In proton or 1H NMR spectroscopy the sample is dissolved in a solvent such as tetrachloromethane, CCl4, or trichlorodeuteriomethane, CDCl3 (also known as deuterated chloroform), because neither solvent contains hydrogen atoms whose signals would swamp proton signals from the sample. However it its worth mentioning that there is often a peak at 𝛅7.26ppm if CDCl3 is the solvent of choice as it is never 100% deuterated in practice and there will be a smidge of CHCl3 present.
There is a method to interpreting a 1H NMR spectrum.
We are looking for:
- the number of peaks / signals (which tells us how many different proton / chemical environments there are in the molecule)
- the position of each peak (this is given by its chemical shift value and referenced against the data sheet to narrow down the types of proton / number of different chemical environments present)
- the intensity of each peak (ratio of protons in each chemical environment)
- the appearance of each peak / the splitting pattern (from which we can deduce the arrangement of protons in the molecule)
NMR data sheets come in two forms – familiarise yourself with the correct one for your syllabus!
1. Number of peaks
The number of peaks simply tells us how many proton environments are present i.e. how many different types of 1H are in the molecule.
2. Position of the peaks
We need to assign each peak a chemical shift, 𝛅, value as read off the spectrum. We can use the data sheet to help us decide what type of chemical environment that proton is experiencing in the molecule – is the proton part of a CH3 or OH group? Bonded to a carbon that is adjacent to an oxygen atom? Bonded to a carbon that is part of a ring system?
- You’ll notice that the higher the 𝛅 value, the closer the proton is to an electron withdrawing group (O, N, Cl).
- We can also use the 𝛅 to determine whether the proton is part of a methyl, methylene or methane group.
If the proton is close to more than one functional group, the effect on the 𝛅 is cumulative.
Explaining 𝛅 of protons in benzenes, alkenes and aldehydes
The question is … why are the 𝛅 of protons in these environments so high?
Benzenes, alkenes and aldehydes all have delocalised ∏ electrons, either as part of a ring system or double bond. When we place these molecules in an external magnetic field, the delocalised ∏ electrons begin to circulate inducing a small local magnetic field which increases the magnitude of the external magnetic field experienced by the protons.
Explaining the 𝛅 of protons bonded to oxygen or nitrogen
You’ll see on the data sheet that protons that are part of an OH or NH group produce a peak that can be anywhere from 0.5 – 12 ppm on the spectrum, and the peak itself often very small and / or broad. This is due to the presence of hydrogen bonding between sample molecules or between the sample molecules and a polar solvent which affects the electron density around the proton, changing the strength of the external magnetic field experienced.
If we run a NMR spectrum after shaking our sample with D2O, the deuterium replaces the proton in the OH or NH group so the peak for this proton disappears – this is confirmation of the presence of an OH or NH proton in our molecule. Compare the spectrum for deuterated ethanol below with the original spectrum at the top of the page and you will notice that the little peak at 𝛅4.9 has disappeared.
Interpreting the NMR data sheet is an art in itself and needs practice. So go practice …
3. Intensity of the peak
The intensity is determined by the total area under the peak (not the height). It tells us the relative number of protons in each chemical / electronic environment and is summarised in the form of an integration curve on the spectrum (see purple line on the spectrum above), although at A level the relative number of protons is simply written on the spectrum by each peak, which is nice.
4. Appearance of the peak – the splitting pattern
You may have noticed that the signal / peak for many protons is actually a doublet, triplet or quartet – the peak has been SPLIT. This splitting pattern is the consequence of spin-spin coupling. The local magnetic field associated with protons on adjacent carbon atoms in the molecule can align with or against the applied magnetic field experienced by a H atom, shifting the signal to slightly higher or lower 𝛅 values.
The number of protons bonded to an adjacent carbon atom determine the splitting pattern of the peak and we can use this to help us untangle the fine structure of a molecule.
Essentially, the number of times a peak is split – 1 = the number of equivalent H atoms on an adjacent carbon.
The more 1H NMR spectra you analyse, the more patterns you will become familiar with …
- 2 H atoms on an adjacent carbon indicates the presence of a CH2 group in the molecule, 3 H atoms means there is a CH3 group present. A -CH2-CH3 fragment is common so we often see a quartet and a triplet on a spectrum.
- a small, broad singlet marked 1H is likely to be a H bonded to either oxygen or nitrogen, but a sharp single marked 1H is from a O=C(H)-O- group.
Let’s pull all this together and look at a couple of spectra …
When we are matching a spectrum to a molecule we need to explicitly mention the four points:
- Number of proton environments = number of peaks
- State the 𝛅 of each type of proton, describing it exactly as the data sheet does (since it is not really possible to write in bold in an exam, underlining the H is a good idea)
- How many protons in each environment
- Explain the splitting pattern
The 1H NMR spectrum for propanoic acid shows three peaks indicating that there are three different proton chemical environments.
The peak at 𝛅1.1 is for a R-CH3 proton. There are 3 equivalent protons in this environment.
The peak at 𝛅2.4 is for a HC-C=O proton. There are 2 equivalent protons in this environment.
The broad peak at 𝛅11.7 is for a O=C-OH proton. There is 1 proton in this environment.
The HC-C=O peak is split into a quartet indicating that there are 3 equivalent H atoms on the adjacent carbon. The R-CH3 peak is split into a triplet indicating that there are 2 equivalent H atoms on the adjacent carbon. Together this is evidence of a -CH2-CH3 fragment in the molecule. The broad peak at 𝛅11.7 is a singlet as the adjacent carbon has no H atoms bonded to it.
Key points:
- it doesn’t matter whether you describe peaks as signals, or protons as H atoms, but if you want to be super sure, check your syllabus and some past paper mark schemes, then use exactly the same terminology 🧐.
- the use of the words ‘adjacent’ and ‘equivalent’ is compulsory 🧑🎓.
- students invariably fail to explain the splitting patterns in 1H NMR spectra exam questions – this is key evidence in your detective work so don’t throw away marks!
- structure your answer as above – it is very easy to talk yourself round in a circle in these questions and still leave something obvious out 😳.
The 1H NMR spectrum for 1-phenylbutan-2-one shows four peaks indicating that there are four different proton chemical environments.
The peak at 𝛅0.9 is for a R-CH3 proton. There are 3 equivalent protons in this environment.
The peak at 𝛅2.4 is for a HC-C=O proton. There are 2 equivalent protons in this environment.
The peak at 𝛅3.6 is for a HC-C=O proton. It is higher than the expected 𝛅 because the protons are also adjacent to the phenyl / aryl / benzyl ring. There are 2 equivalent protons in this environment.
The broad peak at 𝛅7.3 is for a Ar-H proton. There are 5 protons in this environment.
(Ar stands for aryl or benzyl – the protons are bonded to carbon atoms that make up a benzene ring; the peak is broad because in reality these 5 protons are not all exactly equivalent and on some spectra might well show as 3 separate peaks close together – a 2H, a 2H and a 1H).
The HC-C=O peak at 𝛅2.4 is split into a quartet indicating that there are 3 equivalent H atoms on the adjacent carbon. The R-CH3 peak is split into a triplet indicating that there are 2 equivalent H atoms on the adjacent carbon. Together this is evidence of a -CH2-CH3 fragment in the molecule.
The HC-C=O peak at 𝛅3.6 is a singlet as neither adjacent carbon has H atoms bonded to it.
The broad peak at 𝛅7.3 is a singlet as the adjacent carbon has no H atoms bonded to it.
Practice questions
- Combustion analysis of an unknown organic compound gives the empirical formula as C4H8O2 and 1H NMR is shown below. The mass spectrum showed a molecular ion peak at m/z = 88. Deduce the structure of this molecule.
Answer
If we work out the molar mass of C4H8O2 we find that it is 88 g mol-1 so the empirical formula is also the molecular formula.
The NMR spectrum shows that we have 3 different proton environments in the molecule:
- 3 protons in a R-CH3 environment at 𝛅1.2, most likely a methyl group
- 2 protons in a O-CH environment at 𝛅3.9, most likely a -CH2 (methylene) group
- 3 protons in a O=C-CH environment at 𝛅2.1, most likely a methyl group
The peak for the R-CH3 protons at 𝛅1.2 is split into a triplet indicating that the adjacent carbon has two H atoms bonded to it. The peak for the O-CH protons at 𝛅3.9 is split into a quartet indicating that the adjacent carbon has three H atoms bonded to it. Together this suggests an O-CH2-CH3 fragment.
The peak for the O=C-CH protons at 𝛅2.1 is a singlet so there are no H atoms bonded to the adjacent carbon. This methyl group must be out on its own.
The unknown molecule is ethyl ethanoate.
- A student ran an 1H NMR spectrum on a sample she thought to be one of two possible ethers: ethoxyethane or 2-methoxypropane. Use the spectrum shown below to deduce, with reasoning, which isomer was present in the sample.
Answer
The first step is to draw out both isomers so we can predict how many proton environments (and hence peaks) we would expect to see on the 1H NMR spectrum for each isomer.
Ethoxyethane is a symmetrical molecule with two proton environments: O-CH in the range 𝛅3.1 – 3.9 (4 protons in this environment) and R-CH3 in the range 𝛅0.7 – 1.2 (6 protons in this environment). The O-CH peak would be split into a quartet and the R-CH3 peak split into a triplet. This is not what is shown on the spectrum above.
2-methoxypropane has 3 proton environments: R-CH3 at 𝛅1.2 (6 protons in this environment = two equivalent methyl groups), O-CH at 𝛅3.3 (3 protons in this environment = one methyl group) and O-CH at 𝛅3.7 (1 proton in this environment).
The peak for the O-CH proton at 𝛅3.7 is split into a heptet as there are 6 H atoms on the adjacent carbons (the two methyl groups) and the R-CH3 peak at 𝛅1.2 is split into a doublet as there is one H atom on the adjacent carbon.
The isomer in the sample is 2-methoxypropane.
Key point: when making predictions for a peak / signal, state the range from the data sheet for that proton environment, not an exact 𝛅 value.
3. Hexane-2,5-dione and hexane-3,4-dione are isomers. Describe how 1H NMR spectroscopy could be used to distinguish between them.
Answer
Both isomers are symmetrical molecules and both have two proton environments, hence the 1H NMR spectrum for each would show two peaks.
However, the splitting pattern for each isomer would be different. The two peaks in the spectrum of hexane-2,5-dione would be singlets, and at 𝛅2.1 – 2.6 for HC-C=O. The spectrum of hexane-3,4-dione would show a peak at 𝛅2.1 – 2.6 for HC-C=O which would be split into a quartet (the adjacent carbon has three protons bonded to it) and a peak at 𝛅0.7 – 1.2 for R-CH3 which would be split into a triplet (the adjacent carbon has two protons bonded to it).
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