C hem F actsheet Number 113
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High Resolution NMR Spectroscopy To succeed in this to pic 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)
The exact amount of energy required depends on the chemical environment of the hydrogen nuclei. Nuclei with a high electron density surrounding them are shielded from the effects of the magnetic field, so require more energy, while nuclei near to an electron-withdrawing group like chlorine or oxygen are deshielded and need less energy. energy. The NMR machine machine detects the amount of energy needed and scans through a range of energies until all the hydrogen nuclei have flipped.
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
How is an NMR carried out? To carry out an NMR spectrum, the sample is first dissolved in a solvent. The solvent must must not contain any hydrogen atoms, since these would interfere with with the resonance of the sample. The solvents normally used are tetrachloromethane (CCl4), or deuterated solvents like CDCl 3, where the hydrogen atoms have been replaced by deuterium atoms (the 2H isotope of hydrogen), which do not interfere with the magnetic resonance.Spectra are calibrated using tetramethylsilane (TMS) (Fig 3).
What is NMR? NMR stands for nuclear magnetic resonance. It is a powerful technique which gives us detailed information about the structure of compounds. NMR can be used to analyse a number of elements in compounds provided they have an odd mass number; these include 13 C, 19 F and 31 P, but is most commonly used to study hydrogen atoms, 1H. Remember that the nucleus nucleus of a hydrogen hydrogen atom contains a single proton, so we often refer to hydrogen NMR as proton NMR. NMR. NMR is always always used in conjunction conjunction with other techniques, like mass spectrometry and infra-red, but is arguably the most powerful method of analysis available to organic chemists.
Fig 3 Tetramethylsilane CH 3 H 3C
How does NMR work?
CH 3
CH 3
Hydrogen nuclei have a property called spin, which makes them behave like tiny magnets. When you hold the nuclei in a magnetic field, they align themselves with the field, like compass needles all pointing north (Fig 1).
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 measured in parts per million (ppm). (ppm). It indicates the difference in absorbed energy between the TMS protons and the sample protons. The peak for TMS at 0 ppm does not always appear in the spectra, since machines are often internally calibrated, but if there is a peak at 0 ppm, it will be due to the TMS protons.
Fig 1 Hydrogen nuclei without and with a magnetic field applied
No magnetic field applied
Si
d e i l p p a f d o l e n i f o c i i t t c e e n g r a i D m
What does an NMR spectrum tell us? Fig 4 shows the proton NMR spectrum of 1,2-dichloro-2methylpropane. In the spectrum there are two peaks, at 1.6 and 3.7 ppm. There are four pieces of information we can gather from these peaks.
Fig 4 The proton NMR spectrum of 1,2-dichloro-2-methylpropan 1,2-dichloro-2-methylpropanee
If exactly the right amount of energy is absorbed by the nuclei, they will flip and point in the other direction (Fig 2). This is called nuclear magnetic resonance.
H H
Fig 2 Hydrogen nuclei absorb energy energy and their spin flips
C
H
H d e i l p p a f d o l e n i f o c i t t i c e e n g r a i D m
H
10
1
9
C
H C
C
Cl Cl
H
8
7
H
6
5
δ (ppm)
4
3
2
1
0
113 High Resolution NMR Spectroscopy
Chem Factsheet
1. The number of different chemical environments in the compound Firstly, the NMR shows us that there are two different chemical environments for hydrogen atoms in this compound (Figure 5). The six hydrogens in the methyl (CH3) groups are in the same environment since they are the same distance from the chlorine atoms and from the central carbon atom. The two hydrogens in Environment 2 are closer to one of the chlorine atoms and therefore chemically different from Environment 1.
4. How many neighbours each hydrogen atom has Finally, the spectrum gives us information about the neighbouring hydrogen atoms of the nuclei. This comes from the splitting pattern, where the peaks in the spectrum are split into groups of sub-peaks (Table 2). The splitting pattern depends on how many non-equivalent hydrogen nuclei are on the adjacent carbon atoms. Splitting occurs because the magnetic fields of different hydrogen nuclei interact with each other, and the number of sub-peaks is always one more (n+1) than the number of hydrogens (n) on the adjacent carbon atom. You won’t need to explain how this happens, but you do need to understand its effects.
Fig 5 The two hydrogen environments in 1,2-dichloro-2methylpropane H H environment 2
C
environment 1
Table 2 Splitting patterns
H
H H
H
C
C
C
Cl
Cl
H
H
Number of hydrogen atoms on the adjacent carbon atom (n)
2. The relative number of hydrogens in each environment The second piece of information comes from the integration curve, the line above each peak. The height of this line corresponds to the area under the curve and therefore tells us the relative numbers of hydrogen atoms in the different environments. If you measure the height of the two integration curves in Fig 4 you will find that they are in a ratio of 1:3. This matches the structure of the compound, with two hydrogen nuclei in Environment 2 and six in Environment 1. 3. The electron density around the hydrogen atoms The third piece of information is about the nature of each environment. This information is given by the chemical shifts (Table 1). In the exam, you will be given the chemical shifts that you need, so don’t try to memorise them. Looking at Fig 4 again, the peaks at 1.6 and 3.7 ppm correspond respectively to the methyl hydrogens (Environment 1) and the CH2Cl hydrogens (Environment 2), which are deshielded by the closer electron-withdrawing chlorine atom.
Splitting pattern (n+1)
How it looks
Typical structure
0
Singlet
R3CCH
1
Doublet
R2CH-CH
2
Triplet
RCH2-CH
3
Quartet
CH3-CH
Table 1 Typical chemical shifts of a range of hydrogen nuclei Proton
Chemical shift (ppm)
R-CH3
0.8-1.2
R2-CH2
1.1-1.5
R3-CH
1.5
R3-OH
1.0-6.0 Note that the OH hydrogen atom can appear almost anywhere along the chemical shift scale
O R
R R
C
1.0-6.0
Fig 6 No adjacent hydrogen atoms to either environment
CH2R
H
R-CH2-Cl
2.5-4.3
R-CH2-OH
3.3-4.0
C
In the NMR of 1,2-dichloro-2-methylpropane (Fig 4), both peaks are singlets. This means that none of the hydrogens are adjacent to non-equivalent hydrogen nuclei (Fig 6).
C
H
H environment 2
4.4-6.9
R
C
environment 1
H
H H
H
C
C
C
Cl
Cl
H
H
O R
C
H
9.2-9.8 This carbon atom is adjacent to both environments. It has no hydrogen atoms so both peaks are singlets
O R
C
OH
10.0-12.0
2
113 High Resolution NMR Spectroscopy
Chem Factsheet
Answer The displayed formula and NMR spectrum of chloroethane are shown in Fig 9. There are two hydrogen environments: the methyl hydrogens (a) and the CH2Cl 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 CH3) 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 7 shows the proton NMR spectrum of ethyl ethanoate. The three peaks show that there are three different environments for the hydrogen atoms. The integration curves give us a ratio of 2:1:3, and analysis of the splitting patterns (Table 3) leads to a full assignment of peaks (Fig 8).
Fig 7 The proton NMR spectrum of ethyl ethanoate H H
O
C
C O
H
11
10
9
8
H
H
C
C
H
H
7
Fig 9 Chloroethane and its NMR spectrum
H
6
5
4
3
2
1
a
H
b H
H
C
C
H
H
-CH3(a)
-CH 2Cl(b)
Cl
0
δ (ppm)
Table 3 Assigning the peaks in the NMR of ethyl ethanoate Chemical shift
Relative number of hydrogen atoms
Splitting pattern
Number of adjacent hydrogens
10
Assigned to proton group
4.1
2
Quartet
3
b
2.0
3
Singlet
0
a
1.2
3
Triplet
2
c
9
8
7
6
5
4
δ (ppm)
3
2
1
0
An unusual case: Alcohol OH groups The hydrogen atom in alcohol OH groups is unusual in two ways. Firstly, as it says in Table 1, this proton can have almost any chemical shift. The actual shift depends on, among other things, the solvent that the compound is dissolved in. Secondly, at room temperature OH proton peaks do not get split like others, no matter how many adjacent hydrogen nuclei there are. This helps you, therefore, to identify them in the spectra since they are always singlet peaks.
Fig 8 Assigned peaks in ethyl ethanoate a
H O
H
C
b
c
H
H
C
C
H
H
Practice Questions 1. For each of the following compounds, draw the displayed formula, label the different hydrogen environments and predict approximate chemical shifts for each peak in the proton NMR spectrum: (a) 2-chloropropane (b) 1-chloropropane (c) ethanol (d) ethanoic acid (e) 2-chloropentane (f) 3-chloropentane
C O
H
H
Worked example 1. 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 hydrogen atoms in each environment.
2. Sketch the proton NMR spectra you would expect for the following compounds: (a) ethanoic acid (b) propanone (c) ethanal (d) ethanol
3
113 High Resolution NMR Spectroscopy
Chem Factsheet
(b)
Answers
O
Question 1 (a) Environment 2 Cl H 3C C
H3C
Environment 1 = 2.5-4.3 ppm Environment 2 = 0.8-1.2 ppm
CH 3
C
CH 3
H Environment 1 H
H
H 3C C
C
H
H
2
3
(b) 1
(c)
H
Environment 1 = 0.8-1.2 ppm Environment 2 = 1.1-1.5 ppm Environment 3 = 2.5-4.3 ppm
Cl
10
H
H
C
C
H
H
H
2
3
1
8
7
6
5
4
3
2
1
0
4
3
2
1
0
δ (ppm)
Environment 1 = 0.8-1.2 ppm Environment 2 = 3.3-4.0 ppm Environment 3 = 1.0-6.0 ppm
O
9
(c) O
(d) H
H
O
C
C
H3C O
H
2 Cl
H
H
(e) H 3 C C
C
C
CH 3
H
H
H
5
2
3
4
H
Cl
H
H3C C
C
C
CH 3
H
H
H
1
2
3
2
1
(f)
1
H
Environment 1 = 2.0-3.0 ppm Environment 2 = 10.0-12.0 ppm
H
1
C
Environment 1 = 0.8-1.2 ppm Environment 2 = 2.5-4.3 ppm Environment 3 = 1.1-1.5 ppm Environment 4 = 1.1-1.5 ppm Environment 5 = 0.8-1.2 ppm
10
Environment 1 = 0.8-1.2 ppm Environment 2 = 1.1-1.5 ppm Environment 3 = 2.5-4.3 ppm Note that this molecule is symmetrical so has only three different hydrogen environments
9
8
7
6
5 δ (ppm)
(d) H H C
Question 2
H C O
H H
H
(a) O H 3C
C
OH
11
12
10
8
6
4
2
0
10
9
8
7
6 5 δ (ppm)
4
3
2
1
0
Acknowledgements: This Factsheet was researched and written by Emily Perry. Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136
-2
δ (ppm)
4
