2.35H CH3

Alanine

(Ala, A)

O*J\ pKal=2.33

H CH2CH(CHj)2

Leucine

(Leu, L)

pKa2=9.74

H3N+,

O

II

O"y^ pKal=2.32

H CH(CH3)CH2CH3

Isoleucine

(lie, I)

Figure 7-17

Semi-hydrophobicamino acidsThese amino acidsare typically considered tobe hydrophobic in mosttextbooks,but we segregate themfrom otherhydrophobic amino acids, because theydon'tpack as well as the alkyl side chains. These include the aromatic side chains(except for tyrosine) and the amino acids that are more water soluble thanexpected for an aliphatic side chain. Figure 7-18 shows the semi-hydrophobicamino acids proline, phenylalanine, tryptophan, and methionine as theyexist inan aqueous environment at pH = 7.

Semi-Hydrophobic Amino Acids

pKa2=9.44

H3N+.

J\ pKai=2.43H CH,

O"

pKa2= 10.64

h2n!

C

Figure 7-18

Copyright © by The Berkeley Review

O

O"pKal=1.95

H

Proline

(Pro,P)

177

PKa2=9.27

H3N+.

O

pKa2=9.24

H3N+.

O"J\ pKai=2.17

H CH2CH2SCH3

Methionine

(Met,M)Phenylalanine

(Phe, F)

Exclusive MCAT Preparation

Organic Chemistry Nitrogen Compounds Biological

PKa2=8.80

H3N+.

O

Hydrophilic amino acidsHydrophilic amino acids are "water-loving", sothey are water soluble andpreferto be situated in an aqueous environment rather than an organic (Upid)environment. In water, hydrophilic amino acids are found on the exterior of afolded protein. These aresometimes referred toasprotic side chains. Figure 7-19shows the hydrophilic amino acids serine, threonine, cysteine, tyrosine,asparagine, andglutamine asthey exist inanaqueous environment at pH = 7.

Hydrophilic Amino Acids

pKa2=9.13

H3N+.

O

II

^cô-y^ PKal=2.02

H CHo

c oy^ PKal=2.17

H CHn

AO' NH2

Asparagine(Asn,N)

CH,

O NHo

Glutamine

(Gln,Q)|PKa3= 10.07

OH

o

pKa2=9.46 ||O

pKa2=9.10 ||

TyrosineQ (Tyr,Y)

PKa3= 10.33 ||

£ °y ^ pKal=2.19 J\ pKai=2.09 c oJ\ pKal=1.88

H CH2OH H CH(CHj)OH H CH2SH

Serine ThreoninepKa2=8.36

Cysteine(Ser,S) Cnur,T) (Cys,C)

Figure 7-19

Example 7.14Which ofthe following amino acids isLEAST hydrophilic?A. Serine

B. TyrosineC. CysteineD. Leucine

Solution

Hydrophilic is defined as "water-loving," which implies that the compoundshould bewater-soluble. Serine has analcohol side group, soit iswater solubledue to the hydrogen bonding of the side chain. Tyrosine has an alcohol sidegroup aswell, soit too iswater soluble due tothe hydrogen bonding ofthe sidechain. This eliminates both choices A and B. Cysteine has a thiol (SH) sidegroup, so it is watersoluble due to the polarity of its side chain. Leucine has analkylsidechain, so it is hydrophobic. Thebest answeris choice D.


OrgaiHC ChCmiStiy Nitrogen Compounds Biological

Example 7.15Which of the following amino acids contains a benzene ring?

A. Valine

B. TyrosineC. Histidine

D. Isoleucine

Solution

Phenylalanine, tyrosine, and tryptophan are the only amino acids with sidechains that contain a benzene ring. Of those three amino acids, only tyrosine isoffered as a choice, so choice B is the best answer. Histidine has an aromaticring, but it is not a benzene ring.

Example 7.16Which of the following amino acids is classified as hydrophilic?

A. Proline

B. Threonine

C. Valine

D. Phenylalanine

Solution

Proline, valine, and phenylalanine are hydrophobic, because they contain alkylside groups. This eliminates choices A, C, and D. The only choice left isthreonine, choice B, which has an alcohol group as at the end of its side chain.The hydroxyl group forms hydrogen bonds, making threonine hydrophilic.Choice B is the best answer.

Acidic amino acidsAcidic amino acids have side chains that lose a proton from their neutral state.They have three acidic sites, the N-terminal, the C-terminal, and the side chain.In the caseof Glu and Asp, the side chain is a COOH. Cysteine and tyrosinecanalso lose a protonfrom theirneutralsidechain, but it happens at a highenoughpH that we generally think of them as polar more than as acidic. Figure 7-20shows the acidic amino acids aspartic acid and glutamic acid as they exist in anaqueous environment at pH = 7.

Acidic Amino Acids _O O

pK„3 =10.01 || pKa3=9.95 ||H3NV ^C\ H3N\ /C\

^C O" C O"J\ pKai=1.99 y\. PKal=2.13H CH2 H CH2

X^=™ <uO O j zAspartic acid ^v£Ka2 =432

(Asp, D) O O"

Figure 7-20

Glutamic acid

(Glu,E)


OrgaillC ChemiStiy Nitrogen Compounds Biological

Basic amino acids

Basic amino acids have side chains that gain a proton from their neutral state.Thethreeexamples, lysine, arginine, and histidine, havenitrogen-containing sidechains. Like acidic amino acids, basic amino acids have three pKa values. Youmight wish to recall the mnemonic "his lies are basic," which means thathistidine, lysine, and arginine arebasic. Forhistidine, the lower nitrogen in theimidizole ring is the basic site. Figure 7-21 shows the basic amino acids lysine,arginine, and histidine as theyexist in an aqueousenvironmentat pH = 7.

Basic Amino Acids

O O opKa2=9.20 || PKa2=8.99 || PKa3=9.15 ||H3N+\ /C\ I%NV ^C\ H3NV ^^f O- ^fT or ^cT Oy ^ pKal =2.16 y ^ pKal =1.82 y ^ pKal =1.81

H (CH2)4 H (CH2)3 H CH2

pK^ =10.80 +NH, NH

Lysine pKa3 =13.2li ,(^K) H2N+^^NH2 pKa2 =6°5 n=/

Arginine Histidine(Arg,R) (His,H)

Figure 7-21

Amino acids canbecategorized in many ways, not justby theirsidechain, as wehave done here. For instance, let us consider essential and non-essential aminoacids. There are eight to nine essential amino acids, depending on aperson's age.Essential amino acids are the amino acids we cannot synthesize in ourbody, sotherefore wemusttake them in through our diet.

H

Example 7.17Which ofthe following amino acids has acharged side chain atpH=7?A. Valine

B. PhenylalanineC. ArginineD. Cysteine

Solution

Ata pHof7, the side chain for valine isCH3, the side chain for phenylalanine isCH2C6H5, thesidechain forarginine is CH2CH2CH2NHC(NH2)2+, and the sidechain for cysteine isCH2SH. The only charged side group isfound onarginine,which makes choice C the best answer.


OrgaiUC ChCmiStiy Nitrogen Compounds Biological

Example 7.18Which of the following amino acids carries a net positive charge at pH = 8?

A. Threonine

B. ArginineC. PhenylalanineD. Histidine

Solution

At a pH of 8, the amino terminal is protonated and carries a +1 charge and thecarboxyl terminal is deprotonated and carries a -1 charge. The compound existsas a zwitterion if the side chain is neutral. So, to have a net positive charge at pH= 8, the side chain must carry a positive charge. The only amino acid of thechoices to have a positively charged side chain at pH = 8 is arginine, which has aside chain pKa of 13.2. This means that arginine has an overall positive chargeup to the pH where the amino terminal is mostly deprotonated. This makesarginine, choice B, the best answer. At pH = 8, the structures are as follows:

Threonine Arginine Histidine PhenylalanineProtonated q Protonated q Protonated q Protonated q

pKa2>8 || pKa2>8 || pKa3>8 || pKa2>8 ||

H3N+\ /<=v> H3N+\ ^C^ H^ /C^ H3N-^ C^^c o- C O" C O' c oy\ Depro i\ Depro i\ Depro i\ Depro

\. pKal<8 / 5. pKal<8 i\ pKal<8 I % pKal<8H CH H (CH2)3 H CH2 H CH2

A IHO CH3 ProtonaledNH

PK,3>8 I Depro \ i

overall charge =0 overall charge =+1 overall charge =0 overall charge =0

Example 7.19Which of the following amino acids is involved in cross linking within a protein?

A. CysteineB. Threonine

C. Proline

D. Tryptophan

Solution

Two cysteine residues form a disulfide bridge (-CH2-S-S-CH2-) within a proteinor between two separate primary structures (proteinstrands). The formationofthe crosslinkinvolves oxidation of the thiol groups, one of those factsyou shouldknow. The best answer is choice A.

There are some fundamental structural features of proteins that result fromproperties of their component amino acids. Cross-linkingof cysteine residues isone example. Another is structural turns, forced alterations in the structure of aprotein, caused by proline. Proline is cyclic and thus unable to rotate freelyabout its sigma bonds. It is locked into a cyclic conformation, which locks theprotein backbone into a turn.


OrgaillC Chemistry Nitrogen Compounds Biological

Example 7.20Which of the following amino acids is MOSTsoluble in water?

A. Alanine

B. Serine

C. Isoleucine

D. Methionine

Solution

The compound that is most soluble in water of the choices is the hydrophiliccompound. Because the side chain of serine contains a hydroxyl group, it canform hydrogen bonds. This makes serinehydrophilic, whilealanine, isoleucine,and methionine are hydrophobic. Choice B is the best answer.

Example 7.21Whichof the following amino acids has a chiral center in the side chain?A. Proline

B. Threonine

C. TryptophanD. Methionine

Solution

Of the twenty amino acids for which we code, only threonine and isoleucinehave chiral centers in the side chain. The side chain ofthreonine isnaturally R.Thediastereomer ofnaturally occurring threonine (thatdiffers from threonine inthe chirality of the side chain from R to S) is referred to as allo-threonine. Thebest answer is choice B.

Isoelectric Points

In aqueous solution at a pHof7, an amino acid exists as an ionic species. Theamino terminal isprotonated and thecarboxyl terminal is deprotonated. The netresult is zero overall charge (excluding thesidechain in some amino acids), andthe structure is referred to as a zwitterion. The zwitterion is neutral overall butcontains oppositelycharged sites. The zwitterion form of an amino acid, becauseit is a salt, has a high melting point. The pH at which the zwitterion exists inhighest concentration is referred to as the isoelectric pH. The isoelectric pH isobtained by averaging the two pKa values that involve the neutral molecule.Thatis to say thatyouaverage the pKa that takes the molecule from -1 to neutralwith the pKa that takes the molecule from neutral to +1. The molecule is aperfect zwitterion exactly midway between the twopKavalueswhichinclude theneutral species. The isoelectric point coincides with an equivalence point on atitration curve, the first equivalence point ofall amino acids except theones withbasic side chains, where it coincides with the second equivalence point. Tocalculate the isoelectric pH you must decide if the side chain is acidic, basic, orneutral. For amino acids with neutral oracidic sidechains, theisoelectric point isfound by averaging pKai with pKa2- Foraminoacidswith basicside chains, theisoelectric point is found by averaging pKa2 with pKa3. Figure 7-22 shows apictorial overview ofthe structure ofanamino acid asit changes with increasingpH. The R-grouprepresents any neutral side chain.


Organic Chemistry

OH^^r

H R

Overall charge: +1

Nitrogen Compounds

o-fe

H R

Overall charge: 0H R

Overall charge: "1

pl = P^l+PKa2

Figure 7-22

Figure 7-23shows a pictorial overview of the structure of an acidic amino acid asit changes with increasing pH. The side group is specific for aspartic acid, butthe charges are the same with any acidic side chain such as glutamic acid,cysteine, or tyrosine. Acidic side chains lose a proton from their neutral state toform an anion. Acidic side chains contain a hydrogen on either oxygen or sulfur.

Biological

OHP*al

HoN*. pK. HgN*' pKaSL H2B,N,

O" OH

Overall charge: +1O" OH

Overall charge: U

T pKa! +pKa2PI= n

H CHg

cAo-Overall charge: "1

Figure 7-23

Figure 7-24 shows the titration curve associated with full deprotonation of anacidic amino acid, starting from its fully protonated state and proceeding to itsfully deprotonated state, like it is shown in Figure 7-23.

Equivalents base added

Figure 7-24

O" O"

Overall charge: "2

Copyright © by The Berkeley Review 183 Exclusive NCAT Preparation


H,N+,

Figure 7-25shows a pictorial overview of the structure of a basic amino acid as itchanges withincreasing pH. The sidegroup is specific for lysine, but thechargesare the same with any basic side chain such as histidine and arginine. Bydefinition, basic side chains gain a proton to form a cation at low pH values.Basic sidechainscontain a lonepair ofelectrons on nitrogen when neutral.

OHP^l H,N+,PKa3w HoN

(CH2)4

+NH3

Overall charge: +2 Overall charge: +1 Overall charge: U

t_ PKa2 +P*a3

NH,

Overall charge: -1

Figure 7-25

Figure 7-26 shows the titration curve associated with fully deprotonation of anacidic aminoacid, likeshownin Figure 7-25.

^Xa*

0

Isoelectric Point

lV2 2Equivalents base added

Figure 7-26

For calculating the isoelectric point of an amino acid, it is first necessary todetermine the nature of the side chain. The isoelectric point is the average ofpKai and pKa2 for all amino acids except lysine, histidine, and arginine. Theisoelectric point oflysine, histidine, andarginine isanaverage ofpKa2 andpKa3.No matterwhatisoelectric pointyouare determining, you should remember thatit's an average of twoconsecutive pKavalues.

Copyright© by The Berkeley Review 184 The Berkeley Review


Easy Method to Get the Isoelectric Point of a ProteinFirstassume that the protein exists at pH = 1, where all sites on the protein areprotonated. Carboxyl groups are uncharged when protonated, while amino sitescarry a positive charge when protonated. Only the amino terminal and the sidechains of lysine, arginine, and histidinecarrya charge (apositive one)at low pH,so the net charge of the protein is the sum of the lysines,arginines, and histidinesin the protein plus one (for the N-terminal). Considerthe protein in Figure7-27.

0H3N- •Met Thr Ala—

SCH3 CHOH CH3

©Lys

o

Asp Gly His Glu Ser \

NH,

O^ H

OH OH

CH,

HN

Figure 7-27

The protein in Figure 7-27 canbe abbreviated as H6A^+, because when it is fullyprotonated, it carries a +3 charge, and there are six sites from which a protonmay be lost (the N-terminal, the C-terminal, and the side chains of Lys, Asp, His,and Glu). Now consider the number of times the protein must be deprotonatedto reach zero charge. The charge starts at +3, so the fully protonated proteinmust be deprotonated three times to become neutral. This results in pKa3leading to thezwitterion. Figure 7-28 shows anabbreviated titration ofHgA3"1".

CH2OH

3+H6APKal

2+H5AP^

H4A+PK^ PKa^

H3A ^pKdL

H2A" ^pKal

HA"" ^

Figure 7-28

The pi is an average of the two pKa values surrounding the neutral species (thepKa that leads to the zwitterion and the pKa that leads from the zwitterion). Inthis case, the pi is found by averaging pKa3 with pKa4- The shortcut derivedfrom this example is that the total charge at pH = 1 corresponds to the subscriptof the first pKa averaged to determine pi. This is because the pKa leading to theneutral species corresponds numerically to the charge when fully protonated.Consider a protein with a +7 charge at pH = 1. It requires seven deprotonationsto get to the neutral species, so pKa7 is the pKa value that leads to the zwitterion.Figure 7-29 summarizes this example and any generic charge.

Pî+7 ^-=- +6

+q -r +(q-l)

^>+l

>+l

PKa7_0

PKa^

pKa<l0

PKaq+l

pl=pKa7+ pKag

pl=pKaq + pKaq+1

Figure 7-29

The pi is found by summing the number of basic amino acids plus one, andtaking the pKa with that numerical subscript and averaging it with the nextpKa value sequentially up. So, if a protein has two lysines, one arginine andthree histidines, then the pi is an average of pKa7 and pKas- The sample proteinhas a lysine and histidine, so the pi is an average of pKa3 and pKa4.

OH

Biological


OrgailiC ChCmistiy Nitrogen Compounds Biological

The final step is to determine which pKa values are which. For the originalprotein shown, pKai is for the carboxyl terminal (this is always true). Thesequence of the rest must be determined from pKa information. Figure 7-30shows the protein with its pKa values.

© 0 0H3N Met Thr

© © //Ala Lys Asp Gly His Glu Ser \

of

OHPKa =9.7 £ | | £ J* | | Jf

A H ^ ApKa =10.80^ OH 1 Or OH #1CH3 #6 PKa=^ ^^ PKa=4.3

#5 SCH3 CHOH CH3 NH3 ^^ H CH2 .$L CH2OH pKa=2.4

l3 #2 \ im« - #3

Figure 7-30

The assignmentof pKa values is sequential with increasingnumerical value, sothe pi for this protein is found by averaging pKa3 (4.3) and pKa4 (6.1), whichresults in a pi of5.2. This means thatat pH =5.2 it exists as a zwitterion, at pH <5.2 it exists as a cation, and at pH > 5.2 it exists as an anion.

pi „pKa3+pKa4 ,4,3+6.1 _10.4 = 522 2 2*

Example 7.22Which of the following statements is false about amino acids?

A. Thecarboxyl terminalalwayshas the lowest pKa.B. A carboxylic acid side chain always deprotonates before the amino terminal.C. Thesidechain groupof lysine is lessacidic when protonated than the arnino

terminal of lysine.D. The D-isomer ofglycine is thepredominant isomer in humanbeings.

Solution

Thecarboxyl terminal has the lowest pKa valuefor all twentyof the aminoacidsforwhich wecode. This can beverified by looking at theamino acids in Figures7-17 through 7-21. This eliminates choice A. Because thecarboxyl terminal has alower pKa value than the amino terminal, the carboxyl terminal is more acidicthan the amino terminal. By being more acidic, the carboxyl terminal loses aproton more readily than the amino terminal. This means that choice B is validand thus is eliminated. Choice C requires that you look at the amino acids inFigure 7-21. The pKa of the amino terminal of lysine is 9.2. The pKa of the sidechain of lysine is 10.8. The lower pKa is associated with the amino terminal,which implies that the amino terminal is more acidic than the side chain. Thismakeschoice C valid, and thus it is eliminated. Theonlychoice left is answer D.There is no D-isomer ofglycine, because the side group for glycine is a hydrogen,therefore glycine does not contain a chiral center. Without a chiral center, therecan be no D or L configuration associated with the isomer. This confirms that thestatement is false, making choice D the best answer. For the other amino acids,thenaturally occurring form is the L-isomer. This can be recalled by thinking ofthe phrase"L is for life." It canalsobe remembered by thinkingaminoacids arenatural. Eitherway you chooseto recall the meaning of the L label, amino acidsare naturally L. If you are bothered by the word "always" in choiceA, sometimesan always is okay.


OrgaillC ChCmiStry Nitrogen Compounds Biological

Example 7.23What is the isoelectric point for tyrosine?

A. 2.2

B. 5.6

C 6.2

D. 9.1

Solution

Tyrosine is not a basic amino acid, so its isoelectric point is found by averagingpKai and pKa2. The isoelectric point is solved according to the followingmathematics:

j _pKai +pKa2 _2.2+ 9.1 -11.3 = 5652 2 2*

Thismakes choice B the best answer. As a general rule, acidic amino acids havepi values less than 5.5 and basic amino acids have pi values greater than 7.5. Allother amino acids have pi values within the 5.5 to 7.5range.

Example 7.24What is the isoelectric point for lysine?

A. 5.7

B. 6.5

C. 9.2

D. 10.0

Solution

Lysine is a basic amino acid, so its isoelectric point is found by averaging pKa2and pKa3. The isoelectric point is solved according to the following mathematics:

j =pKa2 +pKa3 =9.2 +10.8 =20 = 102 2 2

This makes choice D the best answer. As a general rule, basic amino acids havepi values greater than 7.5, so choices A and B are eliminated before any mathneeds to be done.

Example 7.25What is the isoelectric point for the dipeptide alanine-arginine?

A. 7.5

B. 9.0

C 11.1

D. 11.5

Solution

Alanine is a hydrophobic amino acid and arginine is a basic amino acid, so theisoelectric point for the dipeptide is found by averaging pKa2 and pKa3. Aswritten, the peptide linkage is made from the C-terminal of alanine and the N-terminal ofarginine, so there is no pKa(coOH) for alanine andno pKa(NH3+) f°rarginine. For the dipeptide, pKa2 is the N-terminal of alanine and pKa3 is theside chain of arginine. The isoelectric point is solved as follows:

j _pKa2+pKa3 _ 9.9 +13.2 _23.1 = n 552 2 2'



This makes choice D the best answer. The difficult task in these questions is notdetermining which pKa terms belong in the equation, but rather assigning anumerical value to each pKa term. If you've grown weary of continually turningback to the amino acids in Figures 7-17 through 7-21, then start making someapproximations. Knowing that the N-terminal is pK^ and its value is around 9should tell you that the pi is greater than 9. This eliminates choices A and B,which helps to some extent. From here, you have to know the values. TheMCAT test writers will provide pKa values if the question is this specific.

Example 7.26What is the isoelectric point for the dipeptide histidine-leucine?

A. 5.5

B. 6.1

C. 7.6

D. 9.2

Solution

Histidine is a basic amino acid and leucine is a hydrophobic amino acid, so theisoelectric point for the dipeptide is found by averaging pKa2 and pKa3. Aswritten, the C-terminal of histidine and the N-terminal of leucine make up thepeptide linkage, so pKa3is the N-terminal of histidine and pK^ is the side chainof histidine. The presence of the leucine does not alter the pi from what it wouldhave been with histidine alone. The isoelectric point is solved according to thefollowing mathematics:

pi _PKa2+pKa3 _6.1 +9.2 -15.3 = 7652 2 2'

This makes choice C the best answer. You should have been able to solve thisparticular question without consulting any other data. The side chain ofhistidine has a pKa close to physiological pH, so you know that the pi is anaverage of roughly 7 and 9, which means that the pi is about 8. Only choiceC isclose to that number. Another method for deterrnining the best answer is toconsider that histidine is a basicamino acid. We know that the pi is greater than7.5, but it must be less than the pKa of the amino terminal, given that pKa3corresponds to the amino terminal. Only choice C fits in the 7.5 to 9.2 range.

Example 7.27At what pH is the tripeptide serine-cysteine-isoleucine perfectly neutral?A. 5.3

B. 6.1

C. 8.4

D. 9.4

Solution

The pH at which it is perfectly neutral is the isoelectric point. Of the three aminoacids, none are basic, so the isoelectric point for the tripeptide is found byaveraging pKai and pKa2. The tripeptide has three active protons, the N-terminal of serine, the side chain of cysteine, and the C-terminal of isoleucine.The C-terrninal is always pKai and in this case pKa2 is the side chain of cysteine.The isoelectric point is found using the following mathematics:

! =pKaj_+pKa2_ _ 2.3+ 8.4 -10.7 = 5352 2 2'


OrgaillC CliemiStry Nitrogen Compounds Biological

This makes choice A the best answer. Choices C and D should have beeneliminated, because there are no basic amino acids in the tripeptide. Cysteineisactually an acidic amino acid, but because most cysteineside chains are involvedin crosslinking,we rarely consider their acid-base propertiesin proteins.

Example 7.28Whichof the following tripeptides has the LOWEST pi value?

A. Asp-Lys-AspB. Glu-Lys-GluC. Lys-Asp-LysD. Lys-Glu-Lys

Solution

The lowest pi value corresponds to the most acidic tripeptide. ChoicesA and Bhave two acidic amino acids and one basic amino acid, while choices C and Dhave two basic amino acids and one acidic amino acid. This eliminates choices Cand D. ChoicesA and Beach have one lysine,so both pis are found by averagingpKa2and pKa3- In each tripeptide, there are five active protons (the N-terminal,the C-terminal, and three the side chains). BothpK^ and pKa3correspond to theside chains of the acidic amino acids. In choice A, the acidic amino acid isaspartic acid while in choice B the acidic amino acid is glutamic acid. Asparticacid has a lower side chain pKa than glutamic acid, so the tripeptide in choice Ahas a lower pi than the tripeptide in choice B. Choice A is the best answer.

Example 7.29At pH = 5.0, histidine exists with a net charge of:

A. +2.

B. +1.

C. 0.

D. -1.

Solution

To determine the charge of an amino acid at a given pH, first determine whetherthe sites are protonated or deprotonated. When the pH is lower than the pKa,the site is protonated. When the pH is greater than the pKa, the site isdeprotonated. At pH = 5.0, histidine has a deprotonated carboxyl terminal (pKa= 1.8), a protonated side chain (pKa = 6.1), and a protonated amino terminal (pKa= 9.2). The carboxyl terminal carries a negative charge, while the side chain andamino terminal are both carry a positive charge. The overall charge is therefore apositive one. The best answer is choice B. Histidine at pH = 5.0 is drawn below:

pH <pKa H3N+ J^ pKa =1.8protonated ^syT O" pH>pKa

deprotonated

overall charge = +1 +1 -1 = +1

pKa =6.1 f NHPH<pKa \ /protonated h N1"~~^



Example 7.30At pH = 9.5, tyrosine exists with a net charge of:

A. +1

B. 0

C. -1

D. -2

Biological

Solution

At pH = 9.5, tyrosine has a deprotonated carboxyl terminal (pKa = 2.2), adeprotonated amino terminal (pKa = 9.1), and a protonated side chain (pKa =10.1). The carboxyl terminal carries a negative charge, while the amino terminalis uncharged. The side chain is uncharged when protonated, so the overallcharge is a negative one. The best answer is choice C.

Proteins

Proteins are biologicalpolymers built from amino acids. They are held togetherby covalent bonds in peptide linkages. A peptide linkage is formed when theamino terminal of one amino acid attacks the carbonyl carbon of the carboxylterminal of another amino acid, resulting in the loss of water. Because aminoacids have specific stereochemistry, proteins are highly chiral (on average, theyhave one chiral center for every amino acid in the protein). This large degree ofchirality explains why enzymes are so selective.

Protein Structural Features and Levels

Thestructure of a proteincanbe broken down into levels. The most elementarylevelis the primary structure, which is defined as the sequence (connectivity) ofamino acids within the protein. To break down the primary structure, you mustcleave the peptide bonds. Figure 7-30 shows a tetrapeptide (four- amino acidprotein).

Tetrapeptide (4 amino acid protein)

Peptide linkage (amide bond)

Figure 7-30

In this case, the primary structure is the order in which the four amino acids arearranged from the amino terminal to the carboxyl terminal. The primarystructure of a protein or enzyme is determined by sequencing the protein oneamino acid at a time. This will be discussed later. The interactions between the

amino acids within the protein are responsible for the secondary structure,which is defined as the folding of the amino acids into their natural configurationwithin the protein. The chemical reasons for special features in the secondarystructure are most often attributed to hydrogen bonding, and sometimes to cross-linking through disulfide bridges and kinks and turns caused by the presence ofproline. Someof the structural features of interest in proteins are the oc-helix and



fi-pleated sheets. In the oc-helix, two residues near one another in the protein areheld together by hydrogen bonds. There are 3.6 amino acids per turn and eachturn is 5.4 Ain length. They coil like a phone cord. In 6-pleated sheets, theamino acids strands are not coiled. They are held together by hydrogen bondsfrom thehydrogen on nitrogen to the carbonyl oxygen. Figure 7-31 shows a fi-pleated sheet with antiparallel strands

R74 H

H R„ H O

Figure 7-31

A disulfide bridge is formed when two cysteine residues lose the H from thethiol group to form a sulfur-sulfur bond. Disulfide bridging is most oftenconsidered to be part of the tertiary structure. Because loss of hydrogen resultsin an increase in the oxidation state of sulfur, this process is oxidative, so a cross-linked protein is considered more oxidized than the protein without cross-linking. To break a cystine cross-linkage, one can add a reducing agent such asfi-mercaptoethanol, HSCH2CH2OH. As a side note, cross-linkingis responsiblefor an increase in rigidity for the polymer (polypeptide in this case) due to theloss of both flexibility and entropy. In terms of tnermodynamics, the process ofcross-linking is driven by enthalpy and not entropy (recall that AG = AH- TAS).Figure 7-32 shows the reductive cleavage of a disulfide linkage using thechemical reagent S-mercaptoethanol.

Biological

H Rft H


OrgaiUC Chemistry Nitrogen Compounds Biological

Vasopressin (antidiuretic hormone)S S

I I

Figure 7-32

When any of the molecular interactions that hold together the secondary ortertiary structure of a protein are broken, the protein is said to be denatured (nolonger in its natural form). The overall folding can be changed by theenvironment in which the protein exists. In an aqueous environment, a proteinwill fold in such a way as to expose the hydrophilic side groups to the solvent(water) while nainimizing the exposure of the hydrophobic (alkyl and phenyl)side groups to the solvent (water). In a hydrophobic environment, a protein willfold in such a way as to expose the hydrophobicside groups to the solvent (lipid)while ininimizing the exposure of the hydrophilic (hydrogen bonding) sidegroups to the solvent (lipid). This overall folding of the protein is referred to asthe tertiary structure. Thetertiarystructure is globularfor many proteins.

Example 7.31Taking a protein from a Upid environment and placing it into an aqueousenvironment would most affectwhich of the following?

A. Primary structureB. Secondary structureC. Tertiary structureD. Quaternary structure

Solution

A protein in a Upid environment has its hydrophobic side chains exposed to thesolvent and its hydrophiUc side chains in the inner core. A protein in an aqueousenvironment has its hydrophilic side chains exposed to the solvent and itshydrophobic side chains in the inner core. The conversion from lipidenvironment to aqueous environment breaks apart the internal hydrogenbonding, which denatures the protein. This affects the secondary structure bychanging the structural features. The tertiary and quaternary structures are alsoaffected, but only as a consequence of changes in the secondary structure. Thisquestion calls for the most specificchange. The best answer is choice B.

Copyright ©by The BerkeleyReview 192 The Berkeley Review


Example 7.32Turns are caused by which of the following amino acids?

A. CysteineB. Histidine

C. Proline

D. Glycine

Solution

Because proline is cycUc, it cannot freelyrotate about its sigmabonds, forcing theprotein chain to turn. Choice C is the best answer. The following pictorialrepresentation shows the impact of the cycUc structure of prolineon the turn.

7N

proline residue ^^ Oturn—><*

O /0s

,»>x

Bent peptide containing proline

NH

protein turn / O^f

Blown up view of protein turn

Example 7.33Which of the foUowingresults in a denatured protein?

A. Cleaving disulfide bondsB. SpUtting base pairsC. Breaking alpha bondsD. Forming of beta bonds

Solution

Breaking disulfide bonds requires the reductive cleavage of the S-Sbond. Thisdefinitely changes the tertiary structure, thus denaturing the protein. Basepairing is involved in DNA and RNA, not proteins, so choice B is eliminated.Alpha bonds and beta bonds are throw away answers, so choices C and D areeliminated. Choice A is the best answer.

Unique Biological ProcessA common rule in the chemistry of amino acids is that they exist exclusively as L-stereoisomers in biological systems. An exception to this "aU amino acids are L"rule is found in the bacterial ceU waU. The ceU wall of a bacteriais held togetherby a net-like structure that involves a cross-link between D-alanine and glycine.As a point of biological interest, fi-lactams such as penicillin act in a medicinalfashion by breaking this linkage and destroying the ceU walls of bacteria. Again,theseare not facts to be processed and stored, but they are interestingbiologicalanecdotes that have been presented on previous MCATs. Discussing them hereis simply to provide a Uttle background, so if they show up again, there wiU be asmaUair of famiUarity to them. Figure 7-33shows the formation of the cross-linkin the bacterial cell waU.

Biological



O H O H HII I II I I

R-C— CH2-NH2+ R'-N-C-C-N-C-CO,"|l I z

Terminal Residue of the ** C**3 *-H3pentaglycine bridge Terminal D-Ala-D-Ala unit

O H H H HII III + I

R~C-CH2-N-C-C-N-R' + HoN-C-CO,"II I I 2O CH3 CH3

Gly-D-Ala crosslink D-Ala

Figure 7-33

Gel ElectrophoresisLet us considergel electrophoresis, a biochemistry lab technique based on theidea that charged particles migrate through electric fields. Particles migrateaccording to their charge(which impacts the strength of the electrical force) andtheirsize (which impacts their resistance to flow as they travel through a gel). Apolyacrylamide gel is chosen to offer resistance so that the particles do notmigrate too rapidlyand thus not have the time to separate to a distinguishableamount. Figure 7-34 shows theschematic ofan electrophoresis apparatus.

Figure 7-34

In gelelectrophoresis, cations migrate to thecathode and anions migrate to theanode. This aUows for separation ofcations from anions. To getseparation ofUke charges (cations from cations for instance), the particles must migrate atdifferent speeds. The speed depends on the acceleration, which depends ontheelectric force (F =qE) and the resistive force due to drag (Fdrag)- Equation 7.3shows the relationship ofthe particle's mass andacceleration to itscharge, q, thestrength of theelectric field, E, and theresistive force due to thedrag as it movesthrough the gel.

ma=qE-Fdrag (7.3)

The charge ofan amino acid, polypeptide, or protein fragment depends on thepH of theenvironment and the isoelectric point, pi, of the species. When pH isgreater than pi, the environment is basic, resulting in a negatively chargedspecies. When pH is less than pi, the environment is acidic, resulting in apositively charged species. The bigger the difference between the pH of theenvironment and pi of the species, the greater the magnitude of its charge. Asthe magnitude of charge increases, the electric force increases, and therefore itsacceleration increases. This means that the species can be separatedby theirpivalues. The other factor that influences the migration rate is the size of thespecies. Larger species experience greater resistance as they travel through themedium. As a result, there aretwotypes ofelectrophoresis techniques employedto separate protein fragments.


OrgaillC ChemiStry Nitrogen Compounds Biological

The first type of electrophoresis employs a pH gradient in the gel and thecharged species migrate through the gel until they reach a pH equal to their pi.At thatpoint theyhaveno net charge, so thereisno electric force. This techniqueis referred to as isoelectric focusing. The second typeof electrophoresis involvesadding sodium dodecyl sulfate, SDS, to the protein mixture. The proteinfragments incorporate SDS into their secondary structure, resulting in everyfragment takingon a negative charge. Larger protein fragments canincorporatemore SDS than smaller ones, resulting in a greater magnitude of charge.However, larger speciesare more massive. As a result, the incorporation of SDScreates an environment where aU of the protein fragments have the same mass-to-charge ratio. Given that they are experiencing the same electric field, anyseparation is due to differences in size, where small species migrate faster thanlarger species,because they experienceless resistance due to drag.

Example 7.34What is true of the protein fragment in a mixture that migrates fastest in SDSPAGE and migrates to the highest pH value in a gel with a pH gradient?

A. It is the heaviest fragment in the mixture and is most likely to containaspartic acid.

B. It is the heaviest fragment in the mixture and is most likely to contain lysine.C. It is the Ughtest fragment in the mixture and is most likely to contain aspartic

acid.

D. It is the Ughtest fragment in the mixture and is most likely to contain lysine.

Solution

In SDS PAGE, the speed at which a species migrates depends on the resistance itexperiences moving through the gel. Larger fragments experience greaterresistance, so a species that migrates rapidly through the gel must be small. Thismeans that the fastest fragment must be the Ughtest fragment, so choices A and Bare eliminated. Because the species migrated to a high pH value, it must have alarge pi value. The presence of lysine, arginine, and histidine increases the pi, sothe species must have at least one of those three amino acids. Aspartic acidlowers the pi value, so choice C is eliminated and choice D is the best answer.

Affinity ChromatographyAnother technique that can separate proteins according to their charge is affinitychromatography. We will look specifically at ion-exchange chromatography.Ion-exchange chromatography separates by the affinity of charged proteins foroppositely charged sites on the polymer in the column (stationary phase). Thestationary phase is made of an insoluble polymer, often polystyrene or ceUulose,to which functional groups capable of carrying a charge have been attached.Typical anionic groups include sulfate on sulfonated polystyrene and carboxylateon carboxymethyl-ceUulose. Typical cationic groups include diethylaminoethoxyon diethylaminoethoxy-cellulose. Positively charged columns bind anionicspecies tightly and negatively charged columns bind cationic species tightly. Aprotein carries a positive charge when the solution pH is less than the isoelectricpoint, pi, of the protein. When a mixture of proteins is added to the column andaUowed to migrate down the polymer, selected proteins can be separated fromthe mixture by binding the column. Proteins carrying the same charge as thecolumn elute. To release a bound protein from an ion-exchange column, thepolymer can be washed with a solution of varying pH or a solution of graduallyincreasing salt concentration. Both techniques for releasing the protein coulddenature it irreversibly, so the preferred technique is not always obvious.



Cutting and Sequencing (Primary Structure Determination)Determining the primary sequence of a protein is critical in biochemistry. In thesequencing of proteins, peptide bonds must be broken and component aminoacids identified. To sequence a protein, several reagents are involved.Sometimes it is best to break a large protein into smaUer, more manageablefragments. Each fragment can be sequenced and then the fragments can bereassembled to determine the overaU sequence of the original protein. Thismeans that the original protein must be marked in such a way that the first andlast fragments are easy to identify. We wiU start by determining which aminoacid is first in the protein. The identification of the first amino acid in thesequence can be accomplished by adding 2,4-dinitrofluorobenzene, 2,4-DNFB(Sanger's reagent), to the protein, foUowed by complete hydrolysis using 6 MHCl. Sanger'sreagentbinds the aminoterminalof a protein, thereby labelingthefirst amino acid. Complete acid hydrolysis of the protein then yields all of thecomponent amino acids, of which only one is marked. Figure 7-35 shows bothfree 2,4-DNFB and a complexed amino acid residue.

NO^

NOJ2

2,4-dinitrofluorobenzene

NOJ2

Labeled amino acid residue

Figure 7-35

Sanger's reagent serves tomark thefirst amino acid ina protein and toultimatelyidentify it, but that's just the beginning when sequencing a large proteinwithseveral amino acids. Tobreak the protein into smaUer fragments, biochemistsmimic whatnature does. We employ enzymes that break a large protein intosmaUer fragments by cleaving proteins at specific sites (bonds). The enzymechymotrypsin, which is active in the small intestine, cleaves an amino acid chainat the peptide linkage on the carboxyl side of phenylalanine, tyrosine, ortryptophan. As a point of biological interest, the pancreas actually releaseschymotrypsinogen, an inactive species that gets activated once in the smallintestine. Chymotrypsin wouldcleave the twelve aminoacid protein,Arg-Cys-Gly-Ala-Phe-Thr-Met-Ala-Tyr-Cys-His-Leu, twice, leaving three smallerfragments. The three pieces are: Arg-Cys-Gly-Ala-Phe, Thr-Met-Ala-Tyr, andCys-His-Leu. Reassembling the three fragments canbe doneif the firstfragmentis known from the 2,4-DNFB result. Table 7-1 shows various enzymes andchemical reagents used to break peptidelinkages, their specific activity, and forthe digestive enzymes, their corresponding zymogen and sites of release andactivity.



Enzyme or Agent Cleavage activity Site (if applicable)

Chymotrypsin(digestive enzyme)

C-side of Phe, Trp, and Tyr Released as chymotrypsinogenfrom pancreas into smaU intestine

Clostripain(enzyme)

C-side of Arg

Cyanogen bromide(chemical reagent)

C-side of Met

O-Iodosobenzoate

(chemical reagent)C-side of Trp

Hydroxylarnine(chemical reagent)

Asp—Gly bonds

2-Nitro-5-cyanobenzoate(chemical reagent)

N-side of Cys

Pepsin(digestive enzyme)

C-side of Asp, Glu, and Leu Released as pepsinogen from chiefcells into stomach

Staphylococcal protease(digestive enzyme)

C-side of Asp and Glu (Glu onlyunder specific conditions)

Thermolysin(enzyme)

N-side of Leu, Val, and Ue

Trypsin(digestive enzyme)

C-side of Lys and Arg Released as trypsinogen frompancreas into smaU intestine

Table 7-1

Sequencing is often done in multiple steps. Some methods involve randomhydrolysis foUowed by sequencing of the fragments and analysis of the overlapof information between fragments. There is also a systematic approach wherethe protein is broken down at specific linkages using different reagents and theoverlap between fragments is again analyzed. Consider the foUowing example.

A twenty-seven amino acid protein is treated with various reagents in fourseparate experiments. The results of each experiment are Usted below.1. Treatment of the protein with 2,4-dinitrofluorobenzene and then 6M HCl

yields a histidine residue bonded to 2,4-dinitrobenzene.

2. Treatment of the protein with chymotrypsin yields the followingfragments:a) N-His-Val-Ser-Gly-Ala-Ile-Phe-Cb) N-Leu-His-Gly-Thr-Tyr-Cc) N-Val-Asp-Arg-Cys-Ala-Leu-Cd) N-Val-His-Lys-Glu-Trp-Ce) N-Cys-Ue-Met-Phe-C

3. Treatment of the protein with trypsin yields 3 fragments that are 19amino acids, 5 amino acids, and 3 amino acids in length.

4. Treatment of the protein with thermolysin yields a free histidine residueand a free leucine residue along with the following six fragments:a) N-Ue-Phe-Cb) N-Leu-His-Gly-Thr-Tyr-Cys-Cc) N-Ile-Met-Phe-Cd) N-Val-His-Lys-Glu-Trp-Ce) N-Val-Asp-Arg-Cys-Ala-Cf) N-Val-Ser-Gly-Ala-C

Biological



From the first experiment, the conclusion you should draw is that histidine is thefirst amino acid in the protein, because of the reaction with Sanger's reagent.From the second experiment, we can conclude that leucine is the last amino acidin the protein, because leucine is the only carboxyl terminal amino acid in one ofthe five fragment that would not have been fragmented by chymotrypsin.Combining the conclusion of the first experiment with fragments in the secondexperiment, we know that amino acids 1-7are His, Val, Ser, Gly, Ala, He, and Phesequentially. We also know that amino acids 22-27are Val, Asp, Arg, Cys, Ala,and Leu sequentiaUy. The free leucine from Experiment4 further supports thatleucine is the last amino acid in the sequence. From this point, we wiU label eachfragment from Experiment 2 A through E and use information from the otherexperiments to evaluate the overlap of information and determine the position ofeach amino acid. Thefive fragments from Experiment2 are arbitrarily assignedthe letters A through E as follows:

Fragment A: N-His-Val-Ser-Gly-Ala-Ue-Phe-CFragment B: N-Leu-His-Gly-Thr-Tyr-CFragment C: N-Val-Asp-Arg-Cys-Ala-Leu-CFragment D: N-Val-His-Lys-Glu-Trp-CFragment E: N-Cys-Ile-Met-Phe-C

Because we knowthefirst and lastaminoacids, we know that fragment Ais first,and fragment C is last. This leaves sixpossible combinations for the fragmentsproduced in Experiment 2. The possibiUties are: A-B-D-E-C, A-B-E-D-C, A-D-B-E-C, A-D-E-B-C, A-E-B-D-C, and A-E-D-B-C. From Experiment3, we know thatthe 3-AA fragment must be Cys-Ala-Leu, the last three amino acids. Trypsincleaves the C-sideof Lysand Arg, so the Lys—Glu bond is broken. This leads usto conclude that fragment D must be the fourth fragment, because the fiveaminoacid fragment from Experiment 3 must end with arginine. This narrows thechoices to two: A-B-E-D-C and A-E-B-D-C. From Experiment 4, it canbe inferredthat Phefrom fragment Amustbe bonded at its carboxyl site to eitherLeu, Val,or He. This means that fragment E cannot be second in the overaU sequence,which narrows the answer down to A-B-E-D-C. This makes theoriginal protein:N-His-Val-Ser-Gly-Ala-Ue-Phe-Leu-His-Gly-Thr-Tyr-Cys-Ile-Met-Phe-Val-His-Lys-Glu-Trp-Val-Asp-Arg-Cys-Ala-Leu. Be sure you work through the logic ofsequencing.



Edman's ReagentTo sequence the amino acids in a small polypeptide (twenty to fifty amino acidsmaximum, depending on the amino acids), biochemists employ Edman'sreagent, phenyUsothiocyanate (HsC6N=C=S). This reagent sequentially removeseach amino acid one at a time starting from the amino terminal. This leads to theamino acid sequencing equipment used in most biology labs. Edman's reagentand its reactivity is shown in Figure 7-36.

,N=C=S.

aN=C=S

PhenyUsothiocyanate

Peptide +

NH-peptide

ay. O-

s

11

/ NHN

R

HoN

I

K^ H2N

protontransfer

H R

NH-peptide

OH

Q4.."•HN—-r^

H<Q/ \peptide-HN^\^x ™>

^ + H RH S: (^H

DNH-peptide ;;5=

Example 7.35Hydrolysis of a peptide results in the cleavage of:

A. the carbonyl group.B. a carbon-carbon bond.

C the carboxyl terminal.D. the amide bonds.

Solution

Polypeptides are held together by peptide linkages, which are amide bondsbetween amino acids. To break apart a polypeptide, the amide linkage must besevered. This makes choice D the best answer.

Biological



Example 7.362,4-Dinitrofluorobenzene reacts with the:

A. nitrogen of the amino terminal.B. carbon of the carboxylterminal.C. nitrogen of the amide bonds.D. carbon of the side chains.

Solution

2,4-Dinitrofluorobenzene, known as Sanger's reagent, is an electrophUe thatreacts with the nitrogen of the amino terminal of a protein by way of anucleophUic aromatic substitution reaction. The nitrogen of the amino terminalattacks the benzene ringnucleophilically to displace the fluorine on the aromaticring in a two-step reaction. The best answer is choice A.

Example 7.37The tripeptide Lys-Arg-Ser would NOT:

A. have an overaU positive chargeat pH = 7.B. be water-soluble.

C. from any disulfide bonds.D. havean isoelectric pointgreater thanAsp-Glu-Thr.

Solution

The tripeptide has an amino terminal on lysine (pKa = 9.2), a carboxyl terminalon serine (pKa = 2.2), an ammonium side chain on lysine (pKa = 10.8), and aguanidinium sidechain on arginine (pKa =13.2). At pH = 7, the aminoterminalisprotonated and thus positively charged, the carboxyl terminal is deprotonatedand thus negatively charged, and the two basic side chains are protonated andthus each are positively charged. The overaU charge is therefore positive two.This means that "The tripeptide Lys-Arg-Ser would not have an overall positivecharge at pH = 7"is an invalid statement, so choice A is eliminated. Because allof the sites on the tripeptide are charged, the compound is ionic and thereforehighly water-soluble. This means that"The tripeptide Lys-Arg-Ser would not bewater-soluble" is an invaUd statement, so choice B is eliminated. Because neitherlysine, arginine, norserine contain thiol groups (only cysteine does) there canbeno disulfide cross-Unking associated with this tripeptide. This means that "Thetripeptide Lys-Arg-Ser would not have any disulfide bonds formed" is a vaUdstatement, sochoice Cis thebestanswer. The isoelectric pointfor the tripeptideLys-Arg-Ser is an average of pKa3 and pK^ (the twoside chain pKa values forlysine andarginine), while the isoelectric point for the tripeptide Asp-Glu-Thr isan average ofpKai and pK^ (the carboxyl terminal pKa valueand the sidechainpKa value for aspartic acid). The average ofpKai and pKa2 forAsp-Glu-Thr is alower number than the average ofpKa3 and pKa4 for Lys-Arg-Ser. This meansthat"The tripeptide Lys-Arg-Ser would not havean isoelectric pointgreater thanAsp-Glu-Thr" is an invalid statement, so choice D is eliminated. Basic aminoacids (lysine, arginine, and histidine) always increase the isoelectric point whUeacidic amino acids (aspartic acid and glutamic acid) always decrease theisoelectric point. This question is difficult, because the language is cumbersome.Making use of denotations next to each answer choice like "False,because it ispositive," may prove highly beneficial. You need to not only recaU the subjectmatter when working with these books, but youmust also honeyour test-takingskills.


Organic Chemistry Nitrogen Compounds Section Summary

Key Points for Nitrogen Compounds (Section 7)

Non-biological Nitrogen-Containing Compounds

1. Amines (Contains a central nitrogen with either alkyl groups or hydrogensattached)

a) Basicity (Amines are weak bases)i. The pKb for amines is between 3 and 5, the pKa of protonated

amines is between 9 and 11

U. Typically,2° > 3°~ 1°,because of steric hindrance and H-bondingin. Common reagent in buffers with pH of 10± 1

b) NucleophiUcityi. Good nucleophUes for Sjvj2 reactionsii. Can undergo multiple additions when treatedwith alkylhalides

c) Amines can be synthesized in many ways

i. Gabriel synthesis using alkylation of a phthaUmide followed byhydrolysis

U. Reduction of azides, nitriles, imines, and amidesui. Hofmann rearrangement of an amide

d) Amine reactivity is based on nucleophiUc substitutioni. Amines readily displace a leaving group from alkylhaUdesii. Amines react with nitrite to form diazonium saltsUi. Quaternary amines can undergo Hofmann elimination to yield

terminal alkenes where three of the alkyl groups are methyl groups

2. Imines have nitrogendoublebonded to a central carbon (R2C=NR')a) They reactin a simUar fashion to aldehydes and ketones

i. They are in equUibriumwith enaminesU. Thecarbonis electrophilic and canbe attacked bynucleophilesUi. Oximes and hydrazones form in a similarway as imines

3. Amides have nitrogenbonded to a carbonyl carbon (RC=ONR')a) Amides are neither acidic nor basic

i. Amides are formed from substitution reactions where an amineattacks an ester, anhydride, or acid haUde

u. Amides are also formed from an oxime via Beckmann rearrangementUi. Amide bonds are referred to as peptide bonds when formed between

amino acids

4. Aminoacidscan be synthesizedby severalmethods in vitroa) The alpha carbon has four things attached, an H, CO2H, NH2, and R-

group, sosynthesis routes center around adding these groups tocarboni. Hell-Volhard-Zelinskii synthesis activates the alpha-carbon of a

carboxyUc acid and then adds an amineU. Strecker synthesis adds a cyano group to an imine and then

undergoes hydrolysisUi. Analpha keto acid can beconverted into analpha imino acid, which

is reduced into an amino acidiv. Gabriel synthesis, which is used to synthesize a primary amine can

make an amino acid, where the phthalimide is added to an alkylhaUde with a carboxyUc acid group rather than an alkylhalide


OrgaillC ChemiStiy Nitrogen Compounds Section Summary

Biological Nitrogen-Containing Compounds

1. Amino acids havean amine group on the alpha carbonofa carboxyUc acida) They areclassified according to theirR-group

i. Hydrophobic amino acidshave alkyl side chainsU. Semi-hydrophobic amino acids havefunctional groupsthat are only

slightly water solubleUi. HydrophUic amino acids have highly water soluble side chains,

usuaUy alcohols and amines

b) Side chains can be acidic, basic, or neutral

i. Acidic amino acids contain oxygen, losea proton from their neutralstate, and have isoelectric points less than 5.5

U. Basic amino acids contain nitrogen, gaina proton from theirneutralstate,and have isoelectric points greater than 7.5

Ui. Neutral amino acids do not gain or lose a proton from their neutralstate and they have isoelectric points between 5.5and 7.5

c) Isoelectric points, the pHwhere the charge is0, are found byaveragingthe two pKasleading to and from the zwitterioni. pi is an average ofpKai and pKa2 for all amino acids except Lys,

Arg, His, because 'his lys arebasic." For Lys, Arg, andHis, pi isanaverage of pK^ and pKa3.

U. The piofaprotein isfound easUy byadding the number ofLys, Arg,and His plus one for the N-terminal, which gives the pKa termleading to the zwitterion

Ui. When the pHofthe environment exceeds pi, thespecies isananioniv. When pi exceeds the pHofthe environment, thespecies isancation

2. Proteins are polymers of amino acidsa) Primary structure is the amino acid sequence and peptide linkagesb) Secondary structure involves the interactions ofnearby amino acids

i. An alpha-helix isacoU inthe backbone with 3.6 amino acids perturnU. Beta-pleated sheets involve H-bonding ofamides ofnearby amino

acid moieties

c) Tertiarystructure involvesside chain interactionsof amino acids that arefar apart from one another

i. Beta-pleated sheets involve hydrogen bonding of amides of nearbyamino acid moieties

d) Quaternary structure involves interactions between two differentprimary strands

3. Protein lab techniques center around separation and sequencinga) Gel Electrophoresis

i. Anions migrate toanode andcations migrate tocathode in an EfieldU. Buffered gel gets separation by pi: migrates until pi=pHofgelUi. SDS-PAGE separates bysize: bigger fragments migrate slower

b) Sequencing involves analyzing the primary structure ofa proteini. First treated with urea tobreak H-bonds andfi-mercaptoethanolU. Edman's reagent, Ph-N=C=S,cuts one AA at a time from N-terminalUi. Enzymes can be used to break large proteins into smaller, more

manageable fragments whichcan be sequencedand reassembled


Nitrogen

Compounds

Passages13 Passages

100 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, V, VIII, & XIIGrade passages immediately after completion and log your mistakes.

II: FollowingTask I: Passages II, IV, IX, & XI (30 questions in 38 minutes)Time yourself accurately, grade your answers, and review mistakes.

Ill: Review: Passages III, VI, VII, X, XIII, & Questions 97-100Focus on reviewing the concepts. Do not worry about timing.

^V-

R.E-V-I-E*\AT


Amines and Amino Acids Study Passages

I. Basicity of Amines

II. Amine Basicity and nucleophilicity

III. Effects of Alkyl Groups on Amine nucleophilicity

IV. Imine and Enamine Chemistry

V. Amino Acid Synthesis in vitro

VI. Amino Acid Trivia

VII. Acidity and Basicity of Amino Acids

VIII. Amino Acids and Polypeptides

IX. Physical Properties of Amino Acids

X. Protein Hormones

XI. Enzymatic Cleavage and Sequencing

XII. Peptide Sequencing

XIII. Amino Acid Stereospecificity in Synthesis


Amines and Amino Acids Scoring Scale

Raw Score

84 - 100

66 - 83

47 - 65

34 - 46

1 - 33

MCAT Score

13- 15

10 - 12

7 -9

4-6

1 -3

(1 -7)

(8 - 14)

(15 - 20)

(21 - 27)

(28 - 35)

(36 - 43)

(44 - 51)

(52 - 58)

(59 - 66)

(67 - 73)

(74-81)

(82 - 88)

(89 - 96)

(97 - 100)


The trend for the relative basicity of amines in water is:secondary (2°) > primary (1*) = tertiary (3°) > ammonia. Theorder is explained by competing influences in water. Theelectron donating effect of the alkyl groups supports arelationship of 3" >2° > 1° > ammonia. This prediction issupported by the gas phase basicity of the amines and amineacid/base reactions conducted in an aprotic solvent. Reducedhydrogen bonding caused by the steric hindrance of alkylgroups supports a relationship of ammonia > 1° > 2° > 3°.Differences in hydrogen bonding capacity are demonstrated bythe increasing sharpness of the N-H absorbance seen ininfrared spectroscopy. The sharpest peak is observed with thesecondary amine and gets progressively more broad withprimary amines and ammonia. The two opposing trendscompete to give the observed trend in water. Table 1 showsthe pKa values for some alkyl ammonium chloride salts.

Compound pKa

NH4C1 9.26

(H3Q3NHCI 9.79

H3CNH3CI 10.65

H3CH2CNH3CI 10.71

(H3Q2NH2CI 10.73

(H3CH2Q3NHCI 10.75

(H3CH2Q2NH2CI 10.95

Table 1

The pKb for the conjugate base of any given acid can befound by using the equation pKb = 14 - pKa. This can beuseful for determining whether a compound is charged oruncharged at a given pH value in an aqueous solution.Figure 1 shows the structure of a diamine at varied pH.

H H

W ©NH3

H H

W,N©

H

I.N.

Etx ^^ EtX ^^ Et^ ^^pH = 8.8 pH=10.8 pH=12.8

Figure 1 Diamine at varying pH values

1. In the gas phase, which of the following ammoniumcations is the MOST acidic?

A. NH4+

B. H3CH2CNH3+

C. (H3CH2C)2NH2+

D. (H3CH2C)3NH+

2. What is the approximate pKb of CF3CH2NH2?

A. pKb> 10.71

B. 10.71 >pKb>7

C. 7£pKb>3.29

D. 3.29 >pKb


3. What can be determined about the equilibrium constant(K) for the following acid/base reaction?

H3CNH3+ +NH3 ^ H3CNH2 +NH4+A. K > 1 in both water and gas.B. K > 1 in water; K < 1 in gas.C. K < 1 in water; K > 1 in gas.D. K < 1 in both water and gas.

4. Which of the following amines is the MOST basic?

A. NH3

B. H3CNH2

C. (H3Q2NH

D. (H3Q3N

5. Charged compounds are found to be more soluble inwater than neutral compounds. At which of thefollowing pH values would a tertiary amine be MOSTsoluble in water?

A. 3

B. 5

C. 9

D. 11

6. The pKa for carboxylic acids is between 2 and 5. In apH = 7 buffered aqueous solution, lysine exists inwhich of the following forms?

7.

A. o

H2N,

C.

H3NH

•xH (CH2)4

NH2

O

H (CH2)4

NH2

B. o

H2N\Ao-H (CH2)4

+NH3

D. o

H3N+\Ao-H (CH2)4

+NH3

Why is there no mention in the passage of thebroadness of the IR absorbance of an N-H bond of a

tertiary amine?

A. Steric hindrance hinders N-H bond stretching.B. There is no N-H bond on a tertiary amine.

C. Resonance shifts the absorbance to 3000 cm"1.D. Symmetric molecules exhibit no IR absorbances.

GO ON TO THE NEXT PAGE.


Amines are classified as weak bases, compounds thatpartially hydrolyze water to give OH" and its conjugate acid.Amines are one of the most common types of organic bases.The relative basicity of various amines can be predicted bycomparing the substituents bonded to the nitrogen of theamine. Electron donating groups, when bonded to nitrogen,increase the basicity of the amine. This is to say that thebasicity is directly dependenton the groups attached to theamine. Amines obey standard rules for aqueous phase acid-base chemistry. Reaction 1 is a proton transfer reactioninvolving a primary ammonium cation and ammonia. Theequilibrium for Reaction 1 as written lies to the left.

H3CNH3+ +NH3 « ^ H3CNH2 +NH4+Reaction 1

The equilibrium constant for Reaction 1 is less than 1,because themethyl amine (on theproduct side) is a strongerbase than ammonia (on the reactant side). The equilibriumthus favors the protonation of methyl amine. By knowingthe direction that an equilibrium lies, it is possible todetermine the relative basicity of two compounds bothqualitatively andquantitatively. ThepKa value foranacid (orthe pKb value fora base) canbederived from theequilibriumconstant (Keq) for thereaction and thepKa of theother acid.Table 1 shows the equilibrium constants for a series of acid-base reactions with ammonium, NH4+. All reactions werecarried out at 35°C.

8.

Reactant Base K^ forreaction with NH4H

NH3 1.00

H3CNH2 25.7

H3CH2CNH2 29.5

(H3Q2NH 30.9

(H3CH2Q2NH 50.1

(H3Q3N 3.55

(H3CH2Q3N 32.4

Table 1

Which of the following compounds is the MOST basic?A. H3CNH2

B. H3CCH2NH2C. (H3Q2NH

D. (H3CCH2)2NH

9. Which ofthe following reactions has aKeq of0.030?A. (H3C)3NH+ + NH3 -^=ir (H3C)3N +NH/B. (H3C)3N + NH4+ ^-^ (H3C)3NH++ NH3C. (H3C)2NH2+ + NH3 ^=±^ (H3C)2NH + NH/D. (H3C)2NH + NH4+ ^=i= (H3C)2NH2+ + NH3


10. To separate an amine from an amide, it would be best touse which of the following for extraction?

A. Ether and water at pH = 3B. Ether and water at pH = 7C. Ether and water at pH = 11D. Amides cannot be separated from amines.

11. Which of the following statements is true about therelative basicity of aminescompared to amides?

A. Amines are more basic than amides, because amideshave their lone pair tied up in resonance.

B. Amines are less basic than amides, because amideshave their lone pair tied up in resonance.

C. Amines are more basic than amides, because amidesare more sterically hindered.

D. Amines are less basic than amides, because amidesare more sterically hindered.

12. The highest boiling point is associated with which ofthe following amine compounds?

A. NH3

B. H3CNH2

C. H3CCH2CH2NH2D. (H3Q3N

13. The pKa of trimethyl ammonium chloride,(H3Q3NHCI, is 9.8. Benzene is electron withdrawingby resonance when the substituent has a lone pair ofelectrons, so the pKb of aniline, H5C6NH2, CANNOTbe whichof the following?

A. 3.6

B. 4.8

C. 9.2

D. 10.4

14. Which of the following changes to a primary aminewill NOT result in acompound that yields a lower Keqwhen undergoing proton transfer with ammonium?

A. Replace a hydrogen on the carbon backbone withfluorine.

B. Replace the hydrogens on the carbon bonded tonitrogen witha double bond to oxygen.

C. Replace one of the hydrogens on nitrogen with asecond alkyl group.

D. Replace the alkyl group on nitrogen with ahydrogen.



The nucleophilicity of an amine closely parallels itsbasicity, where as the base strength of an amine increases, itsnucleophilicity also increases. This holds true as a generaltrend with most deviations attributed to steric factors. Table1shows thepKb and log Keq foran SN2-reaction with ethylchloride, H3CCH2CI, for a series of amines.

Amine pKb log Keq

NH3 4.7 1.8

H3CNH2 3.4 2.6

(H3Q2NH 3.2 3.1

(H3Q3N 4.2 0.6

Table 1

One problem with amines as nucleophiles is that theyare capable of undergoing multiple additions. It is difficult toisolate primary amines. To do so, an excess of amine is usedin the reaction. When there is excess electrophile, thereactionproceeds readily to secondary and tertiary amines. Tosynthesize a primary amine, methods other than the reactionof ammonia with an alkyl halide are employed. Figure 1shows alternative routes for synthesizing a primary amine.

Method 1: Gabriel Phthalimide Synthesis

1. O

NH32. NaOH(aq)3. RBr

4. NaOH(aq)/A

•• RNH2

Method 2: Reduction of Nitrogen Compounds

H2(g)/cat.RC""N

RCH=NH

OII

RC-NH2

RCH2N3

H2(g)/Ni

C2H5OH

UAIH4

ether

H2NNH2

Pd/CH3OH

Figure 1 Synthetic routes to form a primary amine

15. Reduction of all of the following types of compoundswill yield a primary amine, EXCEPT:

A. a nitrile.

B. a primary imine.C. a secondary amide.D. an oxime


RCH2NH2

RCH2NH2

RCH2NH2

RCH2NH2

207

16. The IR absorbance for an N-H bond is observed at

approximately:

A. 3200cm"1.

B. 2700cm"1.

C. 1600 cm*1.

D. 1300 cm"1.

17. Which of the following statements BEST explains whymethyl amine is a stronger nucleophile than ammonia?

A. The methyl group is electron withdrawing by theinductive effect, making nitrogen electron poor.

B. The methyl group is electron donating by theinductive effect, making nitrogen electron rich.

C. The methyl group is electron withdrawing byresonance, making nitrogen electron poor.

D. The methyl group is electron donating byresonance, making nitrogen electron rich.

18. Methyl amine, when added to (R)-2-chlorobutane, wouldgive which of the following major organic products?

A. 1-butene

B. 2-butene

C. (R)-2-butyl methyl amineD. (S)-2-butyl methyl amine

19. Which of the following labeled nitrogen atoms is theMOST nucleophilic?

A. Nitrogen aB. Nitrogen bC. Nitrogen cD. Nitrogen d

CN*^H

20. Methyl amine can react with all of the followingmolecules EXCEPT:

A. OB. O

X XH3C^X1 H3C*^ÔCH3

C. D.

H3C Cl.O.

H3C CH3



When ketones are exposed to amines (such as ammoniaor alkyl amines), they can be converted into an imine andwater. In the presence of a small amount of acid, conversionof roughly 99% of the ketone into amine occurs in about anhour for most amines. The reaction has an equilibriumconstant close to one, so the reaction is driven by either theaddition of excess reactant or theremoval of a product. Thereaction proceeds with both primary and secondary amines (inaddition to ammonia), but the secondary aminecannotform aneutral imine. Secondary amines instead form compoundsknown as enamines, rather than form a cationic imine.Figure 1 shows the reactions of a ketone with ammonia, aprimary amine, and a secondary amine respectively.

O

o

R"

+ NH3

+ RNH2

H

N

R"

,RN

XO

XR^©^R

+ N+R2NH "^=^ II

R" R,X^R"Figure 1 Ketone to imine conversions

+ H20

+ H20

+ H20

The ketone can be regenerated by adding water to theimine, as shown by the reverse reaction of the equilibrium.The imine cation formed from the addition ofa secondaryamine to a ketone can undergo the following syntheticallyusefulreaction(Figure 2). •

R^©^RN

ACH2R'

R

CHR'

OH^

R"—X

Rv© RN

R

H+(aq)

ACHR'

IR" R"

Figure 2 Ketone synthesis via an enamine intermediate

Copyright © byTheBerkeley Review®

N

R ^CHR'

RN©^RN

AR CHR'

R"

R

O

ACHR'

208

21. Which of the following conversions CANNOT becarriedout via the enaminesynthesisroute?

A. 3-pentanone into2-methyl-3-pentanoneB. Butanalinto 2-ethylhexanalC. 3-hexanone into 3-heptanoneD. Acetone into 2-octanone

22. What is the product forthefollowing reaction?

H+

X) + H2NCH2CH3

A. B.

H3CH2CN^^x\ H3CH2CHN

xoC. D.

X)

"O(H3CH2C)2N(

©

X)23. What is the product for the following reaction when

carried out under acidic conditions?

•^CO + H20

A. B.

H2N ' ^^ H3CHN^^H3CHN

OC.

H3CH2C

H2N

oD.

24. What is theproduct of thefollowing reaction?

A. A ketoneand a primary amine.B. A ketone anda secondary amine.C. An aldehyde and a primary amine.D. An aldehyde anda secondary amine.


25. To convert 3-pentanone into 4-methyl-3-heptanone,which of the following sequences of reagents should beused?

A. l.R3NH+ 2. H3CCH2I 3.H30+(aq)

B. l.R2NH/H+ 2.H3CCH2I 3.H30+(aq)

C. l.R3NH+ 2. H3CCH2CH2I 3. H30+(aq)

D. 1. R2NH/H+ 2. H3CCH2CH2I 3. H30+(aq)

26. Treatment of an imine with a reducing agent is similarto treatment of a ketone with a reducing agent. What isthe product formed when the imine produced from thereaction of a ketone and primary amine is reduced usingUAIH4 in ether?

A. A primary amideB. A primary amineC. A secondary amine

D. A tertiary amine

2 7. What are the products formed when N-ethyl-3-pentimineis treated with acidic water?

A. Ethanol and 3-pentanone.

B. Ethyl amine and 3-pentanone.C. Ammonia, ethanol, and 3-pentanone.

D. Diethyl ether, ammonia, and 3-pentanone.



There are many naturally occurring amino acids (formallyknown as 2-aminocarboxylic acids) that can be isolated frombiological systems. Amino acids vary only in the alkylgroup attached to carbon 2. Our body can synthesize someamino acids, but others must be consumed in our diet.

Amino acids can be synthesized in vitro using one of threecommon methods. The first method involves the

substitution of the amino group for a bromine functionalgroup on the carbon 2 of an alpha bromo carboxylic acid.This method is known as the Hell-Volhard-Zelinskii

synthesis, and it is shown in Figure 1 below.

O O

R ~x l.Br2/PBr3

OH 2- "2°R

OH

R

O

Br

O

2eqNH3R

OH ^T ~0_ +NH4Br

Br NH3+

Figure 1 Hell-Volhard-Zelinskii synthesis of an amino acid

The second method for synthesizing amino acids in vitroinvolves the addition of a 2-bromodiester to a dicarboxylicimide(Gabriel's phthalimide). The product of this reaction isthen heated in the presence of strong acid in water, whichresults in hydrolysis of both ester sites to form a diacid. Thedicarboxylic imide is completely hydrolyzed under theseconditions, to yield an alkyl ammonium cation. Afterenough timeat an elevated temperature, the diacid undergoesdecarboxylation to form a monoacid, which in this case is anamino acid. This method is known as the Gabriel synthesisand it is drawn in Figure 2 below.

O O

OEt

O O

•OKo o

OEt

OEt

O

°* H+(aq) + 7"0H—T*- H3N+—(

OEt OH

O

Figure 2 Gabriel synthesis of an amino acid


The last method involves the addition of ammonia to an

aldehyde to form an imine. This is followed by the additionof a nitrile anion (C=N") to the imine under acidic conditions.The product is then treated with strong aqueous acid at hightemperature to hydrolyze the cyano group into a carboxylicacid functional group. This synthesis method is known asthe Strecker synthesisand is drawn in Figure 3 below.

•\l.NH3 H3N >—OH

2. NaCN/HCNAHR

3. H30+(aq) /A KR

Figure 3 Strecker synthesis of an amino acid

28. In infrared spectroscopy, the O-H absorbances for D-serineappear at which of the following values?

A. 3200 cm"1 and 3400 cm"1

B. 2700 cm"1 and 3400 cm"1

C. 1700 cm'1 and 3400 cm"1

D. 1700 cm"l and 2700 cm'l

29. Which of the following sequence of reagents wouldyield phenylalanine when added to2-phenylethanal?

2-phenylethanal

A. 1. Cr03/H+(aq)B. 1. NaCN

C. 1. NH3D. 1. 2-aminoethanal

\^ H NH2Phenylalanine

2. NH3

2. H+(aq)/A2. NaCN

2. H+(aq)/A3. H+(aq)/A

30. The following 1HNMR was run in CDCI3 on an aminoacid synthesized by the Gabriel method. Which of thefollowing amino acids could it be?

10

A. Glycine(R = H)B. Alanine (R = CH3)C. Serine (R = CH2OH)D. Phenylalanine (R = CH2C6H5)


31. What is the side product formed during the Gabrielsynthesis of valine?

C0C« "O:CN

CN

3 2. How many units of unsaturation are there in an aminoacid with an aliphatic side chain?

A. 0 units of unsaturation

B. 1 unit of unsaturation

C. 3 units of unsaturation

D. 4 units of unsaturation

33. How many signals do you expect in the •HNMRspectrum for L-cysteine, with a side chain of CH2SH,obtained in a solution with deuterochloroform solvent?

A. 2

B. 3

C. 4

D. 5

34. The pKa for the side chain of histidine in aqueousmedium is 6.1. What is true of its pKa in ahydrophobic environment?

A. It is the same, because log functions cannot bealtered.

B. It would be increased, because the proton leavesmore readily to make the species neutral.

C. It would be decreased, because the proton leavesmore readily to make the species negativelycharged.

D. It would be decreased, because the proton leavesmore readily to make the species neutral.

35. Which of the following amino acids, starting from itsfully protonated state, hasthree equivalence points whentitrated by a strong base?

A. GlycineB. Alanine

C. Leucine

D. Tyrosine



In adult humans, there are nine essential amino acids.

The term essential is applied to amino acids that humansmust obtain through diet, because they are not synthesized inthe cytosol. The essential amino acids include the aromaticamino acids phenylalanine, tryptophan, and histidine, andsome hydrophobic amino acids like valine, leucine, andisoleucine. The other three essential amino acids are

methionine, threonine, and arginine. Of the essential aminoacids, only histidine, threonine and arginine are hydrophilic.The water-solubility of histidine increases at a low enoughpH to protonate the side chain and form a cation.

In neutral water, the carboxylic acid functional group isable to deprotonate and become anionic, while the aminofunctional group gets protonated and becomes cationic. Theamino and carboxyl terminals of all amino acids are chargedin an aqueous environment at or near a pH of 7, causing mostamino acids to exist as zwitterions in neutral aqueoussolution. A zwitterion is the form of the amino acid in

which the overall charge is zero, but the molecule has chargedsites on it. The exceptions to the zwitterion at pH = 7 ruleare the highly acidic and highly basic side chains.

All naturally occurring amino acids have the samechirality in vivo. The natural form for humans is the Inform, which is a result of an S chiral center at carbon numbertwo of the amino acid, with the exception of cysteine.Carbon number two is often referred to as the alpha carbon.Figure 1 shows the generic structure for a typical L-aminoacid where the R refers to the side chain of the amino acid:

o

IIprotonated H^.^ ^y^%

amino terminal C

AH R

L-Amino acid

Figure 1 Generic Amino Acid in Water

Six of the essential amino acids have a hydrophobic sidechain. A hydrophobic side chain reduces the water-solubilityof the amino acid. The amino acid still exists as a zwitterion

in water, as long as the pH is within two units of the aminoacid's isoelectric point. The isoelectric point, pi, can befound by averaging the two pKa values for the amino acidthat include the zwitterion in their reaction. For instance, thepi of glycine is found by averaging pKai and pKa2, becausepKai involves zwitterion formation and pKa2 involveszwitterion consumption.

3 6. There is at least one essential amino acid with each of

the following properties EXCEPT:

A. an aromatic ring.B. a hydrophilic side chain.C. an R-steroecenter at the a-carbon.

D. a cationic side chain at pH = 6.


deprotonatedcarboxyl terminal

*0"

211

3 7. Removing a polypeptide from a lipid environment andplacing it in an aqueous environment would:

A. alter its primary structure by cleaving covalentbonds.

B. alter its primary structure by cleaving hydrogenbonds.

C. alter its secondary structure by cleaving covalentbonds.

D. alter its secondary structure by cleaving hydrogenbonds.

3 8. Before determining the amino acid sequence of a largeprotein, an enzyme is often added, so that the protein is:

A. denatured and turned into its straight chain form.B. cleaved at selected amide bonds and broken down

into smaller fragments.C. coupled with an identical protein in the reverse

order and then transcripted.D. sliced into a cross-section and mounted on a slide

for a better view of the component amino acids.

3 9. Threonine, shown below, has what chirality?

O HO H

A. 2S, 3S

B. 2R, 3S

C. 2S, 3R

D. 2R, 3R

HO CH3

H2N H

40. At a pH of 7.4, the structure of phenylalanine is BESTdescribed as:

A. uncharged.B. an anion.

C. a cation.

D. a zwitterion.

41. What pH would result if you were to mix 10 mL offully protonated alanine with 10 mL of NaOH(aq)solution, where the concentration of the base solution is1.5 times greater than the concentration of the alaninesolution?

A. 2.35

B. 6.11

C. 7.00

D. 9.87


4 2. What is the isoelectricpoint for isoleucine?

pKal=2.3 pKa2 = 9.7

A. 2.4

B. 6.0

C. 7.0

D. 9.7

43. Which of the following amino acids would generate thefollowing titration curve when titrated with 0.10 MNaOH(aq) starting from its fully protonated state?

1.0 2.0 3.0

Equivalents of 0.10 M NaOH(aq) added

A. PhenylalanineB. GlycineC. LysineD. Aspartic acid


Passage VII (Questions 44-51)

An amino acid is given its name based on its havingboth an amine group and a carboxylic acid group. Thegenericstructure for a natural (chirally correct) amino acid isdrawn in Figure 1 below.

O

IIH2N C

C OH

AH R

Figure 1 Generic amino acid

The R represents various groups, each onecorrespondingto a specific amino acid. This is to say that all amino acidsare thesame with theexception of theR group. Thegenericamino acid inFigure 1 is shown as it exists in a hydrophobicenvironment. This is not how it exists in an aqueousenvironment, however. In water, the amino terminal (aminegroup) is protonated and the carboxyl terminal (carboxylicacid group) is deprotonated. Figure 2 shows the aqueousform of a generic amino acid.

O

H3N

^^ ô-

H R

Figure 2 Amino acid in aqueous environment

The pKa values for the protonated amino terminal andthe carboxylic acid group are approximately 9.5 and 2.5respectively. At physiological pH of 7.4, the pKa of theamino terminal is greater than the pH, so it is protonated bythe solution according to the Henderson-Hasselbalch equation.At physiological pH, the pKaof the carboxyl terminal is lessthan the pH, so it is deprotonated by thesolution. Equation1is theHenderson-Hasselbalch equation.

pH = pKa + log [A"J

[HA]Equation 1 Henderson-Hasselbalch Equation

Using theHenderson-Hasselbalch equation, it is possibleto calculate the ratio of the deprotonated form to protonatedform foreach siteon an amino acidat a specific pH.

44. Which of the following changes to n-propanol wouldNOTresult in a lowerpKa than n-propanol?

A. Oxidizing carbon 1 by four electrons, resultingin arc-bond to oxygen instead of two a-bonds tohydrogen.

B. Replacing the oxygen with a sulfur atom.C. Replacing thepropyl group witha phenyl ring.D. Replacing the hydroxyl group with an amine

group.


4 5. What charge does the following dipeptide carry at a pHof 5.0?

O

pKa = 9.2H3N.

O H H2C^

V 1

^ pKa =3.9OH

.OHpKa = 2.0

H

A 1

9 • II

(CH2)4H o

+NH3PKa=10-8A. -1

B. 0

C. +1

D. +2

46. When the pH of solution is greater than the pKa by 1,what is the ratio of the deprotonated form to theprotonated form?

A. 1 : 10

B. 1 :2

C. 2: 1

D. 10: 1

47. How do the three labeled protons on the followingmolecule rank in terms of relative acidity?

0

IIH3N C^ a

C OH

A bH CF2OH

A. Proton a> Proton c > Proton b

B. Proton b > Proton a > Proton c

C. Proton b > Proton c > Proton a

D. Proton c > Proton a > Proton b

4 8. What can be said about the pKa of a hydrogen on sulfur,given that cysteine exists predominantly in thefollowing form at pH = 7.4? Note that the hydrogen onsulfur is the second proton to be lost when cysteine isdeprotonated.

O

H3N

/VH CH2SH

O

A. It is less than 2.5.

B. It is greater than 2.5 but less than 7.0.C. It is greater than 7.0 but less than 9.5.D. It is greater than 9.5.


49. How many times must the fully protonated form ofglutamic acid be deprotonated to form monosodiumglutamate? (For glutamic acid R = CH2CH2CO2H)

A. 0 times

B. 1 time

C. 2 times

D. 3 times

50. At what pH value will the ratio of the +2 to +1 formsof the following molecule be 10:1?

HN-

pKa = 9.2

H (CH2)4

+NH3 pKa =10.8A. 3.0

B. 5.1

C. 7.1

D. 8.2

= 6.1

OH pKa=1.8

51. Given the pKa values for each of the followingcompounds, whichof the compounds is the most basic?

O

,NH3

'^ ÔH^ ÔH -^ A

a apKa= 10.0 pKa = 8.4 pKa = 5.0

B. NH2

aD.

O



Amino acids are the biological building blocks ofnaturally occurring polymers known as proteins. The humanbody codes for twenty amino acids. These twenty make upthe majority of the amino acids found in enzymes, proteinhormones, and other polypeptides. Polypeptides are formedydien amino acids link their respective amino and carboxylJerminals. Table 1 lists the twenty amino acids that havebase codons associated with them:

Amino acid Code R-Group pKasAlanine Ala CH3 2.3/9.7

Arginine Arg (CH2)3NHC(NH2)2+ 2.2/9.5/13.0

Aspartic acid Asp CH2C02- 2.1/3.9/9.8

Asparagine Asn CH2CONH2 2.0/8.9

Cysteine Cys CH2SH 1.8/8.4/10.6

Glycine Gly H 2.3/9.6

Glutamic Acid Glu CH2CH2C02' 2.1/4.3/9.8

Glutamine Gin CH2CH2CONH2 2.2/9.1

Histidine His CH2C3H3N2 1.8/6.1/9.2

Isoleucine He CH(CH3)CH2CH3 2.4/9.7

Leucine Leu CH2CH(CH3)2 2.4/9.6

Lysine Lys (CH2)4NH3+ 2.2/9.2/10.8

Methionine Met CH2CH2SCH3 2.3/9.3

Phenylalanine Phe CH2C6H5 1.8/9.2

Proline Pro —CH2CH2CH2— 2.0/9.7

Serine Ser CH2OH 2.2/9.2

Threonine Thr CH(CH3)OH 2.6/10.3

Tryptophan Trp CH2C8H6N 2.4/9.4

Tyrosine Tyr CH2C6H4OH 2.1/9.1/10.1

Valine Val CH(CH3)2 2.3/9.6 J

Table 1

Each amino acid can be distinguished from the others bythe R-group attached to the alpha carbon, known as its sidechain. The R-groups in Table 1 are shown as they exist atpH = 7.4. Figure 1 shows the structure for a generic aminoacid with a generic R-group.

O

H3N

vc

H R

O

Figure 1 Generic amino acid with natural stereochemistry

52. Which of the following polypeptides will carry apositive one charge when placed in gastric acid (anaqueous solution buffered at pH = 1.5)?

A. Lys-Leu-IleB. Arg-Cys-HisC. Asp-Glu-AsnD. Glu-His-Phe


53. Hydroxyproline, shown below, is found in humanbeings, but it is not coded for in DNA.

What feature is NOT associated with hydroxyproline?A. Hydroxyproline has more than one chiral center.B. Hydroxyproline induces deviations in the secondary

structure of a protein.

C. Hydroxyproline is hydrophilic.D. Hydroxyproline hasa sidechainpKaaround 4.0.

54. Which of the following peptide fragments has theHIGHESTisoelectricpH?

A. Met-Lys-ArgB. Gly-Phe-AlaC. Tyr-Asp-GluD. Leu-Val-He

55. Which of these amino acids contains exactly onechiralcenter?

A. GlycineB. Histidine

C. Isoleucine

D. Threonine

56. Which of these dipeptides requires the lowest solutionpH to exist 100% as a zwitterion?

A. Leu-His

B. Lys-AsnC. Asp-CysD. Tyr-Arg

57. At pH = 7.0, which of the following aminoacids wouldmigrate to thecathode ina buffered electrophoresis gel?A. Alanine

B. Aspartic acidC. Leucine

D. Lysine

58. An isoelectric point less than 5.0 indicates that theamino acid has what type of side chain?

A. An acidic side chain.

B. A basic side chain.

C. A hydrophilic side chain.D. A hydrophobic side chain.



There are twenty amino acids for which human DNAcodes. These twenty amino acids form proteins within thehuman body. The physical properties of the twenty aminoacids are listed in Table 1, shown below:

Amino

Acid

DecompTemp

Water Sol.

@25°C

(8/100 mL)[a]f>5 PKal pKa2 PK33

Gly 233*C 25.2 2.33 9.78

Ala 298°C 16.5 +8.5 2.35 9.87

Val 315eC 8.9 +13.9 2.29 9.72

Leu 293°C 2.4 -10.8 2.33 9.74

lie 284°C 4.1 +11.3 2.32 9.74

Met 2809C 3.4 -8.2 2.17 9.27

Pro 220°C 162.0 -85.0 1.95 10.64

Phe 283°C 3.1 -35.1 2.58 9.24

Trp 289°C 1.1 -31.5 2.43 9.44

Ser 228eC 5.0 -6.8 2.19 9.46

Thr 225'C very sol. -28.3 2.09 9.10

Cys +6.5 1.88 8.36 10.33

Tyr 342°C 0.05 -10.6 2.21 9.12 10.07

Asn 234°C 3.5 -5.4 2.02 8.80

Gin 185'C 3.7 +6.1 2.17 9.13

Asp 270°C 0.6 +25.0 1.99 3.88 10.01

Glu 2478C 0.9 +31.4 2.13 4.32 9.95

Lys 225°C very sol. +14.6 2.16 9.20 10.80

Arg 244°C 15.0 +12.5 1.82 8.99 13.21

His 287eC 4.2 -39.7 1.81 6.05 9.15

Table 1

Amino acids are distinguished by the side chain attachedto the alpha carbon. Within a linear protein, there is oneamino terminal and one carboxyl terminal, so the side chainsaccount for the behavior of each amino acid within a peptide.

59. Which of the following side chains would MOSTLIKELY be found in the hydrophilic pocket of aprotein?

A. —CH2CH2SCH3

B. —CH2C6H4OH

C. —CH2CH2CONH2

D. —CH2SH

6 0. How can the fact that lysine is more water soluble thanleucine best be explained?

A. The side chain of lysine can form hydrogen bonds,while the side chain of leucine cannot.

B. The side chain of leucine can form hydrogen bonds,while the side chain of lysine cannot.

C. The side chain of lysine is non-polar.D. The side chain of leucine is polar.


61. The GREATEST isoelectric pH is found with:

A. alanine.

B. glutamic acid.

C. histidine.

D. phenylalanine.

62. Which of the following structures represents the mo^common form of cysteine in a pH = 7.0 solution?

A. O B. O

H2NÂ» ~ÂOH

H CH2SH

C. O

H3N

OH

H CH2S

D. O

H3N*

x^- "V°-H CH2SH H CH2S

63. Arginine has a lower pKai than alanine, because thecarboxyl terminal of alanine is:

A. more acidic than the carboxyl terminal of arginine,due to the inductive effect of the anionic side chain.

B. less acidic than the carboxyl terminal of arginine,due to the inductive effect of the cationic side chain.

C. more acidic than the carboxyl terminal of arginine,due to the inductive effect of the amino terminal.

D. less acidic than the carboxyl terminal of arginine,due to the inductive effect of the amino terminal.

6 4. Tyrosine best separates from cysteine at a pH equal to:

A. the pKa (Cys> when using ether/water extraction.B. the pKa (Tyr). when using ether/water extraction.C. 7.0, when using ether/water extraction.D. either plcys or P^Tyr. when using ether/water

extraction.

65. At pH = 4.0, what is true for histidine?

A. Most exists with the carboxyl terminal protonated.B. Over 90% exists as a zwitterion.

C. Most exists with the amino terminal deprotonated.D. Over 10% of the side chains are protonated.

66. Which of the following amino acids would bind to acolumn filled with DEAE-cellulose at a pH of 6.20?

A. Alanine

B. Glutamic acid

C. LysineD. Tyrosine



Parathyroid hormone (PTH) and calcitonin (CT) are twoof the three protein hormones involved in mineralhomeostasis. PTH is initially synthesized as a 115-aminoacid inactive polypeptide preprohormone. Preprohormonesare later activated by removal of specific parts of theirprimary structure. The primary structure of parathyroidhormone is shown below:

H2N-

Figure 1 Parathyroid Hormone

Calcitonin is a 32-amino acid peptide hormone thathasits carboxyl terminal in the form of an amide. Bothcalcitonin and PTH are found in several animals, but theprimary sequence varies from species to species. Thereactivity is sometimes altered by the substitution of differentamino acids in the polypeptide.

67. In human PTH, amino acid #46 is alanine. In bovinePTH, amino acid #46 is glycine. What is true whencomparing human PTH with bovine PTH?

A. Thesecondary structure is more disrupted in bovinePTH, because of its bulkier side chain.

B. Thesecondary structure is more disrupted inhumanPTH, because of its bulkier side chain.

C. The tertiary structure is more disrupted in bovinePTH, because of its bulkier side chain.

D. The tertiary structure is more disrupted in humanPTH, because of its bulkier side chain.


68. In PTH, amino acids #1 through #34 make up theactive portion. Amino acids #35 through #84 protectthe structure from proteolysis. Amino acids #35through #43 can best be described as:

A. hydrophilic.B. hydrophobic.C. polar and anionic.D. polar and cationic.

69. In the endoplasmic reticulum, prepro-PTH loses twomethionine residues anda 23-amino acidpeptide to formpro-PTH. Pro-PTHis transferredinto the Golgi regionwhere it is converted into PTH. What occurs in theGolgi region?

A. Gain of an octapeptideB. GainofahexapeptideC. Loss of an octapeptideD. Loss of ahexapeptide

70. Chymotrypsin cleaves at the C-terminus of Phe, Trp,and Tyr. What fragments form when human PTH istreated with chymotrypsin?

A. Three fragments total, of length 22 amino acids, 11amino acids, and 51 amino acids.

B. Three fragments total, of length 27 amino acids, 11amino acids, and 46 amino acids.

C. Four fragments total, of length 27 amino acids, 9amino acids, 16 amino acids, and 32 amino acids.

D. Three fragments total, of length 23 amino acids, 11amino acids, and 50 amino acids.

71. The fact that Mets and Metis oi human PTH can bereplaced with nor-leucine (R = —CH2CH2CH2CH3)withno observed change in reactivityimplies that:

A. sulfur has no effect on reactivity.B. nitrogen has no effect on reactivity.C. sulfurhasa significant effecton reactivity.D. nitrogen hasa significant effecton reactivity.

7 2. Analogs of human-PTH should:

A. be structurally diverse between amino acids #1through #34 compared to human-PTH.

B. be structurally similar between amino acids #1through #34 compared to human-PTH.

C. have a different N terminus and C terminus thanhuman-PTH.

D. be structurally diverse between amino acids #35through #84 compared to human-PTH.


73. Whatcan be concluded from the followingdata?

Fragment % Activity in vitro % Activity in vivo1 - 84 100% 100%

1 - 34 77% 32%

1-31 10% 67%

1 - 28 5% 0.3%

A. The active site probably contains amino acid # 17.B. The active site probably contains amino acid # 30.C. The active site probably contains amino acid # 66.

D. The active site probably contains amino acid # 83.



A researcher was interested in determining the sequenceof two polypeptide fragments she isolated from partialhydrolysis of an enzyme. The first fragment (Fragment I) isa six amino acid fragment which cannot be deciphered usingmolecular weight studies of the component amino acids. Theprimary structure of the second fragment (Fragment II) wasdetermined using an amino acid sequencer. Fragment II hasthe following amino acid sequence from N-to-C terminals:His-Ser-Val-Phe-Ile-Tyr-Phe. An automated proteinsequencer cleaves, isolates, and identifies the componentamino acids one at a time from the amino terminal to the

carboxyl terminal using phenylisothiocyanate. The researcherused the enzymes listed in the Table 1 for analysis.

Enzyme Amino acid Cleavage side

Chymotrypsin Phe, Trp, Tyr Carboxyl

Clostripain Arg Carboxyl

PepsinPhe, Trp, TyrAsp, Glu, Leu Carboxyl

Thermolysin Leu, He, Val Amino

Trypsin Lys, Arg Carboxyl

Table 1

The term "cleavage side" indicates the point at which theenzyme breaks the peptide bond. For instance, thermolysincleaves the peptide bond on the amino (left) side of leucine,isoleucine, or valine. It should be noted that the enzymeitself is a chain of amino acids, but it is present in such asmall concentration that it does not cleave itself. The

enzyme concentration is low because it is a catalyst with asubstantial turnover rate, where a "substantial" turnover rateis defined to be greater than 100/second.

Some small polypeptides can be analyzed according totheir isoelectric point or migration rate in an electrophoresisgel. If the isoelectric point of an amino acid is greater thanthe pH, then that amino acid carries a partial positive chargeand exhibits cationic behavior. Cations migrate to thecathode in gel electrophoresis. Figure 1 shows histidine inits fully protonated state.

PKal = 1.8HO

pKa3 = 9.2 H3N H

pKa2= 6.1

Figure 1 Fully protonated structure of histidine

7 4. What is the approximate isoelectric point of histidine?

A. 4.0

B. 6.1

C. 7.6

D. 9.2


75. In what direction would histidine migrate if it wereadded to an electrophoresis gel buffered at pH = 7.0?

A. It would migrate to the anode.B. It would migrate to the cathode.C. It would not migrate.D. It would denature.

76. If treating a polypeptide with 2,4-dinitrofluorobenzenefollowed by 6 M HCl(aq) results in an alanine residuebound to a dinitrobenzene moiety, then what can beconcluded from this information?

A. The first amino acid in the polypeptide is alanine.B. The second amino acid in the polypeptide is

alanine.

C. The fifth amino acid in the polypeptideis alanine.D. The last amino acid in the polypeptide is alanine.

77. Treatment of FragmentII with thermolysin would yield:

A. 1 fragment.B. 2 fragments.C. 3 fragments.D. 4 fragments.

78. What is the side group on the amino acid that weighs89 grams/moles?

A. —H

B. —CH2OH

C. —CH3D. —SCH3

79. Which of the following polypeptides would be observedafterFragment II is treated withclostripain?A. His-Ser-Val-Phe-Ile-Tyr-PheB. His-Ser-Val-Phe

C. He-Tyr-PheD. His-Ser-Arg

80. If the separate treatment of a protein first with pepsinand then with thermolysin generated the samepolypeptides, then which of the following peptidelinkages is present in the protein?

A. He-Phe

B. Leu-Leu

C. Tyr-ValD. Val-Met


81. Treatment of Fragment I with chymotrypsin yields thefollowing residues: Ser-Val-Phe and Gly-Cys-Gly.Treatment of the six amino acid Fragment I withSanger's reagent (4-dinitrofluorobenzene) binds serine.Which of the following is the sequence for Fragment I?

A. Gly-Cys-Gly-Phe-Val-SerB. Gly-Cys-Gly-Ser-Val-PheC. Ser-Val-Phe-Gly-Cys-GlyD. Phe-Val-Ser-Gly-Cys-Gly



A protein isolated from the saliva of the NortheasternRadish Spider is found to break down muscle fiber in theTuscaloosa Spitting Lizard. Researchers determined theprimary sequence of the protein by carrying out fourexperiments. In Experiment I, the full protein is treated with2,4-dinitrofluorobenzene (2,4-DFNB) followed by hydrolysisof the protein using an acid catalyst. The 2,4-DNFB is astrongelectrophile that accepts electrons from the nitrogen ofthe first amino acid, and in doing so labels it.

The amino acids present in the protein are two alanines,two cysteines, and one each of arginine, glycine, histidine(found to be the first amino acid), isoleucine, leucine, lysine,methionine, phenylalanine, serine, and tyrosine. Followingthe hydrolysis experiment, the researcher treated separatesamplesof the protein with the following reagents:

Experiment II: The protein is first treated with B-mercaptoethanol (breaking any disulfide linkages) followedby treatment with trypsin yields these fragments:

N-Ser-He-Tyr-Ala-C

N-His-Lys-C

N-Met-Cys-Leu-Gly-Ala-Phe-Cys-Arg-C

Experiment III: The protein is first treated with 6-mercaptoethanol (breaking any disulfide linkages) followedby treatment with thermolysin yields these fragments:

N-Ile-Tyr-Ala-C

N-His-Lys-Met-Cys-CN-Leu-Gly-Ala-Phe-Cys-Arg-Ser-C

Experiment IV: The protein is first treated with B-mercaptoethanol (breaking any disulfide linkages) followedby treatment with chymotrypsin yields these fragments:

N-Cys-Arg-Ser-Ile-Tyr-C

free alanine

N-His-Lys-Met-Cys-Leu-Gly-Ala-Phe-C

The researchers were able to use the experimentalinformation to determine the sequence of the amino acids inthe protein. Knowing the first amino acid in the sequencealong with any two of the experiments, the sequence canaccurately be deduced. All disulfide linkages must first bebroken in order to isolate the fragments in their correctsequence.

8 2. What is the role of 6-mercaptoethanol in the first part ofeach experiment?

A. B-Mercaptoethanol serves to bind the first aminoacid in the protein.

B. B-Mercaptoethanol serves to bind the last aminoacid in the protein.

C. B-Mercaptoethanol serves to cleave the proteinfollowing the cysteine residues.

D. B-Mercaptoethanol serves to cleave any disulfidebridges present in the protein.


83. What is the role of 2,4-dinitrofluorobenzene inExperiment I?

A. DNFB serves to bind the first amino acid in theprotein.

B. DNFB serves to bind the last amino acid in theprotein.

C. DNFB serves to cleave the protein following thecysteine residues.

D. DNFBserves to cleaveany disulfidebridges presentin the protein.

8 4. Based on the results of Experiments II, HI, and IV, whatis the LAST amino acid in the protein?

A. Alanine

B. ArginineC. Histidine

D. Serine

85. Based on the results of Experiments n, HI, and IV, allof the following are valid conclusions EXCEPT:

A. chymotrypsin cleaves on the carboxyl side oftyrosine and phenylalanine.

B. trypsin cleaves on the carboxyl side of arginine andlysine.

C. thermolysin cleaves on the amino side of isoleucineand leucine.

D. thermolysin cleaves on the amino side of cysteineand alanine.

86. Why is it a more valid conclusion that thermolysincleaves before leucine and isoleucine, rather than aftercysteine and serine?

A. Thermolysin cannot cleave after molecules involvedin sulfide bridging.

B. Because serine has a protic side chain, it cannotreact with thermolysin.

C. There are two cysteine residues present in theprotein, so if thermolysin cleaved after cysteine, anadditional cut would have been observed.

D. There are two cysteine residues present in theprotein, so if thermolysin cleaved after cysteine,one less cut would have been observed.

87. From the information in Experiments II and III, howmuch of the exact protein sequence is known?

A. It is known only to amino acid two.B. It is known only to amino acid four.

C. It is known up to amino acid eleven.D. The entire sequence of amino acids in the protein is

known.


88. Based on all of the experimental information, what isthe primary sequence of the unknown protein?

A. N-His—Lys—Met—Cys—Leu—Gly—Ala—Phe—Cys-Arg—Ser—lie—Tyr—Ala-C.

B. N-His—Lys—Met—Cys—lie—Tyr—Ala—Leu—Gly—Ala—Phe—Cys—Arg—Ser-C.

C. N-He—Tyr—Ala—Leu—Gly—Ala—Phe—Cys-Arg—Ser—His—Lys—Met—Cys-C.

D. N-Leu—Gly—Ala—Phe—Cys—Arg—Ser—His—Lys-Met—Cys—He—Tyr—Ala-C.


Passage XIII (Questions 89 - 96)

When amino acids are synthesized in vitro, they form asa racemic mixture of the D- arid L-enantiomers. Since the L-

enantiomer is desired, resolution is necessary. For resolvingthe mixture, one of three techniques can be applied.

Technique I: The racemate is first treated with benzoylchloride to form an amide at the amino terminal. The

racemate of N-benzoyl-DL-amino acid is treated with abrucine salt causing the D-amino acid to precipitate fromsolution. The solid is filtered from solution and is treated

with acid to dissolve the brucine-amino acid salt into a new

solution. Upon treatment with aqueous hydroxide, a puresample of the D-amino acid is isolated.

Technique 2: The racemate is first treated with benzoylchloride followed by a strychnine salt, causing the L-aminoacid to precipitate from solution. The solid is filtered fromsolution and is treated with acid to dissolve the strychnine-amino acid salt into a new solution. Aqueous hydroxide isadded to the new solution to isolate the L-amino acid.

Technique 3: The racemate is treated with acetic anhydrideto form N-acetyl-DL-amino acid. This racemic mixture ofacylated amino acids is treated with hog renal acylase,which removes the acetyl group from the L-enantiomer,forming the zwitterion. The D-enantiomer remainsacylated at the N-terminal, making it anionic (due to thedeprotonation of the C-terminal). The two species areseparated using ether-water extraction.

A major drawback to the first two techniques is that onlyone enantiomer is isolated in pure form. The enantiomer thatdoes not precipitate from solution is isolated after the solventis evaporated, which does not result in a pure sample. Thesecond drawback is that the success of the techniques varieswith different amino acids. For instance, they work wellwith alanine, but are difficult to use with tryptophan.Technique 3 has the most universal application. Presentedbelowis an experimentalapplication of Technique 3:

Experiment: A suspension of 12.9 grams of N-acetyl-DL-isoleucine in 0.80 liters of water is buffered to 7.0 in an

aqueous solution, following which 0.0lOg hog renalacylase powder is added. The mixture is stirred for 16hours at 37.0°C (physiological temperature). The mixtureis then acidified with 10.0 mL of concentrated acetic acid,filtered through a frit, and evaporated under vacuum to avolume of roughly 50 mL. Ethanol is added to form L-isoleucine crystals. The crude product is recrystallized froman ethanol-water mixture to yield roughly 3.5 grams(77.1% yield) of optically pure L-isoleucine.

89. Why must the first step in Technique 3 be done in anaprotic solvent?

A. A protic solvent is too inert.B. A protic solvent can react with acetic anhydride.C. To ensure that the N-terminal is protonated.D. To ensure that the C-terminal is deprotonated.


90. When the brucine salt is used to resolve a racemicmixture of alanine, what is likely to occur?

A. L-alanine is isolated in relatively pure form fromprecipitate, while D-alanine is isolated withimpurities from solution.

B. L-alanine is isolated with impurities fromprecipitate, while D-alanine is isolated in relativelypure form from solution.

C. D-alanine is isolated in relatively pure form fromprecipitate, while L-alanine is isolated withimpurities from solution.

D. D-alanine is isolated with impurities fromprecipitate, while L-alanine is isolated in relativelypure form from solution.

91. Which procedure involves an increasein entropy?

A. Separation of enantiomers from a racemic mixture.B. Racemization of one enantiomer.

C. Reaction of a D-amino acid with a diacid.

D. Precipitation of an enantiomeric salt.

9 2. Why is the racemate first treated with acetic anhydride inTechnique 3?

A. To protect the amino terminal.B. To form a bond that will react with hog renal

acylase.C. To make the amino acid cationic.

D. To invert the chiral center.

93. In Technique 3, the filtrate can be treated with HCl(aq)to a pH of 2.0 to crystallize the N-acetyl-D-isoleucine.The crystals can be dissolved and hydrolyzed intosolution of the D-amino acid using 1.0 M HCl(aq).The D-enantiomer can be treated with racemase to

generate a D-L mixture again. Technique 3 can beapplied to this mixture once again to isolate more L-enantiomer. Why does this process become futile afterthree cycles?

A. The percent of the enantiomer obtained from thethird cycle is only 12.5% of the product mixture.

B. The percent of the enantiomer obtained from thethird cycle is only 6.25% of the product mixture.

C. Hog renal acylase is consumed completely by thethird cycle.

D. The acetic anhydride reacts only when the mixturehas a substantial amount of L-enantiomer.


94. In the experiment, why is the solution stirred forsixteen hours?

A. To vaporize the pig renal acylase.B. To agitatethe mixture andenhance reactivity.C. To centrifuge the powder to the bottom of the

flask.

D. To generate a suspension, and thus decrease thereaction rate.

95. Whyis hydroxide anion usedat the end of Technique 1?

A. To remove the benzoyl substituent from the N-terminus.

B. To remove the carboxyl and amino terminalprotons.

C. To remove the cationic brucine salt.

D. To racemize the isolated amino acid.

9 6. What precipitates with strychnine in Technique 2?

A.O

H3N+,O'

R H

C.O

H3N"O"

H R H R


Questions 97 -100 are NOT basedon a descriptive passage

9 7. Which of the following graphs BEST represents the rateof the corresponding nucleophilic substitution reactionrelative to the pH of the solution?

H3CNH2 + CH^Cl (H3C)2NH2+Cr

98. Following neutralization, the product of a reaction ofH3CNH2 and H3CC1 would show which of thefollowing proton NMR peak patterns?

A. Singlet(3H),singlet(3H),septet(1H)B. Singlet (3H), singlet (3H), quartet (1H)C. Singlet (6H), sextet (1H)D. Doublet (6H), septet (1H)

9 9. How many mL of 0.010 M NaOH(aq) need to be addedto take 10 mL 0.010 M lysine from pH = 5.68 (the firstequivalence point) to pH = 10.80 (pK^ for lysine)?

A. 5mL0.10MNaOH(aq)B. 10mL0.10MNaOH(aq)C. 15mL0.10MNaOH(aq)D. 20mL0.10MNaOH(aq)


100. All of the following can be reduce to form aminesEXCEPT.

A. Amides with LiAlH4.

B. Nitriles with HCl/ZnCl2(aq).

C. Imines with H2/Pd.

D. All of the above reactions yield amines.

"GET INTO CHEM!"

1. A 2. C 3. D 4. C 5. A

6. D 7. B • 8. D 9. C 10. A

11. A 12. C 13. A 14. C 15. C

16. A 17. B 18. D 19. C 20. D

21. C 22. A 23. A 24. A 25. D

26. C 27. B 28. B 29. C 30. A

31. B 32. B 33. D 34. D 35. D

36. C 37. D 38. B 39. C 40. D

41. D 42. B 43. D 44. D 45. B

46. D 47. A 48. C 49. C 50. B

51. D 52. C 53. D 54. A 55. B

56. C 57. D 58. A 59. C 60. A

61. C 62. C 63. B 64. D 65. D

66. B 67. B 68. B 69. D 70. D

71. A 72. B 73. B 74. C 75. B

76. A 77. C 78. C 79. A 80. C

81. C 82. D 83. A 84. A 85. D

86. C 87. D 88. A 89. B 90. C

91. B 92. B 93. A 94. B 95. A

96. D 97. C 98. D 99. C 100. B

ES FIN.

Amines and Amino Acids Passage Answers

Passage I (Questions 1-7) Basicity of Amines

1.

2.

3.

4.

5.

6.

7.

Choice A is correct. According to the passage, the most basic amine compound in the gas phase is the tertiaryamine. The weakest base in the gas phase is the base with the least electron donating alkyl groups. This makesammonia the weakest base. Because ammonia is the weakest base, ammonium cation (the conjugate acid ofammonia) must be the strongest acid in the gas phase. This makes choice A correct.

Choice C is correct. If the pKa for CH3CH2NH3+ is 10.71, then the pKb for CH3CH2NH2 must be 3.29. BecauseCH3CH2NH2 is more basic than CF3CH2NH2, due to the electron withdrawing nature of fluorine, the pKb forCF3CH2NH2 must be greater than 3.29. The pKb for CF3CH2NH2 should be only a point or two greater thanthe pKb for CH3CH2NH2, making choice C correct.

Choice D is correct. In both the gas phase and in water, primary amines are stronger bases than ammonia. Thismeans that the stronger base is on the product side of the equation. A favorable reaction has the stronger basereacting to yield the weaker base, which is the reverse reaction as written. Tliis means that reactants are moreabundant than products as written. This makes the equilibrium constant (Keq) less than one in both the gasphase and in water. The best answer is choice D.

Choice C is correct. The donating nature of methyl groups increase the basicity of an amine. According to Table1, the highest pKa is associated with the dimethylammonium cation, so it is the weakest acid. As an acid getsweaker, its conjugate base becomes stronger, so the secondary amine is tiie strongest base. This is choice C.

Choice A is correct. An amine of any substitution, tertiary included, will exist in its protonated state in solutionswith a pH value less than its pKa value. When it is protonated, it is cationic and thus more soluble in waterthan it is in its neutral state. This implies that as the pH of the solution is lowered, the amine becomes moresoluble. The lowest pH of the answer choices is choice A, 3, so choice A is the best choice.

Choice D is correct. When the pH is less than the pKa, the site exists in its protonated state. When the pH isgreater than the pKa, the site exists in its deprotonated state. The pH is seven, which is greater than the pKafor the carboxyl terminal so the carboxyl terminal is deprotonated. When the carboxyl terminal isdeprotonated, it carries a negative charge. This eliminates choice A. The pH is seven, which is less than thepKa for the amino terminal and the amino side chain, so both the amino terminal and the amino side chain areprotonated. When the amino terminal and the amino side chain are protonated, they each carry a positivecharge. This eliminates choices B and C. The best answer is choice D.

Choice B is correct. The passage discusses the N-H absorbance of a secondary amine, but does not mention atertiary amine. This is because all of the hydrogens in a tertiary amine are bonded to carbon. The nitrogen hasthree bonds to carbon, so there is no N-H bond, making choice B the best answer. There is no resonance to speakof, given that there is no 7t-bond, so choice C is eliminated. The tertiary amine may or may not be symmetric, butthat is irrelevant, so choice D is eliminated. Steric hindrance may seem tempting, because a tertiary amine hasmore steric hindrance than less substituted amines. But such an answer is designed to make you jump at sterichindrance without thinking through the question completely, because it is not the reason for no absorbance.

Passage II (Questions 8 -14) Amine Basicity and Nucleophilicity

Choice D is correct. The donating nature of alkyl groups increases the basicity of an amine, although it alsocreates steric hindrance. In this case, the amines are primary and secondary. Secondary amines are more basicthan primary amines, so choices A and B are eliminated. The data in Table 1 confirms that secondary aminesare more basic than primary amines, because secondary amines yield a greater equilibrium constant whenreacting with ammonium than primary amines. Methyl groups are less bulky than ethyl groups, but ethylgroups are better electron donating substituents than methyl groups, so the most basic compound cannot bedetermined using background information. According to the data in Table 1, (CH3)2NH generates a Keq of30.9when reacting with ammonium, while (CH3CH2)2NH generates a Keq of 50.1 when reacting with ammonium.This means that (CH3CH2bNH is a stronger base than (CH3)2NH, so choice D is the best answer.

Copyright © by The Berkeley Review® 223 AMINO ACIDS & AMINES EXPLANATIONS

9. Choice C is correct. Table 1 lists the reactions of neutral amines with ammonium, so for each entry, NH4+ shouldbe found on the reactant side. All of the entries in Table 1 have equilibrium constants greater than 1, so to havean equilibrium constant of 0.03, it must be the reverse reaction of what is shown in Table 1. This means thatNH3 must be found on the reactant side. This eliminates choices B and D. According to the data in Table 1,dimethyl amine ((H3C)2NH) is associated with an equilibrium constant of 50 while trimethyl amine((H3Q3N) is associated with an equilibrium constant of roughly 30. It is in the reverse reaction of trimethylamine that results in an equilibrium constant of 0.03 (about 1/30), so choice C is the best answer.

10. Choice A is correct. Because an amine is a weak base while an amide exhibits no acid-base properties whenadded to water, the two compounds can be separated using acid-base extraction. This eliminates choice D. Theamine can be protonated to form a cationic compound (its ammonium conjugate acid), which is more water solublethan the amide. The extraction therefore will be most successful at a low pH (acidic conditions). The lowest pHof the choices is 3, which makes choice A the best answer.

11. Choice A is correct. Because of the electron withdrawing nature of the carbonyl group through resonance, theelectron pair on the nitrogen in an amide is less available for donating than the electron pair on nitrogen in anamine. The difference in basicity is attributed to resonance and not steric hindrance, so choices C and D areeliminated. An amine is more basic than an amide, so choice A is the best answer.

12. Choice C is correct. The boiling point of a compound is influenced by its molecular mass and its intermolecularforces. The highest boiling point is associated with the compound with both hydrogen bonding and the greatestmolecular mass. Ammonia has hydrogen bonding, but it is very light. Choice A is eliminated. Trimethylamineexhibits no hydrogen bonding, so it does not have the highest boiling point of the choices. This eliminateschoice D. Methyl amine and propyl amine have roughly the same degree of hydrogen bonding, but propyl amineis heavier. As such, the highest boiling point is observed with propyl amine. Choice C is the best answer.

13. Choice A is correct. Given that the pKa of (H3Q3NHCI is 9.8, the pKb of (H3Q3N: must be 4.2. The electrondonating nature of methyl groups increases the basicity of an amine while the electron withdrawing nature ofbenzene decreases the basicity of an amine. Aniline is therefore less basic than trimethyl amine. Because ahigher pKb value is associated with weaker bases, the pKb of aniline can be no less than 4.2. This makes choiceA not possible. Choice A is your choicefor correctness and subsequent happiness.

14. Choice Cis correct. Adecrease in the Keq value for a proton transfer reaction with ammonium indicates that thebasicity of the compound is reduced. The question is thus "Which of the following changes to an amine wouldNOT result in a weaker base?" or better yet, "Which change would result in a compound MORE basic than aprimary amine?" Replacing hydrogen with fluorine would reduce the basicity, because fluorines are electronwithdrawing. Choice A is thus eliminated. Replacing the adjacent (alpha) hydrogens with a double bond tooxygen would convert the compound to an amide and thus reduce its basicity, because oxygen is electronwithdrawing. Choice B is thus eliminated. Replacing one hydrogen with an alkyl group would increase thebasicity, because alkyl groups are electron donating. Choice C is the best answer. Replacing an alkylgroup withhydrogen would reduce the basicity, because alkyl groupsare electron donating. Choice D is thus eliminated.

Passage III (Questions 15 - 20) Effect of Alkyl Groups on Nucleophilicity

15. Choice C is correct. Method 2 in Figure 1 shows the reduction of a nitrile, a primary imine, a primary amide,andan azide into a primary amine, so a nitrile and primary imine, choices A and B, can definitely be reduced into aprimary amine. Choices A and B are eliminated. A secondary amide can be reduced, but R-groups remainconstant, resulting in the formation of a secondary amine. This means that a secondary amide, choice C, cannotbe reduced into a primary amine, so choice C is the best answer. An oxime, R2C=NOH, can be reduced much likean imine can be reduced. The result is a primary amine and water. Choice D is eliminated.

16. Choice A is correct. The N-H bond, like the O-H bond, has an infrared absorbance value above 3000 cm"1. Theabsorbance is broadened due to hydrogen bonding, although not as broad as a hydroxyl absorbance, becauseamines do not form as strong of hydrogen bonds with amines as alcohols form with alcohols. Broadening is notassociated with the question, so we address it for edification purposes only. You simply need to know that N-Hbonds are similar to O-H bonds, so their IR absorbances are close in value. The best choice is answer A.

Copyright © by The BerkeleyReview® 224 AMINO ACIDS & AMINES EXPLANATIONS

17. Choice Bis correct. The methyl group usually donates electrons through what is called hyperconjugation. Thedonating form of the structure is not truly a resonance form in the traditional sense, so the methyl group isconsidered to be electron donating through the inductive effect. This is another case of controversy, so when youare uncertain of what information they are basing their answer on, consult the passage. In this case, which willbe true on occasion, the information is not there. So, from background information, resonance is defined as thederealization of electrons through the 7t-bonding network. The inductive effect is defined as the derealizationof electrons through the a-bonding network. There are no Tt-bonds in an amine, so the best answer cannot involveresonance. This eliminates choices C and D. The best answer must therefore involve the inductive effect. Anelectron donating group would makean amine a better nucleophile, so choice Bis the best answer.

Choice D is correct. The real question here is whether the reaction proceeds by an Sjsjl mechanism or an Sjvj2mechanism. Tiie electrophile is secondary, so there isno easy way to tell. However, on a multiple choice exam,you should let the answers dictate the choices you make. The Sjyjl reaction would result in a racemic mixture,which is not listed as a choice. The answer choices tell us to decide between an Sjsj2 reaction and an E2 reaction.Methyl amine is not a strong enough base to cause elimination, so the reaction must follow an S[\j2 mechanism,eliminating choices A and B. An Sn2 reaction results in an inversion of the chiral center. The reactantelectrophile has R-chirality, so the product should have S-chirality. The best answer is choice D.

Choice C is correct. The most nucleophilic nitrogen is the one with a lone pair that is most readily available forsharing. Electron pair a is the lone pair of an amide nitrogen, so it is tied up in resonance and not available foruse as a nucleophilic lone pair. Choice A is eliminated. Electron pair b is being shared with the benzene (andthe amide carbonyl on the para position of benzene) through resonance, so it is not as available for nucleophilicattack as a standard amine lone pair. Choice B is eliminated. Electron pair c is free from any conjugation(because it has s/?2-hybridization, and sp2-hybridization has the electrons in an orbital perpendicular to the p-orbital of the 71-network). Because of the resonance from nitrogen d, nitrogen c carries a partially negativecharge making it more nucleophilic. Choice C is the best answer. Electron pair d is involved in the aromaticityof the five-membered ring, so it is not readily available for nucleophilic attack. Choice D is eliminated. Thisquestion is analogous to what is observed with histidine. The best choice is electron pair c, so choose C.

20. Choice D is correct. An amine can react with both an acid halide (choice A) and an ester (choice B) to form anamide, so choices A and B are eliminated. It can undergo an S[\j2-reaction with an alkyl halide (choice C), sochoice C is eliminated. Ethers are inert and react with very few compounds. Choice D is the best answer.

18.

19.

Passage IV (Questions 21 - 27) Imine and Enamine Chemistry

21. Choice C is correct. Using the enamine synthesis route, alkyl groups can be added to the alpha carbon. Thesynthesis that cannot be carried out via the enamine procedure laid out in Figure 2 is the one that forms a newcarbon-carbon bond to a carbon other than the alpha carbon. Choice A adds a methyl to carbon 2 (the alphacarbon of 3-pentanone), choice B adds a butyl group to carbon 2 (the alpha carbon of butanal), and choice D addsa pentyl group to carbon 3 (the alpha carbon of acetone). It is in choice C that the new carbon-carbon bond isformed with a non-alpha carbon (carbon 6 of 3-hexanone is not an alpha carbon). Pick choice C for the warm,fuzzy feeling of correct answer picking.

22. Choice A is correct. Water is the side product that is generated in the formation of an imine from a ketone andan amine. The amine reactant is a primary amine, so it will lose both hydrogens and form a bond to the carbonylcarbon. The best answer is choice A.

23. Choice A is correct. The treatment of an imine with acidic water results in the hydrolysis of the imine andsubsequent formation of an amine and a carbonyl. The cyclic nature of the reactant causes the two functionalgroups to be present on the same molecule. Bycleaving the double bond and adding two hydrogens to the nitrogenand an oxygen (carbonyl oxygen) to the carbon, the product is determined rather easily. This shown below. Thebest answer is choice A.


24. Choice A is correct. The imine bond, when cleaved with acidic water, forms a ketone compound and a primaryamine. A quick pictorial route for determining the hydrolysis products formed from the reaction is shown below:

H"

H90

The best answer is choice A.

oPrimary amine

NH, O-0Ketone

25. Choice D is correct. From 3-pentanone to 4-methyl-3-heptanone, the structure has increased its number of carbonsby three. This means that a three carbon chain has been added to carbon 2 (which is equivalent to carbon4 bysymmetry) of 3-pentanone. This eliminateschoices A and B. To carry out this synthesis, a secondary amine mustbe added to the carbonyl reactant (3-pentanone) to form an enamine. This eliminates choice C and makes choiceD the best answer. In the second step of the synthesis, the alkyl halide compound (RX) is a three-carbon chainwith the halide on the terminal carbon. The final step in the synthesis in all four choices is the addition ofacidic water, soit must betrue. Tiie last step converts the nitrogen compound (enamine) into anoxygen compound(carbonyl).

O

1H3CH2CX CH2CH3 i3CH2C

R^ ^RR2NH/Ff N H3CCH2CH2I

CHCH5 H3CH2C

H3CH2CH?C

N

AH+(aq)

CHCHj H3CH2C

O

ACHCH3

i3CH2CH2C

26. Choice C is correct. The imine that is formed from a primary amine and a ketone is the imine shown in thesecond example in Figure 1. Tiie nitrogen of the imine has a single bond to one carbon and a double bond to theother carbon. Upon reduction of the imine, the nitrogen will gain a new a-bond tohydrogen and lose the 7t-bondto the carbon (as the carbonyl oxygen does when the ketone is reduced to a secondary alcohol). The final productisa nitrogen with two alkyl substituents and a hydrogen, which describes a secondary amine. The best answer ischoice C.

27. Choice Bis correct. Questions of this type are best solved by drawing the reaction out. Acidic water causes thehydrolysis of N-ethyl-3-pentimine, as shown below:

_/H2CH3H3CH2C-N=Jo=^

CH2CH3H20

CH2CH3

H3CH2C-NH2 °=\CH2CH3

3-PentanoneEthyl amine

The products are ethyl amine and 3-pentanone, so the best answer is choice B.

Passage V (Questions 28 - 35) Amino Acid Synthesis in vitro

28. Choice Bis correct. D-Serine has two hydroxyl absorbances, because it has two non-equivalent O-H bonds. TheIRabsorbancefor the O-H group of a carboxylic acid is broad and it comes between 2500 cm-1 and 3000 cm"1. TheIRabsorbance for the O-H group of a standard alcohol comes between 3200 cm"1 and 3600 cm"1 and is also broad,although not as broad the carboxylic acid O-H absorbance. The carboxylic acid hydroxyl and alcohol hydroxylare not equivalent in IR spectroscopy. Whether it be by memory orinductive reasoning, pick B. Acarbonyl bond,C=0, absorbs around 1700 cm"1, so choices Cand Dare eliminated. Using inductive reasoning, you can concludethat the O-H bond of an acid is weaker than the O-H bond of an alcohol (which is why it is more acidic).Weaker bonds have lower bond energies (weaker spring constants, if you will), so they areeasier to stretch. Ifthe bond is easier to stretch, then less energy is required to stretch the bond. Less energy can be seen in the IRabsorbance as a lower frequency (lower wave number). This means that the absorbance for an acidic O-H must beless thana standardalcohol, around 3300 cm"1, sochoice Acannot bepossible.


29. Choice Ciscorrect. There are three synthetic routes presented in the passage. This particular synthesis beginswith an aldehyde, therefore it is the Strecker synthesis, as shown inFigure 3. Upon matching the reagents fromFigure 3 with the reagents in the answer choices, it is choice C that is a match. If you chose to evaluate eachstep in each answer choice, the sequence ofproducts from the steps in choice C are an imide, cyano amine andamino carboxylic acid. Following the reaction step by step also leads to the correct choice. In choice A, thecompound would beoxidized into a carboxylic acid and then deprotonated. Inchoice B, the cyano group wouldadd to the carbonyl to form a cyano alcohol which can then form an acid with acidic workup, but not an aminoacid. Without adding nitrogen tothe compound, it isnot possible toform anamino acid. Choice Dhas too manycarbonsin the reactant,so it cannot form the product. Thebest (and correct) choiceis answer C.

30. Choice Ais correct. Every amino acid in a hydrophobic solvent hasa broad signal between 10 and 12 ppmin the1HNMR due to the carboxylic acid hydrogen and a aHNMR signal of twice the intensity between 2.5 ppm and4.5 ppm due to the two hydrogens on nitrogen. Deuterochloroform, CDCI3, is an aprotic, hydrophobic solvent, sothesignal just above 10ppm is attributed to the carboxylic acid protonand thesignal around3 ppm is attributedto the two amino terminal hydrogens. The ^NMR has three signals, in a 1 : 2 : 2 ratio, so the amino acid hasthree types of hydrogens. Because there is only one other signal (the triplet around 2.1 ppm for the alphahydrogens), there is only one other type of hydrogen present on the amino acid molecule. This means that the R-group must be a hydrogen, because any other R-group would introduce additional signals into the 1HNMRspectrum. The only amino acid with two equivalent hydrogens in the side chain is glycine (the side chain is H).Glycine has three types of hydrogens, so choice A is the best answer.

31. Choice B is correct. In the Gabriel synthesis, the amino acid is released when the phthalimide complex ishydrolyzed under acidic aqueous conditions. This means that the side product results from the hydrolysis of thecyclic imide. This makes the side product a diacid, making choice B the best answer. Choice A is not possible,because hydrolysis would cleave the amide bonds. Choice C is too reduced and choice D cannot happen, becausehydrolysis would likely cleave the cyano groups if the pH was correct.

32. Choice B is correct. An amino acid itself has one unit of unsaturation due to the ic-bond of the carbonyl in thecarboxylic acid. The questions centers around the units of unsaturation in the side chain. Aliphatic, bydefinition, refers to an alkyl group with no rings and no rc-bonds. Thus, there is no unit of unsaturation present inan aliphatic side chain. The units of unsaturation for the side chain is zero so the total units of unsaturation forthe molecule is one. Pick B, and you can proudly consider yourself a correct answer picker.

33. Choice D is correct. When L-cysteine is completely uncharged, there are five unique hydrogens present oncysteine. When cysteine exists as a zwitterion, there are only four unique hydrogens present, because thecarboxylic acid proton has been lost. The solvent is deuterochloroform, a hydrophobic solvent because thepresence of deuterium on carbon is aprotic, so cysteine exists in its uncharged state. Consequently, there are fivedifferent hydrogen signals in the ^HNMR. Pick D for optimal results. L-Cysteine is shown below with all ofthe unique hydrogens labeled.

H2N. .CH2SH

OH

o

L-Cysteine

34. Choice D is correct. Histidine has a positively charged side chain when the pH of the aqueous solution is lessthan the pKa value of the side chain This means that the histidine side chain goes from positively charged toneutral upon deprotonation. Because a hydrophobic environment cannot stabilize a positive charge, thehistidine side chain prefers to be neutral in a hydrophobic environment. Histidine will therefore lose the H+more readily in a hydrophobic environment than a hydrophilic environment. Losing H+ more readily results ingreater acidity and thus a lower value for the pKa. Choice A is eliminated, because log functions do changewhen the number changes. The pKa decreases, not increases, when it becomes more acidic, so choice B cannot becorrect. Upon deprotonation of the side chain, the side chain becomes neutral, not negatively charged, so choiceD is a better answer than choice C.


35. Choice D is correct. In order for an amino acid to have three equivalence points, when starting in its fullyprotonated state and being titrated by a strong base, the amino acid must have three acidic protons. In order foran amino acid to have three acidic protons, the amino acid must have an active proton on the side chain at lowpH (i.e., have a side chain that is either acidic or basic). Tyrosine has an acidic side group (a phenol proton hasa pKa around 10) while the R-group for the other three choices (glycine, alanine, and leucine) are all alkylgroups (none of which have an acidic proton). The best answer is therefore tyrosine, choice D.

Passage VI (Questions 36 - 43) Amino Acid Trivia

36.

37.

38.

39.

Choice C is correct. According to the passage, there are three aromatic amino acids, phenylalanine,tryptophan, and histidine. This eliminates choice A. According to the passage, threonine is one of the essentialamino acids. It has a hydroxyl group on its side chain, so there is at least one essential amino acid that ishydrophilic. This eliminates choice B. Essential amino acids are natural amino acids, so they must be L-aminoacids. With the exception of cysteine, L-amino acids have an S-stereocenter at the a-carbon. Consideringcysteine is not one of the essential amino acids listed in the passage, there are no essential amino acids with anR-stereocenter at the a-carbon, so choice C is the best answer. At a pH of 6, arginine carries a positive charge, sochoice D is eliminated. Histidine has a side chain pKa around 6.05, so it will exist in a state where slightlyover half is protonated and cationic.

Choice D is correct. Whenever a protein is removed from a hydrophobic environment and introduced into ahydrophilic environment, the secondary structure of the polypeptide backbone will be altered, because thehydrogen bonds that are intramolecular in thelipid environment cannow be formed with thesurrounding waterin the hydrophilic environment. The result is that the protein will turn itself inside-out as it passes from oneenvironment into another. Pick D, and get another one correct.

Choice B is correct. Enzymes suchas chymotrypsin, trypsin, and thermolysin break the peptide linkages (amidebonds) of specific amino acid residues within a protein. This breaks the protein into smaller fragments, whichcan then be evaluated with greater ease. This is done with large proteins, because sequencing techniques havean upper limit to the number ofamino acids they accurately determine. Before using Edman's reagent, a proteinis denatured using 6-mercaptoethanol and urea, to denature and elongate the protein for easier analysis.However, it isnot an enzyme thatdoes this, sochoice A iseliminated. Enzymes donot cause a protein tocouplewith another protein, and transcription is not associated with protein sequencing, so choice C can be eliminated.Choice Disa throw away answer, because there is no mounting ofa protein onto slides insequencing. The bestanswer is choice B, because a large protein can be accurately sequenced after it has been cleaved into smallpieces.

Choice Cis correct. H is in the front position at both chiral centers, so whatever arc is drawn when prioritizingthe substituents from highest through lowest must be reversed to determine the chirality. In this case carbon 2gets a clockwise arc and carbon 3 gets a counterclockwise arc. Because the molecule should really be viewed fromthe backside, reversing these arcs correctly identifies carbon 2 as an S-center and carbon 3 as an R-center. Thecompound has chirality of 2S, 3R, so choice C is the best answer. The determination is shown below.

HO CH,

For carbon-2, the arc must be reversed forcorrect viewing, so it has S-chirality.

HO H

HO CH,

H.N

For carbon-3, the arc must be reversed for

correct viewing, so it has R-chirality.

Copyright © byThe Berkeley Review® 228 AMINO ACIDS & AMINES EXPLANATIONS

40. Choice D is correct. Phenylalanine has an alkyl side chain, so the only active protons it possesses are for theamino terminal and the carboxyl terminal. The carboxyl terminal has a pKa between 2 and 3, so at a pH of 7.4,the pH is greater than the pKa and therefore the carboxyl terminal of phenylalanine is deprotonated andanionic. The amino terminal has a pKa between 9 and 10, so at a pH of 7.4 the pH is less than the pKa andtherefore the amino terminal of phenylalanine is protonated and cationic. There are no acid/base propertiesassociated with its side chain, so phenylalanine will always have a neutral side group. The result is thatphenylalanine has one cationic site and one anionic site, so it exists primarily as a zwitterion. While it has nonet charge, because it has charged sites, it cannot be considered uncharged. Choice A is eliminated. It has nooverall charge, so it cannot be considered to be an anion or a cation, so choices B and C are eliminated. The bestdescription is a zwitterion, so choice D is the best answer.

41. Choice D is correct. Thisquestionrequires that you understand a titration curvein a conceptual manner. Becausethe base is 1.5 times more concentrated than the alaninesolution and an equal volume of each solution is mixed,the amount of base that has been added is one-and-one-half equivalents. Starting at the fully protonated state,addition of 1.5 equivalents of titrant base takes the alanine solution past its first equivalence point and halfway to the second equivalencepoint of alanine. If you observe the titration curve for alanine, this makes the pHof solution equal to pKa2, which lies exactly halfway between the first and second equivalence points. Thevalue of pKa2 lies between 9 and 10 for most amino acids, so choice D is the best answer. The titration curve foralanine titrated by NaOH is shown below:

Alanine starts

fully protonated

H2A+72 1 lV2

Equivalents base added

Alanine finishes

between the two eq. points.

42. Choice B is correct. For an amino acid with a hydrophobic, alkyl side group, there are only two pKa values. Todetermine the isoelectric point, you must average the two pKa values that are associated with the zwitterion,in this case the only pKa values associated with the amino acid. For isoleucine, the values for pKai and pKa2are 2.3 and 9.7, respectively. The average of the pKa values is 6, so choice B is the best answer.

43. Choice D is correct. Because there are three equivalence points (vertical inflection points) on the titration curve,the amino acid that the titration curve represents must have a side chain with an active proton. An activeproton is associated with either an acidic or basic side group. This eliminates choices A and B, which both haveside chains that contain no active protons. Because both of the first two plateaus in the graph occur below a pHof 7.0, the first two pKa values for the amino acid must be less than 7.0. The middle of the plateau representsthe point at which the pKa is equal to the pH of the solution. So, from the graph, a qualitative determinationof the pKa values can be made. Because the value of pKa2 is less than 7.0, the side chain is a carboxylic acidgroup rather than an amine group. Aspartic acid has a carboxylic acid group on the side chain, which has a pKavalue less than 7.0, so choice D is the best answer. The side chain of lysine is an amine, so its pKa is above 7.0.Choice C is eliminated for this reason.

Copyright © by TheBerkeley Review® 229 AMINO ACIDS & AMINES EXPLANATIONS

Passage VII (Questions 44 - 51) Acidity and Basicity of Amino Acids

44. Choice D is correct. To make the pKa lower, something must be done to the compound to make the proton moreacidic. In choice A, the oxidation of carbon 1 by four electrons results in the conversion of a primary alcohol intoa carboxylic acid, which increases the acidity of the compound through electron withdrawal via resonance.This results in a lower pKa, so choice A is eliminated. Because sulfur is larger than oxygen, exchanging oxygenfor a sulfur allows the proton to be lost more easily. If the proton is lost more easily, then the compound is moreacidic and thus has a lower pKa value. This can be seen when comparing serine (primary alcohol side group) tocysteine (primary thiol side group), where the side chain pKa for cysteine is about 6 less than the side chainpKa for serine. This eliminates choice B. Replacing the propyl group with a phenyl group has the same effectas oxidizing carbon 1 from a primary alcohol into a carboxylic acid, although the resonance effect is not asstrong. Regardless, the compound becomes more acidic, so the pKa decreases. This can be seen when comparingserine (primary alcohol side group) to tyrosine (phenol side group), where the side chain pKa for tyrosine isabout 4 less than the side chain pKa for serine. Choice C is thus eliminated. The only answer choice left ischoice D. By replacing the oxygen with nitrogen, the compound becomes less acidic. Amines are less acidic thanalcohols. Nitrogen is roughly the same size as oxygen, so the effect is not the same as choice B with sulfur.Because nitrogen is less electronegative than oxygen, it "pulls" electrons less than oxygen does. This makes thenitrogen more basic and less acidic than the oxygen. This raises the pKa value making choice D the choice thatwould not decrease the pKa.

45. ChoiceB is correct. When the pH of the solution is greater than the pKa for a proton, that proton is removed bythe solution and the compound exists in its deprotonated state. When the pH is high, the solution is basic andthus deprotonates the proton. This same conclusion can be formed using the Henderson-Hasselbalch equation.When the pH is greater than the pKa, the log term in the H-H equation must be positive, so the concentration ofthe base (deprotonated form) must be greater than the concentration of the acid (protonated form). This meansthat when the pH is less than the pKa, the site retains its proton. Taking the sites one at a time, the carboxylterminal has a pKa of 2.0, so at pH = 5.0, it is deprotonated and carries a -1 charge. The carboxylic acid sidechain on the aspartic acid (right amino acid) has a pKa of 3.9 which is also less than the pH, so it too isdeprotonated and thus carriesa -1 charge. Theamino terminal has a pKa value of 9.2, which is greater than thepH, so it remains protonated and thus carries a +1 charge. The amine side chain on the lysine (left amino acid)has a pKa of 10.8, which is alsogreater than the pH, so it is protonated and thus carries a +1 charge. Thesum ofthe charges is zero,so the best answeris choice B. The structure at a pH of5.0 is drawn below.

°YOH

H fCHol,H OH (CH2)4J

pK, = 3.9

OHPKa= 2.0

NHjPK^lO.8

46. ChoiceD is correct. In order for the pH to be greater than the pKa, the conjugate base (deprotonated form) mustbe in higher concentration than the conjugate acid (protonated form). The question asks for the deprotonated-to-protonated ratio, which is greater than 1. This eliminates choices A and B. Because the pH is one unit abovethe pKa, the log of the base to acid ratiomust equal 1, according to the Henderson-Hasselbalch equation. Thelog of 10is equal to 1,so the ratio of conjugate base to conjugate acid must be 10:1. The best answer is choiceD.

47. Choice A is correct. Because the electron withdrawal associated with resonance is greater than the electronwithdrawal associated with the inductive effect, a carboxylic acid is more acidic than the alcohol group on acarbon with one fluorine atom bonded to it. From the passage, we know that the pKa for the carboxylic acid islower than the pKa for the protonated amine site, so the carboxylic acid (site a) must be the most acidic site.Thismeans that proton a is the most acidic proton on the molecule and only choice A lists proton a as most acidic,so choice A is the best answer.


48. Choice C is correct. Because the sulfur is still protonated at pH = 7.4, the pKa for the proton on sulfur must begreater than 7.4. Because the proton on sulfur is the second proton to be lost (it is lostbefore the proton on theamino terminal), it must be more acidic than the amino proton. The amino proton has a pKa around 9.5, so the Hon sulfur must have a pKa less than 9.5. The pKa for the proton on sulfur lies somewhere between 7.4 and 9.5, sothe best answer is choice C.

49. Choice C is correct. The fully protonated form of glutamic acid starts with a positive one charge, because theamino terminal is protonated carrying a positive one charge and the carboxyl terminal and carboxylic acid sidechain are each protonated and neutral. The compound monosodium glutamate has one sodium cation, sotherefore the glutamate portion must carry a negative one charge. To be converted from a positive one chargeinto a negative one charge, glutamic acid must be deprotonated twice. The best answer is therefore choice C.Drawn below is the step by step conversion from the fully protonated form of glutamic acid to the monoanionform of glutamic acid (glutamate).

O

H3NO™- 1stequivalent

H CH,

H2C .OH

i| net chargeO +1

OK H CH,

H2C

2nd equivalent

OH

i net chargeO O

O

H3NO

H CH,

OH2Cnet charge

O -1

50. Choice B is correct. As drawn, the dipeptide molecule carries a +3 charge when fully protonated. When themolecule is at a pH of 1.8, the two forms of the molecule present in solution are the fully protonated form(carrying a + 3 charge) and the form with the carboxyl terminal deprotonated (carrying a +2 charge). When thepH is 1.8, equal to pKai, these two forms are present in a fifty-fifty ratio. When the molecule is at a pH of 6.1,the two forms of the molecule present in solution are the form with the carboxyl terminal deprotonated (carryinga +2 charge) and the form with both the carboxyl terminal and histidine side chain deprotonated (carrying a +1charge). When the pH is 6.1, equal to pKa2, these two forms are present in a fifty-fifty ratio. In order to have aten fold excess of the +2 species, the pH must be less than pKa2 by a factor of log 10, which equals pKa2 -1. Thiseliminates choices C and D. The pH must be one unit less than 6.1 (pKa2), making choice B, 5.1, the best answer.

51. Choice D is correct. The most basic compound has the lowest pKb value and the weakest conjugate acid. ThepKb value can be found by subtracting the pKa for the conjugate acid from 14. ChoiceA is eliminated, because itis the conjugate acid of choice D. The conjugateacid is always less basic than its conjugate base. Tiie pKfc> valuesfor the three remaining choices can be found from the pKa values given for the sample compounds. For choice B,the pKb value is 14 - 8.4 = 5.6. For choice C, the pKb value is 14 - 5.0 = 9.0. For choice D, the pKb value is 14-10.0 = 4.0. The lowest pKb value is found with choice D, therefore the most basic compound is choice D.

Passage VIII (Questions 52 - 58) Amino Acids and Polypeptides

52. Choice C is correct. At a pH of 1.5, the tripeptide is fully protonated. The protonated amino terminal carries apositive 1 charge and the protonated carboxyl terminal carries no charge. If none of the side chains carry apositive charge, the overall tripeptide carries a positive 1 charge. Only the basic amino acids (histidine,lysine, and arginine) have a positively charged side chain when protonated. This means that the correctanswer choice shall not contain histidine, lysine, or arginine. Only choice C fits this criterion, so the bestanswer is therefore choice C.

53. Choice D is correct. Hydroxyproline has the normal chiral center associated with an amino acid plus it has anadditional chiral center associated with the hydroxyl group of the side chain. This eliminates choice A. Just asproline induces deviations in the secondary structure of a protein (turns and kinks), hydroxyproline causesstructural deviations as well. This makes choice B valid, thus it is eliminated. Alcohols are hydrophilic, sochoice C is valid and thus can be eliminated. The side chain of hydroxyproline is an alcohol. The pKa of analcohol is approximately that of water (15.5), therefore the pKa cannot be around 4.0. If it were around 4.0, thenserine would be expected to have an active side chain proton as well. The best answer is choice D.


54.

55.

Choice A is correct. The highest isoelectric pH is found with the tripeptide with the greatest number of basicside chains. This is true because the basic side chains retain positive charge at higher pH values than the otherside chains. Histidine, lysine, and arginine are the three basic amino acids, and only choice A contains at leastone of the three. Because choice A contains two basic side chains, it has an isoelectric pH that is an average ofpKa3 and pKa4. The four pKa values associated with Met-Lys-Arg are 2.2 (for the carboxyl terminal ofarginine), 9.3 (for the amino terminal of methionine), 10.8 (for the side chain of lysine), and 13.0 (for the sidechain of arginine). The isoelectric pH is an average of 10.8 and 13.0, which is 11.9. This makes choice A thetripeptide with the highest isoelectric point.

Choice B is correct. Glycine has no chiral center, so choice A is eliminated. Isoleucine and threonine each havea chiral center in their side chain resulting in two chiral centers overall, so choices C and D are eliminated.Histidine has all s/?2-hybridized carbons and nitrogens in its ring, so there are no chiral centers on its side chain.Because histidine has only one chiral center, choice B is the best answer.

Choice C is correct. This question is a unique way of asking "which dipeptide has the lowest isoelectric point?"The lowest isoelectric point is found with the dipeptide carrying no basic side chains and preferentially anacidic side chain (such as aspartic acid or glutamic acid). Choice A is eliminated, because histidine is basic,which increases the isoelectric pH. Choice B is eliminated, because lysine is basic, which increases theisoelectric pH. Choice D is eliminated, because arginine is basic, which increases the isoelectric pH. Choice Chas no basic sidechain. The isoelectric point is an average of the carboxyl terminal pKa ofcysteine (1.8) and theside chain pKa of aspartic acid (3.9) which results in an isoelectric point of 2.85.

Choice D is correct. Cations migrate to the cathode, therefore this question is asking for the amino acid thatexists as a cation, or carries a partially positive charge, at a pH of 7.0. To carry a positive charge, partial orfull, the isoelectric pH (pi value) must be greater than the solution pH. This question is therefore asking for theamino acid with the highest pi value (only one can be the best answer, so only one can have a pi value greaterthan 7.0, which makes it the greatest pi value). In this case, the only basic amino acid (which has the highestpi value) is lysine, so choice D is the best answer.

Choice A is correct. The isoelectric pH is an average of the pKa value leading to the zwitterion and the pKavalue going from the zwitterion. To have a pi less than 5.0, the pKa values being averaged must be low numbers.For amino acids with side chains having no active protons, the pi is found by averaging the carboxyl terminalpKa and the amino terminal pKa. This means that the pi is an average of about 2.5 and 9.5, resulting in a piaround 6.0. This eliminates choices C and D. For amino acids with basic side chains, the pi is found byaveraging the side chain pKa and the amino terminal pKa. This means that the pi is an average of a numberbetween 6and 12 andanumber around 9.5, resulting inapi around 9.0. This eliminates choice B. For amino acidswith acidic side chains, the pi is found by averaging the carboxyl terminal pKa and the side chain pKa. Thismeans that the pi is an average of about 2.5 and a number between 3 and 10, resulting in a pi around 3.0. Thismakes choice Athe best answer. To summarize, acidic amino acids have pi values less than 5, basic amino acidshave pi values greater than 7, and amino acids with neutral side chains have pi values around 6.

56.

57.

58.

Passage IX (Questions 59 - 66) Physical Properties of Amino Acids

59. Choice C is correct. The side chain that is most likely to be found in a hydrophilic pocket is the side chain ofthe most water soluble amino acid. The side chains in the answer choices are methionine, tyrosine, glutamine,and cysteine. According to the data in Table 1, the most water soluble amino acid of the choices is glutamine at3.7 grams per 100 mL water. This is predictable, because of the hydrogen bonding of the side chain. The bestanswer is choice C.

60. Choice A is correct. Choices Band Cshould be eliminated immediately, because the alkyl side chain of leucinedoes not form hydrogen bonds and the charged side chain oflysine is polar. Choice D is true, but the fact thatleucine has a nonpolar side chain does not enhance the water solubility of leucine. Choice D is true, butirrelevant, which eliminates it. Because lysine can form hydrogen bonds with the hydrogens on nitrogen, choiceA is valid.

,<SCopyright © by The Berkeley Review 232 AMINO ACIDS & AMINES EXPLANATIONS

61. Choice C is correct. For amino acids with both neutral and acidic side chains, the isoelectric pH is found byaveraging pKai and pKa2- The pi values for the amino acids with both neutral and acidic side chains areconsequently less than 6.0. For basic amino acids, because the pi is an average of pKa2 and pKa3, the value of piis greater than 7.0. The only basic amino acid in the answer choices is histidine, choice C.

62. Choice C is correct. For cysteine, the amino terminal is protonated and carries a positive charge. The carboxylterminal is deprotonated and carries a negative charge. This eliminates choices A and B. Because the pH of thesolution is less than the pKa of the side chain, it is protonated and carries no charge. Tliis makes choice C thebest answer and eliminates choice D.

63. Choice B is correct. The acidity of a compound is measured by its pKa, so the lowest pKa value is associatedwith the most acidic compound. The carboxyl terminal proton is the first to be lost, so it is represented as pKai,meaning that the question is comparing the relative acidity of the carboxyl terminal protons of arginine andalanine. The carboxyl terminal pKa for arginine is lower than carboxyl terminal pKa alanine (according to thevalues in Table 1), so choices A and C are eliminated. Choice B is the better answer, because it is the side chain,not the amino terminal, that differs between arginine and alanine.

64. Choice D is correct. To separate two compounds using extraction in an ether/water biphasic solution, it is bestwhen one is charged and the other is neutral. It is at the isoelectric pH that the species carries no net charge,thus when the pH is equal to the pi of lysine, then lysine is neutral and therefore more soluble in ether than inwater. At this same pH, tyrosine is partially negative and thus water-soluble. When the pH is equal to the piof tyrosine, then tyrosine is neutral and therefore more soluble in ether than in water. At this same pH, lysine ispositive and thus water-soluble. At all other pH values, both lysine and tyrosine are charged. The best answeris choice D. Because of the high water-solubility of lysine in water, the extraction is best carried out at theisoelectric pH of tyrosine.

65. Choice D is correct. At a pH of 4.0, the carboxyl terminal is predominantly deprotonated, because the pH isgreater than the pKa of the carboxyl terminal. This makes choice A false. The isoelectric pH for histidine is7.60, so at a pH of 4.20 (a value less than 6.05), the side chain carries a positive charge, so far less than 90%exists as a zwitterion. This makes choice Bfalse. The pKa of the amino terminal is greater than the pH, so theamino terminal is predominantly protonated. This makes choice C false. Because the pH is far less than thepKa ofthe side chain, the side chain ismostly protonated (more than ten percent). Tiie best answer ischoice D.

66. Choice B is correct. The background information that you have to know is that DEAE-cellulose carries apositive charge at neutral pH, so at a pH of 6.20, the column must be positively charged. Because the columncarries a positive charge, negatively charged amino acids are the ones that bind to the column. The lower thevalue of the pi, the more anionic the amino acid. This means that the amino acid with the lowest pi value isthe one that is most hindered by the DEAE-cellulose column. Glutamic acid has the lowest pi, because the sidechain has a pKa value of only 4.32. This makes choice B the best answer.

Passage X (Questions 67 - 73) Protein Hormones

67. Choice B is correct. The side chain of alanine (CH3) is bulkier than the side chain of glycine (H), thereforehuman PTH hasa bulkier Rgroup than bovine PTH. This eliminates choices A and C. The secondary structure ismore affected by the steric hindrance of the side chain than the tertiary structure, so thebest answer is choice B.

68. Choice B is correct. All of the amino acids from number 35,valine, to number 43, proline, have alkyl side chains,and thus arehydrophobic and uncharged. This eliminates choices A, C, and D. The bestanswer is choice B.

69. Choice D is correct. The polypeptide pre-pro-PTH is 115 amino acids in length. After losing the twenty-threeamino acid segment and the two methionine residues (twenty-five amino acids in total), the resulting pro-PTHcontains ninety amino acids. In the Golgi region, pro-PTH (ninety AA) is converted into PTH (eighty-four AA),so six amino acids must be lost. This means that a hexapeptide is lost, making choice D the best answer.

70. Choice D is correct. Within PTH, there is a tryptophan, Trp, a phenylalanine, Phe, and no tyrosine residues,Tyr. The Trp is in position 23 and the Phe is in position 34. This means that PTH chymotrypsin cleaves after AA#23 and after #34, resulting in three fragments consisting of 23amino acids (AA #1 through AA #23), 11 aminoacids (AA #24 through AA #34), and 50 amino acids (AA #35 through AA #84). Thebest answer ischoice D.


71. Choice A is correct. Choices Band D can be eliminated immediately, because there is no nitrogen in the sidechain ofmethionine. Replacing the sidechain ofmethionine with an aliphatic alkyl group does not increase ordecrease the number of nitrogen atoms in the molecule. Choice C can be eliminated, because no change inreactivity is observed. This makes the best answer choice A. The side chain listed is similar to methionineexcept that the sulfur has been replaced by a methylene group (CH2).

72. Choice B is correct. The active region of human-PTH is given in the passage as being amino acids #1 through#34, so the analogs to PTH should be similar in this same region (the active region). This eliminates choices Aand D and makes choice B the best answer so far. The C-terminal and N-terminal do no affect the active sitereactivity, so choice C is eliminated. The best answer is choice B.

73. Choice B is correct. Choice A canbe eliminated, because reactivity is eliminated when the structure is reducedto twenty-eight amino acids. Choices C and D are poor choices, because both are found in the 35 through 84segment that does not directly affect reactivity, according to the passage. Based on the in vivo reactivity data,the reactivity drops severely when going from 31 to 28 amino acids, so the active site includes an amino acidbetween 29 and 31. Choose B for best results.

Passage XI (Questions 74 - 81) Enzymatic Cleavage and Sequencing

74. Choice C is correct. Because basic amino acids, like histidine, have a +2 charge when fully protonated, theymust be deprotonated twice to reach their neutral state (as a zwitterion). The isoelectric point is found byaveraging the two pKa values that involve the zwitterion. For an amino acid with a basic side group, the pi isan average of the last two pKa values (pKa2 and pKa3). For histidine, the isoelectric point is (6.1 + 9.2)/2 =7.65, making choice C the best answer. Basic amino acids all have a pi greater than 7.0, so choices A and Bshould be eliminated immediately.

Choice Bis correct. To determine the migration direction of an amino acid in abuffered gel, the isoelectric pointmust first be determined. Histidine is a basic amino acid, so the isoelectric point of histidine is found byaveraging its last two pKa values. For all basic amino acids (histidine, lysine, and arginine), the isoelectricpoint (pi) is the average of pKa2 and pKa3. For all other amino acids, the isoelectric point is the average ofpKal and pKa2. The pi for histidine is (6.1 +9.2)/2 =7.65. This value is greater than the pH of the solution, 7.0,so histidine is partially positive (cationic) in asolution with apH of 7.0. Cations migrate to the cathode in gelelectrophoresis, so choice B is the best answer.

Choice Ais correct. Sanger's reagent, 2,4-dinitrofluorobenzene, binds the first amino acid in a polypeptidefragment at its amino terminal. The amino terminal nitrogen substitutes on the aromatic ring for the fluorineatom. This means that the amino acid that binds 2,4-dinitrofluorobenzene must be the first amino acid in thepolypeptide. Because 2,4-dinitrofluorobenzene binds serine in the unknown polypeptide, the first amino acid inthe polypeptide must be serine. Choose Aif you know what's good for you.

Choice Cis correct. Thermolysin cleaves aprotein on the amino side of leucine, isoleucine and valine. FragmentII contains one valine (amino acid #3) and one isoleucine (amino acid #5), neither of which is the first aminoacid. Hence, thermolysin cleaves Fragment II twice, breaking it into three separate pieces. You always checkthat neither valine, leucine, nor isoleucine are the first amino acid in the sequence when using thermolysin. Thebreak for Fragment II caused by thermolysin is as follows: His-Ser I Val-Phe I Iso-Tyr-Phe. Pick Cto score bigon this question.

Choice Cis correct. Without aside group, ageneric amino acid weighs 17 grams (for NH3) +13 grams (for CH) +44 grams (for C02) =74 grams/mole plus the mass of the Rgroup. If the total weight is 89 grams/mole, the massof the side group must be 15 grams/mole. Of the answer selections, only the methyl group has a mass of 15grams/mole. For this reason, you just have to pick C. To determine the answer, you could also have gone througheach answer choice and systematically computed each of their molecular weights if you know the structure ofthe amino acid.

75.

76.

77.

78.

79. Choice Ais correct. According to Table 1, clostripain cleaves at the carboxyl side of arginine. Fragment IIcontains no arginine, so itwill not be cleaved by clostripain, leaving the fragment in one piece. Fragment II staysas His-Ser-Val-Phe-Ile-Tyr-Phe, so choice A is the best answer.

Copyright © byThe Berkeley Review® 234 AMINO ACIDS & AMINES EXPLANATIONS

80. Choice C is correct. Pepsin cleaves at the carboxyl side of phenylalanine, tryptophan, tyrosine, aspartic acid,glutamic acid and leucine. Thermolysin cleaves at the amino side of leucine, isoleucine, and valine. This meansthat if the carboxyl side of either phenylalanine, tryptophan, tyrosine, aspartic acid, glutamic acid or leucineis bonded to the amino side of leucine, isoleucine, or valine, then a peptide linkage exists that will be cleaved byeither pepsin or thermolysin. In such a case, the same fragments would result whether pepsin or thermolysinwere employed. In choice A, the carboxyl side of isoleucine is linked to the amino side of phenylalanine, whichis the opposite of what is needed to lead to equal cleavage by pepsin and thermolysin. Choice A is eliminated.In choice B, the carboxyl side of leucine is linked to the amino side of leucine, and both pepsin and thermolysinreact with proteins containing leucine. The problem with this answer choice is that they cleave on differentsides, so pepsin will generate a fragment ending in leucine and a free leucine while thermolysin generates a freeleucine and a fragment starting with leucine. Choice B is eliminated. Pepsin does not react with either valineor methionine, so no cleavage occurs with choice D when using pepsin. Because thermolysin reacts with proteinscontaining valine, a break will occur before the Val-Met linkage, resulting in fragmentation. This eliminateschoice D. Choice C is the best answer, because pepsin cleaves after tyrosine and thermolysin cleaves beforevaline, so the same break occurs for the Try-Val linkage whether pepsin or thermolysin is used.

81. Choice C is correct. As given in Table 1, chymotrypsin cleaves the carboxyl side of phenylalanine, tryptophanand tyrosine. There are two tripeptide fragments formed when using chymotrypsin, one ending in either Ser orPhe (although by convention we assume it ends in Phe as written) and the other ending in Gly (there is no issuewith the order of the second fragment, because the sequence is the same either backwards or forwards.) The Ser-Val-Phe fragment must be first and it must start with Ser, because it is the only one with either Phe, Trp, or Tyrpresent at the end. This means that the Gly-Cys-Gly fragment is the second of the two fragments. Putting thetwo fragments back together yields the hexapeptide Ser-Val-Phe-Gly-Cys-Gly. This makes choice C the bestanswer. If you didn't solve it that way, there is an easierway. Only choice C has serine as the first amino acidin the sequence. You know serine is the first amino acid in Fragment I from the results of the treatment withSanger's reagent.

Passage XII (Questions 82 - 88) Peptide Sequencing

82. Choice D is correct. The role of the 6-mercaptoethanol is to cleave any disulfide linkages formed between theside chain residues of cysteine. This is stated several times in the passage, at the start of each experiment. Thebest answer is choice D.

83. Choice A is correct. As stated in the passage, the purpose for adding 2,4-dinitrofluorobenzene (2,4-DNFB) wasto bind the amino terminal of the first amino acid in the polypeptide chain. The 2,4-dinitrofluorobenzene onlybinds the first amino acid making it serve as a marker for the first amino acid. This means that the role of 2,4-dinitrofluorobenzene is to identify the first amino acid in the sequence. This makes choice A correct.

84. Choice A is correct. Only histidine appears at the N-terminal in one of the fragments in each of the threeexperiments, Experiments II, III, and IV. Using the data from Experiment II and knowing that histidine is thefirst amino acid (implying that Lys is the second amino acid), the possibilities for the last amino acid in thepolypeptide are either alanine or arginine. The fragments of Experiment III show that the last amino acid iseither serine or alanine (because His is first, and therefore Cys must be fourth). Combining this with theinformation from Experiment II and Experiment III, the last amino acid in the sequence must be alanine. Thismakes choice A your best choice.

85. Choice D is correct. The fragments formed in Experiment IV show that chymotrypsin either cleaves after Tyrand Phe or before Cys and Ala. Because the last fragment (N-His-Lys-Met-Cys-Leu-Gly—Ala-Phe-C) shows apeptide linkage to the amino terminal of alanine still in tack, chymotrypsin must cleave after Tyr and Phe.Choice A is valid. The fragments formed in Experiment II show that trypsin either cleaves after Lys and Arg orbefore Ser and Met. There is no way to determine which is actually the case, so as it is, choice B is not invalid.The fragments formed in Experiment III show that thermolysin either cleaves after Cys andSer or before He andLeu. Because the last fragment listed from the thermolysin cleavage (N-Leu-Gly-Ala-Phe-Cys—Arg-Ser-C)shows a peptide linkage to the carboxyl terminal ofcysteine still in tack, thermolysin mustcleave before lie andLeu. This makes choice D your best choice, because thermolysin does not cleave the amino terminal of alanineand cysteine. This also eliminates choice C.


86. Choice C is correct. Because the last fragment in Experiment III (resulting from the thermolysin cleavage), N-Leu-Gly-Ala-Phe-Cys—Arg-Ser-C, shows a peptide linkage to the carboxyl terminal of cysteine still in tack,thermolysin cannot cleave following cysteine. If it did cleave after cysteine and serine, rather than beforeisoleucine and leucine, then the Cys-Arg linkage in fragment 3 would have been broken, resulting in anadditional fragment. The cleavage has nothing to do with disulfide linkages, so choice A is eliminated. Wehave no information to assess whether thermolysin hasany interactions with protic side chains, so while choiceBmay be true, we cannot choose it. Based on the third fragment, thermolysin must cleave before isoleucine (He)and leucine (Leu). As it is, thermolysin also cleaves before valine (Val). This makes choice C your best choice.

87. Choice D is correct. Histidine is the only amino acid that appears at the N-terminal of one of the threefragments in both Experiment II and Experiment III, so histidine must be the first amino acid. Coupling the factthat histidine is the first amino acid in the polypeptide along with comparison of the fragments fromExperiment II and Experiment III, the fragments can be overlapped and matched to sequence the entire peptide.For instance, Experiment III gives the first four amino acids as His-Lys-Met-Cys. Knowing the third and fourthamino acids in the polypeptide identifies the first two fragments in Experiment II (N-His-Lys-Met-Cys-Leu-Gly-Ala-Phe-Cys-Arg-C), which defines the third fragment in Experiment II by default. The cumulative resultgives the entire polypeptide, so choice D is tiie best answer.

Choice A is correct. Histidine is the only amino acid at the start of a fragment in at least one fragment inExperiments II, III, and IV. This means that the polypeptide must begin with histidine, which eliminateschoices Cand D. Alanine is the only amino acid at the end of afragment in at least one fragment in ExperimentsII, III, and IV. This means that the polypeptide must end with alanine, which eliminates choices Band C. Theonly choice that begins with histidine and ends with alanine is choice A, making it the best answer. This can beverified by matching the fragments from the three experiments, which shows that the correct sequence muststart with His-Lys-Met-Cys and end with Ser-Ile-Tyr-Ala.

Passage XIII (Questions 89 - 96) Amino Acid Stereospecificity in Synthesis89.

91.

92.

Choice B is correct. Choices C and D are eliminated, because in an organic solvent, the N-terminal isdeprotonated and the C-terminal is protonated, resulting in both sites being uncharged. For the experiment towork, the amino terminal must be neutral and nucleophilic. Choice Ais eliminated, because a protic solventwould be reactive, not inert. Aprotic solvent (such as ethanol) can attack acetic anhydride and undergo asubstitution reaction at the carbonyl carbon. The best answer is choice B.

90. Choice Cis correct. Technique 1involves the use of abrucine salt, which forms a precipitate with a D-aminoacid. The precipitate is what is isolated from the mixture in solution in relatively pure form, not the species insolution, so the best answer is choice C. Because of the precipitation of the D-amino acid, rather than the L-amino acid, choices A and B are eliminated.

Choice Bis correct. An increase in entropy involves an increase in randomness. This occurs with aphase changethat increases the volume occupied by the material or a reaction that generates an increase in the number ofmolecules. Choice Ais eliminated, because it involves purification and separation of stereoisomers, whichincreases the order. The reaction of a D-amino acid with an achiral species neither increases nor decreases theentropy, so choice C is eliminated. In choice D, a solute is converted into a solid, which involves a decrease inentropy of the system. This eliminates choice D. The only increase in entropy occurs when one enantiomer isracemized into a mixture of two enantiomers, because they are no longer separate and ordered. This makeschoice B the best answer.

Choice Bis correct. The enzyme acylase cleaves an acyl bond (found in an amide). Individual amino acids haveno peptide Imkages (amide bonds), so amino acids must first be acylated to form a bond that can react withacylase. Acetic anhydride will acylate all of the amino acids, but acylase will only remove the acyl group frommolecules with the correct stereochemistry. This describes choice B. Because the amino terminal is tied up inthe amide bond, it cannot be protonated, and thus the acylated amino acid is only charged at the carboxylterminal. This makes the species anionic, not cationic, which eliminates choice C. The chiral center is notaffected, given that the priorities remain the same, so choice D is eliminated. The amino terminal is notprotected by the acyl group, because there is no reason to protect it. This eliminates choice A, and makes choiceB the best answer.

Copyright © hy The Berkeley Review® 236 AMINO ACIDS & AMINES EXPLANATIONS

93. Choice A is correct. Choice C should be ruled out, because hog renal acylase is an enzyme, and thus it is notconsumed in the process. Choice D can be eliminated, because acetic anhydride is achiral, so it shows nopreference for one enantiomer over another. Tiie original mixture is fifty percent D and fifty percent L. Afterremoval of one isomer and racemization, twenty-five percent of the original product is D and twenty-fivepercent of the original product is L. After the removal of one isomer and racemization again (the thirdpurification), twelve and one half percent of the original product is D and twelve and one half percent of theoriginal product is L. This means that only twelve and one half percent of the desired enantiomer is isolatedfrom the third cycle. After this point, the percent isolated is less than ten percent, and is not worth the effort.The best answer is choice A.

94. Choice B is correct. Stirring will not vaporize (sublime) a solid unless it results in extreme heating. Even withsixteen hours of stirring, not enough energy will be introduced into the system to get a notable amount of phasechange. This eliminates choice A. Stirring implies that the solid does not dissolve well into the solvent, so therole of agitation is not to centrifuge the solid to the bottom, but to increase the collision frequency and thusincrease the reactivity. This eliminates choice C and makes choice B the best answer. Generating a suspensionincreases the surface area for collision (and thus increases the reaction surface), causing the reaction rate toincrease, not decrease. This eliminates choice D.

95. Choice A is correct. The addition of strong base removes the carboxyl terminal and amino terminal protons, ifthere are protons on the amino terminal to be removed. Because the nitrogen is involved in the amide bond withthe benzoyl group, it never gains the proton that an amine group would gain, so there is no hydrogen to bedeprotonated. This eliminates choice B. The brucine salt was broken apart using a strong acid, not a strong base,so choice C is eliminated. Treatment with strong base in an aqueous environment hydrolyzes the amide bond andthereby removes the benzoyl group from the N-terminal. This makes choice A the best answer, an answer youshould pick if you know what is best.

96. Choice D is correct. In technique 2, the racemate is first treated with benzoyl chloride, which N-acylates all ofthe amino acids in solution, both D and L. The strychnine salt is formed from the selective precipitation of theL-amino acid after it has been N-acylated. Choices A and C can be eliminated, because they contain no benzoylgroup attached to the amino terminal nitrogen. Choice Bhas an R-stereocenter at the a-carbon, while choice Dhas an S-stereocenter at the a-carbon. Choice D is the correct answer choice, because S-stereochemistry isassociated with an "L" amino acid. You should recall the mnemonic, "you're either a DoctoR or you LoSe,"where D and R go together and L and S go together.

(Questions 97 -100) Not Based on a Descriptive Passage

97. Choice C is correct. The real question is "what effect does H+ have on the reaction rate when methyl amine isadded to methyl chloride?" The nucleophile is a good nucleophile and the electrophile is small, so the reactionis an Sn2 nucleophilic substitution. In Sn2 reactions, the rate of the reaction depends on both the electrophileconcentration and the nucleophile concentration. This means that the rate of the reaction varies with theconcentration of neutral methyl amine. Methyl amine must have a lone pair available to attack theelectrophile, so if it is protonated, it cannot be a nucleophile. The implies that at lower pH values there is lessneutral amine, so the reaction is slower at lower pH values and faster at higher pH values. The correct graphshould show greater rates at higher pH values, so choices A and Bare eliminated. Because pH is a log scale andit is located on the x-axis, the relationship of the rate and pH is not linear. Choice C is a better answer thanchoice D.

98. Choice D is correct. The final product, after it has been neutralized (deprotonated and returned to an unchargedstate), is (H3Q2NH. This molecule has two unique types of hydrogens, so it should show only two peaks in its1HNMR spectrum. This eliminates choices A and B, which show three signals rather than just two. The twosignals are in a 6:1 ratio, but that is observed in both choice C and choice D, so we are no further along thanbefore. The six like hydrogens split the neighboring signal into a septet (6 neighboring Hs + 1 = 7 peaks) and theone hydrogen on nitrogen splits the neighboringsignal into a doublet (1 neighboring H + 1 = 2 peaks). This meansthat the !HNMR should have a 1H septet and a 6H doublet, which confirms that choice D is a better answerthan choice C, if correct answers are in fact better than wrong answers. Choose D for the warmth and glow ofwisdom.

<B>Copyright © by The Berkeley Review 237 AMINO ACIDS & AMINES EXPLANATIONS

99. Choice C is correct. According to the question, the addition of base takes the amino acid from its firstequivalence point to the point atwhich the pH equals the value of pKa3- This process can be observed using atitration curve, as drawn below. To carry out this conversion would require one and ahalf equivalents of strongbase. When dealing with equivalents, the conversion isbest beviewed conceptually from a titration curve. 1.5equivalents of strong base is equal to 15 mL given that the concentrations of lysine and NaOH are equal andthere are 10 mL ofthe lysine solution present initially. Select choice Cfor an optimal experience.

HoA

10 15 20 25

mL 0.10M NaOH(aq)added

100. Choice Bis correct. This is trivial knowledge, so based on memory, you can say Aand Care valid methods.Should your memory fail you, the reaction must be reduction of the nitrogen containing compound. L1AIH4 andH2/Pd should be familiar reducing agents. In choice B, there is nothing to provide electrons to nitrogen(reduction is the gain of electrons). Zinc has an oxidation state of +2 and chlorine will not give up electrons.Choice B is the best answer.

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238 AMINO ACIDS & AMINES EXPLANATIONS

Section VIII

OrganicChemistryLaboratoryTechniques

by Todd Bennett

Mixture: Amine, Carboxylic acid, Phenol, and Hydrocarbon

Organic solvent I Weak base(aq)I 1

Organic layer Aqueous layer

Organic solvent | Strong base(ag) 10%| HCl(aq)I I Carboxylic acid

Organic layer Aqueous layer

Organic solvent | Strong acid(ag) 10%]HCl(aq)V

PhfenolOrganic layer Aqueous layer

10%]KOH(aq)Amine

[IroHydrocarbon

Separation Techniques

a) Distillation

i.

ii.

iii.

iv.

v.

Vapor PressureDistillation ApparatusSimple DistillationFractional Distillation

Vacuum Distillation

b) Chromatographyi. Mobile Phase and Adsorbents

ii. Thin Layer Chromatographyiii. Rf valuesiv. Column Chromatographyv. Designer Columnsvi. Gas Chromatography (QC)

c) Extraction

i.

ii.

Acid/Base ExtractionAnalyzing Plow Charts

Purification Techniques

a) Recrystallizationi. Solvent choice and Refluxingii. Hot and Cold Filtration

iii. Solvent Washes

Identification Techniques

a) Physical Propertiesi. Melting and Boiling Points

b) Chemical Tests

i. Color of Phase Changesii. Limiting Reagents

c) Derivative Formation

i. Melting Point Evidence

d) Mass Spectroscopy

i. Base and Parent Peaks

ii. Fragmentation

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Laboratory TechniquesSection Goals

Understand and be able to apply the different forms of chromatography.You must be familiar with thin layer chromatography, gas chromatography, high-pressure liquidchromatography, column chromatography, gel chromatography, andbeadchromatography. All ofthe techniques involving chromatography have basically the same function. They afl depend on acompound's affinity for a mobile phase versus its affinity for thestationary phase as it migratesdown the column.

Understand and be able to apply the different forms of distillation.Thereare four formsof distillationto know:simple, fractional, steam,and vacuum. You must knowthe difference in theapparatus and set-up for each of the four. In addition, youmustknow thepurpose ofeach, thesituation where each isbestapplied, andtheadvantages anddisadvantages ofeach of the techniques.

Understand and be able to apply the different types of extraction.Extraction basically involves the separation ofcompounds bytaking advantage ofopposite solubilitybehavior in two immiscible solvents. You must recognize commonsolventmixtures that will bebiphasic, and their solubility properties. Be able todraw a flow chartforstandard extraction andacid/base extraction, which simplyalters the pH of the aqueousphase in a standard extraction.

Understand and be able to apply recrystallization.Recrystallization is applied topurify a solid and form nice long crystals. The crystal form ofacompound is assumed tobe purer than the powder form. Hie technique is a multi-step processwhosedetailsyou should understand.

©"im Be able to deduce structural features using chemical tests.

? Chemical tests involve reagents that selectively react with a small number offunctional groups andundergo either a phase change orcolor change upon doing so. You need tounderstand thelogicbehindchemical tests andbeable tointerpret theresults.

Be able to deduce structural features using mass spectroscopy.You must have a fundamental idea ofhow the instrument operates interms ofion production, ionseparation, and particle detection. You mustknow whata base peakand parentpeakareand beable to identify them on a typical mass spectroscopy graph. You must be able to identify typicalisotopic distributions for chlorine, bromine, hydrogen, andcarbon. You mustbeable to interpretgraphs andindoing soidentifynotonly the original species, butalso beable toidentify thefragmentsthat are'formed during the process.

Organic Chemistry Lab Techniques

Lab TechniquesOnce a compound has been synthesized, it must be separated, purified, andidentified. But let's be pragmatic about things. The average student taking theMCAT doesn't care much about organic chemistry lab techniques,otherwise theywouldbe aimingfor graduate schoolin chemistry and not medicine. Soour goalis to present organic chemistry lab techniques in enough depth to answermultiple choice questions, and no more. Although there aremany techniques forseparation and purification, we shall limit the techniques to distillation,chromatography, extraction, and crystallization. Once purified and isolated, nextyou must identify what you have collected. Taking melting points, boilingpoints, and doingidentification tests goes only so far in identifying a compound.It requires much more once you start dealing with larger molecules. But on amultiple choice exam suchas theMCAT, there isa great deal ofnarrowing downalready done foryou. Asfarasstudying for theMCAT goes, ourconcern is withthesimple compounds, logical analysis, and theprocess ofelimination.

Lab procedures are listed as a topic in the MCAT Student Manual, they appearinthe AAMC sample passages and questions, and have appeared frequently ondifferent versionsof the MCAT over the past few years. Wewillsurvey a few labtechniques, the rationale behind each technique, and their effectiveness undervarious conditions. This section is written from a perspective that assumes youdidonly enough work in yourorganic chemistry lab courses to turn in your labwrite-ups and didn't spend much energy to learn the theory. The importantinformation to extracthere is the theorybehindeachtechnique. Forinstance, theMCAT Student Manual listsboilingpointsand distillation as topics, so distillationis a probable passage topic that can be coupled with theoretical boiling pointquestions. Lab techniques can bebroken down into three categories: separation,purification, and identification.

To separate anything, one component must move away from the other, whetherit be in different directions or at different speeds. There are four separationsperformed in the lab: 1) solid from liquid, 2) solute from liquid, 3) liquid fromliquid, and 4) solid from solid. For each case, there ismore than one techniquepossible. For solid from a liquid, there isfiltration orcentrifugation coupled withdecantation. For solute from liquid, there is distillation, precipitation coupledwith filtration, extraction, chromatography, or ion exchange. For liquid fromliquid, there isdistillation. For solid from solid, there isrecrystallization, densitygradient columns, molecular sieves, acid-base extraction, sublimation, orcolumnchromatography. We shall address some ofthese techniques inthis section.To purify a solid, recrystallization is the most common method. The crystallineform ofa solid is themostpure. The purification ofa liquid is most commonlycarried out via distillation. For purification of hygroscopic organic liquids,distillation from a drying agent (such as magnesium sulfate) is used to removewater from theorganic solvent. The drying agent binds water and theorganicliquid can be boiled away free of any water azeotrope that might be presentotherwise. Specific examples include aromatic solvents suchas toluene.Once a compound hasbeen isolated, it must be identified by techniques such asIR, UV, or NMR spectroscopy, gas chromatography, thin layer chromatography,chemical tests (reactions with reagents that identify specific functional groups),and physical properties (optical rotation, melting point, boiling point, andsolubilities). This section will address each of these techniques except thespectroscopic techniques, which were addressed in section 2.

Introduction


Organic Chemistry Lab Techniques Separation Techniques

Temperature

Separation TechniquesTo separate molecules requires that at least one of the molecules be in motion.We shall use the word flow to describe the motion ofbulk material. The abilityto flow depends on the state of matter, where solids cannot flow, liquids flowdown,gases flow in alldirections (including up),and solutes flow with a solvent.When two materials are in different phases, they can flow in differentdirectionsand therefore can be separated. A common sense view ofseparation techniquesdictates that if you have a mixture of two more components, you must convertthe system into a state where the components are in different phases. Forinstance, distillation can be used to separate a liquid from another liquid byheating the system until one component boils and consequently flows up andaway from the component that remains as a liquid. We shall look at allseparation techniques asaneffort toget the materials toflow indifferent ways.

Distillation

Distillation removes a liquid from either another liquid or from solute byexploiting their boiling point differences. Upon heating, the most volatilecomponent converts to a gas more readily than the less volatile components,although there is a small amount of a less volatile component that vaporizes too.Indistillation, after the vapor escapes from the surface ofa liquid it can travel upthe distilling column. The vapor collides with the inner walls of the distillingcolumn, where it condenses. The amount ofcondensation depends on the walltemperature. Some of the condensed liquid drips into the distilling flask andsome reevaporates. Every time the vapor condenses and reevaporates, it goesthrough a cycle ofpurification. The design ofthe distilling apparatus issuch thatby the time it reaches the top of the distilling column, ithas gone through enoughpurification cycles that nearly 100% ofthe vapor is the more volatile component.

Vapor Pressure Diagram and Vapor RatioThe separation of two liquids via distillation is based on the vapor pressure ratioofthe components in solution. Tiie vapor pressure of a material depends on itsheat of vaporization, AHvaporjZation/ and its percent composition in solution. AsAHvap0rjzatjon increases, it isharder for the molecules to escape solution, so thevapor pressure is reduced. As the species' mole fraction in solution decreases, ittakes up a smaller percentage of the surface area, and therefore cannot have asmany molecules escape per unit area. This too reduces the vapor pressure. Toaddress this phenomenon, Figure 8-1 shows the vapor pressure as a function oftemperature for two hypothetical components ina mixture oftwo liquids.

3:1

1:1

Init. Sol'n

Vapor Ratio (More : Less)

9:1 27:1

3:1 9:1 27:1

1st Condensate 2nd Condensate 3rd Condensate

Solution Ratio (More : less)

Figure 8-1



Figure 8-1 demonstrates that over three evaporation-condensation cycles, thesystem becomesricher and richer in the more volatile component until the ratiois roughly 27 :1, which equates to a mixture that has 96.4% of the more volatilecomponent. This value gets larger, approaching 100%, with each subsequentevaporation-condensation cycle. As the surface area in the distilling columnincreases, the number of cycles increases, so the purity increases. For mixtureswhere the ratio of vapor pressure of the more volatile component to the vaporpressure of the lessvolatile component is small, moresurface area is required togetgoodseparation. This iswhat is referred toasfractional distillation.

Distillation ApparatusHeat is added at the lowest point, because heat rises through the solution andhomogenizes thesystem through thermal circulation. As liquid at the surface ofthe solution in the distilling pot evaporates, the vapor rises up the distillingcolumn. At the top of the distilling column, the vapor can flow down the sidearm, which is enveloped by a cooling jacket, resulting in condensation of thevapor. The condensed liquid drips from the side arm into the collection flask.Figure 8-2 showsthebasiccomponents of a distillation apparatus.There is an opening betweenthe side tube and the collection flask, whichallowsthe system to vent any build up of pressure that takes place as the liquidvaporizes. As the vapor forms, it pushes the air that is initially present in thedistilling apparatus out throughthe venting hole. Thesystem canbe sealedanda dryingtubecanbe fitted to the collection flask to prevent moisture from the airfromentering the collection flask and condensingwith the distilled product. Anice bath around the collection flask is used when the distillate is highly volatile.

Distillation

Column

DistillingPot

Thermometer

Warm water out

Cold water in

Figure 8-2

Collection

Flask

Simple Distillation versus Fractional DistillationDistillation can be carried out under conditions that are referred to as eithersimple distillation or fractional distillation. Fractional distillation employs adistilling column with more surface area than simple distillation. Simpledistillation is used with liquidshaving a largedifference in boiling points,whilefractional distillation is used with liquids that have a small difference in boilingpoints (less than 30°C in difference, although that is not an absolute number.)Simple distillation is also used to remove a solvent from a solute. Simpledistillation is faster than fractional distillation and it generates a higher yield,while fractional distillation leads to greater distillate purity.



OrgaillC ChCmiStry Lab Techniques Separation Techniques

Fractional distillation is thesameas simple distillation for the most part with theexception beingthat thedistillation column is filled with an inert solid to providemoresurface areaon which thevaporscancondense and then re-evaporate. Thisserves to re-distill the mixture as the vapor climbs up the distillation column.Most often, the column is packed with glass beads, because glass is both inertand cheap. The columnmust be fitted with a trap in the inner core so that theglass beads do not fall through the column into the distillation flask. Fractionaldistillation allows theliquids tobeseparated more efficiently and thus ina purerfashion. The apparatus in Figure 8-2 shows the set up for simple distillation,becausethe distillation columnis empty.

Example 8.1Which ofthefollowing mixtures canbeseparated usingsimple distillation?A. Alpha-D-glucopyranose frombeta-D-glucopyranoseB. Butanol from diethyl etherC. 1-Pentanol from 2-methyl-l-butanolD. Acetone from propanal

Solution

Simple distillation separates a solvent from a solute or a liquid from a liquidwhen they have a large difference in boiling point. The largest difference inboiling pointis found withchoice B, an alcohol and etherofequalmass. ChoiceA is two diastereomeric solids and choices C and D are of two liquids withroughlyequalboiling points. They cannotbe separatedby simpledistillation.

Example 8.2Which of the following materials could be used to pack the column in thefractional distillation of acetic acid from trichloroacetic acid?

A. Paraffin wax

B. Magnesium turningsC. Aluminum mesh

D. Glass beads

Solution

The material chosen must provide surface area, remain a solid at elevatedtemperatures, and be inert. Paraffin waxhas a low meltingpoint, so choice A iseliminated, because the waxwillmeltand drop into the distillation flask. Metalscan oxidize in the presence of acids, resulting in metal oxides that can neutralizethe acids, so choices B and C are eliminated. The best answer is choice D,because the glass beads are inert and will remain solid at the elevatedtemperatures used in distillation.

Vacuum Distillation

Vacuum distillation involves attaching the distillation apparatus to a vacuumpump at a point after the condensate has been collected. The purpose is toreduce the pressure in the apparatus and in doing so lower the boiling point.You should recall that the boiling point is defined as the temperature at whichthe vapor pressure equals the atmospheric pressure, so lowering atmosphericpressure lowers the vapor pressure required to boil, which results in a lowertemperature needed to boil. This is done when the compounds have a highboiling point, higher thantheflash pointor decomposition point.


OrganiC ChCmiStiy Lab Techniques Separation Techniques

ChromatographyChromatography is the separation of two or more components in a mixture byexploiting their differences in solubility in a migrating solvent and their affinityfor a polymer. Both factors play a significant role in the separation process, sothey must both be considered. The various chromatography techniques are allcentered around a sort of swimming race between the components in the mixtureto some arbitrary finish line. What is meant by "race to the finish" is thatdifferent components exhibit different rates of migration across the surface of astationary phase, referred to as the adsorbent, depending on their solubility in themobile phase, (the solvent), and their affinity for the stationary phase. The solventis migrating through the adsorbent and the components are traveling with thesolvent to a varying extent. As the solvent moves, the components are towedalong with it in most cases. The more soluble the component is in the solvent,the further it is pulled by the solvent through the adsorbent. Likewise, thegreater the affinity for the polymer, the slower the component migrates.

Mobile Phase (Solvent) Properties versus Stationary Phase PropertiesBecause separation in chromatography depends on relative affinities, we shouldconsider the nature of typical materials. The mobile phase is one of severalpossible solvents,ranging from nonpolar (hexane) to polar/protic (ethanol), or insome instances a mixture of two solvents. When mixtures are used in columnchromatography, it usually starts with one pure solvent followed by differentmixtures which gradually become richer in a second solvent. Typical solventpairs include ether-hexane and methanol-methylenechloride.

Thestationary phase is one of two materials,aluminaor silica gel. Bothof thesepolymers are polar, so in the absence of any other information, we know thatpolar materials have a higher affinity for the stationary phase than nonpolarmaterials. As a consequence, polar species have slower migration rates thannonpolar species in chromatography. How much slower depends on thematerial and solvent. In addition, the activity of a polymer can be affected by thesolvent, where the solvent binds the silica gel or alumina and thereby reduces theinteraction of the migrating solute with the polymer. For instance, as aluminagainsmoisture, it is less able to bind other materials, so it is said to be less active.Alumina comes with a Brockmann activity rating from I to V, where V representshighly moist alumina which exhibits low activity. Anhydrous alumina is morepolar than anhydrous silica gel.

The only thing that can be said with certainty in chromatography is that polarspecies in nonpolar solventshave slow migrationrates whilenonpolar species innonpolar solvents have a fast migration rate. The questions on the MCAT willlikely stress thispoint, whereyoucansimply refer to theidea"like dissolves like"to answer them.

Thin Layer Chromatography (TLC)Thin layer chromatography, TLC, is carried out on a smallscale to identify thenumber of components or type of componentsin a mixture. It involves spottingsmallaliquots of sample in a line parallel to the baseof the platenear the bottomof a vertical, rectangular plate with either silica gel or alumina on its surface.Thisplate is then placed into a container that can be fitted with a lid and solventis added to the container until it is just below the level of the spots. It isimportant to not have the solvent level in the container above the spots,otherwise they can dissolve into solution. Due to capillary action, the solventslowly migrates up the plate, interacting with each of the spots as it travels. Inorder to prevent the solvent from evaporatingfrom the surface of the plate, a lidis employed to maintain a closed system. Some chemists place a piece of filter



paper on the inside wallof the jar,not touching the plate, to provide additionalsurface area from which the solvent can evaporate. This keeps the chambersaturated withvapor. Once thesolvent isnearly to the top of the plate, the plateis removed from the container andthe top ofthe solvent front is marked using apencil.

The platecanbe developed in oneofseveral ways, suchas spraying ninhydrin onthe surface, adding it to a closed chamber with iodine crystals, or shining UVlight on the plate. In some way, the spots must be made visible for evaluation.Once it is determined where the spots ended, one can ascertain such informationas how many components were in the mixture and whether the componentswere nonpolar, semi-polar, or highly polar. By initially spottinga plate with aknown material you suspect to be the unknown compound in the originalsample, the spots of the unknown composition can be compared to the knownspot to determine identity of one of the unknown compounds in the originalsample. A TLC set Up is shownin Figure8-3.

Lid-

Thin Layer Chromatography

Jar-

TLC plate-(with silica gel or alumina) Sample

(spotted onto plate)

Solvent top

Figure 8-3

Example 8.3Thin layer chromatography can be used todoall ofthefollowing EXCEPT:A. determine the number ofcomponentsin a mixture.B. distinguish the relative abundance of enantiomers.C. determine thebestsolvent for column chromatography.D. monitor the progress of a reaction.

Solution

TLC separates the components in a mixture by their relative affinities for thesolvent and theadsorbent. Asa result, thenumberofcomponents in themixturecanbe ascertained from thenumber ofspotson theplate. This eliminates choiceA. Because column chromatography uses the same solvent and adsorbent asTLC, theresults ofTLC can serve as a preview ofcolumn chromatography, andthus beused to test potential solvents. This eliminates choice C. Samples ofareaction can be taken and analyzed by TLC to determine the appearance ordisappearance ofeithera reactant or product. Thiseliminates choice D. There isno quantitative capability with TLC and enantiomers travel at the same rate, sotheycannotbeseparated, making choice Bis the bestanswer.


OrgaiUC ChCmiStry Lab Techniques Separation Techniques

Example 8.4If the sample on the plate were submerged below the top of the solvent, then thesample would:

A. migrate more rapidly than expected.B. dissolve into solution rather than climb the plate.C. react with the plate.D. cause the plate to dissolve into solution.

Solution

If the sample on the plate is placed into a flask at a level that is lower than the topof the solvent, then the compound is free to dissolve into the solution, and thus itwill not migrate up the plate. This eliminates choice A and makes choice B thebest answer. The plate has an inert adsorbent, so no reaction takes place. Thiseliminates choice C. The plate does not dissolve into solvent, and even if it did,the height of the plate wouldn't matter. Choice D is eliminated.

Rf ValuesIn TLC, we often measure an Rf value for the compounds. We can think of an Rfvalue as a "ratio of the fronts," where the distance a spot travels is compared tothe distance that the solvent travels. The distance that the solvent travels is

measured from the line on which the spots start, where the solvent first interactswith the sample, to the line where the solvent front finishes. The distance thespot travels is measured from the center of the spot when it starts to the center ofthe spot when it finishes. Some spots deform and spread out as they climb, so itis best to use the center of the spot as reference. The distance the componenttravels is divided by the distance the solvent travels to yield its Rf value. Thecomponent with the greatest Rfvalue is the component that has the least affinityfor the stationary phase (adsorbent) and is most soluble in that solvent. Therangefor Rfvalues is:0 ^ Rf< 1,becausethere is a chance the spot doesnot travelat all or that it travels almost at the solvent front. The Rf value is a physicalconstant associated with a solute/solvent/adsorbent combination. The basicidea of all forms of chromatography is separation by relative affinity.

Example 8.5What is NOT true about the Rfvalue for a compound?

A. It cannot be greater than 1.0.B. It has no units.

C. It is different in different solvents.

D. It is always the same on the same adsorbent.

Solution

The spot cannot travel faster or farther than the solvent carrying it, so the Rfvalue cannot be greater than 1.0,eliminatingchoice A. Because it is calculatedbydividing a distance by a distance, the units cancelout. This means that Rfvalueshave no units, which eliminates choice B. Because solubility is different in eachsolvent, the impact of each solvent varies from solvent to solvent. As a result, theRf value differs with each solvent, which eliminates choice C. Because themigration rate depends on the affinity of the compound for both the adsorbentand the solvent, the distance it travels in a given amount of time depends on theadsorbent and solvent. The Rf value varies with adsorbent, making choice D nottrue about the Rfvalue for a compound and therefore the best answer.



Figure 8-4 shows theTLC plates at thestart and aftercompletion of the migrationand developmentwith iodinecrystalsfor two separate TLC experiments. In onecase the solvent is nonpolar while in the other case, the solvent is polar. Thenature of the compound can be inferred from the relative Rf values obtained for aseries of solvents. Because a solute dissolves best into a solvent of "like"nature,the higher the value of Rf, the more that the solute is like that solvent.

•Solvent frontstopped

Ispot•dsoiv

^ R,=-

After

Nonpolar Solvent

"spot

'solvento o o o

Before

Figure 8-4

Solvent frontstopped

dsolvent ^

d

spot

Rf = "spot

•'solvent

After

Polar Solvent

Based on Rfvalues, you should be able to make predictionsas to the best solventfor column chromatography. The best separation occurs when the Rf valueshave the greatest multiplicative (notadditive) difference. Forinstance, Rfvaluesof 0.2 and 0.1 separate better than Rfvalues of 0.3 and 0.2.

Column ChromatographyColumn chromatography is applied to separate bulk quantities of product. Itoperates on thesame principles as thin layer chromatography: thesolubility ofacompound in thesolvent as it flows down the column versus the affinity of thecompound for the adsorbent in thecolumn. If the compound migrates quicklydown the column, wecanconclude that it must experience a minimum attractionto the silica gel or alumina (stationary phase) in the column. In columnchromatography, the "race to the finish line" is to the bottom of the column andthe components in the mixture separate according tohow quickly they finish.To set up a column, it is first packed with a plug (glass wool works well)followed bya thin layer ofsand (to form a flat surface), then a large layer ofsilicagel, followed again bya thin layer ofsand, to protect the top ofthe silica gel fromdisruptions in itsflatness. The top andbottom of thesilica gel mustbe flat andparallel to one another, so that components all travel the exact same distancethrough the silica gel no matter where they start. We assume that thecompounds have no affinity for the sand or glass. Solvent is added to thecolumn to saturate it. Once the top of the solvent is flush with the top of thesand, the sample can be added.

The sample is dissolved into minimal solvent, which is poured gently into thetop of the column. The solvent slowly flows down the silica filled glass columnto an opening at thebottom, taking the compounds at different ratesas it goes.Samples ofthe solution are then collected in increments thatareanalyzed for anycomponents theymaycontain. Faster migrating compounds reach the bottomofthe column first, so they have shorter elution times. Elution time is defined as thetime it takes a component to travel the length of thecolumn and drip out. Likethe Rf value, elution time is a measurement from the experiment. A large Rfvalue corresponds to a low elution time, given that the compound has a fastmigration rate in each case. Eachaliquot of solvent collected from the column isanalyzed by some technique (usually spotting on a TLC plate and then eitherillumination with a UV lamp or treatment with I2 crystals) for the presence of



components. The aliquots containing the component are consolidated and thesolvent is driven off via evaporation. Figure 8-5 shows a column at differentstages, where the bands separate as they migrate down the column.

Solvent is added so itwill flow down the column

&^

shi: Sand layer

Slowest

elution

f?^2?


Silica

Gel

Sand layer

Fastest

elution1st aomponent hasreached the bottom

t=0 t = l t=2 t = 3

Figure 8-5

Example 8.6Column chromatography relies on which force to separate compounds?

A. GravityB. Gas flow

C. Magnetic attractionD. Centrifugal

Solution

The solvent migrates down the column, because gravity is pulling the solventdown the column. The solute compounds are pulled along by the solvent as itmigrates down the column. They also experience some affinity for the silica gel,which slows some solutes more than others. But the driving force behind theoverall flow of solution down the column is gravity. This makes choice A thebest answer. Gas flow is associated with gas chromatography, not columnchromatography, so choice B is eliminated. There is no magnetic field or circularmotion associated with column chromatography, so choices C and D cannot betrue. You should make the correct choice and select choice A.

Designer Polymers (Chiral Columns, DEAE-Cellulose, and Affinity Columns)Given that the migration rate through a column depends on the affinity of acompound for the polymer, there are some columns designed with specificmolecules in mind. While the application of such columns is more common inbiochemistry than organic chemistry, the basic idea is still rooted in thefundamentals of column chromatography. An extreme case of designing apolymer is observed in biochemistry, where an antigen column is used to bindand isolate antibodies from a solution. One problem encountered in such anexperiment involves how to release the antibody from the column. Onceeverything else has eluted from the column, the column can be treated with a



Et2HN

OH

varying pH buffer until the antigen and antibody denature enough to no longerbind one another. One problem with changing the pH of the column is thechance that the proteins are irreversibly denatured. In such cases, the antibodymay instead be released by adding a salt solution of gradually increasing ionconcentration or by adding free antigen to compete with the bound antigen forthe antibodies.

In organic chemistry, chiral columns have become popular in recent years,because they are capable of selecting for one enantiomer over another. If thedesired product is a compound with a hydroxyl group on a stereogenic carbonwith S-chirality, then a column with R-alcohols on the polymer will bind andthereby hinder the flow of the S-enantiomer through the column. As a result, theS-enantiomerhas a greater elution time than the R-enantiomer, so they separateas they traverse the length of the column. The fundamental difference betweencolumn chromatographyin biochemistry and organicchemistry is that in organicchemistry, the aim is to have all of the compounds elute at different times wherein biochemistry the goal is to have all but one compound elute.

The most common designer polymers on the MCAT to date are the adsorbentsused in ion exchangechromatography. Three with which you should be familiarare sulfonated-polystyrene, carboxymethyl-cellulose (CM-cellulose), anddiemylaminoethoxy-cellulose (DEAE-cellulose). The first two carry negativecharges at neutral pH, while DEAE-cellulose carries a positive charge at neutralpH. Proteins with high pi values, ones that are rich in lysine, arginine, orhistidine, will bind CM-cellulose columns,allowing them to be isolated. Proteinswith low pi values, ones that are rich in aspartic acid or glutamic acid, will bindDEAE-cellulose columns, allowing them to be isolated. Typically withsulfonated-polystyrene columns, the protein mixture is added at low pH so all ofthe amino acids and small peptide residues bind the column. The pH of thecolumn is gradually increased,releasing amino acids according to increasing pivalues. Figure8-6shows DEAE-cellulose and sulfonated-polystyrene.

DEAE-Cellulose

OH HOH2C OH HOH2CQ

OSO, OH

Et2HN

Sulfonated-polystyrene

OSO, OH OH OH

Figure 8-6

OH

OSO,



Gas Chromatography ApparatusGas chromatography, GC, works by first vaporizing a sample (mixture ofcomponents) into the gas phase and then forcing that organic vapor through apacked column using elevated pressure from a cylinder of an inert gas (usuallyhelium and sometimes argon). The machine measures retention time on thecolumn by recording collisions as gas molecules leave the end of the long, coiledcolumn and strike the detector. The collisions are converted to peaks (like a bargraph), which can be integrated to determine their relative abundance. A greaterarea for the signal indicates a greater quantity of material present. Thecompounds in the sample must be highly volatile at 200°C and inert with regardsto silica gel or wax for GC analysis to work. Figure 8-7 shows the design for astandard gas chromatography machine and a standard chart output for amixture of three components.

Inert carrier gas(Usually He)

Injection port, through which asample may be added by syringe

Packed

column

Detector

A carrier gas, such as helium, is added to create a positive gauge pressure that forces thevaporizedsample through the packed coiledcolumn and into the detector. The detectorreads collisions. The chart graphs intensity of collisions versus time. This means that noidentification can be obtained directly from gas chromatography. The retention timedepends on the nature of the column and the structure of the compound.

Figure 8-7

The relative quantities of the components can be determined by integrating thesignalsassociated with the compounds. GC output is continuous,so the graphsare actually read from right to left, where signals on the right representcomponents with smaller elution times than components on the left. Todetermine the purity of a product mixture (and relative abundance ofcomponents), the graph can be read directly. The purity of the product mixturecan be observed directly in the total number of peaks that are seen on the graph.To ascertain information about one of the components in a mixture, a puresampleof that component can be added to a gas chromatography apparatus andthe retention time for that particular component can be determined. Analternative way to identify a component involves the addition of a small sampleof that compound to the mixture to enrich the sample in that component. Thatcomponent corresponds to the signal that increases its relative size. Thistechnique is referred to as spiking the sample with a component. Gaschromatography is used as both an identification tool for the reaction mixtureand as a tool for measuring the kinetics of a reaction. By measuring the samplemixture at given intervals, the growth of a peak can be monitored to determinethe rate of formation of a product. Figure 8-8 shows a typical graph obtainedfrom the GC apparatus. GC peaks are relatively sharp as the graph shows.




Retention time

Figure 8-8

Example 8.7Gas chromatography does NOT work with:A. alcohols.

B. aldehydes.C. ketones.

D. fatty acids.

Solution

In this case, the component that cannot work is the component that cannotvaporize in the gas chromatographer. The temperature of the vaporizingchamber isoften higher than 200°C, a highenough temperature tovaporize mostorganic liquids, but a fatty acid cannot vaporize at that temperature. The fattyacid is likely to melt and decompose at this temperature, so it cannot beidentified usinggaschromatography techniques. Choice D is the best answer.

Analysis of GC DataThe MCAT is apt to askdata analysis questions rather than details of operationfor theGC apparatus. It is imperative thatyoube able to interpret graphs suchas the one in Figure 8-8. GC analysis focuses on quantitative aspects of themixture rather than specifics about the structures. Given that the compoundsvaporize instantaneously upon addition to a GCmachine, the boilingpoint doesnot affect the elution time. However, because many columns are often slightlypolar, affinity for polarspecies is greater than affinity for nonpolar species. Asecond factor that dictates themigration rate is mass. Heaviergasesmoveslowerandhave longer elution times. Heavier gases andpolargases typically havehighboiling points, so although boiling pointhas nothingto do with elution time, thefactors thatraise boiling point also increase theelution time in a GC. This iswhymany chemists say that molecules with lower boiling points come off of a GCcolumn first.

Extraction

Extraction works based on solubility and it is typically employed to takeadvantage ofdrastic differences in thesolubilities ofcomponents in twodifferent(immiscible) solvents. To start, wefirst need to define partitioning. A compoundhas a different solubility in every solvent. The ratioof its maximum solubility inone solvent compared to its maximum solubility in another solvent is known asthepartition coefficient. In cases wherethe compoundcan go into eithersolvent, itwill selectbetweenthe solvents in a ratio equal to the partition coefficient. Whenthe solventsystem involves two immiscible liquids, then the solute must split


OrganiC ChemiStry Lab Techniques Separation Techniques

between the two solvents. Some components will dissolve primarily into one ofthe solvents, while the remaining components will dissolve primarily into theother solvent. This means that the components of a mixture can be separated ifthey have differing partition coefficients. By decanting one solution from theother, the components are then separated from one another in their solute formaccording to their relative solubilities in the two different solvents. From thispoint, you must remove the component from the solvent in which it is nowdissolved. This can be done via distillation or controlled evaporation.

Extraction does not always involve two immiscible solvents, although that ismost often the case. A solid component can be removed from a mixture of solidsby adding a solvent in which it is highly soluble and the other compounds arerelatively insoluble. This is how many natural products are isolated from plants.

Acid-Base Extraction

When one of the two immiscible solvents is water, the properties of the solutescan change when the solution pH changes. Acid-base extraction in essenceseparates compounds by their pKa values. By using a biphasic solvent mixture(heterogeneous system), selective protonation or deprotonation (depending onthe compound) will either increase or decrease the water solubility of thecompound, depending on whether an ion is generated. Loss of charge results inan increase in preference for the organic layer while a gain of charge results in anincrease in the preference for the aqueous layer. Once extracted into the waterlayer, the species can be returned to a neutral state by adjusting the pH, and inmost cases it will crash out of solution. Figure 8-9 shows the flowchart for theacid-base extraction of a mixture of n-butylamine (CH3CH2CH2CH2NH2),butanoic acid (CH3CH2CH2CO2H), and ethyl benzene (CH3CH2C6H5).

Mixture: Butylamine, Butanoic acid, Ethylbenzene

50.0 mL ether | 50.0 mL 10%NaOH(aq)

Layer I (organic) Layer II (aqueous)

10% HCl(aq) 10% HCl(aq)

Layer III (organic) Layer IV (aqueous) Butanoic acid

evapDrate 10% NaOH(aq)

T TEthylbenzene Butylamine

Figure 8-9

In the first extraction, aqueous sodium hydroxide solution deprotonates butanoicacid to form butanoate, a carboxylate anion (which is highly water soluble due toits negative charge). Butylamine and ethylbenzene remain in the organic layer.The butanoic acid can be "crashed out" of solution by adding acid to regeneratethe carboxyUc acid. The carboxyUc acid form is not very water soluble, so itprecipitates out of the water. Butyl amine is removed from the ether layer byextracting it with acidic water, which forms butyl ammonium cation, a highlywater soluble species. Butyl ammonium cation can be "crashed out" of solutionby adding strong base to the aqueous layer. The ether layer contains theremaining ethylbenzene, which can be isolated by evaporating the ether away.


OrganiC ChemiStry Lab Techniques Separation Techniques

Example 8.8To separate benzoic acid (C6H5CO2H) from toluene (C6H5CH3), it would be bestto use a mixture of an:

A. alcoholand water at pH = 11.B. alcohol and water at pH = 3.C. ether and water at pH = 11.D. ether and water at pH = 3.

Solution

To carry out an extraction, the two solvents cannot be miscible in one another.Alcohols are typicaUy miscible in water, so choices A and B are eliminated.Benzoic acid can be deprotonated in basic solution, where it becomes its anionicconjugate base (benzoate) which is readily soluble in water. Toluene dissolvesinto the ether rather than water, so the best answer is choice C.

OH ^ -^^~ ^ -CH,a

more soluble in more soluble in insolubleether than water water than ether in water

SeparatoryFunnel and Partitioninginto the Biphasic LayerThe physical actofseparation is mostoftencarried out in a separatoryfunnel. Aseparatory funnel is an upside-down, pear-shaped flask with a fritted glasscapon top and a stopcock controUed valve on the bottom. The two layersnaturaUyseparate, with the less dense layer assuming the top. The system is mixed andshakentoensure complete interaction of the twolayers. Onething to note is thatthe aqueous layer has a meniscus, so the layerwith the mostcontact against theglass walls is the aqueous layer. To increase the differences between the twolayers, the aqueous layer is often saturated with salt. Such a solution is referredto as a brine solution.



FlowChart with an Ether, Amine, Phenol, and Carboxylic AcidOn the MCAT, perhaps the most difficult task when it comes to extraction wiUbeanalyzing a flow chart. Due to the time limitations of the exam, the chart cannotbe too exhaustive, so we shall consider a four-compound mixture. Figure 8-10shows a typical four-compound flow chart. It is a goodgeneral rule that a weakbaseshould be added before a strong base, so as to isolate the stronger acid. Ifyou were to add the sodium hydroxide first, it would extractboth the phenoxideand the carboxylate into the aqueous layer.

OCH NHCH3

H,C


N02 O

OH

Organic Layer 0.10MNaHCO3(aq)

Organic Layer

H,C

distill awayfrom residue

OCH3 NHCH3 CH2CH3

H3C^ ^^ ^*^ ^^ ÔH

Organic Layer

OCH3 NHCH3

O.lOMNaOH(aq)

acidify andfilter

+NH2CH3

basify andseparate

Figure 8-10

Special Cases including Esters, Anhydrides, and Acid HalidesAs is true with all lab techniques, the conditions must be inert for the techniqueto be effective. Because extraction involves both acidic water and basic water,there is the potential for hydrolysis. As such, extraction cannot be used to isolatewater-sensitive compounds such as acid anhydrides, acid halides, and esters.Thesecompounds can be extracted from a pulp into an organic solvent, but theycannot undergo acid-base extraction.

N02 O

acidify andfilter

N02 O

OH


OrganiC ChemiStry Lab Techniques Purification Techniques

Purification TechniquesMany of the separation techniques we have discussed so far can also be used aspurification techniques. Column chromatography is useful for purification,because the bands separates nicely from one another. Depending on thecompounds, the isolation of a few grams of highly pure material is common incolumn chromatography. Distillation, when done multiple times, can generate ahighly pure liquid. But rather than describe chromatography and distillationagain with the attention to detail that makes for purification, we shaU onlydiscuss recrystallization. Yournumber one goal is to zero in on topics that havea good probability of showing up on the MCAT, so we shall considerrecrystallization in the purification section.

Recrystallization of a Crude Product MixtureA solid product may be separated from solid impurities through selectiveprecipitation, which organic chemists refer to as recrystallization.RecrystaUization involvesdissolving the soUd into hot solvent, hot filtering outthe insoluble soUd impurities, and then slowly cooling the solution to precipitatethe purified crystals. Byadding a minimal amount of solvent in which the soluteis saturated at high temperature but insoluble at a lower temperature, one mayfirst dissolve a product mixture leaving solid impurities in solution to be filteredout. Upon cooling the solution, the product precipitates out of solution in acrystalline form free of the impurities from before. The crystals are filtered atlow temperature, to prevent any of the material from dissolving back intosolution, to avoid any reduction in the yield.

Solvent Choice (Dissolving Solute into Minimal Refluxing Solvent)In recrystallization, solvent choice is essential. Because the soUd is dissolved intorefluxing solvent, it is ideal that the desired material is highly solubleat elevatedtemperatures. The term refluxing describes a system where the solvent isboiling, but it is condensing on the waUs of the flask and dripping back intosolution. You may recall that acid refluxis a conditionwhere your gastricfluidsclimb up your esophagus only to fall back into your stomach. Reflux is theclimbing and returning process. In order to maximize the amount of crystalcollected, the desired materialshould be insoluble (or minimaUy soluble) in thesolvent at lower temperatures. The solvent can be a homogeneous mixture oftwo solvents to meet desired solubility properties. By mixing solvents, it ispossible to change the overallpolarityof the solvent and thus change the overallsolubility of thesolute at bothhigh temperatures and low temperatures.

Example 8.9If a solid is too soluble in water at room temperature, it would be best to addwhich of the foUowing solvents?A. Hexane

B. Ethanol

C. Diethyl etherD. Tetrahydrofuran

Solution

The key to this question is that the solvent you mix with water must reduce theoverall polarity and hydrogen bonding capacity of the solvent mixture and besoluble in water. All of the choices are less polar and less protic than water, butof the choices, only ethanol is soluble in water. Choice B is the best answer.


OrganiC ChemiStiy Lab Techniques Purification Techniques

Decolorizing(Adding Charcoal to Solution to Bind Colored Impurities)In some instances, the purification can be enhanced by adding a material topreferentially bind someor aU of the impurities. Onesuchcase occurs when youhave colored impurities and a colorless target crystal. By adding activatedcharcoal, charcoal that is free of impurities, the colored species will bepreferentially bound by the surface of the charcoal and can thus be filtered out.Charcoal binds aU organic materials, but because of its it-network and thepresence of conjugated ^-networks in colored species, charcoal has a higheraffinity for colored organic molecules than colorless organicmolecules.

Hot Filtering (to Remove Insoluble Impurities, i.e.,boiling chips)Filtration is employed to removesolids fromliquids. It is a laboratory techniquepatterned after everyday activities, such as collecting spaghetti in a colanderwhile running water over it. In organicchemistry, a finemicroscopic net is usedto separate the soUd from the Uquid by allowing the Uquid to flow through whilehindering the flow of soUds, which are larger than the pore size of the filter. ThesoUds are thus coUected by the filter. Filtration canalsobe used to separate largesolid particles from small solid particles by setting up two filtering sites ofdifferent size. The first filtration should be of a larger pore size than the secondfiltration so that the larger solid may be collected first and thereby separatedfrom the smaller solid, which will then be collected in the second filter. This issometimes referred to as sievefiltering.

Filtering typically depends on gravity to force the liquid to flow through thefilter, although it can be expedited by a pressure difference between theatmosphere above the filter and the atmosphere below the filter. Gravityfiltration allows large insoluble solid particles (crystals) to be removed from thesolvent (mother liquor). As the solution flows through the filter, the solidparticles are collected in the pores of the filter. Vacuum filtration uses a filter inconjunction with a vacuum, so that the liquid flows faster. As with aU types offiltration, insoluble soUd particles are removed from a solvent. The Uquid flowsthrough the filter due to both gravity and the pressure difference between aboveand below the filter, created by the vacuum below the filter.

Example 8.10Which of the foUowing mixtures can be separated using gravity filtration?A. Ethanol from diethyl ether.B. Sodium chloride from water.

C. Butane from dichloromethane.

D. OxaUc acid from hexane.

Solution

In this question, you must rely on experience and memory. In choice A, bothcompounds are liquids, so filtering is pointless. In choice B, the salt is completelysoluble in water, so the salt is actually in solute form. It must be a solid, notsolute, to be filtered, so choice B is eliminated. Butane is a liquid as isdichloromethane, as you may recaU from lab experiences. This eliminates choiceC, leaving choice D as the only remaining choice, so it better be correct. Hexaneis a Uquid and aU carboxylic acids, besides formic acid and acetic acid, are solidsat room temperature. Acids do not readily dissolve into hydrocarbon solvents,so oxalic acid exists as a solid and not as a solute in hexane. Choice D iscomprised of a soUd in a Uquid, so it is the best answer.

Copyright © by The BerkeleyReview 257 Exclusive MCAT Preparation

Organic Chemistry Lab Techniques Purification Techniques

Crystallization, Cold Filtering (to Collect Crystals), and Solvent WashThe final step in a recrystallization procedure is to coUect and isolate the targetcrystals. If the solution is slowly cooled from its refluxing state, it wiU precipitatelong, pure crystals. These crystals are coUected using filtration, but with coldsolvent, so as to prevent the loss of product to dissolving. The crystals at thispoint still have a residue of solvent with soluble impurities on their surface. Thecrystals must lose this "dirty solvent" to be pure. The dirty solvent can beremoved by washing the crystals with a solvent in which the impurities aresoluble, but the target compound is not. It is optimal to use a solvent with a lowboiling point, so it can readily evaporate away from the solid. The rinsingsolvent dissolves the impurities as it passes across the crystals. A commonexample is the removal of water from glass by rinsing the glass with acetone.Youno doubt have done this in lab a thousand times (okay, maybe only three orfour times, but 1000is so much more impressive). The acetone does not dissolvethe glass, but it does dissolve the water and rinses it away. The residual amountof acetone left behind on the glass can evaporate away leaving behind dry(anhydrous) glassware. Solvent washes are often carried out on the solidcoUected from filtration. It's similar to use rinsing off with water once we havelathered up in the shower. If we didn't rinse the lather away, the water wouldevaporate and leave a soap film on our skin. We opt to rinse in a pure, volatilesolvent in which we are insoluble, at least most of us do.

Example 8.11Thebest choice with whichto rinse sodium acetate in order to remove propanolresidue is:

A. hydrochloric acid.B. propanol.C. diethyl ether.D. water.

Solution

The goal of rinsing a product is to remove the dirty solvent (in this casepropanol) from the surface of the desired product (in this case H3CC02Na)without reacting with the product nor dissolving the product. Choice A iseliminated, because HCl wiU protonate the sodium acetate to form acetic acid.Choice B is out, because you cannot add propanol to remove propanol (thatwould be the same as washing a dirty shirt in mud to remove the dirt). Choice Dis eliminated, because sodium acetate can dissolve into the water. The bestchoice is the only choice left, choice C. Diethylether will dissolve the propanolwithout dissolving the sodium acetate. The propanol can be rinsed away withthe ether and only residual ether wiU be left behind in the sodium acetate. Theether can easilyevaporate away from the sodium acetate crystals leaving behindpure sodium acetate.



The last aspect of lab we shaU consider are the identification techniques. Once acompound is isolated, it is important to be able to characterize the material.Compounds are most easily identified by physical properties. IR, UV-visible,and NMRspectroscopy have already been discussed, so we shallconsidermostlychemical diagnostics in this section. By fingerprinting compounds with theirphysical properties, it is possible to distinguish one from another. Chemical testsfor specific functional groups can also be employed. In some instances, acompound can be converted into a new compound (derivative), and the physicalproperties of the derivative can be used to help identify the original compound.

Physical PropertiesIt is common to verify a compound or ascertain its purity from its melting point.A rapid change from a solid into a Uquid (occurring over two degrees or less) isindicative of a pure compound. Sharp boiling points are not as obvious. Toidentify a solid, one can mix an unknown with the compound they suspect it is.A sharp melting point confirms that the unknown is in fact that compound, whilea broad melting range indicates that the compound is something else. This is themixed melting point test. The name comes from mixing the unknown and knownto form a homogeneous mixture. Other physical properties to compare includesolubilities and optical activity. Whatever physical difference compounds mayhave can be exploited to distinguish one unknown from either a secondunknown or a comparable standard compound (known). Table 8-1 shows thephysical properties for some common organic molecules.

Substance Density (g/mL) Melting Point Boiling Point

Acetone 0.79 -95°C 56°C

Benzophenone 1.15 48°C 306°C

Chloroform 1.49 -64°C 61°C

Cyclohexane 0.78 68C 81°C

Methanol 0.79 -98°C 65°C

Naphthalene 1.15 94°C 288°C

Stearic acid 0.85 708C 291°C

Table 8-1

Example 8.12According to the data in Table 8-1, which of the following compounds is a soUdat room temperature and sinks to the bottom of the flask when added to water?

A. CyclohexaneB. Chloroform

C. NaphthaleneD. Stearic acid

Solution

The compound is insoluble in water and denser than 1.00grams/mL, because itsinks, and doesn't dissolve, in water. Incorrect density eliminates choices A andD. Chloroform has a melting point of -64°C, so it is a Uquidat room temperature.This eliminates choice B. Naphthalene, choice C, is a solid at room temperature,because it has a melting point of 94°C. Choice C is the best answer.



OrgailiC Chemistiy Lab Techniques Identification Techniques

Example 8.13If a student suspects that the compound they have is L-tyrosine, what shouldthey add to the sample for a mixed melting point?A. L-TyrosineB. D-TyrosineC. XyloseD. Glycine

Solution

To support the notion that an unknown compound is L-tyrosine using a mixedmelting point study requires mixing the unknown with a pure sample of L-tyrosine. If the unknown compound is in fact L-tyrosine, then the mixture is infact pure L-tyrosine. This would result in a sharp melting point at 114°C, themelting point for L-tyrosine. If the unknown compound were not L-tyrosine, themelting point range would be broad and would faU outside of 114°C. The bestanswer is choice A. D-tyrosine, although having the same melting point as L-tyrosine, won't work, because the enantiomers interact with each otherdifferently than they do with themselves. Pure enantiomers typicaUy do notpack into crystals as weU as racemic mixtures, so theyhave lower meltingpoints.

Example 8.14A meltingpoint rangefrom 31.5°C to 38.5°C supports which conclusion?A. The compound is very pure.B. Thecompound is highly reactive at low temperatures.C. Thecompound wiU boilat a temperature less than 138.5°C.D. The compound has impurities.

Solution

The melting point range does not indicate anything about the reactivity of acompound. As such, no conclusion can be made about the compound'sreactivity, so choice Bis eliminated. Knowing a compound's melting point doesnot giveanyhint as to itsboiling point. This eliminates choice C. A rangeof 7°Cis broad in termsofmeltingpoint, and a broad melting point range is associatedwith an impure compound. Theonly reasonableanswer is choice D.

Example 8.15TodistinguishD-leucine fromL-leucine, it is best to compare their:A. specific rotations.B. solubiUty in ethanol.C. melting points.D. odors.

Solution

D-leucine and L-leucine areenantiomers, so theyhave opposite optical rotationsof equal magnitude. To distinguish the two enantiomers from one another, it isbest to compare their specific rotations. Their solubility and melting pointsshould be identical, because of their structural similarities (enantiomers have thesame physical properties.) Their odors may or may not be different, because theolfactory lobes are chiraUy sensitive, but we cannotdetermine for sure that theyhave different odors. We know forsure that they rotate plane-polarized light inopposite directions, so the best answer is choice A.


OrganiC Chemistry Lab Techniques Identification Techniques

Example 8.16The difference in solubiUtybetween 4-nitrobenzoic acid and glyceraldehyde (2,3-dihydroxypropanal) in water is most dissimilar with their solubiUty difference in:A. benzene.

B. ether.

C. 5% acid solution.

D. 5% base solution.

Solution

Nitrobenzoic acid gets deprotonated in a basic solution and the anion that isformed is more soluble in water than the original acid. Glyceraldehyde has noacidic or basic sites, so its solubiUty is not affected by the pH of the solution. Forthis reason, nitrobenzoic acid shows greater solubiUty in the 5% base solutionthan pure water while glyceraldehyde shows roughly the same solubility inwater as 5% base solution. The best answer is choice D. The important point onthis question is that you are looking for differences in solubility in water andanother solvent, not differences between 4-nitrobenzoic acid and glyceraldehydein their solubiUty in benzene, ether, acidic solution or basic solution. This was infact a reaUy poorly worded question, which happens from time to time on theMCAT.

Example 8.17Butanol can be distinguished from butanone by comparing aU of the foUowingEXCEPT their:

A. boiling points.B. freezing points.C. densities.

D. infrared spectra.

Solution

Butanol and butanone have different boiling points and melting points, becausean alcohol is polar and protic while a ketone is simply polar. This eliminateschoices A and B. The infrared spectra for both compounds differ greatly, becausebutanol has a broad peak around 3500 cm"1 that butanone does not have andbutanone hasa sharp peak around 1700 cm"1 thatbutanol does nothave. Thesetwo peaks can be used as features that will differentiate the two compounds.This eliminates choice D. The two compounds have different densities, but thequestion does not allow for aUfour choices to be eliminated. You are looking forthe "best" answer. Both compounds have roughly equal masses and are liquidsat room temperature, so their densities are not going to be that different.Although aUof the choices work, the best answer is choice C.


Organic Chemistry Lab Techniques Identification Techniques

Chemical Tests

Chemical tests involvereactionsthat are selectivefor very few functional groups.The ideal chemical test is positive for only one functional group, but this is rare.Most chemical tests have multiple functional groups that give positive results.For a chemical test to be effective,a few things must be true.

1) The reagent must react with very few functional groups.

2) The chemicaltesting reagent must undergo some visible change, such asa color change or phase change, so it can be detected.

3) The testing reagent must be the limiting reagent, so that any excess of thetesting reagent wiUnot interfere with the reading.

We shall focuson the tests associatedwith compounds other than carbohydrates.Table8-2shows four commonchemicaltests used in organic chemistry.

Compound Tested For Reagent Positive

Methyl ketones I2(s)/OH-(aq) YeUow precipitate1° and 2° alcohols, aldehydes K2Cr04/H2S04 Color change: orange to greenAlkenes Br2/CCl4 Color change: brown to clearUnsaturated fats IBr/CCl4 Color change: purple to clear

Table 8-2

Do not memorize any of these tests. Know the concepts behind a chemical test,but expect that the passage wiU provide the details about a specificchemical test.

Example 8.181-Iodohexane is distinguished from3-iodo-3-methylpentane MOST readilyby:A. reaction with ammonia.

B. molecular mass.

C. water-solubiUty.D. the iodoform test.

Solution

1-Iodohexane and 3-iodo-3-methylpentane have the same molecular weight, sochoice B is eliminated. Alkyl halides are not very water soluble, so choice C iseliminated. The iodoform test is for methyl ketones, not alkyl iodides, so choiceD is eliminated. The difference between them is that 1-iodohexane is a primaryalkyl iodide and 3-iodo-3-methylpentane is a tertiary alkyl iodide. The primaryalkylhaUde undergoes Sn2substitution with ammonia, makingchoice A best.

Example 8.19What makes KMn04 in basicwater not useful as a chemical test agent?A. It reacts with too many compounds.B. It undergoes both a phase change and color change.C. It is too inert and requires a high concentration.D. It is a gas with low solubiUty in water.

Solution

Permanganate can oxide many compounds, including alkenes, alcohols, andaldehydes. Although it has a distinct color change, it reacts with too manycompounds for a conclusion to be drawn. Choice A is the best answer.



Derivative Formation

Derivatives are products formed from the reaction of a given reagent with anunknown compound. Derivatives are made to be used as a diagnostic for anunknown. The additional physical measurements associated with the newlyformed derivative can be used to identify both the derivativeand indirectly theunknown. They are used to support and confirm an identity for an unknown.The functional group on the unknown compound must be known to decide whatderivative should be synthesized. Common derivatives include 2,4-dinitrophenyUiydrazones for ketones or aldehydes, benzylesters for alcohols, andosazones for sugars. The ideal derivative is a soUd that can easily be identifiedby its melting point. An example of a 2,4-dinitrophenylhydrazine derivative isshown in Figure 8-11.

Aldehyde

O

A-R R'

Ketone

NOj

Phenylhydrazine derivative

R' H Np2

Phenylhydrazine derivative

Example 8.20The limiting reagent of a reaction has a molecular mass of 128 grams per moleand the product has a molecular mass of 160 grams per mole. If 0.20grams of thelimiting reagent generates 0.20 grams of product, then what is the percent yield?

A. 100%

B. 89%

C. 80%

D. 75%

Solution

Percent yield is found by taking the actual yield (in either grams or moles) anddividing it by the theoretical yield (in either grams or moles). In this case, it iseasier to solve using moles rather grams. Either way you choose, you must becertain that the units cancel out. The mathematics is as follows:

0-2 g

160 /mole - 7160 -128 =64 =_8_ =081/ 160 80 10

%yield = actual molestheoretical moles 0.2 g.

^lo,e "/128The yield is 80%, choice C. The percent yield for a reaction must be calculatedusing the limiting reagent. Most numbers should result from easy calculations.




Example 8.21Todistinguish2-hexanone from3-hexanone, it is best to compare:A. their boiling points.B. their reactivitywith an oxidizingagent.C. their alcohol solubiUty.D. the melting points of their phenylhydrazine derivatives.

Solution

The two compounds (2-hexanone and 3-hexanone) have different boilingpoints,but only slightly different, because they are both six carbon ketones. Thiseliminates choice A. Ketones are inert with mostoxidizing agents, so choice Biseliminated. Ketones have sirmlar alcohol solubUities, so choice C is invalid. Todistinguish compounds, derivatives aremade. Phenylhydrazine derivatives of2-hexanone and 3-hexanone have different melting points, because they do notpack the same in the solid phase. This makes choice D the best answer. As apoint of interest, the iodoform test could work too, but it is not a choice.

Example 8.22Which of the foUowing physical properties of the 2,4-dinitrophenylhydrazinederivative ofa ketone isbestto useasan identity diagnostic?A. DensityB. Molecular weightC. Boiling pointD. Melting point

Solution

Phenylhydrazones are soUds, soit isnotfeasible tocompare their boiling points.This eliminates choice C. Their densities andmolecular masses arenot going tobe asuseful indistinguishing, because most wiU have similar values for densityand molecular mass, or close enough that no strong conclusion can be drawn.This eliminates choices A and B. Melting point is the distinguishing physicalproperty between derivatives. This makes the best answer choice D.

Example 8.23To distinguish an alcohol from an ether of approximately the same molecularmass, allofthefollowing can beemployed EXCEPT comparing their:A. boiling points.B. water solubiUty.C. rate of evaporation.D. index of refraction.

Solution

An alcohol exhibitshydrogen bonding while ether does not. This increases theboiling point of the alcohol, which also makes it harder to evaporate than anether. Analcohol hasa higher boUing pointand slower rate ofevaporation thanan ether of roughly equal mass, which eliminates choices A and C. Hydrogenbonding also increases the watersolubility of an alcohol of roughly the samenumber of carbons as an ether. This eliminates choice B. The index of refractionmay very well be different for an ether and an alcohol, but it is not predictablelikeboiling pointand water solubility. This makes it hard to distinguish whichcompound is the ether and which is the alcohol, so choice D is the best answer.



Mass SpectroscopyThelast laboratory technique we shaU address is mass spectroscopy. Because themechanics of the device are covered in physics, we shall focus on the data andinterpretation only. At the level of the MCAT, there are only three things weshould understand about the data. First, we can ascertain the molecular mass ofa compound with great accuracy from the parent peak (peak of highest mass).This canhelp us to distinguishalkanes fromalkenes and alkynes ofequalcarboncount. Second, we can determine if the compound has either chlorine or bromine(from the distribution of their isotopes). Chlorine has two abundant isotopes,35C1 and 37C1, in a ratioof approximately 3:1. Therefore, when weseesome ofthehighermasspeaksin a 3 :1 ratiofor theMand M+2 signals in a grouping, wecan conclude that the peak combination must be due to chlorine. Bromine hastwo abundant isotopes, 79Br and 81Br, in a ratio of approximately 1 : 0.98. Apeakratio of 1 : 1 for the M and M+2 peaks in a group generally indicates thepresence ofbromine in thecompound. Lastly, wecandetermine themass of thealkyl groups attached to either a carbonyl carbon or a heteroatom. From thedifference is mass between the parent peak and the base peak (the tallest peak,which is assigned an intensity of 100 on the y-axis ofmass spectroscopy graphs),we can ascertainthe alkylgroups that have fragmented offof the originalcation.

The one fact we need from physics is that the machine detects ions, and in ourcase,cations. In some cases, the cation is also a free radical,but that is irrelevantto the operation of the machine. The basic ideas behind the device are thatcharged particles move in a circular path when traveling perpendicular to linearmagnetic fields and that heavier particles traverse a circular path of greaterradius thanlighter particles. Themachine accelerates thecations and thencausesa 90° turn, which allows particles of different mass to spread out before theystrike a detector. Figure 8-12 showsthe massspectroscopy graphforbutanone.

100—1

80-

§ 60-|

40-

20-

Base

peak

t—i—r48 50 52 54 56

m


M+Parent

peak

TT68 70 7214 16

T18

T22

ll T ^-A tW24 26 28 30 32 34 36 38 40 42 44 46 58 60

T62

~i—r64 66 74

Molecular Mass (amu)

Figure 8-12

The mass of the most abundant molecule of butanone is 72.0575 g/mole, so thepeak at 72 is due to the radical cation formed when a nonbonding (lone pair)electron was ionized from oxygen. In ionizinga hydrocarbon, a bonding electronis lost from a carbon-hydrogenbond. An electronis ionizedwhen a molecule isbombarded with high energy incident particles, usually electrons. Becausecations and free radicals are unstable, the species rearranges and fragments,



sheering off pieces of the molecule. The mass spectrophotometer detects anycharged fragments that splinter off, sothe peaks that are observed correspond tothemoststable cationic species thatrearrangement and fragmentation canform.The peaks themselves represent fragments and the difference between peakscorrespond touncharged fragments that sheered off. InFigure 8-12, thepeaks ofinterest arefound at 72,57,43, and29. The peak at 72 is theparentpeak, so it isattributed to the cationic compound. The peaks at 57 and43 correspond to theacyUum ions formed when the parent compound loses a methyl group and ethylgroup respectively. The peakat 29ispossibly due to an ethylcation.

Example 8.24In massspectroscopy, thematerial being analyzed shouldbe:A. at lowpressureand in the gasphase.B. at high pressure and in the gas phase.C. at lowconcentration and in theUquid phase.D. at high concentration and in the Uquid phase.

Solution

Because the particles must travel independently ofone another in the apparatus,they must be in the gas phase. This eliminates choices C and D. In order tominimize the number of peaks, the material isat low pressure, sothat the highlyreactive free radical and cationic particles donotcolUde andundergo reactions.If there are too many molecules, they can combine, which would lead to morecompUcated data. Thepressure should be low, so choiceA is the best answer.

Example 8.25Which of the following peak combinations would imply that chlorine waspresent on the molecule?

A. B. C.

+ + +-f U +131132 133134 135 131132 133134 135 131132 133134 135 131132 133134 135

Solution

As mentioned earlier, chlorine hastwo abundant isotopes thatcome in a ratio ofapproximately 3 :1 and differ in mass by 2. We shaU ignore the peak at 135 tointerpret the ratios, because that peak does not vary from choice to choice.Choice Ashows peaks ina2:1: 2:1 ratio, which innoway can beinterpretedas3:1. This eliminates choice A. Because the peaks that differ inmass by2areofroughly comparable height, it ismore likely thatthe species contains bromine.Choice Bshows peaks ina 1:1:2 :2 ratio, which innoway can beinterpreted as3:1 infavor ofthe Ughter peak. This eliminates choice B. Choice Cshows peaksina 9:4:3:1.3 ratio which shows a3:1 ratio for the 133-to-131 peaks anda3:1ratio for the 134-to-132 peaks. This makes choice C the best answer. Choice Dshows peaks in a 9 : 3 : 2 : 8 ratio, which in no way can be interpreted as 3 : 1.This eliminates choice D. Although it may seem arbitrary as tohow the heightratiowasdetermined, only choice C shows peaks differing by 2 in mass with aheightratioof3:1 in favor of thelighter one.


Organic Chemistry Lab Techniques Section Summary

Key Points for Lab Techniques (Section 8)


1. Based on different flow directions or different flow speedsa) DistiUation

i. Converts the most volatile component into a gas so it can flow up thedistilling column and away from the mixture

ii. Vapor pressure differences are maximized at the boUingpoint of theleast volatile component

Ui. The distillation apparatus is comprised of a distilling pot, a distillingcolumn, a thermometer, a condensing side arm, and a collection flask

b) Simple distillation has minimal surface area in the distilling columni. Maximizes yield at the expense of purityu. It is employed when the boiling points of the components differ by a

large amount.

c) Fractional distiUation has significant surface area in the distilling columni. Maximizes purity at the expense of yieldu. It is employed when the boiling points of the components differ by a

smaU amount (less than 30"C).

d) Vacuum distillation is employed when boiling points are highi. Carried out in a closed system where the internal pressure is actively

reduced

e) Chromatography separates by relative affinities for a mobile versusstationary phasei. Mobilephase is the solvent which travels across the adsorbentii. Adsorbents (the stationary phase) are polar polymers through which

or across which migration occursiii. Thin layer chromatography provides a flat surface across which the

solvent migrates, taking solute with itiv. Rf values are the ratio of the solute migration distance to the solvent

migration distance in thin layer chromatographyv. Column chromatography is a separation and/or purification

techniquewhere a mixtureseparatesas it travels down a columnvi. Elution time is defined as the time it takes from when the solutes first

interact with the adsorbent to when they exit the base of the columnvu. Gas chromatography (GC) involves the vaporization of a mixture

which migrates through a hot column and is pushed along by aninert carrier gas

f) Extraction involves the removal of a component from a mixture byselective solubiUty in a new solventi. Partitioning describes the distribution of a compound between two

solvents

U. Partition coefficient is the numerical ratio of the solubility of acompound in one solvent versus another solvent

iii. Acid/base extraction involves partitioning between an organicsolvent and water, where the pH is varied to enhance the solubilityof compounds that can form ions when protonated or deprotonated

iv. The yield is better when carried out in three consecutive extractionsof small volume rather than one extraction of large volume

v. Extraction flow charts involve tracing the pathway of variouscomponents in a mixture when they travel through the steps


Organic Chemistry Lab Techniques Section Summary

Purification Techniques

a) Recrystallization purifies a solid by dissolving it and then crystallizing itin a slow, methodical fashion

i. The compound must be highly soluble at high temperatures andminimally soluble at low temperaturesThe dissolving step is carried out under refluxing conditions tooptimize the solubilityA decolorizing step is carried out when there are colored impuritiesanda colorless compound. Activated charcoal is typically usedHot and cold filtration are used depending on whether thecompound must stay in solution or remain crystallizedSolvent washes are employed to rinse away residual solvent thatmay contain impurities

11.

in.

iv.

v.


a) Physical properties are a quick way toverify the identity ofa compoundi. Melting points and boiling points are the common physical

properties usedas a diagnostic for a compound's identityb) Chemical tests involve reagents that react with a minimal number of

functional groupsi. Coloror phasechangesindicatea positive testii. The test agent must be the limiting reagent in the reaction

c) Derivatives are formed to provide additional evidence as to the identityof a compound

i. Melting points are themost common physical property measured fora derivative

d) Mass spectroscopy converts a molecule into a cationic radical and thenaccelerates, deflects, and separates the cationic species that are formedfrom fragmentation

i. Base and parent peaks represent the most stable molecular ion andtheionic form of the original compound respectivelyFragmentation occurs when the unstable species formed uponionization undergoes chemical processes in an effort to form a morestable speciesBromine oncompound results in a 1 : 1peak ratio for the high massfragments while chlorine oncompound results in a 3 : 1 peak ratiofor the high mass fragments

n.

in.

"Chemistry, it's not just for breakfastanymore!"


Laboratory

TechniquesPassages

13 Passages

100 Questions

Suggested schedule:

I: After reading this section and attending lecture: Passages I, IV, VI, & XGrade passages immediately after completion and log your mistakes.

Following Task I: Passages II, VII, IX, & XI (29 questions in 37 minutes)Time yourself accurately, grade your answers, and review mistakes.

Review: Passages III, V, VIII, XII, XIII, & Questions 94 - 100Focus on reviewing the concepts. Do not worry about timing.

II:

III:

BerkeleySpecializing in MCAT Preparation

I.

II.

III.

IV.

V.

VI.

VII.

VIII.

IX.

X.

XI.

XII.

XIII.

Distillation and Separation (1-7)

Fractional versus Simple Distillation (8 - 14)

Steam Distillation of natural Products (15-21)

Caffeine Extraction Experiment (22 - 28)

Partition Coefficient Experiment (29 - 35)

Acid/Base Extraction Experiment (36 - 42)

Extraction and Thin Layer Chromatography Experiment (43 - 49)

Thin Layer and Column Chromatography (50 - 56)

Column Chromatography (57 . 63)

Gas Chromatography (64 - 71)

Recrystallization (72 - 79)

Qualitative Analysis (80 - 86)

Synthesis and Extraction (87 - 93)

Questions not Based on a Descriptive Passage (94 - 100)

Lab Techniques and Spectroscopy Scoring Scale


84 - 100 13- 15

66-83 10- 12

47 -65 7-9

34-46 4-6

1 -33 1 -3


A student is given a mixture of two liquids to separate.There are three possible compositions for the mixture: I)equal parts by volume of Z-l,2-dichloroethene (b.p. 60"C)and E-l,2-dichloroethene (b.p. 488C), 2) equal parts byvolume of 3-pentanone (b.p. 102°C) and 2-hexanone (b.p.128°C), and 3) equal parts by mass of diethyl ether (b.p.34*C) and di-n-propyl ether (b.p. 90°C). Because the studentis uncertainof the components, he uses fractional distillationto separate the liquids. Fractional distillation is chosen whenthe boiling points of the components are within 30°C of oneanother. The distillate is collected in 10-mL aliquots. Figure1 shows the distillation apparatus used by the student. Theapparatus contains the following components: boiling flask,distilling column (with optional packing), thermometer,condenser with cooling sleeve, and collection flask.

Distillation

Column

DistillingPot

Thermometer

Figure 1 Distillation apparatus used in the experiment

The boiling flask is held at a constant temperature andthe distilling column is insulated with glass wool. Afterabout ten minutes, the first drop of liquid appears in thecollection vial. A total of three samples (of roughly 8 mLeach) were collected before the distilling pot was removedfrom the heat source. A small portion of liquid remained inthe distilling flask after the heat source was removed. Oncethe system cooled back to room temperature, there wereapproximately 10 mL of solution still remaining in thedistilling flask.

1. Whendistilling a liquid with a boiling point of 56.6"C,you should use an apparatus with:

A. two openings to the atmosphere, one before thecollection flask and one after the collection flask.

B. one opening to the atmosphere before thecondenser.

C. one opening to the atmosphere after the collectionflask.

D. no opening to the atmosphere, so no vapor canescape.


For the experiment, the thermometer read 49.2°C as thefirst aliquot was collected. The compound beingcollected is most likely which of the following?

A. E-l,2-dichloroethene

B. Z-l,2-dichloroethene

C. Diethyl ether

D. 3-pentanone

A boiling chip, when added to the distillation flask,prevents bumping (the sudden pop of vapor in thesolution) by:

A. providing surface area from which to boil.

B. causing even distribution of the heat.

C. reducing intermolecular forces.

D. increasing intermolecular forces.

The majority of the vapor from the boiled liquidcondenses on which of the following areas?

A. Cold glass in the condenser

B. Cold glass in the collection flask

C. Warm glass in the condenserD. Warm glass in the collection flask

To separate a dissolved solid (solute) from a liquidsolvent in which it is completely soluble, it is best touse:

A. gravity filtration.

B. vacuum filtration.

C. simple distillation.

D. fractional distillation.

Simpledistillation would workbest with which of thefollowing mixtures?

A. H3CH2COCH2CH3 and H3CH2CCOCH2CH3

B. H3CH2COCH2CH3 and E-1,2-dichloroethene

C. H3CH2CH2COCH2CH2CH3 andH3CH2CCOCH2CH3

D. H3CH2CH2CH2CCOCH3 andH3CH2CCOCH2CH3

Fractional distillation is employed to separate:

A. a gas from a liquid.

B. a liquid from a gas.

C. a solid from a liquid.

D. a liquid from a liquid.



A student intended to separate 10.0 mL of toluene from10.0mL of heptane. The boiling point of heptane is 98.4'Cand the boiling point of toluene is 110.6°C. To separate thetwoliquids, the studentcould use either simple or fractionaldistillation. Simple distillation involves distillation througha short, hollow distilling column and it is chosen when thesubstance to be separated has a boiling point that is at leastthirty degrees lower than any other component in themixture. Simple distillation takes less time than fractionaldistillation andgenerates a higher yield. In theseparation ofheptane from toluene, it would be impractical to use simpledistillation, because their boiling points are too close.

Fractional distillation differs from simple distillation inthat the distillation column has more surface area. This isaccomplished either by using a longer column or bypackingthe column with an inert material onto which the vapors cancondense. The advantages of fractional distillation are thattheproduct is purerand the vapor pressure is more constant,which reduces bumping in the solution. Fractionaldistillation does not recover all of thecomponent, however,and it can take upto five times as long as simple distillation.Table 1 lists the boiling points of some organic liquids.

LiquidBoilingPoint

LiquidBoilingPoint

Diethyl ether 34.6°C Octane 125.7'C

Pentane 36.rc n-Pentanol 138.0°C

THF 65.4°C Nonane 150.8°C

Hexane 68.9°C Anisole 158.3°C

Isopropanol 82.3°C Decane 174.1'C

8.

9.

Table 1

Over time, the effectiveness of fractional distillation isdiminished. How can this best beexplained?

A. Over time the distilling column heats up, and asthe temperature increases, the amount ofcondensation and reevaporation decreases.

B. Over time the distilling column heats up, and asthe temperature increases, the amount ofcondensation and reevaporation increases.

C. Over time, the solution becomes poor in the morevolatile component, so azeotropes increase.

D. Over time, the solution becomes poor in the morevolatile component, so azeotropes decrease.

When separating the following molecule pairs usingdistillation, which pair requires the longest distillingcolumn?

A. Diethylether and pentanol.B. Tetrahydrofuran (THF) and anisole.C. Octane andpentane.D. Isopropanol and toluene.


10. Which of the following is NOT an advantage offractional distillation oversimple distillation?

A. Thecomponents thatarecollected arepurer.B. The fractional distillation procedure is faster than

simple distillation.

C. Fractional distillation allows for the mixture tocontinually re-distill throughout the distillationcolumn.

D. Fractional distillation provides a more constantvapor, thus reducingthe tendency of the solution to"bump".

11. Why is distillation bulb (at the base) placed in a sandbathin a heating mantle ratherthanplaced overa flame?

A. The sand serves as a thermal insulator.

B. Thesandis less reactive withthe glass than theair.C. The sand bath keeps the vapor pressure of water

low by absorbing moisture.

D. The sand bath allows for uniform distribution ofheat around the distillation bulb.

12. What is the role of coppermesh whenit is inserted intoa distilling column?

A. The copper mesh serves as a filter to collect anysolid particles that have become airborne.

B. The copper mesh serves to neutralize any acidicvapor that forms.

C. The copper mesh allows for an even distribution ofheat dueto theelectrical resistance of thecopper.

D. The copper mesh serves to provide additionalsurface area onto which vapors may condense andthen re-evaporate in fractional distillation.

13. Which of the following pairs of compounds is bestseparated using simple distillation?

A. Diethylether and heptane.B. Tetrahydrofuran (THF) and hexane.

C. Octaneand pentanol.D. Nonane and anisole.

14. Which stereoisomers would MOST likely show thesame boiling point?

A. 2R,3S-dichloropentane and2R,3R-dichloropentaneB. 3R-bromocyclopentene and3S-bromocyclopenteneC. D-Glucose and D-mannose

D. 3R-ethyl-trans-decalin and3S-ethyl-trans-decalin



Because molecules can exist in the vapor state attemperatures below the compound's boiling point, it ispossible to carry out distillation at temperatures below theboiling point of the components in the mixture when themixture does not form an ideal solution. A common

technique for carrying this out is steam distillation. Steam isdirected to the distilling flask, resulting in a temperature of100°C during the course of distillation. The solution slowlyevaporates, and is collected in the same fashion as standarddistillation. Because of the presence of water vapor in theimmediate atmosphere, the distillate is rich in water. Unlikethe distillate in simple or fractional distillation, the distillatein steam distillation must be further purified. To simplifyfurther separation, steam distillation is often employed withoils that are immiscible in water.

A student was assigned the task of separating citral (b.p.= 229°C) from lemon grass oil. Citral is a 10-carbon terpenethat is a precursor in the commercial synthesis of Vitamin A.It is isolated in conjunction with neral, which varies at thedouble bond between carbons 2 and 3. Citral and neral are

shown in Figure 1 below.

O CH3 CH3

Citral

Figure 1 Citral and neral, components of lemon grass oil

After steam distillation, the mixture is approximately90% water, 9% citral, and 1% neral. To isolate the organicproducts, the mixture is extractedusingdiethyl ether and driedusing anhydrous magnesium sulfate.

15. What compounds are best separated using steamdistillation?

A. Organic liquids with boiling points over 100°CB. Organic liquids with boiling points under 100'CC. Organic solids with melting points over 100°CD. Organic solids with melting points under 100°C

16. Neral and citral are best described as:

A. conformational isomers.

B. geometrical isomers.

C. optical isomers.

D. structural isomers.


17. Standard distillation of lemon grass oil at 229°C couldresult in all of the following EXCEPT:

A. decomposition of citral.

B. hydrogenation of citral.

C. oxidation of citral.

D. polymerization of citral.

18. When ether is added to the distillate, what is observed?

A. Water dissolves into the ether layer while citralsinks to the bottom of the flask.

B. Water dissolves into the ether layer while citralfloats to the top of the solution.

C. An ether layer forms on top of the water layer, andcitral dissolves into it.

D. An aqueous layer forms on top of the ether layer,and citral dissolves into it.

19. What is the purpose of adding anhydrous magnesiumsulfate to the ether extract?

A. To isomerize neral into citral

B. To reduce the aldehyde into a primary alcohol

C. To remove any residual water in the ether layer

D. To selectively bind neral, leaving behind pure citral

20. What is the approximate boiling point of neral?

A. 78°C

B. 100'C

C. 147°C

D. 2176C

21. Whatproperty is expectedfor VitaminA?

A. A lower melting point than citral

B. A high water solubility

C. A high vapor pressure at room temperature

D. A high affinity for lipids



A studentattempts to removecaffeinefrom tea leaves bya series of extraction procedures. Initially, 2.0 grams of tealeaves are added to 10 mL of boiling water for 20 minutes.The aqueous solution is then extracted three times withexactly 5.0 mL of methylene chloride(CH2CI2) during eachof the threeextractions. The 15.0mL of methylene chloridesolution is collected and consolidated into one flask. Thesolvent is removed through evaporation by flushing vaporfrom the flask using a constant stream of nitrogen gas. Thetemperature of the solution is maintained at 60°C by partiallyimmersing the flask in a water bath.

After the solvent is completelyremovedfrom the flask, awhite residue remains at the bottom. This white residue issublimed under vacuum and 28milligrams of a white powderare isolated. The melting point and infrared spectroscopy dataof the white powder matches that of caffeine (m.p. = 238 *C).The student concludes that the isolated compound is in factpure caffeine. Following sublimation, the bottom of theflask has no white powder remaining, just a waxy residue.Thestructure of caffeine is shown in Figure 1 below.

The caffeine could have been removed from the organicsolvent by treatment with a strong acid such ashydrochloricacid. The acid can protonate the caffeine to form the ionicsalt complex, which is relatively insoluble in organicsolvent.

22. From the values presented in the passage, what is theapproximate percentage by massof caffeine in tea?

A. 1.4%

B. 2.8%

C. 14%

D. 28%

23. Based on the information in the passage, caffeine haswhich of the following solubility properties?A. It is soluble in both water and CH2CI2, but more

soluble in CH2CI2.

B. It is soluble in both water and CH2CI2, but moresoluble in water.

C. It is soluble in water, butinsoluble in CH2CI2.D. It is soluble in CH2CI2, but insoluble in water.


2 4. The flow of nitrogen gas serves to:

A. reduce the partial pressure of the solvent vapor somore solvent can evaporate.

B. increase thepartial pressure of thesolvent vapor somore solvent can evaporate.

C. reduce thepartial pressure of the solvent vapor soless solvent can evaporate.

D. increase thepartial pressure of thesolvent vapor soless solvent can evaporate.

25. Which procedure would you useto determine thepurityof the caffeine?

A. Adding it to methylene chloride to test itssolubility.

B. Adding it to water to test its solubility.C. Mixing the compound with caffeine and observing

the melting point for a sharprange.D. Mixing the compound with an organic solid other

than caffeine andobserving the melting pointfor asharp range.

26. Why is the methylene chloride added in three 5-mLportions rather thanone 15-mLportions?

A. Methylene chloride is too unstable to be added in abig portion.

B. 15mL of methylene chloride weighs more than 10-mL of water,so the organic layer wouldsink to thebottom of the flask.

C. Less caffeine can be extracted by three smallextractions than one large extraction.

D. More caffeine can be extracted by three smallextractions than one large extraction.

27. Methylene chloride was added to the aqueous solutionto:

A. extract all of the caffeine while leaving behind thewater-soluble impurities.

B. extract most of the caffeine while leaving behindthe water-soluble impurities.

C. extract all of the caffeine along with the water-soluble impurities.

D. extract some of the caffeine along with the water-soluble impurities.

28. Sublimation is employed to purify a:

A. solid from other solids.

B. solid from liquids.C. solid from gases.D. liquid from solids.



A student wishes to determine the partition coefficient of4-methylaniline between the water layer and diethyl etherlayer in a biphasic mixture. The student decides to dissolve aknown mass of 4-methylaniline into a 1:1 mixture byvolume of water and anhydrous diethyl ether. In the firsttrial, the student uses 10 mL of water, 10 mL of anhydrousdiethyl ether, and 1.0 gram of 4-methylaniline. This mixtureis placed into a separatory funnel and shaken. The lowerlayer is removed, without removing any of the upper layer.A drop of water is added to the lower layer, and it is found todissolve completely into the solution. The student concludesfrom this that the upper layer, still in the separatory funnel,is the organic (diethyl ether) layer. The student removes theupper layer into a 25 mL Erlenmeyer flask and adds to this0.5 grams of anhydrous calcium chloride, which forms a layerof powder on the bottom of the flask.

The flask is swirled gently for two minutes and then sitsfor five minutes. The ether layer is removed from the flaskby decantation using a warm pipette and is added to a 10.50gram vial. The calcium chloride remaining in the Erlenmeyerflask is rinsed with 5.0 mL of anhydrous diethyl ether. This5.0 mL portion is removed by decantation using a warmpipette and added to the vial. The vial is heated mildly underan exhaust fan until the diethyl ether evaporates awaycompletely. The mass of the vial and dried 4-methylanilineis 11.25 grams. The student assumes that the missing 4-methylaniline from the original 1.0 g sample must havedissolved into the water layer. By dividing the grams whichdissolved into the ether layer by the grams that dissolved intothe water layer, a partition coefficient is calculated.

To verify that the answer is valid, the student consults abook of physical constants and finds that the solubility at25°C for 4-methylaniline in water is 3.8 grams/100 mL andthe solubility at 25°C for 4-methylaniline in diethyl ether is22.8 grams/100 mL. From these values, the studentconcludes that the experimental partition coefficient is toolow. This deviation is most likely attributed to the loss ofsolid in the transfer steps.

29. From the experimental values using 1.0 gram in amixture of 10 mL each of water and diethyl ether, whatis the partition coefficient?

A. 2.5

B. 3.0

C. 4.0

D. 6.0

30. From the literature values for solubility, what is thepartition coefficient?

A. 2.5

B. 3.0

C. 4.0

D. 6.0


31. A second student attempted this same experiment using3.0 grams 4-methylaniline in a mixture of 10 mL eachof water and diethyl ether. Which of the followingstatements predicts the results?

A. The partition coefficient was found to be the sameas with the 1.0-gram experiment.

B. The partition coefficient increased from the valuedetermined in the 1.0-gram experiment.

C. The partition coefficient decreased from the valuedetermined in the 1.0-gram experiment.

D. The partition coefficient could not be determinedbecause the 3-methylaniline didn't dissolvecompletely.

3 2. How can the 4-methylaniline that adhered to the pipettewall during transfer be recovered?

A. The pipette could be rinsed with water.

B. The pipette could be rinsed with diethyl ether.

C. The pipette could be heated to a high temperature.

D. The pipette could be cooled to a low temperature.

3 3. If the ether used was NOT anhydrous, how would thisaffect the calculated partition coefficient?

A. It would lower the value, because initially therewould be more than 10 mL of water and less than

10 mL of ether.

B. It would raise the value, because initially therewould be more than 10 mL of water and less than

lOmL of ether.

C. It would lower the value, because initially therewould be more than 10 mL of ether and less than

10 mL of water.

D. It would raise the value, because initially therewould be more than 10 mL of ether and less than

10 mL of water.

34. All of the following solvents be used in place of diethylether EXCEPT:

A. tetrahydrofuran,THF.

B. ethanol.

C. cyclohexane.

D. methylene chloride.

3 5. Which of the following is a desired property of theorganic solvent?

A. Highly miscible in water

B. Highly dense

C. Solute is highly soluble in the solventD. Its boiling point is less than room temperature



Twostudents wereeachpresented with a mixture ofequalparts benzoic acid, resorcinol, and meta dinitrobenzene. Tohelp in separating the three compounds into their pureindividual components, the students had available 10%sodium bicarbonate solution, 5% hydrochloric acid solution,10% acetic acid solution, and 5% sodium hydroxide solution.Each student mapped out a scheme in the form of a flow chartto separate the three compounds. Student I chose first to addsodium hydroxide solution and ether to separate thecompounds. Student II chose first to add sodium bicarbonatesolution and ether to separatethe compounds. Eachscheme,along with the structures of each compound and their pKavalues, are shown in Figure 1 below.

O^ ÔHpKa =4.17 ?H pKa =9.89

Benzoic Acid

j>H pKa =9.89 NO2

?« 6. NO2Resorcinol meta-Dinitrobenzene

Proposed Scheme for Student I

Ether layerAqueous sodiumhydroxidelayer

Ether layer

(Aqueous sodiumbicarbonate layer Tube 1

1

1

Tube 2 Tube 3

Proposed Scheme for Student II

Ether layerAqueous sodiumbicarbonatelayer

Ether layer

(Tube 5

Aqueous sodiumhydroxide layer Tube 4

ITube 6

\

Figure 1 Reaction schemes for Students I and II

The ether layer and aqueous layer are immiscible in oneanother, sothey can beseparated using decanting techniques.To isolate the solute from theetherlayer, theethersolvent isallowed to evaporate. To isolate the components from theaqueous layers, the solutions are neutralized and filtered. Thetwo students did not get thesame results. Themelting pointranges for the three solids isolated by Student II are sharperthan the melting point ranges for the three solids isolated byStudent I.


36. Which of the following organic compounds could beused in place of ether in this acid/base extraction?

A. Ethanol (b.p. = 79"C)

B. 1,4-Dichlorobenzene (m.p. = 53°C)C. Naphthalene (m.p. = 94°C)

D. Chloroform (b.p. = 61°C)

3 7. What solution should be added to Tube 4 to cause aprecipitate to fall out of solution?

A. 10% sodium bicarbonate solution.

B. 5% hydrochloric acid solution.

C. 10% acetic acid solution.

D. 5% sodium hydroxide solution.

3 8. What can be said about Tube 2 and Tube 5?

A. Tube 2 contains dinitrobenzene, while Tube 5contains resorcinol.

B. Tube 2 contains dinitrobenzene, while Tube 5contains both resorcinol and benzoic acid.

C. Tube 2 contains resorcinol, while Tube 5 containsdinitrobenzene.

D. Tube 2 and Tube 5 both contain dinitrobenzene.

39. What is true about the pK values for the conjugatebases of the benzoic acid and resorcinol?

A• pKb(sodium benzoate) >pKb2(sodium resorcinoxide)and pKa1(resorcinol) > pKa2(resorcinol)

B• pKb2(sodium resorcinoxide) > pKb(sodium benzoate)and pKai(resorcjnoi) > pKa2(resorcinol)

C. pKb(sodium benzoate) > pKb2(sodium resorcinoxide)and pKa2(resorcinol) > PKal (resorcinol)

D• P*M>2(sodium resorcinoxide) > pKb(sodium benzoate)and pKa2(resorcinol) > PKal (resorcinol)

40. What canbeconcluded about the twoschemes presentedin the passage?

A. The scheme for Student II worked while the schemefor Student I did not work.

B. The scheme for Student I worked while the schemefor Student II did not work.

C. Both the scheme for Student I and the scheme forStudent II worked.

D. Both the scheme for Student I and the scheme forStudent II did not work.


41. If 3-methylaniline were present in the mixture forStudent n, where it be found?

A. Exclusively in Tube 4

B. Exclusively in Tube 6

C. Evenly split between Tubes 5 and 6

D. Predominantly in Tube 5

42. Which of the following compounds cannot be isolatedusing acid-base extraction techniques?

A. 4-Ethylcyclohexanone

B. 3-Bromophenol

C. 3-Nitrobenzoic acid

D. Ethyl benzoate



Organic compounds can be isolated from plants througha process known as extraction, a common industrial practiceused to isolate natural products for medicine, preservatives,dyes, and flavoring agents. Trimyristin, for instance, can beisolated from nutmeg beans by grinding the beans into a pulpand then adding diethyl ether to the ground pulp. Trimyristindissolves readily into diethyl ether. In ordinary extractionprocedures, the solution is isolated by either decanting itaway from the suspension or by filtering the solid impuritiesout of the suspension by passing it through cotton. In bothprocedures, the desired compound remains dissolved in theorganic solvent. When left to sit, the solvent will slowlyevaporate, leaving behind residue of the compound. In thecase of trimyristin, precipitation can be expedited by addingmethanol to the solution. The purity of the extract may betested by one of several spectroscopic techniques (including*HNMR and IR), by thin layer chromatography, or byevaluating the melting point of the powder.

Trimyristin is a fatty acid triglyceride, derived fromglycerol and three fourteen-carbon fatty acids. Like all esters,it is sensitive to acid-base extraction techniques. In anexperiment, students attempt to extract trimyristin fromnutmeg beans. The solvent chosen for extraction wastetrahydrofuran, THF. After filtering through cotton andadding methanol to solution, a white powder formed on thebottom of the flask. It was isolated by passing it through aBuchner funnel fitted with filter paper. A small sample ofthe powder was dissolved into three drops of dichloromethane,and the new solution was spotted onto a TLC (thin layerchromatography) plate in three places. The plate was placedinto a beaker with minimal dichloromethane at the bottom.

The beaker was sealed and the solvent proceeded to climb theplate. Before the solvent reached the top of the plate, theplate was removed from the solvent and the top of the solventfront was marked. The plate was set into an iodine chamberfor developing. Spots appeared and were analyzed. Figure 1shows a rough sketch of what appeared:

•Solvent

front

stopped

>«solvent

o or*

o

d pot">

o

dspoM•

fle o

1spot|

Before After

Dichloromethane Solvent

y

Rf =Jspot

dsolvent

Figure 1 Thin layer chromatography results of extract


The three spots resulted in Rf values of 0.19, 0.59, and0.83. The compound with an Rf = 0.59 (referred to asCompound II) was the most abundant species of the threecompounds that were isolated from the crude product mixtureusing column chromatography techniques. The other twoproducts werepresent in smallerquantities.

43. A spot that shows an Rf value of 0.72 in a nonpolarsolvent like hexane will likely show what Rf value in apolar solvent like ethyl acetate?

A. 1.84

B. 0.72

C. 0.13

D. -0.21

44. What solvent should be added to the beans to removeany naturally occurring sugar? A typical sugar has acarbonyl group and all other carbons have an alcoholgroup (CH2OHCH(OH)CH(OH)CH(OH)CH(OH)CHO)A. A long-chain alkane.

B. A cyclic ether.

C. A short-chain alcohol.

D. A long-chain ketone.

45. Why should the initial height of the solvent in thebeaker NOT be above the initial height of the spots onthe TLC plate?

A. The spots would not migrate up as fast if they aresubmerged below the level of the solvent.

B. Thespots would become saturated and thus migratetoo rapidly up the plate.

C. The spots would migrate in a radial manner ratherthan a linearmanner, if submerged below thetopofthe solvent.

D. The spots would dissolve into solution ifsubmergedbelow the top of the solvent.

46. If two compounds in a mixture have respective Rfvalues of 0.58 and 0.29 in hexane, and 0.07 and 0.21 inethanol, how are theybestseparated?

A. By column chromatography, using hexane in ashort column.

B. By column chromatography, using hexane in along column.

C. By column chromatography, using ethanol in ashort column.

D. By column chromatography, using ethanol in along column.


47. Which of the following solvents would show a similarresult as a TLCsolvent as carbon tetrachloride (CCI4)?

A. Acetone(CH3COCH3)

B. Propanal (CH3CH2CH=0)

C. Ethanol (CH3CH2OH)

D. Pentane (CH3CH2CH2CH2CH3)

48. Which of the following values is NOT possible for theRf value in a TLC analysis?

A. 0.00

B. 0.35

C. 0.70

D. 1.05

4 9. How can THF and trimyristin best be described?

A. THF is a nonpolar solvent and trimyristin is ahydrophobic compound.

B. THF is a nonpolar solvent and trimyristin is ahydrophilic compound.

C. THF is a polar solvent and trimyristin is ahydrophobic compound.

D. THF is a polar solvent and trimyristin is ahydrophiliccompound.



Chromatography operates based on the partitioning of acompound between a mobile phase and a stationary phase. Ifthe compound has a higher affinity for the eluting solvent(mobile phase) than the adsorbent (stationary phase), then itwill travel quickly. One type of chromatography is thin layerchromatography, which is done on a small scale to analyzethe components of a mixture. It can also be done on a largerscale in a column, where the length of the column is at leastten times the diameter of the column. Columns are capableof separating up to approximately 10 grams of material,whether it is a liquid or solid.

In thin layer chromatography, the adsorbent is bound to arectangular plate made of either glass or plastic, depending onthe adsorbent. The sample is spotted near the bottom of theplate and solvent is allowed to climb up the plate throughcapillary action. The solutes in the spots migrate with thesolvent up the plate, but at varying rates depending on theirrelative partitioning between eluting solvent and adsorbent.In column chromatography, the same material is added to thetop of a column packed with adsorbent and a thin layer ofclean sand at the top. The sand prevents the adsorbent fromdistorting when solvent is poured into the column.

There are two common adsorbents, alumina (AI2O3)andsilica gel (Si02). Alumina comes is three forms, acidic,basic, and neutral. The neutral form is given a BrockmannActivity rating from I to V, where I is the most active interms of binding a solute. The activity can be reduced by theaddition of water, because water binds the alumina, therebyreducing the surface area to which the solute can bind. ABrockmann Activity rating of III refers to alumina that is 6%water by mass.

Solvents are rated according to their polarity, from leastpolar to most polar. As a solvent becomes more polar, it hasa greater affinity for a polar solute. Table 1 shows solventsaccording to elution rate for a nonpolar solute from fastest toslowest across alumina. Often, the affinity of polar solutesfor the adsorbent exceeds all other affinities.

Rank Solvent Rank Solvent

1 n-Hexane 8 Tetrahydrofuran

2 Petroleum ether 9 Dioxane

3 Cyclohexane 10 Ethyl acetate4 Ligroin 11 Dimethyl sulfoxide

5 Carbon disulfide 12 2-Propanol

6 Ethyl ether 13 Ethanol

7 Dichloromethane 14 Methanol

Table 1

The general elution order for solutes in nonpolar solventsis: alkanes > alkenes > dienes > aromatic hydrocarbons>ethers > esters > ketones > aldehydes > amines > alcohols >phenols > carboxylic acids. Rf values can be predicted fromthe relative elution rate and the solvent rankings by applyingthe simple principle: "like dissolves like."


50. Which combination has the fastest elution time througha column filled with alumina?

A. Benzoic acid solute in n-pentane

B. Glycerol solute in petroleum ether

C. o-Xylene solute in ligroin

D. Glycine solute in ethanol

51. For two compounds with Rf values that are different bya factor of 2, what is true in column chromatography?

A. The elution time is 4 times greater for thecompound with the larger Rf value than thecompound with the smaller Rf value.

B. The elution time is 2 times greater for thecompound with the larger Rf value than thecompound with the smaller Rf value.

C. The elution time is 4 times greater for thecompound with the smaller Rf value than thecompound with the larger Rf value.

D. The elution time is 2 times greater for thecompound with the smaller Rf value than thecompound with the larger Rf value.

5 2. Using TLC, which solvent would give salicylic acid thegreatest Rf value?

A. n-Hexane

B. Ligroin

C. Methanol

D. 2-Pentanol

53. What is true of alumina with a Brockmann Activityrating of IV?

A. It has 4% water by mass.

B. It has a greater affinity for solute than alumina witha Brockmann Activity rating of III.

C. It is rich in silicon atoms.

D. It can be formed by adding more than 6 grams ofwater to 94 grams of alumina with a BrockmannActivity rating of I.

5 4. How does a sudden change in solvent from n-pentane tomethanol in an alumina column harm the effectiveness

of the column?

A. The methanol endothermically binds the alumina,forming vapor pockets in the column.

B. The methanol exothermically binds the alumina,forming vapor pockets in the column.

C. Methanol is immiscible in hexane, so it dropsthrough the column too fast.

D. Methanol is immiscible in hexane, so it dropsthrough the column too slow.


55. Why does adding water to alumina raise its BrockmannActivity rating?

A. After water binds the surface of alumina, there arefewer sites on the alumina for solute to bind.

B. Water causes the alumina to contract, reducing itssurface area for the binding of solute.

C. Water makes thenonpolar alumina more polar.D. Wateroxidizes the alumina into a newspecies that

is more hydrophobic than alumina.

56. Alumina would likely have the highest affinity forwhich of the following solutes?

A. 1,3-Cyclohexadiene

B. CyclohexanolC. Ethyl butanoate

D. 3-Methylpentanal


Passage IX (Question 57 - 63)

Benzylic hydrogens are unusually reactive forhydrogenson an s/^-hybridized carbon. They have pKas around 23-25,considerably lower than 45-50 for alkyl hydrogens. They arealso more susceptible to oxidation than alkyl hydrogens.Figure 1 shows the oxidation of toluene into benzoic acid, atypical oxidation reactionfor benzylic protons.

[O]

Figure 1 Oxidation of the benzylic hydrogens of toluene

A chemist decided to oxidizefluorene into fluorenone byaddingsodiumdichromate, Na2Cr207, in acetic acid solventThe reaction is shown in Figure 2 below.

Figure 2 Oxidation of fluorene into fluorenone

Because both fluorene and fluorenone are solids at roomtemperature, separation cannot be done by distillation. Forseparating solids, extractionor chromatography are typicallyemployed. If the sample is 10 grams or less, then columnchromatography works well. The chemist chose to separatethe productmixture using column chromatography. Prior tocarryingout the column chromatographyexperiment, she rana series of thin layer chromatography trials to determinewhich elution solvent and adsorbent to use in the column.

The adsorbent in each trial is silica gel. Two spots areobserved in every trial. One spot is yellow in color, and bothare UV active. Table 1 shows the Rf values for each of thetwo compounds in various solvents used in the TLC trials.

SolventRf for Yellow

CompoundRf for Colorless

compound

Liquid I 0.18 0.37

Liquid II 0.24 0.33

Liquid m 0.19 0.51

Liquid IV 0.41 0.55

THF 0.31 0.42

Table 1

57. Fluorenone should have the greatest affinity for whichof the following solvents?

A. Acetone

B. Cyclohexane

C. Dichloromethane

D. Methanol


58. Why does fluorene elute from the column beforefluorenone when cyclohexane is theeluting solvent?

A. Fluorene has a higher affinity for silica gel thanfluorenone.

B. Fluorene is lighter than fluorenone, so it travelsfaster.

C. Fluorene has a higher affinity for cyclohexanethanfluorenone.

D. Fluorene is smaller than fluorenone, so it can passmore easily through the pores in silica gel.

59. According to the data in Table 1, what eluting solventis best for separating fluorenone from fluorene?

A. LiquidII

B. Liquid m

C. Liquid IV

D. THF

60. When eluting a mixture of biphenyl, ethyl benzoate,and toluic acid from an alumina column using ligroin asthe eluting solvent, the sequence off the column is:

A. biphenyl first, ethyl benzoate second, and toluicacid last.

B. toluic acid first, ethyl benzoate second, andbiphenyl last.

C. ethyl benzoate first, biphenyl second, and toluicacid last.

D. biphenyl first, toluic acid second, and ethylbenzoate last.

61. Once the silica gel is set in the column, why shouldyou wait to add the sample until the solvent is flushwith the top of the silica gel?

A. So that the sample does not float on the solventand never get to the silica gel

B. So that the sample does not adhere to the glass ofthe column

C. So that all of the sample starts traveling down thecolumn at the same time

D. So that top of the column can dry out before thesample interacts with the silica gel

6 2. Why is a thin layer of sand added to the top of the silicagel?

A. To prevent the sample from reaching the silica gel

B. To reduce the polarity of the gel

C. To keep the top of the silica gel flat when moresolvent is poured into the column

D. To bind sandphilic solutes


63. Why do nonpolar solutes elutebefore polarsolutes fromcolumns containing silica gel, even when a polarsolvent is used?

A. Polar compounds do not follow the "like dissolveslike" rule.

B. The silica gel is highly polar, which hinders themigration of polar solutes.

C. Nonpolar species become more polar when thesolvent is polar.

D. Polar compounds aggregate and form largemolecules that do not migrate quickly because ofsteric hindrance.


Passage X (Questions 64-71)

A researcher carried out a two-step reaction using theketone 3,3-dimethylcyclopentanone as the reactant. Thereaction sequence is shown in Figure 1 below.

Reaction 2

Figure 1 Two-step reaction of 3,3-dimethylcyclopentanone

The reaction sequence involves deprotonating an alphahydrogen and subsequently addingan electrophile to form abond to the alpha carbon. Because the reactant contains twoalpha carbons, there are two possible structural isomerproducts. The two methyl groups on carbon three offer sterichindrance, so one alphacarbon is lesssterically hindered thanthe other alpha carbon.

Thereaction wascarried outat different temperatures. Atlower temperatures, the product mixture favored the lesssterically hindered product. This is often referred to as thekinetic product of thereaction. Thethermodynamic productresults from the formation of the more stable bond. In thereaction of Figure 1, there is no distinct thermodynamicproduct. Table 1 shows the percentage of each species insolution following thereaction at various temperatures.

Trial I:

Temp

-33°C

% Reactant

25

% Product A

66

% ProductB

9

Trial H: 0°C 20 60 20

Trial HI: 15°C 16 56 28

Trial IV: 30°C 13 53 34

Trial V: 50°C 10 50 40

Trial VI: 78°C 7 48 45

Table 1

Figure 2 is the gaschromatography chartproduced whenthe product mixture from Trial III is analyzed. Thecompounds were passed through a Carbowax-12 column thatispartially polar. The identity of each peak was obtained byspiking theproduct mixture with a sample of oneof the purecomponents. Spiking is the term given to adding a smallportion of one of the components to the product mixture.The peak that increases isdeduced tobelong to the compoundthat has been spiked.

Copyright © byThe Berkeley Review® 282

Time

Figure 2 Gas chromatography results for Trial HI

Gas chromatography can be used to quantitativelyanalyze the composition of mixtures by evaluating the areaunder each peak. The relatives areas under each curve can beused to determine the percent composition of the mixture.

64. If the product mixture from Trial m were spiked withthe reactant, which of the peaks in the example wouldgrow?

A. Peak 1

B. Peak 2

C. Peak3

D. Without knowing the composition of the column,it cannot be determined.

Peak 2

Peak 3

65. If the column in the gas chromatography machine werepacked with a nonpolar polymer, what would be trueabout the retention times of polar and nonpolarcompounds?

A. Nonpolarcompounds would have longer retentiontimes, because they bind the column less tightlythan polar compounds.

B. Nonpolar compounds would have longer retentiontimes, because they bind the column more tightlythan polar compounds.

C. Nonpolar compounds would have shorter retentiontimes, because they bind the column less tightlythan polar compounds.

D. Nonpolar compounds would have shorter retentiontimes, because they bind the column more tightlythan polar compounds.

66. Which of the following relationships accurately depictsthe relative retention times of the components?

A. Product A > Product B > Reactant

B. Reactant > Product A > Product B

C. Reactant > Product B > Product A

D. Product B > Product A > Reactant


67. According to Table 1, what can be concluded aboutincreasing the temperature of the reaction?

A. It makes the reaction more favorable.

B. It favors the formation of the kinetic product.C. It favors the retention of reactants.

D. It makes the reaction less favorable.

68. For the compounds associated with the crude productmixture from Trial IV, gas chromatography separatesprimarily by:

A. the size of molecule.

B. the mass of the heaviest atom in the molecule.

C. the charge of the compound.

D. the affinity for the column.

69. The best gas to use as a carrier gas (to propel thevaporized components of the mixture through thecolumn) is:

A. carbon dioxide.

B. hydrogen.

C. helium.

D. water.

7 0. The major product from the reaction in Trial II can beisolated using which of the following techniques?

A. Column chromatography

B. Simple distillation

C. Vacuum sublimation

D. Acid-base extraction

71. Why must a strong base be used in Reaction 1 inFigure 1?

A. Because alpha hydrogens are highly acidic, withpKa values around 2-5.

B. Because alpha hydrogens are highly acidic, withpKa values around 17-20.

C. Because alpha hydrogens are weakly acidic, withpKa values around 2-5.

D. Because alpha hydrogens are weakly acidic, withpKa values around 17-20.



Recrystallization is the laboratory process of dissolving asolid into a minimal amount of solvent, and thenprecipitating the solute from solution in a highly purecrystalline form. The procedure includes some filtering stepsto remove solid impurities (such as a boiling chips orcharcoal that may have been added) and ultimately to isolatethe desired crystals. The general procedure for therecrystallization process is listed below:

I. Add the impure solid mixture into a minimalamount of solvent at room temperature to dissolvethe compound partially.

II. Heat the solvent to its boiling point in a refluxingsystem. If not all of the solid has dissolved, add hotsolvent drop by drop until the solid is completelydissolved.

III. Once the solid is completely dissolved, hot-filter thesolution, avoiding any precipitation of the solute.

IV. Slowly cool the filtered solution until it reachesroom temperature, and then place the container intoan ice bath. Crystals should form slowly duringthis step of the procedure.

V. Filter the collected crystals from the solvent atreduced temperature.

VI. Rinse the crystals with a volatile solvent in whichthey are insoluble and allow them to air dry.

The procedure is generic and can be applied to any solidthat is soluble in an organic solvent. If the solid is toosoluble in the solvent, the quantity of solute that isrecrystallized is minimized. If the solid is not solubleenough in the solvent, the quantity of solvent is so large thatimpurities are dissolved as well. The ideal solvent has thesolid barely soluble at room temperature and fully soluble atthe boiling point of the solvent. The solvent is refluxed sothat it exists at its highest temperature (boiling point), butdoes not evaporate away.

Charcoal is a decolorizingagent that is sometimesaddedto the first solution. Charcoal has a high affinity for coloredorganic molecules, because both charcoal and organicchromophores are rich in rc-bonds. Charcoal is onlyadded tothe solution when the desired compound is colorless and thesolution is colored.

The purity of the crystals can be quickly tested bylooking at the melting point range. A narrow rangecorresponds to purecrystals. As a general rule, largercrystalsare purer than smaller crystals.

7 2. Which of the following crystals are the purest?

A. Small cubic crystals formed by rapid cooling.B. Small rectangular crystals formed by slow cooling.C. Long cubic crystals formed by slow cooling.D. Long rectangularcrystals formed by rapid cooling.


7 3. The ideal solvent for recrystallization should have whichof the following properties?

I. A low affinity for the solid lattice

II. The solid should be highly soluble in the solvent atall temperatures

m. Impurities should be highly soluble in the solvent

A. I only

B. Honly

C. I and Honly

D. I and HI only

7 4. The purity of the crystal CANNOT be verified by whichof the following?

A. Density of the crystal

B. Melting point of the crystal

C. Mass of the crystal

D. Index of refraction of the crystal

7 5. What is the role of hot-filteringthe solution in Step HI?

A. To filter out any soluble impuritiesB. To filter out any insoluble impuritiesC. To filter out the compoundD. To reduce the amount of solvent in the flask

76. What is the purposeof using the ice bath in StepV?

A. To increase the amount of crystals formed byincreasing the solubility of the solid

B. To decrease the amount of crystals formed byincreasing the solubility of the solid

C. To increase the amount of crystals formed bydecreasing the solubility of the solid

D. To decrease the amount of crystals formed bydecreasing the solubility of the solid

77. What is a potential problem if the crystals are formedtoo rapidly?

A. They grow to be too large.

B. They trap impurities from solution in their latticestructure.

C. It does not allow enough time for impurities toform in the lattice.

D. It prevents solventfrombeingincorporated intothelattice structure.


7 8. What is NOT true about recrystallization?

A. The final crystals are purer than the starting solid.

B. The process requires knowing the boiling point ofthe solvent.

C. It is best carried out in a highly volatile solvent.

D. It results in less than 100% recovery of the originalsample.

7 9. To which solution should activated charcoal be added

during a recrystallization experiment?

A. A solution with colorless impurities and a colorlesstarget crystal

B. A solution with colorless impurities and a coloredtarget crystal

C. A solution with colored impurities and a colorlesstarget crystal

D. A solution with colored impurities and a coloredtarget crystal



A researcher has chemicals organized according tomolecular mass. In one section there are three compoundswhich say only 100 grams/mole on their respective bottles.In an attempt to identify the three unknowns, the researcherlabels the bottles Compound A, Compound B, andCompound C. 1.00 gram of each compound is isolated andpurified.

A 1.00 mg sample of each compound was completelyoxidized. For all three compounds, 2.64 mg CO2 gas and1.08 mg H2O liquid was collected after complete oxidation.No nitrogen, sulfur or halides were found in any of thecompounds. No analysis for the oxygen content wasconducted due to the difficult nature of analyzing oxygencontent. Calculations show that there are exactly six carbonsand twelve hydrogens in each of the three compounds,implying that the three unknowns are isomers. After theformula analysis was complete, the researcher subjected eachof the three compounds to standard chemical tests, assummarized in Table 1.

Cr03/H+ Br2(CCl4) (N02)2C6H3C0C1

Compound A turned

green

remained

brownprecipitate formed

Compound B remained

orange

remained

brown

no precipitateformed

Compound C remained

orange

turned

clear

no precipitateformed

Table 1

It should be noted that the organic product from thereactionof Compound A with Cr03/H+ does not turn litmuspaper red. Compound B, when treated with I2 and KOH,forms a yellow precipitate and acidic solution (a positiveresult for the iodoform test for methyl ketones). From themolecular mass and carbon count, it can be inferred that thecompounds contain one oxygen. The chemical tests indicatethat each compound is either an alcohol, an ether or acarbonyl compound. A positive test with bromine in carbontetrachloride indicates the presence of an alkene. Each ofthese three types of functional group reacts differently, sopredictions can then be made about further reactions that theisomers can undergo, in order to further support their identity.

8 0. Which of the following IR absorbances would be thekey peaks in Compound A and Compound B?

A. Compound A: broad absorbance at 3500 cm"1Compound B: sharp absorbance at 1710 cm"1

B. Compound A: sharp absorbance at 1710 cm"1Compound B: broad absorbance at 3500 cm"1

C. Compound A: sharp absorbance at 2980 cm"1Compound B: sharp absorbance at 1710 cm"1

D. Compound A: sharp absorbance at 1710 cm"1Compound B: sharp absorbance at 2980 cm"1


81. A compound that turnsclear withBr2/CCl4 and hasonedegree of unsaturation CANNOT be:

A. a ketone.

B. an alkene.

C. an ether.

D. an alcohol.

8 2. Which of the following types of compounds would losea peak inits JHNMR when D2O is added?

A. Alcohol (ROH)

B. Ketone (RCOR)

C. Ether (ROR)

D. Aldehyde (RCHO)

83. Which of the three compounds would undergo a colorchange when treated with KMn04, considering KMn04can oxidize alkenes, selected alcohols, and selectedcarbonyl compounds?

A. Compound A

B. Compound C

C. Compounds A and C

D. Compounds B and C

84. (N02)2C6H3C0C1 is NOT a good reactant to test forwhich of the following?

A. Alcohol (ROH)

B. Amine (RNH2)

C. Ether (ROR)

D. Thiol (RSH)

85. Compound A, by turning green with Cr03/H+ and notforming a compound that turns litmus red, is mostprobably:

A. a primary alcohol.

B. a secondary alcohol.

C. a tertiary alcohol.

D. an ether.

86. A positive iodoform test can be associated with whichof the following?

A. A yellow precipitate confirming a methyl ketone.

B. A yellow precipitate confirming an aldehyde.C. A black precipitate confirming a methyl ketone.D. A black precipitate confirming an aldehyde.


Passage XIII (Question 87 - 93)

A student starts with 138 mg of 1,4-dimethoxybenzene(138.166 grams/mole) and treats it with 1.32 mL anhydrousHNO3 (1.50 g/mL, 63.012 g/mole) dissolved into 2.61 mLanhydrous H2SO4. This mixture is then heated to 35°C fortwenty minutes and then left to cool to room temperature.As the mixture is being heated, a brown gas appears in thereaction flask. To this mixture is added 5.0 mL of water

previously cooled to 0°C. This results in a white crystallineprecipitate forming instantly in the solution upon addition ofthe water.

The white solid is filtered from solution using a Hirschfunnel and washed with three aliquots of 1.0 mL of coldwater. The product is then isolated, dried and weighed. Themass collected for the isolated crude product (assumed to be2,5-dimethoxynitrobenzene (183 g/mole)) is 91.5 mg. Themelting point range is measured to be from 72.5°C to 74°C.

A small sample of the product is then pulverized and adrop of mineral oil is added to the pulverized sample. Themull (mixture of the mineral oil drop and pulverized solid) isstirred and added to the face of a salt plate for analysis usinginfrared spectroscopy. Key peaks in the HI show that thecompound contains a nitro group, a C-O single bond, and abenzene ring. Based on the infrared data and the meltingpoint of the crude product, the student concludes that thecrude product is 2,5-dimethoxynitrobenzene. The literaturevalue for the melting point of 2,5-dimethoxynitrobenzene is74 - 758C. The error in the observed melting point isattributed to thermometer error and not an error in thereaction.

8 7. What is the percent yield for this reaction?

A. 25%

B. 50%

C. 75%

D. 100%

88. What can be deduced from themelting point range andthe IR data for the crude product?

A. The product is still wet with solvent.

B. The productis free of any significant impurities.C. The product is impure.

D. The starting material was isolated rather than theproduct.

8 9. Filtering is used to separate a:

A. solid from other solids.

B. solid from liquids.C. solid from gases.D. liquid from other liquids.


9 0. What is the identity of the brown gas formed in the firstpart of the reaction?

A. N2

B. CO

C. N02

D. S02

91. Which of the following is a possible cause for thethermometer error?

I. Air has leaked into the thermometer resulting inreduced vacuum within the thermometer.

H. A divot exists on the inside wall of the

thermometer between the 45 and 55 °C mark.

HI. A divot exists on the inside wall of the

thermometer between the 85 and 95 °C mark.

A. I only

B. I and Honly

C. I and HI only

D. I, H, and IH all explain the thermometer error.

92. To further purify the product mixture, what type oflaboratory technique should be carried out?

A. The crude product mixture could be dissolved intohot solvent and then recrystallized from the solventas it cools.

B. The crude product mixture could be filtered a secondtime with the same solvent.

C. The crude product mixture could be heated to liquidform and then distilled via fractional distillation.

D. The crude product mixture could be treated withnitric acid and sulfuric acid again to react anyunreacted starting material.

93. Why is the addition of water carried out at a lowtemperature?

A. To increase the solubility of the product.B. To decrease the solubility of the product.C. To increase the acidity of the solution.D. To terminate the reaction by protonating the nitro

group.


Questions 94 through 100 are NOT based on adescriptive passage.

94. Ifa pure sample ofa compound with S-stereochemistryhas a specific rotation of -55°, then what is thecomposition of a mixture of that compound and itsenantiomer that exhibitsa specific rotationof+22"?

A. 50% S-enantiomer with 50% R-enantiomer

B. 33% S-enantiomer with 67% R-enantiomer

C. 30% S-enantiomer with 70% R-enantiomer

D. 20% S-enantiomer with 80% R-enantiomer

95. Thefollowing TLCplateandcolumn correspond to thesame components within the mixture and the samemobile phase (solvent).

Column

:—m BandC

\

BandB

Band A

TLC Plate

Which correctly correlates the spot to the band?

A.Band A = •; Band B =0 ; Band C = O

B.Band A = O;Band B = 0; Band C = •

C.Band A= Q; Band B = •; Band C = O

D.Band A = •; Band B =O;Band C = Q

96. Fractional distillation differs from simple distillation inall of the following ways EXCEPT:

A. The distilling column in fractional distillation hasmore surface area than the distilling column insimple distillation.

B. Fractional distillation is chosen over simpledistillation when the boiling points are close forthe components in the mixture.

C. Fractionaldistillation generates a higher purity thansimple distillation.

D. Fractional distillation generates a greater yield thansimple distillation.


97. Which of the following compounds, during an acid-baseextraction, would be extracted into0.10M NaOH(aq)?

A. 4-Methyl benzoicacidB. 3-Ethyl anisole

C. Ethyl benzoate

D. N-Methyl aniline

98. In which of the following solvents or solutions is 4-hydroxybenzoic acid MOST soluble?

A. O.lOMHCl(aq)

B. O.lOMKOH(aq)

C. Water

D. Diethyl ether

99. Which of the following lab techniques will NOT workfor the task listed?

A. Using distillation to separate two liquids.B. Using chromatography to separate two solutes.C. Using extraction to separate two gases.

D. Using crystallization to separate two solids.

100. Butanone shows all of the following EXCEPT:

A. high solubility in acetone.

B. a negative Jones test for oxidation.

C. a mass spectroscopy peak at 57 amu.D. a specific rotation of ocd = 41.2°.

1. C 2. A 3. A 4. A 5. C 6. A

7. D 8. A 9. D 10. B 11. D 12. D

13. A 14. B 15. A 16. B 17. B 18. C

19. C 20. D 21. D 22. A 23. A 24. A

25. C 26. D 27. B 28. A 29. B 30. D

31. D 32. B 33. A 34. B 35. C 36. D

37. B 38. D 39. C 40. A 41. D 42. D

43. C 44. C 45. D 46. C 47. D 48. D

49. A 50. C 51. D 52. C 53. D 54. B

55. A 56. B 57. A 58. c 59. B 60. A

61. C 62. C 63. B 64. A 65. B 66. D

67. A 68. D 69. C 70. A 71. D 72. C

73. D 74. C 75. B 76. C 77. B 78. C

79. C 80. A 81. A 82. A 83. C 84. C

85. B 86. A 87. B 88. B 89. B 90. C

91. B 92. A 93. B 94. C 95. A 96. D

97. A 98. B 99. C 100. D

STOP! START GRADING.

Laboratory Techniques Passage Answers

Passage I (Questions 1-7) Distillation and Separation

1. Choice C is correct. No matter what the boiling point of the liquid, no distillation should ever be done in aclosed system, otherwise pressure will build up (as T increases, sodoes P if there is constant V) and the systemwill eventually explode. This eliminates choice D. The venting of the system must occur after the vapor hasbeen condensed and collected, otherwise the vapor will escape and nothing will be collected. This eliminateschoices A and B and makes choice C the best answer.

2. Choice A is correct. A temperature of 49.2°C indicates that the boiling point of the compound beingcollected isnear49.2°C The compound has tobeone ofthe six given in the passage, so you must scan the first paragraph forthe compound with a boiling point closest to 49.2 °C. The best choice for the identity of the compound is E-1,2-dichloroethene, which has a boiling point of 48°C. This makes choice A correct. Tliis question might seem sostraight forward that you get nervous and think you're missing something. Keep in mind that the MCAT hasarange of question difficulties and that straight forward questions appear.

3. Choice A is correct. When a boiling chip is added to the distillation flask, it sinks to the bottom, which is thehottest point in the solution (heat is added at the bottom of the solution.) Because vapor must escape from asurface and there is minimal surface area at the point where the solution is hottest, the solution can superheat(reach a temperature above its boiling point). Because heat rises (hotter solutions are less dense and thereforemore buoyant than colder solutions), the superheated solution will migrate to the top of the solution, at whichpoint it has surface area from which to evaporate. Because the solution is beyond its boiling point, it rapidlyevaporates (bumps), causing a splash. To avoid this problem, the system needs more surface area at the bottomof the boiling flask. A boiling chip is added to provide a rough surface on which vapor can collect and build upnear the bottom of the flask. This is the kinetic theory explanation of bumping. The best explanation is choiceA. Choices C and D should have been eliminated, because the boiling chip does not interact with the moleculesin solution other than providing a surface area for them to collect on. Intermolecular forces are neither enhancednor diminished by a boiling chip. The boiling chip is stationary, so it does not help in the homogenization of thesolution. This eliminates choice B.

4. Choice A is correct. The majority of the vapor collects on cold glass, as opposed to warm glass, which eliminateschoices C and D. The condenser is cooled so as to condense the vapor. The condensed vapor then collects andtrickles into the collection flask in droplets. The actual condensation occurs in the condenser, however, whichmakes choice A the best answer.

5. Choice C is correct. Filtration can not be used when there is a solute, because the compound is fully dissolved intosolution and cannot be collected in the filter. This eliminates choices A and B. A liquid may be separated from adissolve solute by distilling (evaporating) the liquid away. This makes choice C or choice D the best answer.Because the boiling points are very different (the solute is normally a solid at room temperature, so it has tomelt before it can boil), there is no need to employ fractional distillation. The sample is best distilled usingsimple distillation, because the boiling points are drastically different. Choice C is your choice.

6. Choice A is correct. Simple distillation works best when the two compounds have the greatest difference inboiling points. Choice A is composed of two compounds which have boiling points of 34°C and 102°C, a differenceof 68°C. Choice B is composed of two compounds which have boiling points of 34°C and 48°C, a difference of14°C Choice C is composed of two compounds which have boiling points of 90°Cand 102°C,a difference of 12°C.Choice D is comprised of two compounds which have boiling points of 128°C and 102°C, a difference of 26°C. Thelargest difference in boiling points between the components is observed with the compounds in choice A. Thisquestion is really a nomenclature question. Youmust translate from formula in the question into the IUPACnameand boiling point from the text of paragraph 1.

7. Choice D is correct. Fractional distillation is employed to separate two liquids with relatively close boilingpoints. This means that it is employed to separate a liquid from a liquid. This eliminates choices A, B, and Cand makes choice D the best answer.

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Passage II (Questions 8 -14) Fractional versus Simple Distillation

8. Choice A is correct. Fractional distillation depends on the condensation and re-evaporation of the more volatilecomponent in the mixture. Each cycle of condensation/re-evaporation further purifies the vapor so that the firstcondensate off of the column is rich (relatively pure) in the more volatile component. Over time, the innersurface of the distillation apparatus heats up so that the amount of condensation on the inner walls is reduced,and the components condense and re-evaporate less. This does not segregate the less volatile component as wellas when the walls are cooler. The net result is that over time, the distillate becomes less pure in terms of themore volatile component present in the vapor. The best answer is choice A. Azeotropes should have an equalbearing on the mixture whether the distillation is simple or fractional. Tiie surface area does not affect theazeotropic mixture.

9. Choice D is correct. A long distilling column is used in fractional distillation, which is chosen when the boilingpoints of the components in the mixture are close in value. Close is defined as boiling points within 30°C of oneanother. In this question, you are asked for the best choice, which for fractional distillation is the pair ofcomponents with the closest boiling points. In choice A, diethyl ether (b.p. = 34.6°C) and pentanol (b.p. =138.0°C) have a difference in boiling points of over 100°C. Simple distillation works with choice A. In choice B,tetrahydrofuran (b.p. = 65.4°C) and anisole (b.p. = 158.3°C) have a difference in boiling points of nearly 100°C.Simple distillation will work fine for choice B. In choice C, octane (b.p. = 125.7°C) and pentane (b.p. = 36.1°C)have a difference in boiling points of nearly 90°C. Simple distillation will work just fine in choice C as well. Inchoice D, isopropanol (b.p. = 82.3°C) and toluene (b.p. = 110.6°C) have a difference in boiling points of under30°C. Simple distillation will not work well, so fractional distillation should be used with choice D.

10. Choice B is correct. The major advantage of fractional distillation over simple distillation is that thefractional distillation procedure allows components to be distilled and collected in a purer manner. Fractionaldistillation can always be used no matter how close the boiling points of two substances may be. Choice A isvalid. Fractional distillation works by providing surface area from which the vapor can continuously condenseand re-evaporate. Choice C is thus valid. The drawback to fractional distillation is the time required to carryout the distillation and the loss of the vapor that condenses throughout the apparatus. Fractional distillationtakes a larger amount of time than simple distillation, thus choice B is an incorrect statement. The best answeris choice B.

11. Choice D is correct. Just as a water bath is used for better heat transfer (in both cooling and heating), a sandbath is chosen to transfer heat into the flask most efficiently. The sand bath covers the outside surface of theglass, and transfers heat into the system through the glass. The sand serves to conduct heat (not insulate), sochoice A is eliminated. Sand is not reactive with glass, but neither is air. Choice B may or may not be a truestatement, but either way, it is not the best answer. Sand is not particularly hygroscopic, so it will not affect thevapor pressure of water, eliminating choice C. The best answer involves heat transfer, so choose D for optimalperformance.

12. Choice D is correct. The role of the copper mesh in the distilling column is to increase the surface area ontowhich the vapor may condense so as to collect more condensate in the column. The condensate can evaporate onceagain from the copper mesh, further distilling the condensed liquid. The copper mesh does not filter anything,it is inert, and it does not distribute heat any better than the vapor does. Copper mesh just allows for fractionaldistillation to take place. The best answer is choice D.

13. Choice A is correct. Simple distillation is chosen when the boiling points of the components in the mixture arefar apart. Far apart is defined as boiling points that are more than 30°C apart from one another. For thisquestion, you are asked for the best choice. The best choice for simple distillation will be the two componentswith the greatest boiling point difference. In choice A, diethyl ether (b.p. = 34.6°C) and heptane (b.p. = 98.4°C)have a difference in boiling points of over 60°C. Simple distillation will work fine with choice A, but you mustlook at the other choices to see if there is a difference greater than 63.8°C. In choice B, tetrahydrofuran (b.p. =65.4°C) and hexane (b.p. = 68.9°C) have a difference in boiling points of only 3.5°C. Simple distillation will notwork with choice B, thus fractional distillation must be used. Choice B is eliminated. In choice C, octane (b.p. =125.7°C) and pentanol (b.p. = 138.0°C) have a difference in boiling points of about 12°C. Simple distillation willnot work with choice C. Choice C is thus eliminated. In choice D, nonane (b.p. = 150.8°C) and anisole (b.p. =158.3°C) have a difference in boiling points of only 7.5°C. Simple distillation will not work in this case, thuschoice D is eliminated. Choice A is the best answer.

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14. Choice B is correct. The same boiling point for two stereoisomers is observed when the two stereoisomers areenantiomers, as they have the same functional groups and the same steric hindrance. They show the sameintermolecular forces, which results in the same boiling point. The two molecules in both choice A and choice Dare diastereomeric pairs, therefore both choice A and choice D can be eliminated. Diastereomers do not showthe same intermolecular forces, so they likely show different boiling points. Glucose and mannose are C-2epimers of one another, which is another way of saying that they are diastereomers. The only pair ofenantiomers in the choices are 3R-bromocyclopentene and 3S-bromocyclopentene, making choice B the bestanswer. When a compound has only one chiral center, it cannot have a diastereomer, because diastereomersrequire at least two chiralcenters. Choices A, C, and D have multiple chiral centers.

Passage III (Questions 15 - 21) Steam Distillation of Natural Products

15. Choice A is correct. Steam distillation involves boiling a mixture ofcomponents in the presence ofwater, so theboiling point is more important than the melting point. This eliminates choices C and D. Water boils at 100°C,so adding steam to help distill a mixture of organic compounds is best done when the boiling point of eachcomponent is greater than 100°C. This makes choice A the best answer.

16. Choice B is correct. Both compounds contain ten carbons and have three units of unsaturation, so they areisomers. All of the answer choices are isomers of some kind, so that didn't help much. They have the sameconnectivity (and therefore the same root for their IUPAC names), so they are not structural isomers. Thiseliminates choice D. Neither citral nor neral has a chiral center, so they are not optical isomers. Thiseliminates choice C. They cannot readily contort or rotate about a bond to inter-convert between one another, sochoice Ais eliminated. Conformational isomers must be able to rotate or flip about sigma bonds to interchangebetween one another, like the gauche and anti conformations of acompound like butane. Citral and neral vary atthe positioning around the double bond between carbons 2and 3, so they are cis-trans isomers. Cis-trans isomersare more correctly defined as geometrical isomers, so the best answer is choice B.

Choice Bis correct. Standard distillation of lemon grass oil at 229° exposes the compounds in lemon grass oil toextreme heat, which can decompose, destroy, or denature acompound. Decomposition can certainly occur at hightemperatures, as is seen with most biological molecules, so choice Ais eliminated. High heat in the presence ofoxygen gas results in the oxidation of most organic compounds, so choice C is valid, and therefore eliminated.Citral can react with itself at the 7t-bond carbons, so at high temperature, polymerization is a possibility. Thiseliminates choice D. Without asource of hydrogen, hydrogenation is not a viable possibility, so hydrogenationis not a likely result when heating a citral-neral mixture. This makes choice B the best answer.

Choice C is correct. Ether is chosen to extract the organic components from the aqueous layer, because it isimmiscible in water and it has a high affinity for organic molecules. The immiscibility of water in ethereliminates choices A and B. Citral dissolves into the ether layer, which favors choice C over choice D. Thewording is not perfect, so we must also consider that ethers float on water, given that their densities are lessthan 1.00g/mL. This supports choice C as the best answer.

Choice Cis correct. The term anhydrous provides a big clue here, because it refers to something without water,which implies that water must have been removed from it, otherwise the fact that it is anhydrous is irrelevant.The best answer is that an anhydrous salt is added to the ether layer to absorb, and thereby remove water.Choice C is the best answer. Isomerization of neral into citral is done with either light orheat, but an insolublesalt has no bearing on the isomerization. Choice A is eliminated. Sulfate, SO42-, is rich in oxygen andmagnesium dication, Mg2+, is poor in electrons, so neither could be a reducing agent. Neither can reduce analdehyde into an alcohol, so choice Bis eliminated. Selectively binding a compound such as neral might soundlike something a chemist would want to do, but there is no reason to believe that magnesium sulfate could dothat. Choice D is eliminated.

Choice D is correct. Neral only varies from citral in the positioning ofone double bond, so in solution, it exhibitssimilar intermolecular forces as citral. While the melting point may be impacted by the cis versus trans positionof the double bond (given that packing depends on steric hindrance), the boiling point is less impacted. Theboiling point of citral is 229°C, so the boiling point ofneral should be close to that value. Choice D is the bestanswer. Given that steam distillation is employed, the boiling point of neral is above 100°C, so at the veryleast, choices A and B should have been eliminated.

17.

18.

19.

20.

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21. Choice D is correct. Vitamin A is made from citral, so it is likely a hydrophobic aldehyde too. It is larger thancitral, so it will have a higher melting point, eliminating choice A. It is hydrophobic, so it has a low watersolubility, eliminating choice B. Citral is a liquid with a high boiling point, so it has a low vapor pressure.Vitamin A is likely the same, so choice C is eliminated. Being a lipid, Vitamin A has a high affinity forlipids, which makes choice D the best answer. Vitamin A is a lipid-soluble drug.

Passage IV (Questions 22 - 28) Caffeine Extraction Experiment

22. Choice A is correct. The percentage of caffeine in the tea leaf is the mass of the extracted caffeine divided bythe mass of the tea leaves multiplied by 100%. The amount extracted is 0.028 grams (28 milligrams) and theoriginal sample of tea leaves weighed 2.0 grams, so the value is 0.028/2 x 100% = 0.014 x 100% = 1.4 %. Choice Ais the best answer.

23. Choice A is correct. Because caffeine is extracted with both hot water and methylene chloride, it must besoluble in both hot water and methylene chloride. This eliminates choices C and D. In the second extraction,methylene chloride is used to remove the caffeine from the water, so caffeine must be more soluble in methylenechloride than water. This eliminates choice B and makes choice A the best answer.

24. Choice A is correct. A stream of nitrogen, when flowing across the surface of a liquid, will force the gases presentabove the surface away, thus reducing the partial pressure of the vapor above solution. If the air above thesolution is saturated with vapor before the nitrogen gas stream, it will not be saturated afterwards, allowingnew vapor to form. The solution continues to vaporize (evaporate) to equilibrate with the vapor present abovesolution, but because the vapor is continually being removed, it can never reach equilibrium. Eventually, theliquid evaporates away completely. This is best explained as choice A.

25. Choice C is correct. A quick and easy test for the identity of a solid is the mixed melting point test. By adding asmall amount of the compound you believe the sample to be, it is possible to observe the melting point for themixture of the known and unknown compounds. If they are the same compound, then the melting point will besharp and exact. If the two compounds are not the same, then the melting point will be broad. This means thatto prove that an unknown is caffeine, the unknown is best mixed with caffeine for a melting point measurement.This makes choice C the best answer.

26. Choice D is correct. Methylene chloride is common as a solvent, so it must be a stable compound. Even if it wereunstable, the quantity should not be the factor that triggers its instability. Choice A is eliminated. Choice Baims for the test taker who is asleep at the wheel, so to speak. The top and bottom layers have to do withrelative densities, not relative masses. Less dense species float in more dense mediums, no matter how massivethey are. Choice B is eliminated. The goal of extraction is to isolate the product, so hopefully any method orprocedure would aim to maximize what is isolated. This makes choice D a better answer than choice C,although that logic does not explain why it is true. This is where you have to keep in mind that it's astandardized exam, and not an exam of what you know. Using test logic to choose D in this case is perfect,because you save time. A mathematical example is shown here to support the answer, but it is unnecessary toconsider during the actual exam.

Consider a solute that splits evenly between two solvents. In one case, 15 mL of Solvent A is used with 10 mLsolvent B. This would result in 60% of the solute being extracted into Solvent A. In a second approach, the 15 mLof Solvent A is added in three 5-mL portions. In portion 1, 33.3% goes into Solvent A and 66.6% remains inSolvent B. In portion 2, based on the percentage of the original material, the partitioning ratio of Solvent A toSolvent B is 22.2% to 44.4%. In portion 3, based on the percentage of the original material, the partitioning ratioof Solvent A to Solvent B is 14.8% to 29.6%. Adding up the percentages extracted into Solvent A from the threeseparate extractions, we get: 33.3% + 22.2% + 14.8% = 70.3%.

Three separate smaller extractions yields 70.3% while one larger extraction yields 60%. Three smallextractions do in fact generate a better result than one big extraction, so choice D is the best answer.

27. Choice B is correct. Initial treatment of the tea leaves with hot water dissolved the water-soluble products,including caffeine. Methylene chloride is added to this solution to remove organic substances that are onlypartially soluble in water at lower temperatures. Because caffeine is water soluble at elevated temperatures, itis still partially soluble at a lower temperature, so not all of the caffeine can be extracted into the methylenechloride layer. The best answer is choice B.


28. Choice A is correct. Sublimation is the physical process of converting a solid directly into a gas, so choices C andD are not valid answers. As a laboratory technique, it is used to convert one solid into a gas, while leavingbehind the other solids. The gas can migrate away from the other solids and then be collected on a cold surfaceas a pure sample of the solid. Purifying a solid from a liquid does not work, because the liquid can vaporize,resulting in more than one compound in the gas phase. This eliminates choice B. The best answer choice is A.

Passage V (Questions 29 - 35) Partition Coefficient Experiment

29. Choice Bis correct. Exactly 0.75 grams of the original 1.0 gram sample was isolated from the organic layer. It isassumed that the other 0.25 grams of sample were dissolved into the aqueous layer. The partition coefficient isthe amount that dissolves into the organic layer divided by the amount that dissolves into the aqueous layer,which in this case is 0.75/0.25 = 3.0. This makes choice B the best answer.

30. Choice D is correct. As stated in the passage, the solubility of4-methylaniline in water at 25°C is 3.8 grams/100mL and in diethyl ether at 25°C is 22.8 grams/100 mL. The partition coefficient is the solubility of the sample indiethyl ether divided by the solubility of the sample in water. The partition coefficient is found as follows:

31.

32.

33.

22.8100

3.8

Choice D is the best answer.100

_ 22.8 -228-114 - 6xl93.8 38 19 19

= 6

Choice D is correct. As stated in the passage, the solubility of 4-methylaniline at 25°C in water is 3.8 grams/100mL and the solubility of 4-methylaniline at 25°C in diethyl ether is 22.8 grams/100 mL. This means that themaximum amount of 4-methylaniline that can dissolve into 10 mL diethyl ether is 2.28 grams. The maximumamount of 4-methylaniline that can dissolve into 10 mL water is 0.38 grams. The maximum total mass betweenthe two solutions is 2.66 grams. Adding 3.00 grams exceeds this amount, so not all of the 3.00 grams of 4-methylaniline can dissolve into the two solutions combined. The undissolved portion will not partition betweenthe two solvents, but instead forms a precipitate at the bottom of the flask. This precipitate makes the lowerlayer appear to have more 4-methylaniline than it actually has dissolved into solution. This makes choice Dthe best choice. Had the compound been fully soluble, the partition coefficient should have been the same,assuming any errors remains the same in magnitude.

Choice Bis correct. Because 4-methylaniline is more soluble in the diethyl ether than water, the pipette couldbe rinsed with diethyl ether to wash off any residue of 4-methylaniline. Heating and cooling the pipette ispointless once the compound has adhered to the walls of the pipette. Water is a poor choice to remove theadhered solid, because 4-methylaniline is relatively insoluble in water. The residue on the glass was lost fromthe diethyl ether layer, so if it is recovered, it is best recovered into the diethyl ether layer. Pick choice Bforbest results.

Choice Ais correct. If the ether were not anhydrous, itwould contain some water impurity. This implies that inthe 10 mL of ether, there is actually a small portion of water and therefore less than 10 mL of ether. Thiseliminates choices C and D, because they indicate an excess (a value greater than 10 mL) of ether. With lessthan 10 mL of ether, less 4-methylaniline than expected would dissolve into the ether layer. This means that asmaller amount of 4-methylaniline would be isolated from the ether layer. This would lower the numerator,and thus lower the value for thecalculated partition coefficient. This makes choice A the best answer.

34. Choice Biscorrect. Diethyl ether is used because itis immiscible in water and the organic compounds are solublein it. Any solvent that is used in place of diethyl ether must have the same properties. Tetrahydrofuran, THF,is a cyclic ether, so it has the same properties as diethyl ether. This eliminates choice A. Cyclohexane andmethylene chloride are both organic solvents that are immiscible in water, so choices C and D are eliminated.Ethanol is miscible inwater, so it cannot be used in lieu ofdiethyl ether. This makes choice B the best answer.

35. Choice Cis correct. For two layers to form, the organic solvent must be immiscible in water, eliminating choiceA. The density is not important, as long as its not the same density as water. As long as it has a densitydifferent than water, higher or lower, the system will have a layer on top and on bottom. This eliminateschoice B. The solute should be soluble in the organic layer, so that itcan be removed from the aqueous layer.Choice Cis the best answer. If its boiling point is less than room temperature, then it will exist as a gas at roomtemperature. To beuseful asa solvent, itneeds tobea liquid at room temperature. This eliminates choice D.

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Passage VI (Questions 36 - 42) Acid/Base Extraction Experiment

36. Choice D is correct. Ether is chosen as a solvent, because it is immiscible in water and it is a liquid at roomtemperature. The organic solvent used in lieu of ether must also be immiscible in water and a liquid at roomtemperature. Choices Band C are eliminated, because they are not liquids. Choice A (ethanol) dissolves intowater, so choice A is eliminated. Only choice D remains. Chloroform is an immiscible liquid, so choose D.

37. Choice B is correct. To extract the contents of Tube 4 from the organic layer, a basic aqueous solution was used, soTube 4 must be at a high pH. This means that the solute is likely in its anionic form. To remove the solute(dissolved anion) from solution, the solution must be neutralized. To neutralize a basic solution, an acid isadded. This eliminates choices A and D. The acid should be strong, so it can fully react with the base insolution. The best answer is choice B, hydrochloric acid.

38. Choice D is correct. Tubes 2 and 5 contain the component that does not react with either the strong base or theweak base to become more water soluble. Dinitrobenzene has no acidic protons, given that all of its hydrogensare bonded to carbon, so it will remain neutral and organic soluble during the entire separation. It is not possibleto convert dinitrobenzene into a charged species with just acid and base reactions. The best answer isdinitrobenzene for both Tube 2 and Tube 5, so choice D is the best answer.

39. Choice C is correct. The second proton is by definition less acidic than the first proton, so the second pKa isgreater than the first pKa. Choices A and Bare eliminated. The value of pKb can be found by subtracting thepKa for the conjugate acid from 14. Because the pKa for benzoic acid is lower than the pKa for the first proton ofresorcinol, the pKb for the conjugate base of benzoic acid (benzoate) must be greater than the pKb for theconjugate base of the first acidic proton of resorcinol (pKb2)- The first proton lost by a diprotic acid is the secondone gained by the fully deprotonated form of the conjugate base. This makes choiceC the best answer.

40. Choice A is correct. Because the melting points for the compounds isolated by Student II showed a narrowerrange than the compounds isolated by StudentI, the productsfrom thescheme ofStudent IImust have beenpurerthan the compoundsfrom the scheme ofStudent I. Thus, the scheme for Student II must have worked better thanthe scheme for Student I. The scheme for Student II isolated the stronger acid first (Tube 4), the weaker acidsecond (Tube 6), and the neutral species in the organic layer (Tube 5). It can be concluded that the scheme forStudent II worked while the scheme for Student I did not. The best answer is choice A.

41. Choice D is correct. 3-Methylaniline is an amine, making it a basic compound. Basic compounds dissolve intowater under acidic conditions. However, in the reaction scheme for Student II, only base is added to the waterand not acid. As such, the amine is not protonated, and remains in the organic layer for both extraction steps.The result is that it starts and finishes in the organic layer, which finishes in Tube 5. A small amount maydissolve into water, so the best answer is choice D, predominantly in Tube 5.

42. Choice D is correct. Acompound can beseparated using extraction techniques as long as it is not reactive in any ofthe mediums. 4-Ethylcyclohexanone is a large ketone, which is not reactive with the ether layer, acidic water,or basic water. 4-Ethylcyclohexanone can be isolated using acid-base extraction techniques, so choice A iseliminated. 3-Bromophenol is a mild organic acid, which is not reactive with the ether layer or acidic water,and undergoes reversibledeprotonation in basicwater. 3-Bromophenol can be isolated using acid-base extractiontechniques, so choice B is eliminated. 3-Nitrobenzoic acid is a mild organic acid, which is not reactive with theether layer or acidic water, and undergoes reversible deprotonation in basic water. 3-Nitrobenzoic acid can beisolated using acid-base extraction techniques, so choice C is eliminated. Ethyl benzoate is an ester. Esters canundergo hydrolysis under acidic aqueous conditions or saponification (base catalyzed hydrolysis) under basicaqueous conditions. Becauseesters react under the conditions used in acid-base extraction, ethyl benzoate cannotbe isolated using acid-base extraction techniques. Choice D is the best answer.

Passage VII (Questions 43 - 49) Extraction and Thin Layer Chromatography Experiment

43. Choice C is correct. An Rf value of 0.92 implies that the compound is highly soluble in the solvent and that thecompound has a very low affinity for the surface of the TLC plate. A change in solvent from nonpolar to polarwould show a reduction in solubility, and thus a reduction in the Rfvalue. The Rfcannot be negative, so choice Dis eliminated. The Rf value cannot be greater than 1, so choice A is eliminated. The only value greater than zerobut less than 0.92 is choice C, making it the best answer.


44' Sfvinl ?SICT? Be,CaUSe 3SUglr C°ntainS 3largG nUmber °f hydr0Xy{ SrouPs' *wiI1 be most solub1^ in asolvent with a hydroxyl group, such as an alcohol. This is based on the idea that "like dissolves like." Analkane, ether, and ketone do nothave a hydroxyl group, so the best answer is choice C.

45' r^ttZ^f 'T'1; ^ ^^ ?rind?le bGhind thin kyer chromatography is having the spots migrate up theplate at different rates by exploiting their differences in attraction to the solvent and affinity to the TLC plateSpotting should be done in away that allows the spots to migrate. If the spots are placed below the top of thesolvent (submerged into the solvent), they can dissolve into the solvent. This makes choice Dthe best answerChoice Ashould be eliminated, because the rate of migration does not depend on the placing of the solvent andspots. The spots are not going to become saturated, so they will not rise up the plate too rapidly. Choice Biseliminated. The spots will dissolve in all directions, so radial migration is an incorrect answer. Choice Ciseliminated. Choice D is better than the other pooranswers.

46. Choice Cis correct. For column chromatography, the best separation results from the solvent that shows thegreatest difference in Rf values in terms of ratio. In hexane, the Rf values are different by a factor of 2 while inethanol, the Rf values are different by afactor of 3. Do not subtract the Rf values to determine the separation,the Rf values must be divided. Abetter separation is observed with ethanol than hexane, so choices Aand Bshould be eliminated. Because the Rf values are small, the column should be short, so that less solvent isrequired and the components do not remain on the column for any unnecessary time after separation has beenestab ished. Amixed-solvent of hexane and ethanol would not work well, because the solutes will travel fasterand the ratio of the Rf values will decrease only slightly. Choice C is the best answer.

Choice Dis correct. Carbon tetrachloride is anonpolar solvent, so it shows similar solubility properties (andthus similar TLC results) as other nonpolar solvents. Acetone and propanal both contain the carbonyl functionalgroup, which makes them polar. This eliminates both choices Aand B. Ethanol is not only polar, but it also iscapable of forming hydrogen bonds with solute particles. Choice Ccan thus be eliminated. Because pentane is ahydrocarbon, it is nonpolar. Pentane will behave most like carbon tetrachloride, so the best answer ischoice D.

Choice Dis correct The Rf value is defined as the distance that aspot travels up aTLC plate (dspot) divided bythe distance that the solvent travels up aTLC plate (dsolvent)- Because the spot migrates due to its affinity forthe solvent, the spot can never migrate a greater distance than the solvent migrates. This means that thedsolvent will always be agreater value than the dspot. This consequently means that the Rf value can never begreater than or equal to 1.00. Choice D, 1.05, is not possible. An Rf value of 0.0 simply means that the spot doesnot dissolve mto that particular solvent, and therefore it does not move. An Rf value of 0.0 is possible.

49. Choice Ais correct. Trimyristin is a fatty acid triglyceride, meaning it is a triester. Long chain esters arehydrophobic, because they cannot form hydrogen bonds and they are relatively nonpolar. This eliminateschoices Band D. THF is an ether, making it hydrophobic. Because of the oxygen in the ring, THF is a mildlypolar solvent, although it is aprotic. Tiie best answer is choice A.

47.

48.

Passage VIII (Questions 50 - 56) Thin Layer and Column Chromatography

50. Choice Cis correct. Alumina is a polar adsorbent, so it hinders polar solutes more than nonpolar solutes.Nonpolar solutes come off the column first, resulting in a fast elution time. Benzoic acid is hydrophilic andpolar, eliminating choice A. Glycerol is also polar, so choice Bis eliminated. Glycine is polar and hydrophilic,so despite the presence of ethanol as the eluting solvent, it is still hindered by the alumina. Choice D iseliminated. Anonpolar compound like o-xylene exhibits little affinity for the alumina, so it travels quicklydown the column. Using anonpolar solvent like ligroin speeds itup even more, so choice Cis the best answer.

Choice Dis correct. The compound with the larger Rf value climbs aTLC plate faster, so it travels through acolumn faster as well. This means that it should have a smaller elution time. This eliminates choices A and B.The Rf values differ by afactor of two, so the relative migration distance in a given time differs by a factor oftwo. This means that their rates also differ by a factor of two, meaning that their times for traveling a givendistance shall also differby a factor of two. This makes choice D the best answer.

51.

52. Choice Cis correct. Salicylic acid is highly polar, as stated in the passage. The greatest Rf value corresponds tothe greatest migration, which is observed in the most polar solvent. Hexane and ligroin are nonpolar, so choicesAandBare eliminated. Methanol is more polar than pentanol, sochoice Cis the best answer.

Copyright © by TheBerkeley Review® 294 LAB & SPECTROSCOPY EXPLANATIONS

53. Choice D is correct. The Brockmann Activity rating for alumina has to do with the amount of water bound to thecolumn. According to the passage, a rating of III refers to a column that is 6% water by mass. A rating of IV hasmore water than a column with a rating of III, so choice A is eliminated. Because there is so much water bound toalumina, there is less binding of solute to the column. Having more water bound with a rating of IV than a ratingof III, a rating of IV represents a lower affinity than a rating of III. This eliminates choice B. Alumina is madeof aluminum oxide, so there is no silicon present. This eliminates choice C. A Brockmann Activity rating of III iscaused by 6% water by mass, so a rating greater than III results from more than six percent water by mass. Giventhat a rating of I is perfectly anhydrous, a rating of IV requires more than 6 grams of water per 100 grams ofcolumn, so choice D is the best answer.

54. Choice B is correct. Methanol has a high affinity for alumina, so the binding of methanol to the alumina isexothermic. This eliminates choice A. When methanol is added to the column, heat pockets can form, whichresults in the formation of vapor pockets in the column. These vapor pockets can crack the gel, resulting in abreak in uniformity for the pathway of solute through the column. Choices B is the best answer. Methanol ismiscible in hexane, but even without knowing such a fact, there is no reason to assume that rate at whichmethanol migrates through the column has any dependence on the miscibility of the two solvents. Thiseliminates choices C and D.

55. Choice A is correct. A higher Brockmann Activity rating represents a greater amount of water on the alumina,which corresponds to a lower affinity for polar molecules. The lower affinity is due to the fact that water istaking up space that the polar molecule would bind if it was open. This makes choice A the best answer. Waterdoes not cause the contraction of alumina. If anything, it may dissolve alumina with such a high affinity forwater. Choice B is eliminated. Alumina is polar, so although the addition of water may affect the polarity,choice C is incorrect, because alumina is polar to begin with, not nonpolar. Water does not oxidize alumina,given that aluminum is fully oxidized in AI2O3. This eliminates choice D.

56. Choice B is correct. Alumina is polar, so the solute for which it has the highest affinity must also be polar.Choice A, 1,3-cyclohexadiene, is nonpolar and hydrophobic, so it is eliminated. Choice C, ethyl butanoate, isan ester, so while it has a dipole, it is classified as nonpolar species and hydrophobic. Choice C is eliminated.Choice B, cyclohexanol, is polar and able to form hydrogen bonds, so it has a high affinity for alumina. ChoiceD, 3-methylpentanal, is an aldehyde, so it polar and should also exhibit an affinity for alumina. A proticspecies has a greater affinity for alumina than an aprotic species, so choice B is a better answer than choice D.

Passage IX (Questions 57 - 63) Column Chromatography

57. Choice A is correct. Because like dissolves like, fluorenone should be most soluble in another ketone. Assolubility increases in the solvent, the affinity is said to be higher. The only ketone in the answer choices isacetone, so choice A is the best answer.

58. Choice C is correct. Fluorenone is more polar than fluorene, so independent of solvent, fluorene should elute fromthe column first. The fact that a nonpolar, hydrophobic solvent is being used makes the difference in migrationrate even greater. Fluorene is a nonpolar species, so it has a low affinity for silica gel. This eliminates choice A.Fluorene is in fact lighter than fluorenone, but in a column, much like free fall, the mass of the species does notfactor into its migration rate except for the steric hindrance associated with larger, more massive molecules.Choice B is eliminated. Fluorene does in fact have a higher affinity for cyclohexane then fluorenone does, sochoice C is a valid statement. Fluorene is smaller than fluorenone, but there are no pores in silica gel thatseparate by molecular size (molecular sieves do that), so choice D is eliminated. Choice C is the best answer.

59. Choice B is correct. The best solvent is the one that allows the two species to separate by the most. This isdetermined by comparing the Rf values of fluorene and fluorenone in each of the solvents. The greatest ratiorepresents the best solvent. For Liquid II, the ratio is 0.33 to 0.24, which is less than 2. For Liquid III, the ratiois 0.51 to 0.19, which is slightly greater than 2.5. For Liquid IV, the ratio is 0.55 to 0.41, which is less than 2.For THF, the ratio is 0.42 to 0.31, which is slightly over 1. Liquid II is best, so choice B is the best answer.

60. Choice A is correct. Alumina is a polar adsorbent and ligroin is a nonpolar solvent, so nonpolar species will exitthe column before polar species. Of the three compounds, biphenyl is nonpolar and hydrophobic, so it will elutefrom the column first. This eliminates choices B and C. Toluic acid is polar and hydrophilic, so it will elutefrom the column last. This makes choice A the best answer.


61.

62.

63.

Choice C is correct. When the solvent is flush with the top of the silica gel, all of the solute is also flush withthe top of the silica gel. This means that the solute will enter the gel at the same time, resulting in sharperbands going down the column. Choice Cis the best answer. The sample must dissolve into the solvent to beeffective, so it cannot float, making choice Aincorrect. Whether the solvent is flush with the top of the silicagel does not impact whether or not the solute adheres to the glass. If that is an issue, it will result in problemsas the sample migrates down thecolumn, no matter where thesolvent level starts. Choice Bis eliminated. Thesilica gel should not dry out, so choice D is incorrect. The best answer is in fact choice C.

Choice Cis correct. Although the world might be a better place with words like "sandophilic" in it, there is nosuch term. This unfortunately makes choice Dincorrect. The sample must reach the silica gel, so choice A iseliminated. The sand does not interact with the silica gel, so ithas no impact on the polarity of the gel. ChoiceBis eliminated. The role of the sand at the top of the silica gel is ensure that the top of the silica gel stays flat,so all solute travels the same distance through the column no matter where itstarts. Without the sand, pouringsolvent into the system could disturb the flat surface. Choice C is the best answer.

Choice Bis correct. Silica gel is a polar adsorbent, so its affinity for polar species is high while its affinity fornonpolar species is low. The question states that when using silica gel, nonpolar species elute first, no matterwhat solvent is used. This simply means that the affinity of the adsorbent must outweigh the affinity for thesolvent. Polar compounds do follow the "like dissolves like" rule, so choice Ais eliminated. Nonpolar speciesremain nonpolar when polar solvent is added, so choice C is eliminated. Polar compounds do not form largemolecular networks any more than nonpolar molecules would, so choice D is eliminated. Choice B is the bestanswer, because silica gel is highly polar and it thereby hinders polar species by binding them as the traveldown a column.

Passage X (Questions 64 - 71) Gas Chromatography

64. Choice A is correct. According to the data in Table 1, the reactant is in the lowest concentration in Trial III(16%), so Peak 1 must be due to reactant. In GC graphs, the peak with the smallest area correlates to the lowestconcentration. Upon adding reactant, spiking the mixture with reactant, Peak 1should be the one to grow. Thisis done to confirm the identity of a peak. The best answer is choice A.

65.

66.

Choice Bis correct. If the column ispacked with a nonpolar material, then nonpolar compounds will bind to thecolumn more so than the polar compounds. This allows the polar compounds to travel through the column withless resistance. Polar compounds will thus have shorter elution times. The best answer is choice B.

Choice D is correct. Figure 1 represents the output for Trail III at 15°C. According to Table 1, the relativeconcentrations at 15°C are Product A > Product B> Reactant. Peak 2 is the largest of area, so Peak 2 must beassociated with Product A. Peak 3 is the second largest,so Peak 3 is associated with Product B. Peak 1 must bedue to the reactant. The first peak to appear has the shortest retention time, so the reactant has the lowestretention time. This eliminates choices Band C. Product Bcomes off last, soProduct Bhas the longest retentiontime. The best answer is choice D.

67. Choice A is correct. By definition, an increase in temperature favors the formation of the thermodynamicproduct, so choice Bis eliminated. Choices C and Dare the same answer, but worded differently. There cannotbe two best answer choices, so choices C and Dare both eliminated. The MCAT writers love to place the sameanswer twice with slightly different wording. From Table 1 in the passage, it can be seen that the amount ofreactant decreases with increasing temperature, so the reaction is proceeding more towards products. Thisimplies that the reaction is more favorable with increasing temperature. The best answer is choice A.

68. Choice D is correct. Gas chromatography separates by differences in migratory rates of vaporized organiccompounds. Although size and mass are variables that affect the migratory rate of a gas, the primarydifferences in retention times in this example can be attributed to the attractive forces between the packingmaterial in the column and the compounds. Ifmass and size were the primary factors, then good separationwould not be possible for the two products from the reaction in the passage. The charge ofthe compound wouldhave aneffect ifcharged compounds could easily vaporize. Because charged compounds do not readily vaporize,the gas chromatography apparatus does not distinguish charged species. Although the passage doesn't mentionit, background knowledge should tell you that ionic compounds (like NaCl) do not vaporize easily. The bestanswer is choice D.


69. Choice C is correct. The best gas to propel through the gas chromatography apparatus is an inert gas that doesnot react with any of the compounds. Nucleophiles can react with carbon dioxide, so choice A is eliminated.Hydrogen gas can react with air, so hydrogen gas should not be used. This eliminates choice B. Helium is theonly inert gas of the choices, sohelium should bechosen. The best answer ischoice C. Water is too reactive.

70. Choice A is correct. The major product from Trial II is the less hindered compound. It has no acid-baseproperties, so acid-base extraction is not viable. Choice Dis eliminated. The boiling point of the major productis likely to be close to the boiling point of the minor product, so only fractional distillation would work, notsimple orvacuum distillation. Choices Band Care eliminated. Choice Ais the best answer by default.

71. Choice D is correct. Astrong base isused to deprotonate a very weak acid. This eliminates choices AandB. ThepKa for a hydrogen on a carbon that is alpha to a carbonyl is around 19, so the best answer is choice D.Sometimes trivia is necessary to solve a problem.

Passage XI (Questions 72- 79) Recrystallization

72. Choice C is correct. Longer crystals are indicative of higher purity. Impurities disrupt the crystal lattice, socrystals cannot grow as large. This eliminates choices Aand B. Slow gradual formation of crystals yields thefinest, purest crystals. This makes choice C the best answer.

73. Choice D is correct. The ideal solvent should not bind to the lattice structure of the crystals, otherwise it couldincorporate as an impurity in the crystal lattice as it precipitates from solution. This makes Statement I valid.The ideal solvent for recrystallization should have the solid insoluble at low temperatures and highly solubleat elevated temperatures. This eliminates Statement II. The impurities should be highly soluble in the solventso that they do not precipitate out and possibly get trapped in the crystals. This makes Statement III valid.Statements I and III are both valid, so choice D is the best answer.

74. Choice Cis correct. Physical properties can be used to determine the purity ofa compound. If the density of thecrystal is precise, it implies that there are no impurities in the crystal. The melting point of the compound(crystalline ornot) can be used to determine the purity. Abroad melting point range implies that the compoundis not pure. The mass of the crystal has nothing to do with its shape and purity. This makes choice Can invalidstatement, and therefore the best answer. The index of refraction is another of the physical properties thatdepends on the lattice structure. The index of refraction varies with impurities. The best answer is choice C.

75. Choice Bis correct. By keeping the solvent and glassware hot, the original solid remains in solution during thefiltration. The purpose of filtering at this point is to remove any solid (insoluble) impurities that are present, sothe target compound should remain insolution for highest yield. This makes choice Bthe best answer choice.

76. Choice C is correct. The ice bath serves to cool the solution to 0°C. The solubility of a solute decreases as thetemperature of the solution decreases, causing more solid to crystallize (precipitate) out of solution. Overall,the purpose of the ice bath is to cool the solvent to decrease the solubility and thus increase the amount ofcrystals formed. Pick C for correctness.

77. Choice Bis correct. If crystals form too rapidly, then impurities (such as solvent molecules) can get trapped inthe crystal lattice. If an impurity happens to collide with the surface of the crystal, it has time to escape if thecrystals are formed slowly. However, if the next layer of the crystal forms too quickly, the impurity cannotescape. The best answer is choice B.

78. Choice C is correct. Ideally, the crystals formed from recrystallization are purer than the original solid,otherwise recrystallization would be pointless. Choice A is a valid statement. Considering that the solventmust be heated to itsboiling point to maximize solubility (and thus minimize the amount ofsolvent required), itis important to know the boiling point of the solvent. This makes choice B valid. If the solvent is highlyvolatile, then it cannot be raised to a very high temperature (the more volatile the liquid, the lower its boilingpoint), thus more solvent is required to fully dissolve the solid. The more solvent required at maximumtemperature, the more solvent present when the solution is cooled. The more solvent present whenrecrystallizing, the less solid that will crystallize out from solution. This reduces the percent recovery, makingchoice C the incorrect statement. It is never possible to recover 100% of a material after it has been dissolvedinto solution. A small portion always remains soluble. The best answer is choiceC.

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79. Choice C is correct. Charcoal binds colored organic molecules. If impurities are colored, they can bind thecharcoal and the charcoal can be filtered away, taking the bound impurities with it. This means that charcoalis employed when there are colored impurities. This eliminates choices A and B. If the target molecule iscolored, then charcoal can bind it, and using charcoal would decrease the yield. This means that the desiredcompound must be colorless, making choice Cthe best answer. To be achemistry warrior supreme, pick C.

Passage XII (Questions 80 - 86) Qualitative Analysis

80. Choice A is correct. Because Compound A is oxidized by chromic acid, it must be either an alcohol or analdehyde. Dinitrobenzoylchloride reacts readily with nucleophiles (such as alcohols) to form a derivativethat is a solid at room temperature. The formation of a precipitate therefore confirms that Compound A is anucleophile, so the aldehyde choice is not possible. Although you may not know what these reagents dospecifically, you should be able to deduce the purpose of each test. The molecular formula for Compound A(C6H12O) indicates that the compound must be either a carbonyl, ether, or alcohol. The ether choice iseliminated, because it would not be oxidized by the chromic acid. The carbonyl choice is eliminated for thereasons stated above. The compound must be an alcohol. An O-H stretch is found at3500 cm'1. Tiie peak isbroaddue to hydrogen bonding. Because Compound Bdoes not change color in CrC>3/H+, it must eitherbe an etherorketone. Because it shows a positive iodoform test, it is a methyl ketone. If you did not know that, you candeduce that the ether possibility is out, because ethers are inert and will not test positive with I2 and KOH.The key peakin a ketone would be theC=0 peak at 1710 cm"1. This makes choice A the bestanswer.

Choice Ais correct. This question has more depth than meets the eye. According to the formula, the compoundhas one degree of unsaturation, which can be either a ketone 7t-bond, an alkene 7t-bond, ora ring. The reagentadded (Br2 in CCI4) reacts with an alkene 7t-bond, therefore an alkene functionality must be present in themolecule. The ketone is not possible for the structure, because the C=0 ofa ketone would require a second degreeof unsaturation and given that the one unit of unsaturation present in the compound has already been used (bythe alkene). There can be no carbonyl present in the compound. The positive test confirms that there is analkene present, so it does not show that it cannot be an alkene. Choice B is thus eliminated. The ether andalcohol functionalities are possible, because the molecule could be an alkene (for which the bromine in CCI4 testis positive) with an ether or alcohol functionality. The positive test only shows that the compound cannot be aketone, thus you mustpick Ayou correctness trooper you!

Choice Ais correct. An acidic hydrogen (protic hydrogen) will exchange a proton for deuterium in the presenceof D2O. This results in the loss of one signal (broad peak) in the proton NMR spectrum. The only compound witha protic hydrogen is the alcohol, choice A. An ether should have been eliminated immediately, because ethershave no acidic protons. Aketone and aldehyde have alpha hydrogens that can be removed by strong base, butthey do not readily exchange in water without the presence of a strong base or strong acid. This question is bestsolved by selecting the most acidic compound of the choices. Aproton on the alcohol oxygen is more acidic thanthe hydrogen on an alpha carbon. No matter what your path to wisdom, finish off beautifully by choosinganswer choice A.

Choice Ciscorrect. Because Compound Awas oxidized by the chromic acid, itcan be oxidized by the KMnC>4 aswell. Both chromic acid and potassium permanganate are strong oxidizing agents. The other compound ofinterest isCompound C, which turned the bromine solution clear (confirming the presence analkene double bond).Alkenes also react with permanganate solution, forming a vicinal diol with syn orientation. This is stated inthe passage. Both Compound Aand Compound Ccan react with aqueous potassium permanganate (KMnC>4(aq)),resulting in the purple permanganate solution changing color upon reaction. This makes choice C the bestanswer.

Choice C is correct. Acid chlorides are highly reactive electrophiles (because the chloride is a good leavinggroup and the carbonyl carbon iselectron poor). Acid chlorides can react with even the weakest ofnucleophiles.Alcohols, amines, and thiols are all nucleophiles, so they all will react with the dinitrobenzoyl chloride((N02)2C6H3C0C1) kicking out the chloride leaving group. Thus the best choice for a compound that will notreact with (N02)2C6H3C0C1 is the ether. The best answer is choice C. Based on the inert nature of ethers ingeneral, it is safe to pick the ether, even if you know nothing about acid halide reactivity. In other words, forquestions that ask "Which will not react?", they are really asking "Which is the least reactive species?"

81.

82.

83.

84.

®Copyright © by The Berkeley Review 298 LAB & SPECTROSCOPY EXPLANATIONS

85. Choice B is correct. Choices C and D can be eliminated,because tertiary alcohols and ethers cannot be oxidizedby chromic acid (Cr03 in H+). The color change observed from the treatment of Compound Awith chromic acidshows that the compound can be oxidized. A primary alcohol (choice A) can be eliminated, because whenprimary alcohols are oxidized, they form carboxylic acids which would turn litmus paper red when added tothe paper. For this reason, the secondary alcohol is the best choice. Feel correctness euphoria with choice B. Beeuphoric and pick it.

86. Choice A is correct. This is perhaps a question of trivial knowledge to a point. If youhaveyour chemical testsmemorized, then this question is a no brainer. The MCAT provides a great deal of information, therefore werecommend that you focus onextracting information via critical analysis of the passages and tables, rather thanmemorization. The color of the precipitate is given in the passage as yellow. This eliminates the choices CandD which list the precipitate as being black. The passage also states that the test confirms the presence of amethyl ketone, which outright gives choice Aas the correct answer. This is a free question for just reading thepassage. If you did not read the passage, the aldehyde should be eliminated, because aldehydes can beoxidized, and Compound Bdid not react with chromic acid. If Compound Bwere an aldehyde, it would haveshowed a positive test with the chromic acid (a color change from orange to green). From the information listedin Table 1 and knowledge that the precipitate is yellow, the best choice is answer A.

Passage XIII (Questions 87 - 93) Synthesis and Extraction

87.

88.

89.

90.

91.

Choice Bis correct. The limiting reagent is 1,4-dimethoxybenzene, not nitric acid, so the percent yield should becalculated based on 1,4-dimethoxybenzene. The percent yield for the reaction is the actual yield (91.5 mg 2,5-dimethoxynitrobenzene) divided by the theoretical yield (183 mg 2,5-dimethoxynitrobenzene) based on the 1.0mmole of the limiting reagent (1,4-dimethoxybenzene). Because the actual yield is only half of the theoreticalyield, the percent yield is fifty percent. The best answer is thus choice B. The math is shown below.

% yield =moles product

91.5 mg,183 g/mole _ 0.5 mmolesproduct

moles limiting reagent 138mg.138g/mole

1.0 mmoles limiting reagent= 50%

Choice B is correct. The melting point of the product has a range of 1.5 °C, which is a small enough range todeduce that the product is relatively pure. Because the melting point is sharp (a sharp melting point range isgenerally regarded as two degrees or less) and close to that of the literature value for the product, the bestanswer is choice B. Be a Bsupporter! There isno mention ofa broad OH peak in the passage, so it is unlikelythat it is wet. Choice A is eliminated.

Choice Bis correct. Filtering collects (and thus removes) the solid particles from a solution ormixture. Filteringis most commonly used inorganic lab for the separation of solids from liquids. The best answer ischoice B.

Choice Cis correct. Both nitrogen gas and carbon monoxide gas are colorless, so choices Aand Bare eliminated.You should know the color of these two gases from your everyday experiences inlife. Nitrogen gas makes up themajority of our air, and air is transparent. Carbon monoxide is given off with the exhaust from combustionengines. If you are aware of your trivia, sulfur dioxide is colorless to faint yellow in color, which eliminateschoice D. If you didn't know this, you hopefully thought of smog, a brown gas containing a nitrogen/oxygenmolecule of some kind (nitrogen dioxide). Because smog is comprised of anitrogen/oxygen compound, and smog isbrown, it is a logical conclusion to select NO2, choice C. Choice C is the bestanswer.

Choice Bis correct. Theerror in the thermometer resulted in a reading that was too low (the experimental valueof 72.5 to 74 is less than the literature value of 74 to 75). This means that the mercury did not climb as high as itshould have in the inner column. The presence ofsome air above the mercury column (resulting from a reducedvacuum) hinders the ability of the mercury column to rise, thus Statement I results in a reading lower thanexpected. Adivot present between 45°C and 55°C results in the mercury column filling the divot at temperatureswhere the top of the mercury column is above the divot (temperatures greater than 55°C in Statement II). Thiscauses the mercury column to not rise as high. This results in a reading that is too low. If the divot is above themercury level, then it will have no effect, because the mercury has not risen high enough to fill the divot. Thismeans that Statement II is valid but Statement III is invalid. Choice B is the best answer.

Copyright © by The Berkeley Review*1 299 LAB & SPECTROSCOPY EXPLANATIONS

92. Choice Ais correct. Filtering the compound again would not purify the crystals, it would simply reduce theyield, so choice Bis eliminated. Distilling the liquid (formed from melting) would be impractical given thehigh temperature and probability of deposition of the gas into solid throughout the distilling column, so choiceCis eliminated. Addition of more sulfuric acid and nitric acid could further react with the product rather thanthe starting material, so choice Dis eliminated. Recrystallization is the ideal laboratory technique to employin the purification of a solid. The best answer ischoice A, so you better pick thatA.

Choice Bis correct. The addition of cold water causes the crashing out (rapid crystallization) of product fromsolution. This means that solubility decreased, not increased, eliminating choice Aand making choice Bthe bestanswer. The solution already has two strong acids, a mixture of nitric and sulfuric, so the addition of water willnot increase its acidity of the solution. This eliminates choice C. Water will not protonate a species that isdeprotonated in the presence of strong acid, so choice D is eliminated. You should recall that lowertemperatures result in reduced solubility, so choice B is the best answer choice.

93.

Not Based on a Passage Questions 94 -100 Organic Chemistry Lab Techniques94.

98.

99.

Choice C is correct. If the S-enantiomer has a specific rotation of -55°, then the R-enantiomer has a specificrotation of +55°. The net rotation is positive, so there must be more of the R-enantiomer than the S-enantiomer.This eliminates choice A. Solving this is going to require math, so it is easier to start with an answerchoice andwork back to the specific rotation. For plugging in the numbers, choice C is best, because only one calculationneeds to be done. The value is either going to be too high (implying that choice Bis correct), right on the mark(making choice C correct), or too low (implying that choice D is correct). A mixture of 30% of the S-enantiomerand 70% of the R-enantiomer has a40% excess of the R-enantiomer. A100% R-enantiomer solution has aspecificrotation of +55°, so 40% excess of R yields a rotation of 40% of +55°, which is 22°. Choice C is the best answer.

95. Choice Ais correct. The solute that elutes first from the column is the fastest. In a TLC experiment, the fastestsolute climbs the plate the most, resulting in the highest spot. This means that the dark circle corresponds toBand A. This eliminates choices Band C. By the same reasoning used to ascertain the dark spot, the hollowspotcorresponds to Band C,making choice Athebest answer and thus, themost pickable answer.

96. Choice D is correct. Fractional distillation is used to separate liquids with boiling points that are close to oneanother. This is achieved by increasing the surface area in the distilling column. This makes choice A a validstatement, thereby eliminating it. The boiling points are close to one another in fractional distillation, whichmakes choice B a valid statement, thereby eliminating it. Fractional distillation is more selective, so thepurity is high, buta great deal of residue is left behind in the distilling column, resulting in a lower yield withfractional distillation. This makes choice D the best answer.

97. Choice Ais correct. The compound that gets extracted into sodium hydroxide solution is the organic compoundthat becomes an ion when treated with hydroxide. Neutral acids become anions when they are deprotonated, sothe answer is an organic acid. The only organic acid is choice A, a carboxylic acid. Choice A is the best answer.

Choice B is correct. Carboxylic acids are hydrophilic, although the aromatic ring helps with its lipidsolubility. It should be somewhat soluble in diethyl ether. Despite the two hydrophilic groups onhydroxybenzoic acid, it is only slightly soluble in water, because of the benzene ring. Carboxylic acids arerather easily deprotonated, so they can be converted into an anion and made to be more water soluble. Thismakes a compound like benzoicacid highlysolublein a basicenvironment. The best answer is choice B.

Choice C is correct. Distillation involves boiling a liquid away from a mixture, so it works with two liquidsthat have different boiling points and don't form an azeotrope. Choice A is eliminated. Chromatographyinvolves solutes migrating down a column at different rates, so it works for solutes. Choice B is eliminated.Crystallization dissolves and selectively precipitates a solid from a mixture of solids. This eliminates choiceD. Extraction separates solids and liquids, but not gases. Choice C is thebestanswer.

100. Choice D is correct. Butanone should be highly soluble in other ketones such as acetone, so choice A iseliminated. Ketones don't oxidize with chromium (VI), so they do show a negative Jone's test. Choice Biseliminated. Butanone has a molecular mass of 72 grams per mole, so if it loses a methyl group during, it canform a fragment with a molecular mass of 57 amu. Choice C is eliminated. There are no chiral centers inbutanone, so it cannothave optical activity. This makes choice D the best answer.


Writing a book of this nature ended up being a more difficult project than I first imagined back in 1990.For every one hundred questions you see in this book, about three hundred had to be written. A greatdeal of time went into selecting which questions served the purpose of conceptual analysis, developingtest-taking skills, honing table-reading skills, testing memory, and building confidence. But aschallenging as it was at times, it has been a wonderful experience. This particular edition is theculmination of fourteen years of writing and rewriting. Along the path I have incorporated thefeedback and input of roughly 5000students. I have tried my best to include many of the examples thathave proven useful in lecture. I have tried to keep in mind the sole purpose of this book in your life: tohelp you achieve a higher MCAT score in the biological sciences section and offer test-taking tips thatare universally applicable. It is imperative that you keep that in mind while working through ourmaterials. Your goal is to learn information and strategies that allow you to quickly pick the best offour choices when asked a multiple-choice question. Completing this book would have never beenpossible without the support of many special people in my life. I would like to thank:

Cecile Santos: You have been an inspiration to my teaching. I see the love you have in your eyes for thestudents you teach and it is awe-inspiring. You have been my sounding board for teaching ideas andmuch of the text in this book. Without you, I could never have done this.

Cambria and Madison: You fill my heart with so much love. You keep me balanced at times I need itmost. Your laughter and smiles keep my spirits high and I am so lucky you came into my life. You arethe greatest children a father could ever want.

Our Wonderful Teachers: Over the years I have gleaned so many great ideas from you. It has been anhonor working alongside so many of you. While so many of you have been helpful in my path, I wouldbe remiss if I didn't specifically mention six very special teachers to me (although we have had manymore.) Anthony Gopal showed me about the power of enthusiasm and how demonstrations and real lifeexamples can drive classroom learning. Moin Vera showed me how simple genius and fundamentalconcepts can make even the hardest topic seem easy. Derek Welsbie showed me how simple logic,when applied with a little bit of information, can make every question easier to understand. SamBiggers showed me how bonding with the class and empathizing with their anxieties can build a bondthat helps the entire team do better. Kim Gordy showed me the importance of silliness in gaining thestudents' interest and trust and how analogies can power learning. Cecile Santos showed me theimportance of engagement and open dialogue with her open forum discussion in lecture where everystudent is involved. You are all incredible teachers and I learn something from you every time I watchyou in action. Thank you!

Dale Schmidt: You have been incredibly supportive over the years, picking up the slack when it came.Your work ethic and attention to detail is second to none. I am always sure that if you did it, it wasdefinitely done correctly.

Kimberly Gordy and Doug Coe: You spent so much time combing through my books to help them flowbetter. I can't thank you enough, because I am not an easy person to work with when it comes to editing.You managed to not only make this book better, but you made me realize the need for a different styleand voice at many junctures.

Our Incredible Students: For the books I have written, I owe my biggest thank you to the thousands ofstudents who have used them to help prepare for their MCAT. Some of you are well into your practiceand this is a distant memory. Others of you have just recently finished your exam and are working yourbutt off to make it into medical school. To each of you who offered some feedback, told me about a typo,asked me why a question was worded a certain way, or just simply said what you thought, I am deeplygrateful. This book, and the others I have written, is as much a part of you as it is a part of me. This isour project and I am so happy you were a part of its development.

Copyright © byThe Berkeley Review® 301 LAB & SPECTROSCOPY EXPLANATIONS

TfuPERKEJLEYrjR.F..V.|Ê-W'M

PERIODIC TABLE OF THE ELEMENTS

2

He

4.0

1

H

1.05

B

10.8

6

C

12.0

7

N

14.0

8

O

16.0

9

F

19.0

10

Ne

20.2

3

Li

6.9

4

Be

9.0

13

Al

27.0

14

Si

28.1

15

P

31.0

16

S

32.1

17

ci

35.5

18

Ar

39.9

11

Na

23.0

12

Mg24.3

19

K

39.1

20

Ca

40.1

21

Sc

45.0

22

Ti

47.9

23

V

50.9 52.0

25

Mn

54.9

26

Fe

55.8

27

Co

58.9

28

Ni

58.7

29

Cu

63.5

30

Zn

65.4

31

Ga

69.7

32

Ge

72.6

33

As

74.9

34

Se

79.0

35

Br

79.9

36

Kr

83.8

37

Rb

85.5

38

Sr

87.6

39

Y

88.9

40

Zr

91.2

41

Nb

92.9

42

Mo

95.9

43

Tc

(98)

44

Ru

101.1

45

Rh

102.9

46

Pd

106.4

47

Ag107.9

48

Cd

112.4

49

In

114.8

50

Sn

118.7

51

Sb

121.8

52

Te

127.6

53

I

126.9

54

Xe

131.3

55

Cs

132.9

56

Ba

137.3

57tLaT138.9

72

Hf

178.5

73

Ta

180.9

74

W

183.9

75

Re

186.2

76

Os

190.2

77

Ir

192.2

78

Pt

195.1

79

Au

197.0

80

Hg200.6

81

Tl

204.4

82

Pb

207.2

83

Bi

209.0

84

Po

(209)

85

At

(210)

86

Rn

(222)

87

Fr

(223)

88

Ra

226.0

89 RAc§

227.0

104

Rf

(261)

105

Db

|(262)

106

Sg(263)

107

Bh

(262)

108

Hs

(265)

109

Mt

(266)

110

Uun

(269)

111

Uuu

(272)

112

Uub

(277)

+

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

(145)

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

71

La

175.0

§

90

Th

232.0

91

Pa

(231)

92

U

238.0

93

Np(237)

94

Pu

(244)

95

Am

(243)

96

Cm

(247)

97

Bk

(247)

98

a

(251)

99

Es

(252)

100

Fm

(257)

101

Md

(258)

102

No

(259)

103

Lr

(260)

Specializing In MCAT Preparation

Documents

The Berkeley Review MCAT Organic Chemistry Part 2