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INTRODUCTION TO STRUCTURE DETERMINATION

There are so many organic compounds in existence that it can be difficult to establishexactly what compounds, or what combinations of compounds, are present in a given

sample. Chemical tests can be used to distinguish between different functional groups,

and melting point and boiling point data can provide some further information as to theidentity and purity of a compound, but they can generally only confirm suspected

structures and can not usually be used to identify new compounds. Chemical techniques

also require fairly large quantities of sample and are not always effective at distinguishingbetween similar compounds.

Chemists have developed a number of other, more precise, analytical techniques to

determine the exact structure of organic compounds. Knowledge of three of thesetechniques is required for AQA Alevel Chemistry. These are mass spectrometry, infra-red spectroscopyand nuclear magnetic resonance nmr! spectroscopy.

As infrared spectroscopy is covered at A!level, this topic is concerned with a moredetailed consideration of mass spectrometry and nmr spectroscopy.

At Alevel, these analytical techniques must be used to determine the structure of organic

molecules containing carbon, hydrogen and oxygen only and generally containing no

more than six carbon atoms. The molecule being analysed will be an al"ane, al"ene,alcohol, ether, carbonyl, carboxylic acid or ester.

#sually, information is given which enables the empirical formula to be deduced before

the analysis begins. This is usually given in the form of composition by mass$

%g A compound contains &'.'( carbon, )*.*( oxygen and +.( hydrogen.

-ole ratio$ C &'/0 1 *.*

2 +./ 1 +.

3 )*.*/+ 1 *.*

!implest whole number ratio$C *.**.* 1 /

2 +.*.* 1 0

3 *.**.* 1 /

!o empirical formula 1 C203

The molecular formula of the molecule, and its structure, can then be deduced from one

or more of the analytical techniques.

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MASS S"ECTROMETR#

a4 5ntroduction

The technique of mass spectrometry was used at A!level to determine the relative

abundances of different isotopes in a sample of an element, to deduce relative atomicmasses and to deduce molecular formulae.

-ass spectrometry can also be used to determine the structure of organic molecules. 5nsome respects molecules behave in a similar way to atoms in a mass spectrometer, but

there are important differences$

6hen a molecule is ionised, one electron is removed. As molecules generally containpaired electrons only, the result is the formation of a species with an unpaired electron, or

a free radical.

%g ethane$

C CH H

HH

HH

xoxxxx

o

ooo

x x

xo C CH H

HH

HH

xoxxx

x

o

ooo

x x

xo +

or C02+7C02+89:9 e

The resulting species is thus a positi$ely c%arged ionand also a free radical. 5t is "nown

as the molecular ion;or parent ion4.

Two possible things can happen to a molecular ion when it is formed$

it can pass intact through the mass spectrometer and onto the detector, being

detected as having an m< ratio of which < 1 / and m 1 relative mass of

molecular ion.

it can brea" up into two smaller, more stable species, one of which is a

positively charged ion and one is a free radical. This is "nown as

fragmentation. This will result in the detection of species with m< ratioswhich are less than that of the molecular ion.

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The result is that a number of different pea"s are seen in the mass spectrum of an organic

molecule$

the pea" with the largest m< ratio corresponds to the molecular ion, and this

m< ratio corresponds to the relative molecular mass of the molecule.

the pea"s with smaller m< ratios result from fragmentation of the molecular

ion. These pea"s can be used to deduce more information regarding thestructure of the molecule, because different molecules fragment in different

ways and some fragments are more stable than others.

%g mass spectrum of ethanol$

The mass spectrum of ethanol contains several pea"s$

the largest m< ratio in the mass spectrum is &+. This is therefore the

molecular ion pea" which means that the molecule has a relative molecularmass of &+.

the other pea"s with smaller m< ratios result from fragmentation of the

ethanol molecule. The most abundant fragment ions appear to have relative

masses of &) and */, and there are less abundant fragment ions with masses of0 and 0=

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&! fragmentation

>uring fragmentation, the free radical molecular ion is bro"en up. As the molecular ion

contains an odd number of electrons, one of the species into which it fragments will

contain an even number of electrons ;this will be a positive ion4 and the other will containan odd number of electrons ;this will be a free radical4.

%g consider the case of ethane$

+CCH H

HH

HH

xoxxx

o

ooo

x x

xo CH

H

H

ox

xx

oo

xC H

H

H

xo

ox

xo

This is the free radical molecular

ion

This species "eeps

the unpaired electronand is an unchargedfree radical

;not detected in massspectrometer4

This species loses the

electron and is a positivelycharged ion

;detected in massspectrometer4

This process can be represented by the equation$

7C02+89:7C2*8

:9 7C2*89

3nly the charged species is detected as the neutral free radical is neither accelerated nordeflected.

The process of fragmentation can thus be represented by the following general equation$

R'( ?99 @:

9:is the molecular ion and is detected ;assuming not all of them fragment4?9is the fragment ion and is detected

@:is the fragment radical and is not detected

c! sta&le ions

Bot all possible fragments are detected some of the fragments form stable ions and

these are more li"ely to be formed.

The most stable cations are carbocations ;eg C2*9, C2*C20

9and C2*C2C2*94 and

acylium ions ;eg 2C39, C2*C39, C2*C20C3

94. These are thus the fragments most li"elyto be detected, and will give the most intense pea"s in a mass spectrum.

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The main pea"s in the mass spectrum of an organic compound will thus be the molecular

ion pea" and the pea"s corresponding to the most stable ions which can be formed from

fragmentation of the molecular ion.

%g butanone$ C2*C20C3C2*

%xpected in the mass spectrum of butanone would be$

m< ratio 5on responsible %quation to show formation of ion

0 7C2*C20C3C2*89:

/) 7C2*89 7C2*C20C3C2*8

9:7C2*899 7C20C3C2*8

:

0= 7C2*C2089 7C2*C20C3C2*8

9:7C2*C20899 7C3C2*8

:

) 7C2*C20C389 7C2*C20C3C2*8

9:7C2*C20C3899 7C2*8

:

&* 7C3C2*89 7C2*C20C3C2*8

9:7C3C2*899 7C2*C208

:

The real mass spectrum of butanone is shown below$

The most abundant pea" is &*, followed by 0 and 0=. The pea"s at /) and ) can also be

accounted for. ;5t is rarely possible to account for all the pea"s in a mass spectrum4

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d! using mass spectra to determine structure

The molecular ion pea" enables the relative molecular mass of the compound to be

deduced.

The other pea"s give the masses, and hence the possible identities, of the fragments.

These gives important clues to the structure of the unfragmented molecule.

%g A pea" at 0= suggests the presence of C2*C209

A pea" at &* suggests the presence of C2*C20C209, C2*C2C2*

9or C2*C39

A pea" at ) suggests the presence of C2*C20C39or C2*C3C20

9

The presence ;or absence4 of these fragments gives important information as to whichstructural features are present ;or not present4 in the molecule. This is particularly useful

for distinguishing between isomers$

%g C&2/' butane gives pea"s at /), 0=, &* and )Dmethylpropane gives pea"s at /), &* and )D only ;no pea" at 0=4

%g C*2+3 propanal gives pea"s at /), 0=, &* and )D

propanone gives pea"s at /), &* and )D only ;no pea" at 0=4

%g C02+3 ethanol gives pea"s at /), 0= and &+

-ethoxymethane gives pea"s at /) and &+ ;no pea" at 0=4

Thus the presence ;or absence4 of specific fragments enables the structure to bedetermined.

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N)M)R S"ECTROSCO"#

a! Introduction

!ome nuclei have magnetic properties. 5f these nuclei are subEected to a strong magnetic

field, they can either align themselves in the same direction as the magnetic field ;a lowenergy state4 or in the opposite direction ;a highenergy state4. 6hen these nuclei are then

subEected to radio waves with a range of frequencies, each nucleus can absorb the

frequency which corresponds to the difference in energy between its lowenergy state andits highenergy state and switch from one to the other. The absorption can be detected and

converted into a spectrum. The frequency of the radiation absorbed by the nuclei varies

depending on the type of nucleus and the arrangement of electrons around that nucleus,

and can thus be used to provide important structural information about the molecule. Thistechnique for structure determination is "nown as nuclear magnetic resonance ;nmr4

spectroscopy.

There are a number of different types of nmr spectroscopy, depending on the type ofnucleus being investigated. Two techniques are required at Alevel$

proton nmr spectroscopyinvestigates the absorption of radiation by nuclei of

hydrogen atoms ;/24 and is thus a technique for obtaining information about

the number and arrangement of hydrogen atoms in a molecule car&on-*+ nmr spectroscopyinvestigates the absorption of radiation by

nuclei of carbon/* atoms ;/*C4 and is thus a technique for obtaining

information about the number and arrangement of carbon atoms in a molecule

&! proton nmr spectroscopy

A number of important pieces of information can be deduced from analysis of a protonnmr spectrum$

5dentical hydrogen atoms all absorb radiation at the same frequency and socontribute to the same pea". The number of different pea"s, therefore, tells

you the num&er of different en$ironments of %ydrogen atomsin the

molecule.

The area under each pea" is related to the intensity of the absorption and gives

information about the num&er of %ydrogen atoms of t%at typein the

molecule. The relative intensity of the pea"s is "nown as the integration

factorand is generally given as a simplest whole number ratio ;as it is not

always easy to see from the spectrum4

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The frequency of the radiation absorbed depends on the environment of the

hydrogen atom and gives information about the position of t%ose %ydrogen

atoms in t%e molecule relati$e to ot%er car&on atoms and functional

groups. The actual frequency of the absorption is difficult to measureF instead

the frequency is measured relative to a standard. 6hat is measured is thedifference between the frequency absorbed by the sample atoms and the

frequency absorbed by the standard, expressed as a fraction of the frequency

absorbed by the standard. This fraction is "nown as the c%emical s%iftand isgiven the symbol G.

G 1 f;sample4 f;standard4

f;standard4

The value of G is generally very smallF typically //' x /'+. 5t is thereforeexpressed in parts per million ie 0.' x /'+is equivalent to 0.' ppm. The

chemical shift is generally greater if the hydrogen atoms are closer to

electronegative atoms, but is also relatively high for hydrogen atoms close to

al"ene and arene groups. 2ydrogen atoms with a large chemical shift are saidto be des%ielded, and hydrogen atoms with a low chemical shift are said to be

s%ielded. The pea" resulting from the standard always has ;by definition4 achemical shift of

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A typical proton nmr spectrum loo"s li"e this$

Irom this spectrum, it is clear that there are four pea"s. The chemical shift can be read

from the hori

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c! car&on-*+ nmr spectroscopy

Carbon/* nmr spectroscopy wor"s in a similar way to proton nmr spectroscopy, but it ismuch simpler$

5dentical carbon atoms all absorb radiation at the same frequency and socontribute to the same pea". The number of different pea"s, therefore, tells

you the num&er of different en$ironments of car&on atomsin the molecule.

The area under the pea"s does not correspond exactly to the intensity of the

absorptions, so it is not possible to conclude the number of carbon atoms

present in each environment.

The frequency of the radiation absorbed depends on the environment of the

carbon atom and gives information about the position of those carbon atoms in

the molecule relative to other atoms and functional groups. This is measured

in the same way as in proton nmr spectra, using c%emical s%ift.

Jea"s in the carbon/* nmr spectrum are not split, so it is not possible todeduce information about the number of adEacent carbon atoms in each carbon

environment

A typical carbon/* nmr spectrum loo"s li"e this$

http://en.wikipedia.org/wiki/File:Vanillin_13-C_NMR_spectrum.PNG

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Irom this spectrum, it is clear that there are eight pea"s. This means that there are eight

different types of carbon atom in the molecule, each with a different chemical shift.

c! "reparing samples for proton nmr analysis

efore a spectrum can be analysed, it is necessary to produce a spectrum and preparing asample for proton nmr analysis requires some consideration$

i4 Choosing a suitable solvent

5t is necessary to dissolve the sample in a solvent before it can be analysed. The problem

with most solvents is that they themselves contain hydrogen atoms which absorb

radiation, interfering with the spectrum produced by the sample. 5t is therefore necessaryto use a solvent which contains no hydrogen atoms. The usual choices are CCl&and

C>Cl*. > is the symbol for deuteriumF this is an isotope of hydrogen containing one

neutron in its nucleus. 5t has no magnetic properties and does not interfere with a proton

nmr spectrum.

ii4 Choosing a standard

5n addition to the solvent, a standard ;or reference4 molecule must be added to calibrate

the spectrum. The chemical shift is then measured relative to this standard as describedabove. The standard normally used is tetramethylsilane ;T.-.!.4 which has the following

structure$

Si

C

C

C

C

H

H

H

HH

H

H

H

H

H

H

H

This substance has a number of advantages as a standard$

5t produces a single, singlet pea" which is very intense ;there are /0 2 atoms,all identical4 which ma"es the pea" easy to identify.

The 2 atoms are highly shielded ;as the !i is an electropositive atom,releasing electron density onto the 2 atoms4 so the pea" is generally at a

significantly lower frequency than that found in most organic molecules,

meaning that it does not interfere with the other pea"s and can be easilydistinguished from them.

5t is cheap and nontoxic.

The large singlet which arises in proton nmr spectra at G 1 ' as a result of the T-!

standard is usually erased from the spectra before it is produced ;so you never see it4.

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d! Using proton nmr spectra in structure determination

Jroton nmr spectra are very useful for wor"ing out the structure of organic molecules,particularly if the molecular formula is "nown. The information from a spectrum can be

bro"en down into four parts$

i4 the number of pea"s

%ach pea" corresponds to one set of identical hydrogen atoms. The number of pea"stherefore gives the number of types of hydrogen atom present.

5n some molecules, all the hydrogen atoms are identical. These molecules give only one

pea" in the spectrum.%g ethane, propanone, methoxymethane

C

H

H

C

H

H

H H H C

H

H

C

O

HC

H

H

C C

H

H

H

H

H

H O

5n some molecules, there are two types of hydrogen atom. These molecules will give two

pea"s in a proton nmr spectrum.

%g butane, pentan*one, ethanal, methanol

C C

H

H

C

H

H

H C

H

H

H

H

H

C

O

H

CH3

C OH

H

H

H

-ost molecules, however, give more than two pea"s$

utanone gives three pea"s$

HC C

H

H

C

H

H

H C

H

H

O

these H atomsare identical these H atoms

are identical

these H atomsare identical

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utanal gives four pea"s$ %thanol gives three pea"s$

H

H H

C

H

H

CC

H O

HCH

H H

HH

CC OHH

%ach of the 2 atoms within the same circle are identical and contribute towards the same

pea".

-ethylpropanal gives three pea"s$

H

H

H

C

H

H

CCH

O

H

C

H

these six H atomsare all identical

this H atomis unique

this H atomis unique

ii4 integration factor

The integration factor indicates the number of identical hydrogen atoms corresponding to

that pea". -olecules which give the same number of pea"s can be distinguished because

the integration factor of each pea" is different, corresponding to a different distribution of2 atoms in the molecule.

%g. -ethylpropanal and butanone ;C&2D34 both give three pea"s in their proton nmrspectra$

H

H

H

C

H

H

CCH

O

H

C

H

these six H atomsare all identical

HC C

H

H

C

H

H

H C

H

H

O

2owever the integration factors of the three pea"s in methylpropanal will be +$/$/, but inbutanone they will be *$*$0. The two molecules can thus be distinguished by their proton

nmr spectra because the integration factors of the pea"s are different.

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iii4 chemical shift

The chemical shift depends on the environment around the 2 atomF in particular it isrelated to the proximity of electronegative atoms. The nearer a hydrogen atom is to an

electronegative atom, the more deshielded it will be and the greater the chemical shift.

Ior example, 2 atoms in al"anes tend to have low chemical shifts ;G1 ' 04, but 2 atomsattached to 3 atoms in carboxylic acids tend to have high chemical shifts ;1 /'/04.

All 2 atoms not within one carbon atom of a functional group will have a chemical shiftbetween ' and 0 ppm.

The following table summarises the important chemical shifts for 2 atoms in common

environments close to a functional group$

environment Type of molecule Chemical shiftppm

2CC13 carbonyl 0.' 0.)

2C3 alcohol or ether *.* &.'32 alcohol '.) ).'

2C1C al"ene &.+ ).=

2C13 aldehyde = /'

32 acid /' /0

The chemical shift data provides useful information in deducing the environments

responsible for a particular pea" and identifying functional groups$

Consider the proton nmr spectra of two un"nown compounds$

/.

The pea" at G 1 =. suggests an 2 atom in an aldehyde group.

The pea" at G 1 0.) suggests 2 atoms on a C adEacent to a carbonyl groupThe pea" at G 1 /./ suggests 2 atoms on a C not adEacent to a functional group

5ntegration factors ;/ for G 1 =., 0 for G 1 0.), * for G 1 /./4 would suggest propanal$

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H

H

H

C

H

CCH

O

H

9.7 (1 H)

2.5 (2 H)

1.1 (3H)

0.

The pea" at G 1 *.+ suggests an 2 atom on a C adEacent to a C3 bond

The pea" at G 1 0.* suggests 2 atoms on a C adEacent to a carbonyl group, but as this is

inconsistent with the structure it could also suggest an 2 atom bonded to 3 in an alcoholThe pea" at G 1 /.+ suggests 2 atoms on a C not adEacent to a functional group

The pea" at G 1 '.= suggests 2 atoms on a C not adEacent to a functional group

5ntegration factors ;0 for G 1 *.+, / for G 1 0.*, 0 for G 1 /.+ and * for G 1 '.=4 would

suggest propan/ol$

C C C OH

HHH

H

HHH

2.3 1H

3.7 2H

1.! 2H".9 3H

Chemical shift data is thus useful for identifying functional groups and for deducing the

position of other 2 atoms on the molecule.

iv4 coupling

As was discussed earlier, the absorption of radiation by hydrogen atoms is affected by the

presence of hydrogen atoms on adEacent carbon atoms ;provided they are not involved in

hydrogen bonding4. These hydrogen atoms on adEacent carbon atoms cause a splitting ofthe pea" according to the ;n9/4 rule as described earlier.

This is very useful for positioning the hydrogen atoms in different environments relative

to each other.

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Consider a C2* group. This will account for a pea" with integration factor *.

The splitting of this pea", however, will depend on the hydrogen atoms on the adEacentcarbon atom$

C C

HH

H

HH

2ere n 1 0 so a triplet will be observed

C C

OH

H

HH

or

C C

HH

H

HO

2ere n 1 / so a doublet will be observed

C C

CH

H

HO

or

C

H

H

H

O

2ere n 1 ' so a singlet will be observed

Consider a C20 group. This will account for a pea" with integration factor 0.

CC

H

H

H

H

OC

or

CCC

H

H

H

H O

C

2ere n 1 0 so a triplet will be observed

C C

HH

H

HH O

C

or

C C

HH

H

HH

O

2ere n 1 * so a quartet will be observed

Thus the splitting of the pea"s gives much useful information about the splitting patterns

in the molecule.

%g Consider the following proton nmr spectrum$

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The pea" at G 1 /.' is a triplet ;0 2 atoms on adEacent carbons4, with integration factor *.

this suggests that it is caused by a C2* group adEacent to a C20 groupThe pea" at G 1 0.& is a singlet ;' 2 atoms on adEacent carbons4, with integration factor *.

this suggests that it is caused by a C2* group attached to 3 or C13

but if it were attached to 3 it would have a chemical shift of *.* &.' so it is probably caused by a C2*group attached to C13

The pea" at G 1 0./ is a quartet ;* 2 atoms on adEacent carbons4, with integration factor 0.

this suggests that it is caused by a C20 group attached as C2*C20C13 orC2*C203

but if it were attached to 3 it would have a chemical shift of *.* &.'

so it is probably caused by C2*C20C13

!o the molecule is butanone$

HC C

H

H

C

H

H

H C

H

H

O

1." 3Hn # 2 soit is a tri$let

2.1 2Hn # 3 soit is a quartet

2.% 3Hn # " soit is a sin&let

2ydrogen bonded atoms do not contribute to coupling$

%g consider the proton nmr spectrum of ethanol$

C C OH

HH

H

HH

2.7 1H' sin&let

3.7 2H' quartet1.2 3H' tri$let

The pea" at G 1 /.0 is a triplet, with integration factor *. 5t is the C2*bonded to the

C20.

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The pea" at G 1 *. is a quartet. 5t is the C20 bonded to the C2*. 5t does not couple

with the 2 attached to the 3, even though it is on an adEacent carbon atom, because this 2atom is involved in hydrogen bonding.

The pea" at G 1 0. is a singlet, with integration factor /. 5t is the 2 bonded to the 3. 5tdoes not couple with the two C20 hydrogen atoms because it is involved in hydrogen

bonding.

Two important general points in particular should be noted$

5f a triplet with integration factor * and chemical shift ' 0 is present, as

well as a quartet with integration factor 0, then C2*C20 is almostcertainly present

!inglet pea"s with integration factor / strongly suggest that an 32 bond

is present

e! Using car&on-*+ nmr spectra in structure determination

Carbon/* nmr spectra are also useful for wor"ing out the structure of organic molecules,particularly if the molecular formula is "nown. 2owever they give less information about

a molecule than proton nmr spectra.

i4 the number of pea"s

%ach pea" corresponds to one set of identical carbon atoms. The number of pea"s

therefore gives the number of types of carbon atom present.

5n some molecules, all the carbon atoms are identical. These molecules give only one

pea" in the spectrum.%g ethane, methoxymethane, cyclohexane, ben

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ii4 chemical shift

The chemical shift depends on the environment around the carbon atomF in particular it isrelated to the proximity of electronegative atoms. The nearer a carbon atom is to an

electronegative atom, the more deshielded it will be and the greater the chemical shift.

Ior example, carbon atoms in al"anes tend to have low chemical shifts ;G 1 ' &'4, butcarbon atoms attached to 3 atoms with a double bond tend to have high chemical shifts

;G 1 /+' 00'4.

All carbon atoms not directly attached a functional group ;ie singly bonded to only

carbon or hydrogen atoms4 will have a chemical shift between ' and &' ppm4. The nearer

an electronegative atom, the higher the chemical shift

The following table summarises the important chemical shifts for 2 atoms in common

environments close to a functional group$

environment Type of molecule Chemical shiftppmC3 alcohol, ether, ester )' ='

C1C al"ene, arene =' /)'

C13 carboxylic acid,

ester

/+' /='

2C1C Aldehyde, "etone /=' 00'

%g consider the following carbon nmr spectrum of an alcohol with molecular formula

C&2/'3$

The spectrum gives three pea"s, so there must be three different carbon atoms. As thereare four carbon atoms in the molecule, two of them must be identical.

3f the possible isomers of C&2/'3$

utan/ol and butan0ol would give four pea"s ;all four C atoms are different4-ethylpropan/ol would give three pea"s ;two of the C atoms are identical4

-ethylpropan0ol would give two pea"s ;three of the C atoms are identical4

The spectrum is therefore most li"ely to come from methylpropan/ol.

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USIN T.E ANA/#TICA/ TEC.NI0UES TOET.ER

5f all the different spectra are available, then the most effective analytical strategy is touse them together. There are a number of useful steps to deducing the structure$

5f composition data is available$

/. #se the composition data to wor" out the empirical formula.

0. #se the mass spectrum to deduce the relative molecular mass, and hence wor" outthe molecular formula.

*. #se the infrared spectrum to chec" for the presence of C13 and 32 bonds, and

hence identify the functional group.

&. -a"e a list of the possible isomers consistent with the molecular formula andfunctional group.

). #se the number of pea"s in the proton nmr spectrum to deduce the number of

hydrogen environments. %liminate the isomers inconsistent with this number.

+. #se the integration factors in the proton nmr spectrum to wor" out the number ofhydrogen atoms in each environment. %liminate the isomers inconsistent with this

distribution.. Compare the splitting of the pea"s in the nmr spectrum with the expected splitting

patterns of the remaining possible isomers. %liminate the isomers which do not

give this splitting pattern.D. 5f necessary, compare the chemical shifts of the pea"s in the nmr spectrum with

the expected values. %liminate the isomers which are inconsistent with the

chemical shifts.

5f composition data is not available, the molecular formula must be found by a trial and

error method.

/. >educe the relative molecular mass from the mass spectrum.

0. >educe the functional groups present from the infrared spectrum and hence

establish the li"ely number of oxygen atoms present.*. Add up the integration factors in the proton nmr spectra. The number of hydrogen

atoms present is an integral multiple of this number.

&. #se the relative molecular mass and the information on hydrogen and oxygen

atoms to deduce the molecular formula.

A wor"ed example of combined analysis is given on the following page ;no composition

data available4.

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#se the following spectra to deduce the structure of A$

-ass spectrum$ Jroton nmr spectrum$

-olecular ion pea" 1 DD three pea"s$-ost intense fragments$ &*, 0=, &), +/, ' G 1 /.*, triplet, integration factor *

G 1 0.', singlet, integration factor *

G 1 &./, quartet, integration factor 0

5nfrared spectrum$ sharp absorption at //) cm/, no broad absorptions between /)''and *)'' cm/

ANS1ER2

Irom mass spectrum, rmm 1 DDIrom infrared spectrum, C13 present

Irom proton nmr spectrum, sum of integration factors 1 D

!o D, /+ or 0& hydrogen atoms present and at least one oxygen5f one 3 atom present, remaining mass 1 +0 and cannot ma"e this using D/+0& 2 atoms

5f two 3 atoms present, remaining mass 1 )+F can ma"e this from & carbons and 2

hydrogens!o li"ely molecular formula 1 C&2D30.

Irom infrared spectrum, no 32 is present, so it is not a carboxylic acid.

!o is probably an ester.Jossible structures$

-ethyl propanoate, ethyl ethanoate, propyl methanoate, methylethyl methanoate

Irom proton nmr spectrum$

There are * hydrogen environments so it cannot be propyl methanoate ;which has &4.

The integration factors are *$*$0 so cannot be methylethyl methanoate ;which has the

ratio +$/$/4.The coupling ;*2 triplet, 02 quartet, *2 singlet4 is consistent with both remaining

possible structures.

ut the *2 singlet has a chemical shift of 0.', which is consistent with C2*C3 and not

C2*3 ;for which the chemical shift would be *.* &.'4

The 02 quartet has a chemical shift of &./, which is consistent with C2*C203 and notC2*C20C3 ;for which the chemical shift would be 0.' 0.)4

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!o it cannot by methyl propanoate and therefore the structure is ethyl ethanoate.

Iull explanation of spectra$

5nfrared spectrum$

CH3 C

O

O CH2 CH3

asors at 1715 cm)1

Jroton nmr spectrum$

CH3 C

O

O CH2 CH3

2."' 3H' sin&let

%.1' 2H' quartet

1.3' 3H' tri$let

-ass spectrum$

CH3 C

O

O CH2

CH3

stale carocations*ith m, # 15

stale carocations*ith m, # 29

stale ac-lium*ith m, # %3

7C2*C33C20C2*89:is the molecular ion pea"

7C2*C33C20C2*89:7C2*C38

99 73C20C2*8:accounts for the pea" at m< 1 &*

7C2*C33C20C2*89:7C2*C208

99 7C2*C338:accounts for the pea" at m< 1 0=

7C2*C33C20C2*89:7C2*8

99 7C2*C33C208:accounts for the pea" at m< 1 /)

7C2*C33C20C2*89:7C2*8

99 7C33C20C2*8:accounts for the pea" at m< 1 /)

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C.ROMATORA".#

Chromatography is a technique by which different compounds in a mixture can beseparated and then analysed. There are two main types of chromatography$

t%in-layer c%romatograp%y5n thinlayer chromatography a liquid solvent is allowed to flow up a piece of

chromatography paper or a TLC plate. The mixture is placed in a small area on

the paper and allowed to flow up the paper with the solvent. >ifferent substancesmove at different speeds, and the distance travelled by that substance compared to

the solvent can be used to identify the substance.

gas-li3uid c%romatograp%y5n gasliquid chromatography a gaseous mixture is allowed to flow through a

column lined with a solid. Holatile liquids can also be vaporised and then allowed

to flow through the column. >ifferent gases ta"e different amounts of time to flow

through the column, and this can be used to identify the gas ;or volatile liquid4

The mixture of substances moving through the column ;or up the plate4 is called the

mo&ile p%ase. The substance lining the column ;or the plate4 is "nown as the stationary

p%ase.

The speed at which each substance moves through the column or up the plate depends on

its relative solubility in the two phases.

5n gasliquid chromatography, a substance that is strongly attracted to the stationaryphase will move slowly through the column and ta"e a long time to pass through the

column. A substance which is not strongly attracted to the stationary phase will move

more quic"ly through the column and ta"e less time to pass through the column.