High Resolution NMR SpectroscopyNumber 113

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ChemFactsheetwww.curriculum-press.co.uk

To succeed in this topic you need to understand:• the structure of atoms• the functional groups of the organic compounds studied at AS

and A2• that chemists need to analyse compounds (for example through

mass spectrometry and infra-red)

After working through this Factsheet you will be able to:• understand the importance of NMR as an analytical technique• predict and interpret high resolution proton NMR spectra

What is NMR?NMR stands for nuclear magnetic resonance. It is a powerfultechnique which gives us detailed information about the structureof compounds. NMR can be used to analyse a number of elementsin compounds provided they have an odd mass number; theseinclude 13C, 19F and 31P, but is most commonly used to studyhydrogen atoms, 1H. Remember that the nucleus of a hydrogenatom contains a single proton, so we often refer to hydrogen NMRas proton NMR. NMR is always used in conjunction with othertechniques, like mass spectrometry and infra-red, but is arguablythe most powerful method of analysis available to organic chemists.

How does NMR work?Hydrogen nuclei have a property called spin, which makes thembehave like tiny magnets. When you hold the nuclei in a magneticfield, they align themselves with the field, like compass needles allpointing north (Fig 1).

Fig 1 Hydrogen nuclei without and with a magnetic field applied

No magnetic field applied

Dir

ecti

on

of

app

lied

mag

netic

fiel

d

If exactly the right amount of energy is absorbed by the nuclei, theywill flip and point in the other direction (Fig 2). This is called nuclearmagnetic resonance.

Fig 2 Hydrogen nuclei absorb energy and their spin flips

Dir

ecti

on

of

app

lied

mag

netic

fiel

d

The exact amount of energy required depends on the chemicalenvironment of the hydrogen nuclei. Nuclei with a high electrondensity surrounding them are shielded from the effects of themagnetic field, so require more energy, while nuclei near to anelectron-withdrawing group like chlorine or oxygen are deshieldedand need less energy. The NMR machine detects the amount ofenergy needed and scans through a range of energies until all thehydrogen nuclei have flipped.

How is an NMR carried out?To carry out an NMR spectrum, the sample is first dissolved in asolvent. The solvent must not contain any hydrogen atoms, sincethese would interfere with the resonance of the sample. The solventsnormally used are tetrachloromethane (CCl

4), or deuterated solvents

like CDCl3, where the hydrogen atoms have been replaced by

deuterium atoms (the 2H isotope of hydrogen), which do notinterfere with the magnetic resonance.Spectra are calibrated usingtetramethylsilane (TMS) (Fig 3).

Fig 3 Tetramethylsilane

CH3

Si

CH3

CH3H

3C

All the hydrogen atoms in TMS are in the same, highly shielded,environment and so will produce a single peak in the NMR spectrum.The horizontal axis of the spectrum is called the chemical shift (δ),which is measured in parts per million (ppm). It indicates thedifference in absorbed energy between the TMS protons and thesample protons. The peak for TMS at 0 ppm does not always appearin the spectra, since machines are often internally calibrated, but ifthere is a peak at 0 ppm, it will be due to the TMS protons.

What does an NMR spectrum tell us?Fig 4 shows the proton NMR spectrum of 1,2-dichloro-2-methylpropane. In the spectrum there are two peaks, at 1.6 and 3.7ppm. There are four pieces of information we can gather from thesepeaks.

Fig 4 The proton NMR spectrum of 1,2-dichloro-2-methylpropane

δ (ppm)

C HH

H

C CC

H H

H H

HClCl

10 9 8 7 6 5 4 3 2 1 0

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113 High Resolution NMR Spectroscopy

1. The number of different chemical environments in the compoundFirstly, the NMR shows us that there are two different chemicalenvironments for hydrogen atoms in this compound (Figure 5).The six hydrogens in the methyl (CH

3) groups are in the same

environment since they are the same distance from the chlorineatoms and from the central carbon atom. The two hydrogens inEnvironment 2 are closer to one of the chlorine atoms andtherefore chemically different from Environment 1.

Fig 5 The two hydrogen environments in 1,2-dichloro-2-methylpropane

2. The relative number of hydrogens in each environmentThe second piece of information comes from the integrationcurve, the line above each peak. The height of this linecorresponds to the area under the curve and therefore tells usthe relative numbers of hydrogen atoms in the differentenvironments. If you measure the height of the two integrationcurves in Fig 4 you will find that they are in a ratio of 1:3. Thismatches the structure of the compound, with two hydrogennuclei in Environment 2 and six in Environment 1.

3. The electron density around the hydrogen atomsThe third piece of information is about the nature of eachenvironment. This information is given by the chemical shifts(Table 1). In the exam, you will be given the chemical shifts thatyou need, so don’t try to memorise them.Looking at Fig 4 again, the peaks at 1.6 and 3.7 ppm correspondrespectively to the methyl hydrogens (Environment 1) and theCH

2Cl hydrogens (Environment 2), which are deshielded by the

closer electron-withdrawing chlorine atom.

Table 1 Typical chemical shifts of a range of hydrogen nuclei

Chemical shift (ppm)Proton

R-CH3

0.8-1.2

R2-CH

21.1-1.5

R3-CH 1.5

R3-OH 1.0-6.0

Note that the OH hydrogenatom can appear almostanywhere along the chemicalshift scale

C

O

CH2RR

1.0-6.0

R-CH2-Cl 2.5-4.3

R-CH2-OH 3.3-4.0

4.4-6.9C CH

RR

R

9.2-9.8

C

O

OHR

C HH

H

C CC

H H

H H

HClCl

environment 1

environment 2

C

O

HR

10.0-12.0

4. How many neighbours each hydrogen atom hasFinally, the spectrum gives us information about theneighbouring hydrogen atoms of the nuclei. This comes fromthe splitting pattern, where the peaks in the spectrum are splitinto groups of sub-peaks (Table 2). The splitting pattern dependson how many non-equivalent hydrogen nuclei are on theadjacent carbon atoms. Splitting occurs because the magneticfields of different hydrogen nuclei interact with each other, andthe number of sub-peaks is always one more (n+1) than thenumber of hydrogens (n) on the adjacent carbon atom. Youwon’t need to explain how this happens, but you do need tounderstand its effects.

Table 2 Splitting patterns

0 Singlet R3CCH

1 Doublet R2CH-CH

2 Triplet RCH2-CH

3 Quartet CH3-CH

Number ofhydrogen atomson the adjacentcarbon atom (n)

Splittingpattern(n+1)

How it looks Typicalstructure

In the NMR of 1,2-dichloro-2-methylpropane (Fig 4), both peaks aresinglets. This means that none of the hydrogens are adjacent tonon-equivalent hydrogen nuclei (Fig 6).

Fig 6 No adjacent hydrogen atoms to either environment

C HH

H

C CC

H H

H H

HClCl

environment 1

environment 2

This carbon atom is adjacent to both environments. It has no hydrogenatoms so both peaks are singlets

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113 High Resolution NMR Spectroscopy

Fig 7 shows the proton NMR spectrum of ethyl ethanoate. Thethree peaks show that there are three different environments for thehydrogen atoms. The integration curves give us a ratio of 2:1:3,and analysis of the splitting patterns (Table 3) leads to a fullassignment of peaks (Fig 8).

Fig 7 The proton NMR spectrum of ethyl ethanoate

11 10 9 8 7 6 5 4 3 2 1 0

H

δ (ppm)

Table 3 Assigning the peaks in the NMR of ethyl ethanoate

Chemical Relative Splitting Number of Assignedshift number pattern adjacent to proton

of hydrogen hydrogens groupatoms

4.1 2 Quartet 3 b

2.0 3 Singlet 0 a

1.2 3 Triplet 2 c

C

H

H

H

C

O

O C C H

HH

HH

C

H

H

H

C

O

O C C H

HH

HH

Fig 8 Assigned peaks in ethyl ethanoate

cba

Worked example1. Draw the displayed formula of chloroethane.2. State the number of different hydrogen environments found

in chloroethane.3. Predict approximate chemical shifts for the different

environments.4. Predict the splitting pattern of the NMR peaks.5. Sketch the NMR spectrum of chloroethane, including

integration curves to show the relative numbers of hydrogenatoms in each environment.

AnswerThe displayed formula and NMR spectrum of chloroethane areshown in Fig 9. There are two hydrogen environments: the methylhydrogens (a) and the CH

2Cl hydrogens (b). Using Table 1 the

approximate chemical shifts are 1.1-1.5 and 2.5-4.3 ppm respectively.The peak at about 1.4 ppm (corresponding to the CH

3) is split into a

triplet by the neighbouring CH2Cl hydrogens, while the peak at 3.5

ppm (corresponding to the CH2Cl) is split into a quartet by the

methyl group.

Fig 9 Chloroethane and its NMR spectrum

δ (ppm)10 9 8 7 6 5 4 3 2 1 0

H C

H

H

H

C Cl

H

Hb

a

-CH2Cl(b)

-CH3(a)

An unusual case: Alcohol OH groupsThe hydrogen atom in alcohol OH groups is unusual in two ways.Firstly, as it says in Table 1, this proton can have almost any chemicalshift. The actual shift depends on, among other things, the solventthat the compound is dissolved in. Secondly, at room temperatureOH proton peaks do not get split like others, no matter how manyadjacent hydrogen nuclei there are. This helps you, therefore, toidentify them in the spectra since they are always singlet peaks.

Practice Questions1. For each of the following compounds, draw the displayed

formula, label the different hydrogen environments and predictapproximate chemical shifts for each peak in the proton NMRspectrum:(a) 2-chloropropane(b) 1-chloropropane(c) ethanol(d) ethanoic acid(e) 2-chloropentane(f) 3-chloropentane

2. Sketch the proton NMR spectra you would expect for thefollowing compounds:(a) ethanoic acid(b) propanone(c) ethanal(d) ethanol

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113 High Resolution NMR Spectroscopy

Answers

Question 1(a)

C

H

Cl

CH3

H3C

Environment 2

Environment 1

Environment 1 = 2.5-4.3 ppmEnvironment 2 = 0.8-1.2 ppm

(b)H

H3C

H

C

H

H

C Cl

1

2 3

Environment 1 = 0.8-1.2 ppmEnvironment 2 = 1.1-1.5 ppmEnvironment 3 = 2.5-4.3 ppm

(c)H

C

H

C

2 3

H

C

H

O

H

H

H

1

Environment 1 = 0.8-1.2 ppmEnvironment 2 = 3.3-4.0 ppmEnvironment 3 = 1.0-6.0 ppm

(d) O

C

H

C

2

H

C

H

OH

H

1

Environment 1 = 2.0-3.0 ppmEnvironment 2 = 10.0-12.0 ppm

(e)

Cl

H3C

H

C

H

H

C

1

H

H

C CH3

2 3 4

5

Environment 1 = 0.8-1.2 ppmEnvironment 2 = 2.5-4.3 ppmEnvironment 3 = 1.1-1.5 ppmEnvironment 4 = 1.1-1.5 ppmEnvironment 5 = 0.8-1.2 ppm

(f)H

H3C

H

C

Cl

H

C

1

H

H

C CH3

2 3 2

1

Environment 1 = 0.8-1.2 ppmEnvironment 2 = 1.1-1.5 ppmEnvironment 3 = 2.5-4.3 ppmNote that this molecule issymmetrical so has only threedifferent hydrogen environments

Question 2

(a)

C

O

OHH3C

12 10 8 6 4 2 0 -2

δ (ppm)

(b)

C

O

CH3

H3C

10 9 8 7 6 5 4 3 2 1 0

δ (ppm)

(c)

C

O

HH3C

10 9 8 7 6 5 4 3 2 1 0

δ (ppm)

(d)

Acknowledgements: This Factsheet was researched and written by Emily Perry.Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF11NU. ChemistryFactsheets may be copied free of charge by teaching staff orstudents, provided that their school is a registered subscriber. No part of theseFactsheets may be reproduced, stored in a retrieval system, or transmitted, inany other form or by any other means, without the prior permission of thepublisher. ISSN 1351-5136

11 10 9 8 7 6 5 4 3 2 1 0δ (ppm)

CC CH

H

H H

H

HO