Chapter 24 The Organic Chemistry of Carbon

In 1828, F. Wohler synthesized urea from ammonium chloride andsilver cyanate. The compound synthesized was identical to theurea generated by animals. Previous to this time it was thoughtthat only life forms were capable of producing organic compounds.Wohler demonstrated that organic compounds could be made frominorganic compounds.

Important Terms❏ organic compounds ❏ structural formulas❏ saturated hydrocarbons ❏ unsaturated hydrocarbons❏ hybridization ❏ structural isomers❏ functional groups ❏ IUPAC❏ geometric isomers ❏ cis and trans❏ optical activity ❏ chiral❏ SN1 and SN2 ❏ alkene❏ primary carbon cation ❏ secondary carbon cation❏ carbanion ❏ radical❏ Markovnikov's Rule ❏ electrophilic❏ nucleophilic ❏ aromatic❏ ortho, meta, para ❏ carboxylic acid❏ alcohol ❏ amine

The Saturated Hydrocarbons: Alkanes and CycloalkanesA hydrocarbon is a compound that contains only hydrogen andcarbon. An important aspect of carbon is its ability to form chainsand branches with other carbon atoms.

The different ways that carbon can bond, makes it important for usto be able to distinguish between compounds. For example, giventhe formula C4H10, there are two compounds with this sameformula.

H

C

H

H

H

C

H

H

C

H

H

C

H

H

H

C

H

H C

H

H

C

H

H

H

CH H

Structural formulas indicate the actual bonding arrangement fororganic compounds. They indicate how and to what the atoms are

Chapter 24

bonded to. There are two ways of writing structural formulas, theexpanded or full structural formula, and the line or semi-condensedformula. In the semi-condensed formula, he number of linesrepresent the type of bonding between carbon atoms. A single linebetween two atoms means a single bond, a double line refers to adouble bond while a triple line refers to a triple bond. These twoformula types are illustrated for ethane and propane. below. Oneshould be able to write both expanded and condensed structuralformulas.

Ring systems, like structural formulas, are generally abbreviated. Ahydrocarbon ring is represented by a polygon where each vertexrepresents a carbon with suitable hydrogen to satisfy bondingrequirements. Consider cyclopropane. This compound has thegeneral formula of C3H6. It is classified as an cyclic alkane , it fitsthe general formula of CnH2n. The general structure is a trianglewhere each vertex represents a carbon atom bonded to two othercarbon atoms in the ring. Each carbon atom of the ring has twohydrogens bonded to it to satisfy bonding requirements.

Expanded Structural, Condensed, and Molecular Formulas.

CC C

H

H

H

H

HH

H

H

C H

H

H

C

H

H

H

CH3--CH3 or H3C-CH3 or CH3CH3 C2H6

CH3-CH2-CH3 or H3C-CH2-CH3 or CH3CH2CH3 C3H8

The Organic Chemistry of Carbon

Cyclopropane can be represented using any of the followingstructures.

2

2

2CHCH

CH

H

H H

H

H H

C

C

C

The compound cyclobutane, C4H8, is a four-membered ringsystem which can be represented by a square.

HH

H

H H

H

H H

C

C C

C CH2

CH2

CH2

CH2

Cyclopentane, C5H10, is a five-membered ring system and can berepresented by any of the following structures.

CH2

CH2

CH2

CH2

CH2

HH

H

H H

H

H H

C

C C

C

C

H H

Structural or Positional IsomersStructural or positional isomers have the same number of atomsbut exhibit different bonding arrangements. This usually results incompounds that have different physical properties and sometimesdifferent chemical properties as well.

Chapter 24

Example 24.1. Draw all of the structural isomers for C6H14.

For C6H14 the longest continuous chain contains six carbon atomsand yields

CH3-CH2-CH2-CH2-CH2-CH3 Structure I

Now, write five carbons in a row and number the carbons. Thesixth carbon atom may be placed on carbons 2, 3, or 4.

C C C C C

C

1 2 3 4 5C C C C C

C

1 2 3 4 5C C C C C

C

1 2 3 4 5

Structure II Structure III Structure IVStructures II and IV are identical since the -CH3 group is in thesecond position from the end. Remember that when counting thelongest continuous chain, you should start from one end, andcount, then start from the other end and count, to be sureduplication does not occur.

C C C C C

C

1 2 3 4 5

6

Do not make the common error of thinking a carbon in the up ordown position near the end or beginning of a chain is not part ofthe longest continuous chain. Number the longest chain basedupon the continuous connection of carbon atoms.

We can place the proper number of hydrogens around each carbon.

Writing four carbons in a chain, and adding two CH3 groups oncarbon number 2 gives structure IV. We can have a CH3 group oncarbon 2 and a CH3 group on carbon 3 to give structure V.

The Organic Chemistry of Carbon

C 3H C C

H

C 3H C 3H

H

C 3HC 3H C C

C 3H

C 3H

C 3HH2

Structure IV Structure VThere are a total of five structural isomers for C6H14.

The Unsaturated Hydrocarbons: Alkenes and AlkynesSingle, double, and triple bonds have been found between carbonatoms. The table below indicates the relationship among bonding,structure, and hybridization schemes.

Geometry and Hybridization Schemes for Carbon.Bonds Geometry Hybridization Angles (HCC)CH3-CH3 tetrahedral sp3 109.5°

CH2=CH2 planar sp2 120.0°CH CH linear sp 180.0°

Alkanes are saturated hydrocarbons with the general formula ofCnH2n+2 and each carbon atom has sp3 hybridization withtetrahedral site symmetry around each carbon; that is, a H-C-Cangle of 109.5°. Alkenes are unsaturated hydrocarbons that havethe general formula CnH2n. This type of compound contains adouble bond between carbon atoms, has trigonal planar sitegeometry, and the carbon atoms of the double bond have sp2hybridization. The approximate angle around the carbon atomsinvolved in the double bond is 120°. Alkynes have the generalformula, CnH2n-2, are unsaturated hydrocarbons with a carboncarbon triple bond, sp hybridization for the carbon atoms of thetriple bonds, linear site geometry, and an angle of 180°.

Example 24.2. Identify the following compounds as beingsaturated or unsaturated hydrocarbons and also identify the class ofthe hydrocarbons (alkanes, alkenes, or alkynes):

a) CH3-CH2-CH3 b) CH3-CH=CH-CH3 c) CH3-C C-CH3

Chapter 24

Compound a has all single bonds as well as single carbonhydrogen bonds. Hence, it is a saturated hydrocarbon, an alkane.This compound also has the molecular formula of C3H8 which isof the form CnH2n+2.

Compound b contains a double carbon-carbon bond and thereforethis compound is unsaturated. The presence of the double bondindicates that the compound is an alkene. It has the molecularformula of C4H8 which coincides with the general formula of analkene, CnH2n.

Compound c has a triple carbon carbon bond. It is an unsaturatedhydrocarbon and falls into the classification of an alkyne. Themolecular formula for this compound C4H6 agrees with that of analkyne, CnH2n-2.

Example 24.3. Indicate the hybridization and site geometry foreach carbon atoms in the following structures:

CH2 C CH CH32CH

O

C HC 3H

2CH

O

C HC 3H123

Carbon atom number 3 has four single bonds to it. It has sp3hybridization and tetrahedral site symmetry. The same is true forcarbon 2. Carbon 1 has sp2 hybridization and planar sitesymmetry. It has one double bond and two single bonds.

CH2 C CH CH31 2 3 4

Carbon atom 1 has one carbon-carbon double bond and two CHsingle bonds. It exhibits sp2 hybridization and has trigonal planarsite symmetry. The same situation is found for carbon number 3.Carbon number 4 has sp3 hybridization. It has four single bondsand exhibits tetrahedral site symmetry. Carbon atom 2 is a specialcase for it has sp hybridization. This arises because the p orbitalson carbon number 2 that are used in the p bonding to carbons 1 and3 are perpendicular to each other.

The Organic Chemistry of Carbon

Geometric Isomers are isomers that have very similar chemicalproperties but different physical properties. This arises because thecompounds have the same basic framework for their carbonstructures. The requirements for geometric isomers are fairlystraight forward:

1. Compound must have restricted rotation. a. Double bond b. Ring structure

2. No two groups attached to the same carbon atom of thedouble bond may be identical.

Examples of geometric isomers (structures I and II) are givenbelow for 2-butene.

C 3H

C

HC

H

C 3HI

cis-2-butene

C 3H

CH

C

H

C 3H

II

trans-2-butene

C 3H

C

H

C

HC 3H

III

2-methyl-1-propene

Structure III is not a geometric isomer for it contains identical CH3groups on the same carbon atom of the double bond. It alsocontains two H's on the other carbon atom of the double bond.

Geometric isomers are often referred to as being cis- or trans-. Cis- means on the same side of the double bond while trans- means onthe opposite side of the double bond. Note that in the case of thecis- isomer you pass through the double bond and arrive on thesame side of the double bond plane.

Example 24.4. Indicate if the following ring systems are examplesof geometric isomers. If so indicate if they are of the cis - or trans-type.

Cl

Cl

ClClCl

Cl

a) b) c)

Structure a has two chlorine groups on the same carbon atom.Thus it is not a geometric isomer. It is named as 1,1-dichlorocyclopropane.

Chapter 24

Structure b has the two chlorine atoms on opposite sides of the ringtherefore it is a trans- geometric isomer. It is named as trans-1,2-dichlorocyclopropane.

Structure c has the two chlorine atoms on the same side of the ring.This is a cis geometric isomer. It is named as cis-1,2-dichlorocyclopropane.

Some Reactions of Alkanes and Alkenes

Example 24.5. Indicate a simple chemical test that woulddistinguish between an unsaturated hydrocarbon and a saturatedhydrocarbon. Use the following general structures to illustrateyour answer.

Unsaturated hydrocarbons undergo electrophilic addition reactionswith bromine. This reaction is usually performed in the dark toavoid photochemical substitution of bromine. The disappearanceof the dark red color associated with bromine is indicative of apositive reaction.Unsaturated hydrocarbon

+dark

red color

C

Br

R C

Br

H

R Rcolorless productcolorless

Br2R

R

C C

H

H

Saturated hydrocarbon

+dark

red colorcolorless

Br2 no reactionC

H

R C

H

H

R H

The Organic Chemistry of Carbon

Example 24.6. Identify the major product (I or II) of the followingelectrophilic addition of HBr across the carbon carbon doublebond.

+ BrH or C

Br

C HC 3H

H

C 3HC 3H

II

C

Br

C HC 3H

H

C 3HC 3H

I

C C

H

C 3H

C 3H

C 3H

Markovnikov's rule states that the major product in an electrophilicaddition reaction of H-X across an unsymmetric double bond willbe the one where hydrogen adds to the carbon atom of the doublebond that already contains the most hydrogens directly bonded toit.

+ BrH orC C

H

C 3H

C3

H

C3

H

C C HC 3H

H

C 3HC3

H

3° carbonium ion

+C C HC 3H

H

C3

HC 3H

2° carbonium ion

+

Structure I is the major product. Structure II is the minor product.

The Aromatic Hydrocarbons and Their DerivativesBenzene has the shape of a hexagonal ring with the molecularformula of C6H6 and all of its carbon carbon distance the same(0.139) nm compared to single carbon carbon bond distance of0.154 nm and a double bond distance of 0.133 nm). Benzene doesnot undergo electrophilic addition nor does it readily add hydrogenor bromine. This stability is attributed to its unique bondingsystem that involves the overlap of p orbitals which form acontinuous p electron cloud of electron density around the ring.The resonance contributing structures of benzene can be written intwo forms

Chapter 24

CH

C

C

CC

CH H

H

H

H

CH

C

C

CC

CH H

H

H

H

A resonance hybrid structure of benzene is representedas a full circle inside of the hexagonal ring to representthe overlap of electron density. This is a short-handnotation for benzene that is commonly used.Compounds with benzene-like structures are classified

as aromatic hydrocarbons. Some common examples of aromaticsare naphthalene, anthracene, and substituted benzenes.

X

benzene naphthalene anthracene substituted benzene

Substituted benzenes are formed by electrophilic substitutionreactions. Such reactions require the presence of a Lewis catalyst.

Lewis Acid+ E-Nu + H-Nu

H E

Some typical mono- and di- substituted benzenes are given below:

CH3 Cl NH2 OH

toluene chlorobenzene aniline phenol

The Organic Chemistry of Carbon

Example 24.7. Draw the structures for ortho-, meta-, and para-isomers of dichlorobenzene.

ortho-dichlorobenzene

ClCl

meta-dichlorobenzene

Cl

para-dichlorobenzeneCl

Cl Cl

The terminology of ortho-, meta-, and para- is used only fordisubstituted benzenes.

Functional GroupsSpecific arrangements of elements that act as reactive sites within amolecule are called functional groups. Classification of reactionsthat occur in organic chemistry are often by functional groups. Forexample, the structure R-OH is a general structure for an alcohol.R-O-R' is a general structure for ethers, where R represents anyattached system. One should be able to recognize and name thecommon functional groups. Knowledge of the general structure ofcommon functional groups is a key concept in chemistry.

Example 24.8. Identify the functional group present in each of thefollowing structures

C 3H 2CH

O

C HC 3H2CH

H

C 3H Na) b) c) C 3HC 3H CH 2CH 2 O

a) The N-H linkage is attached to two carbon atoms and onehydrogen atom. This is the structure for a secondary amine. Thename of this compound is ethylmethyl amine.

Chapter 24

b) The carbonyl functional group with a hydrogen atom attachedto it identifies this compound as an aldehyde. The IUPAC namefor this compound is propanal.

O

C HR

c) This compound has the general structure R-O-R, it is an ether.The name for this compound is propyl methyl ether.

Organic NomenclatureThe variety of functional groups and the early development oforganic chemistry have led to an interesting situation for namingorganic compounds. Many organic compounds are identified bytheir common name. The International Union of Pure and AppliedChemistry (IUPAC) has rules for naming organic compounds andthere is a systematic method.

To begin with, we first determine the longest continuous carbonchain and then determine the root name. The root name is derivedfrom the name of the compound having that number of carbons inits longest carbon chain. You must memorize the names of thefirst twelve hydrocarbons. The following table provides thechemical formula, IUPAC name, the root name and the number ofcarbon atoms in the chain.

IUPAC Names and Roots.Chemical Formula IUPAC Name Root AtomsCH4 methane meth 1CH3-CH3 ethane eth 2CH3-CH2-CH3 propane prop 3CH3-(CH2)2-CH3 butane but 4CH3-(CH2)3-CH3 pentane pent 5CH3-(CH2)4-CH3 hexane hex 6CH3-(CH2)5-CH3 heptane hept 7CH3-(CH2)6-CH3 octane oct 8CH3-(CH2)7-CH3 nonane non 9CH3-(CH2)8-CH3 decane dec 10CH3-(CH2)9-CH3 undecane undec 11CH3-(CH2)10-CH3 dodecane dodec 12

Alkyl (Al-keel) groups are formed by removing a hydrogen fromthe alkane; for example, removing a hydrogen from methane H3C-

The Organic Chemistry of Carbon

H yields the methyl group H3C-. The alkyl group is named afterits corresponding hydrocarbon by removing the suffix ending aneand adding the ending yl. Some of the more common alkyl groupsare listed on the next page.Simple Alkyl Groups.______________________________________________________Group Name Group Name______________________________________________________

CH3- methyl CH3-CH2- ethyl

CH3-CH2-CH2- n-propyl CH3-CH2-CH2--CH2- n-butyl

CH3-CH- isopropyl CH3-CH-CH2- isobutyl| |

CH3 CH3

CH3|

CH3-C- t-butyl CH3-CH-CH2-CH3 sec-butyl| |

CH3

F- fluoro Cl- chloro Br- bromo I- iodo______________________________________________________

Example 24.9. Give the IUPAC name to the following chemicalcompounds:

a) CH3-CH-CH2-CH3 b) CH3-CH-CH2-CH3| |

CH2-CH2-CH3 CH2Cl-CH-CH3

c) CH3-CBr2-CHBr-CH3

Chapter 24

For compound a, the longest carbon chain contains six carbonatoms and it is numbered to give the methyl group the lowestpossible index number.

H

C 3H C

CH 2 CH 2 C 3H

CH 2 C 3H123

4 5 6

H

C 3H C

CH 2 CH 2 C 3H

CH 2 C 3H

123

4 5 6

3-methylhexane rather than 4-methylhexaneFor compound b, the numbering scheme is selected to give thelowest set of index numbers.

HC 3H C

CH 2 C 3H

C 3H

1 2

3 4

CHCl

HC 3H C

CH 2 C 3H

C 3H1 2

34CHCl

1-chloro-2,3-dimethylbutane rather than 4-chloro-2,3-dimethylbutane

For compound c, you need to number the carbons so that thelowest set of index numbers for the substituents is obtained.

C 3H CBr 2 CHBr C 3H1 2 3 4

C 3H CBr 2 CHBr C 3H1234

2,2,3-tribromobutane rather than 2,3,3-tribromobutane

Example 24.10. Give the appropriate chemical structure for thefollowing compounds:a) 2,2-dichloropentane b) 1,2-dibromo-2-methylpropanec) 1,1,2-trichloroethane d) 2,3-dimethylhexaneFor compound a, the parent hydrocarbon is a five-memberedcarbon chain (pent) that is saturated (all single bonds) (ane).Writing this carbon sequence out and numbering the carbons yields

The Organic Chemistry of Carbon

Cl1 2 3 4 5 |C-C-C-C-C Inserting the chlorines results in C-C-C-C-C andcompleting the structure |

Cl

Cl|

with hydrogen gives CH3-C-CH2-CH2-CH3|

ClFor compound b, 1,2-dibromo-2-methylpropane, the parenthydrocarbon is a three-membered chain (prop-) that is saturated(ane). Writing this carbon sequence out, numbering, inserting thesubstituents, and then the hydrogen yields

Br Br Br Br H| | | | |

C-C-C C - C - C H - C - C - C - H| | | |

CH3 H CH3 H

For compound c, the parent hydrocarbon is a two-membered chain(eth) that is saturated (ane). Writing the carbon sequence out,numbering, inserting the substituents, and then adding thehydrogen results in the following structure

1 2C C C C

ClCl

Cl

C C

ClCl

Cl

H H

H

or C CH 2HCl 2Cl

For compound d, 2,3-dimethylhexane, the parent hydrocarbon is asix-membered chain (hex) that is saturated (ane). Writing thiscarbon sequence out, numbering, inserting the substituents, andthen adding the hydrogen yields

Chapter 24

C C C C C C1 2 3 4 5 6

C 3H CH CH

C 3H C 3H

CH 2 CH 2C 3HC C C C C C

C 3H

C 3H

Alcohols: Some Reactions, Isomers, and Boiling PointInformation

The number of organic compounds is in the billions and thesecompounds are usually placed in specific categories based upon thetype of functional group present. Once you are able to classifyreactants according to functional groups, you are able to learnreaction chemistry of specific functional groups.

Example 24.11. Determine the number of structural isomers forC4H10O. Please draw only alcohols and ethers. There are fourstructural isomers with the alcohol functional group.

C 3H CH 2CH 2CH 2 OH C 3H C

H

CH 2C 3H

OH

C 3H C

H

CH 2C 3H

OH C 3H C C 3H

C 3H

OH

For the ethers there are three structural isomers.

C 3H C 3HCH 2CH 2O C 3HC 3H CH 2CH 2 O C 3HC 3H C

H

C 3H

O

There are a total of seven structural isomers for C4H10O that areethers or alcohols.

Substitution ReactionsSN1 refers to a unimolecular nucleophilic substitution reactionwhile SN2 refers to a bimolecular nucleophilic substitutionreaction. Both SN1-type and SN2 -type reactions are usuallydetermined by experimentation. An SN1 reaction rate dependsonly upon the concentration of the substrate while an SN2 reactionrate depends upon the concentration of the substrate (organicreactant) and the nucleophile (Lewis base). Remember thatmechanisms are the intermediate steps in a reaction. The rate ofchemical reaction is determined by the slow step in the mechanism.

The Organic Chemistry of Carbon

For the unimolecular nucleophilic substitution reaction between(CH3)2CHI + OH- ---> (CH3)2CHOH + I-

the formation of a carbon cation intermediate is the rate controllingstep. Experimental evidence indicates that SN1-type reactions arefavored by the formation of tertiary carbon cations over secondarycarbon cations which in turn are more favorable than primarycarbon cations. Carbon cations are identical to carbonium ions.

C 3H

CC 3H

H

Islow

OH

fast

C 3H

CC 3H

H

IOH +

C 3H

CC 3H

H

I+ +

For the bimolecular nucleophilic substitution reaction betweenmethyl iodide and hydroxide ion, H3CI + OH- ---> H3COH + I-

the formation of a five-membered coordinate intermediate is therate-controlling step. Experimental evidence indicates that SN2-type reactions are favored by carbon compounds that would formprimary or secondary carbon cations rather than tertiary carboncations.

+

slow fastC OH

H H

H

I +C

H

HH

OHI

transition state

C

H

H

H

I OH

This reaction involves a collision between two species, the methyliodide molecule and the hydroxide ion. Carbocations orcarbonium ions are positively charged ions that exist for a shortperiod of time and are known as intermediates. You cannot bottleand store reaction intermediates. The general representation of thethree different types of carbonium ions are shown below

Chapter 24

C +

H

R

H

C +

R

R

H

primary carbon cation

C +

R

R

R

secondary carbon cation tertiary carbon cation

1° 2° 3°

R refers to any type of attached group. Two hydrogens and one Rgroup indicates a primary structure, one hydrogen and two Rgroups indicates a secondary structure, and three R groupsindicates a tertiary structure. The type of intermediate structureformed in a reaction often governs the nature of the productsformed. If the intermediate has a negative charge it is a carbanion.If the intermediate does not have a charge, it is neutral, and is afree radical.

Example 24.12. Identify the following structures as beingprimary, secondary, or tertiary carbon cations, carbanions, orcarbon free radicals.

C 3H

C

H

C 3H

C 3H

C

H

+C 3H C

C3

H

C3

H

C 3H

C 3H

C

H

CH 2 +a) b) c)

Structure a has a positive charge. It is a carbon cation. The carbonwith the charge has two groups attached to it and only onehydrogen. It is a secondary carbon cation.

Structure b has a negative charge. It is a carbanion. The carbonwith the charge has two groups attached to it and only onehydrogen atom.

Structure c has a positive charge. It is a carbon cation. The carbonwith the charge has three groups attached to it. It is a tertiarycarbon cation.

Oxidation and Reduction ReactionsOxidation and reduction reactions were encountered in Chapters 2and 19. In the field of organic chemistry, oxidation is usuallyassociated with the addition of oxygen, the removal of hydrogen or

The Organic Chemistry of Carbon

the conversion of alcohols to carbonyls. Reduction refers to theaddition of hydrogen, the removal of oxygen or the conversion ofcarbonyls to alcohols. Primary alcohols are oxidized to aldehydesor acids. Secondary alcohols are oxidized to ketones. Tertiaryalcohols do not undergo mild oxidation.

Example 24.13. Predict the products of the following oxidationand reduction reactions

C 3H C

O

H

H

H

K2Cr 7O2

SO42H

∆C 3H C

OH

H

C 3Hb)

+ K2Cr 7O2

SO42H

∆a)

C 3H C

O

Hd) + H2∆

K2Cr 7O2

SO42H

∆c) C 3H C

OH

C 3H

C 3H

Reactions a, b , and d are oxidation-reduction reactions. Tertiaryalcohols do not undergo mild oxidation. Therefore, no productsare formed in reaction c.

Carboxylic Acids and Their DerivativesCarboxylic acids are weak organic acids that have the carboxylicfunctional group. The lower molecular weight carboxylic acidsare very water soluble while the heavier acids are water soluble.The acidity of these acids is general represented by the followingequilibrium system.

+O

C HOR H2O+

O

C OR H3O+

Chapter 24

Recall that the acid dissociation constant for such an equilibriumsystem is

O

CC 3H O H3O+

O

C HC 3H O

K =a

Ka values for organic acids are usually less than 10-3. Meaningthat at equilibrium most of species in solution are the unreactedorganic acid molecules. There are relatively few ions present.

Organic acids react with bases to produce a salt and water,however, since it is a weak acid, an equilibrium system isestablished.

+ H2O

O

C O Na+

RNaOH+O

C HOR

You should know the following four types of reactions involvingcarboxylic acids:Ionization (Carboxylic acid + water ∆ carboxylate ions +hydronium ion)

+O

C HOR H2O+

O

C OR H3O+

Neutralization (Carboxylic acid + base ∆ salt + water)

+ H2O

O

C O Na+

RNaOH+O

C HOR

Amide Formation (Carboxylic acid + amine ----> amide +water)

+O

C HOR H2O+ N

H

R

O

CR N

H

RH

∆

The Organic Chemistry of Carbon

Acid Anhydride Formation (Carboxylic acid + Carboxylic acid-----> acid anydride + water)

+

+ H2O+

O

C HOR

O

C HOR

O

C OR C

O

R

Η+

∆Grignard ReagentsGrignard reactions involve the attack of an alkyl halide towards acarbonyl group. Grignard reactions produces alcohols fromaldehydes and ketones.

Example 24.14. Predict the major product A and B in thefollowing Grignard reaction.

+ A BC 3H C

O

H C 3H MgBrether H+

H2O

Addition of a Grignard reagent across a carbonyl bond of analdehyde results in the formation of a secondary alcohol becauseGrignard reagents tend to follow the general scheme fornucleophilic addition across a carbon oxygen double bond.

C 3H C

O

H+H+

H2OC 3H MgBr

etherC 3H C C 3H

O MgBr

H

+C 3H C C 3H

H

OH

MgBrHO

Optical Activity and Optical IsomersOptical isomers exhibit identical chemical and physical propertiesbut differ from each other in the direction that they rotate planepolarized light. The criteria for recognition of optical isomers isthe compound is nonsuperimposable upon its mirror image.Optical isomers that are nonsuperimposable mirror images of eachother are referred to as enantiomers. Below are a pair ofenantiomers.

Chapter 24

A

BD

EE

DB

Amirror

In about 99 % of the encountered cases for optical isomers, we findthat carbon has four different groups bonded to it. A carbon atomthat has four different groups bonded to it is often referred to asbeing a chiral or an asymmetric center. Carbon atoms with twoor more identical groups bonded to it are referred to as beingachiral or symmetric centers. Only chiral or asymmetric centersare associated with optical activity; the rotation of plane polarizedlight.

Example 24.15. Indicate the chiral centers in each moleculebelow.

Br

C 3H C

Cl

H

Br

C C

OHOH

OH

H H

Cl

C C

OHOH

C C 3H

OH

C 3H

Cl Cl

a) b) c)

a) The starred carbon atom is a chiral center. It has four differentgroups bonded to it and has a nonsuperimposable mirror image.

Br

C 3H C

Cl

H

Br

C C

OHOH

OH

H H

Cl

C C

OHOH

C C 3H

OH

C 3H

Cl Cl

a) b) c)* * * *

b) The starred carbon atom is a chiral center. It has four differentgroups bonded to it . This compound is optically active.c) Compound c has two chiral centers which are indicated by thestarred carbon atoms. The central carbon atom is not a chiralcenter because it has two identical groups bonded to it; that is,"CH3-C(OH)Cl-".

Chapter 25 Polymers: Synthetic and Natural

Polymers are large molecules formed from small molecules.Polymers can be long straight chains or can be highlybranched. Examples of synthetic polymers include nylon,polyester and polypropylene. Examples of natural polymersinclude proteins, carbohydrates and deoxyribonucleic acids.

Important Terms❏ polymer ❏ linear polymer❏ branched polymer ❏ crosslinked polymer❏ addition polymer ❏ condensation polymer❏ thermoplastic ❏ proteins❏ thermoplastic ❏ amino acids❏ peptide ❏ carbohydrates❏ monosaccharides ❏ disaccharides

Addition PolymersAddition polymers are formed by addition of smallmolecules called monomers into long chains. There is little orno crosslinking between chains. These polymers can floweasily when heated and can be molded into a variety of shapes;thus we use the term thermoplastic to describe thesematerials. Addition polymers can be recognized by therepeating unit which always has the same formula as themonomer from which the polymer is formed. An example isthe monomer ethylene CH2=CH2, which when it forms thepolymer polyethylene uses -(CH2CH2)- as the repeating unit.

Addition polymers can be made from free-radicalpolymerization, ionic polymerization and coordinationpolymerization. Free-radical polymerization proceeds by achain-reaction mechanism. Ionic polymerization proceedsthrough ionic intermediates. Coordination polymerizationproceeds through a transition metal catalyst.

Chapter 25 277

Example 25.1. Given the following monomers, write aplausible structure for the polymer:a) CH=CH2 b) CH=CH

| | |Cl Cl CH3

a) -CH-CH2-CH-CH2- b) -CH-CH-CH-CH-| | | | | |

Cl Cl Cl CH3 Cl CH3

Condensation PolymersWhen condensation polymers are formed, a small moleculesuch as water condenses out. The repeating unit in acondensation polymer is smaller than the monomer. Onceformed, condensation polymers contain extensivecrosslinking and attempts to change the shape of this polymerwill be unsuccessful. These materials are called thermosetplastics.

Peptides and ProteinsProteins are formed from the polymerization of monomerscalled amino acids. Amino acids contain both an amino (-NH2) group and a carboxylic acid (-CO2H) group. There are 20common amino acids used to synthesize proteins. Yourtextbook will provide the structures of these amino acids.

Proteins are formed by the reaction of the -CO2H end of oneamino acid with the -NH2 end of another to form an amide.This -CONH- amide bond is known as a peptide bond. Asmore and more amino acids link together, a protein is formed.

By convention, proteins are listed from the N-terminal aminoacid to the C-terminal amino acid end. The order in whichamino acids are linked together gives rise to the specificcharacteristics of the protein. When the amino acid ofaspartene is linked to phenylalanine, in the ASP-PHE orderthe artificial sweetener aspartame if produced. If these

278 Polymers: Synthetic and Natural

amino acids are linked in the PHE-ASP order the peptideformed is not sweet.

Example 25.2. Draw the structure for the dipeptide GLY-TYR

From your textbook, locate the structures of GLY and TYR.

Draw the structures of GLY and TYR. Link these two aminoacids by a peptide bond.

N

H

H

C C

OH

H

N

H

C

H

C HH

OH

C

O

OH

25.11 The Structure of ProteinsProtein structures can be described on several differentlevels. The primary structure is the sequence of amino acidslisted one at time in order of appearance. The secondarystructure is how the protein folds back on itself, two of thesesecondary structures are α-helix and β-pleated sheets. Thetertiary structure of proteins is produced by the interaction ofthe amino acid side chains. In some proteins a quaternarystructure can be described between the individual polypeptideside chains.

Carbohydrates: The Disaccharides andPolysaccharidesCarbohydrates are the primary food source for living systems,plants and animals. Carbohydrates include simple sugars

Chapter 25 279

such as glucose, fructose and sucrose as well as polymers ofthese sugars such as starch, glycogen and cellulose. Threeimportant classes of carbohydrates include monosaccharides,disaccharides and polysaccharides.

Monosaccharides are divided into two groups the aldoses orketoses, depending upon if they contain an aldehyde orketone functional group.

The dissacharides are formed by condensing a pair ofmonosaccharides. Maltose, lactose and sucrose are examplesof disaccharides. Common polysaccharides include starchand glycogen. Poly- saccharides serve two primaryfunctions, that of storage of energy and as mechanicalstructure for cells.