Everything you need to know about C-13 NMR spectroscopy

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13C NMR signals are considerably weaker than 1H NMR signals because the natural abundance of 13C is 1.1%.

We also know that magnetically active nuclei such as 1H and 13C will split each others’ signals (spin-spin coupling) if they are in close proximity which can make interpreting a 13C NMR spectrum messier and trickier than it needs to be. 13C NMR spectra are produced in such a way that any splitting is removed and the spectrum is often described as being ‘proton decoupled’.

Consequently, all the peaks / signals on a 13C spectrum are singlets and it is the chemical shift, 𝛅, of these peaks that tells us about the chemical environment of individual carbon atoms in the molecule. As with 1H NMR spectra, the 𝛅 is measured against a reference signal from TMS (tetramethylsilane).

A typical 13C NMR spectrum looks much like the one below πŸ™ƒ.

  • the 𝛅 of 13C atoms is much larger than that of protons so the x-axis typically runs from 0-200 ppm (compared with 0-12 ppm for 1H NMR).
  • the area under the peak is NOT necessarily proportional to the number of carbon atoms in a particular chemical environment but the height of a peak may be indicative of the number of hydrogen atoms bonded to the carbon. 13C in CH3 and CH2 groups usually have taller peaks than 13C bonded to one or no hydrogen atoms.
  • 13C atoms bonded to electron withdrawing groups such as oxygen, nitrogen, chlorine or bromine give peaks with a higher 𝛅.
  • 13C atoms that are part of a methyl group have lower 𝛅 than those in methylene or methine groups.

Analysing a 13C NMR spectrum is done in much the same way as with a 1H NMR spectrum. The first thing is to familiarise yourself with the data sheet for your exam board.

The number of peaks on the spectrum tells us the number of different carbon chemical environments we have in our molecule. We can read the 𝛅 of each peak off the spectrum and then compare against the data sheet to identify each of these carbon chemical environments. All this detective work is used to confirm the structure of a molecule, or we might be asked to predict the range of 𝛅 for different carbon atoms in a known molecule.

Let’s have a look at a few key examples …

There are 4 peaks on the 13C NMR spectrum for 3-methylpentane indicating that there are 4 unique carbon environments in the molecule. The molecule is symmetrical so carbon atoms in the methyl groups at either end are equivalent and give a single peak at 𝛅11. Similarly, the carbon atoms in the two methylene, CH2, groups are equivalent and give us a single peak at 𝛅30. There is a peak at 𝛅20 for the carbon in the branch methyl group and a peak at 𝛅39 for the carbon in the middle of the chain.

You’ll notice that on the data sheet the 𝛅 range for all these carbon atoms is simply 0-50 ppm / 5-40 ppm – they are in a C-C chemical environment. Without the additional information on the spectrum it would be difficult to assign a particular peak to a particular carbon in the molecule other than using the general trends discussed above (as in CH3 carbons tend to have a lower 𝛅 than CH2 and CH) to make some tentative suggestions.

There are 3 peaks on the 13C NMR spectrum for methoxyethane indicating that there are 3 unique carbon environments in the molecule.

There is a peak at 𝛅68 for a C-O carbon and a peak at 𝛅58, also for a C-O carbon. Read the data sheet carefully – the key here is that both of these carbons are bonded to oxygen, NOT that they are part of a CH2 / CH3 group.

Examiners love a benzene ring! You’ll notice that because this is a substituted benzene, the carbons in the ring are NOT equivalent. There are 4 peaks on the spectrum indicating 4 unique carbon environments. Each of the peaks is due to the presence of a carbon atom in a benzene ring, with 𝛅 in the range 110-160 ppm.

We number the carbons from the chlorine in a clockwise fashion – carbons 2 and 6 are equivalent (either side of the carbon that is bonded to the chlorine) giving a single peak and carbons 3 and 5 are equivalent, also giving a single peak. This is useful information (even it seems obvious) so tell the examiner 😊.

There are 6 peaks on the 13C NMR spectrum for 1-phenylpropan-1-one indicating that there are 6 unique carbon environments in the molecule.

The peak at 𝛅200 is for the C=O carbon (ketone functional group).

The four peaks at 𝛅128-137 are typical of carbon atoms in a benzene ring. These six carbon atoms are not all in the same chemical environment. Carbon atoms 2 and 6 are equivalent giving a single peak. Carbon atoms 3 and 5 are equivalent giving a single peak. Carbon atoms 1 and 4 are unique, each responsible for one of these four peaks (in an exam, I’d draw out the molecule and number the carbon atoms on the benzene ring just to make it nice and clear).

The peak at 𝛅34 is for a R-C=O-C– carbon (AQA data sheet) / is for a C-C carbon (OCR data sheet). The peak at 𝛅9 is for a C-C carbon.

In summary, your answer needs to cover two things:

  1. link the number of peaks on the spectrum to the number of carbon environments (again, obvious but so many students fail to put this in their answer).
  2. use the data sheet to identify the type of carbon for each and every peak. giving both the 𝛅 as read off the spectrum (or the range of 𝛅 if making predictions) and the carbon environment (is it a C-C carbon or a C=O carbon?).

Practice questions

  1. 4-hydroxycyclopent-1-ene and cyclopentanone are isomers with the molecular formula C5H8O. Explain, with reference to the data sheet, how 13C NMR spectroscopy could be used to distinguish between the two.

  1. Pentane, C5H12, reacts with Br2 in the presence of high energy UV radiation to give a mixture of three straight chain isomeric products with the molecular formula C5H11Br. Deduce the structure of the isomer that is responsible for the 13C NMR (proton decoupled) spectrum below. Explain your reasoning, with reference to the data sheet.

  1. Predict the number of peaks that would be expected in the 13C NMR spectrum for propylpropanoate, CH3(CH2)2OCOCH2CH3.  Use the data sheet to identify the chemical shift for each peak. 

4. A carboxylic acid with a molecular formula C5H9O2Cl has a 13C NMR spectrum with 4 peaks. Draw a possible structure for this molecule.Β 

5. The following molecules are isomers with the formula C5H10O.Β 

Which isomer is responsible for the 13C NMR spectrum shown below? Explain your reasoning.

6. A diol has the empirical formula C3H6O. The mass spectrum shows a molecular ion peak at m/z 116 and the 13C NMR spectrum is shown below. Suggest a structure for the molecule, explaining your reasoning with reference to the data sheet.Β 

7. (a) Draw a structure for an isomer of C6H12Cl2 which has two peaks in its 13C NMR spectrum.

(b) Draw a structure for an isomer of C6H3Cl3 which has four peaks in its 13C NMR spectrum.Β 

Answers (using the second / AQA data table)

  1. Both isomers have 3 unique carbon environments and so the 13C NMR spectrum for each will show 3 peaks.

Carbon atoms 1 and 2 are equivalent in 4-hydroxycyclopent-1-ene, and will give a peak with a 𝛅 90-150 ppm for a C=C. Carbon atoms 3 and 5 are also equivalent, giving a peak at 𝛅 5-40 ppm for a C-C. Carbon atom 4 will give a peak at 𝛅 50-90 for a C-O.

Carbon atoms 3 and 4 are equivalent in cyclopentanone, and the peak will be at 𝛅 5-40 ppm for a C-C. Carbon atoms 2 and 5 are also equivalent and will give a peak at 𝛅 20-50 ppm for a C-C=O. Carbon atom 1 will give a peak at 𝛅 190-220 ppm for a C=O.

  1. The three isomeric products have the following structures:

There are 5 unique carbon environments in 1-bromopentane and 2-bromopentane so each spectrum will show 5 peaks.

3-bromopentane has 3 unique carbon environments as it is a symmetrical molecule so its spectrum will show 3 peaks. Carbon atoms 1 and 5 are equivalent and will give a peak at 𝛅 5-40 ppm for a C-C. Carbon atoms 2 and 4 are equivalent and will also give a peak at 𝛅 5-40 ppm for a C-C. Carbon atom 3 will give a peak at 𝛅 10-70 ppm.

The spectrum belongs to 3-bromopentane.

3.

4.

5. Isomers (a) and (d) are symmetrical with three unique carbon environments so their spectra would show 3 peaks, not 5. Neither spectrum would show peaks for C=C environments at 𝛅115 ppm and 𝛅138 ppm.Β 

Isomer (c) has two carbon atoms bonded to an oxygen so the spectrum would show 2 peaks 𝛅50-90 ppm for the C-O environment. The spectrum above shows only one peak in this range which is consistent with isomer (b). The spectrum has two peaks for R-CH at 𝛅15 ppm and 𝛅31 ppm which is consistent with the two carbon atoms in the -CH2-CH3 fragment of isomer (b).Β 

The spectrum belongs to isomer (b).Β 

EXAM TIP: another example of a watertight answer, explaining why the spectrum cannot belong to isomers a, c or d as well as explaining why it does belong to b. Don’t give the examiner an excuse for not giving you the marks when you know what you are talking about!Β 

6. The molecular formula of the diol is C6H12O2. The 13C NMR spectrum shows that the molecule has three unique carbon environments as there are three peaks on the spectrum. This would suggest that there are 2 equivalent carbon atoms in each environment.

The peak at 𝛅63ppm is for C-O (the carbon atoms bonded to the hydroxyl / OH groups). The peak at 𝛅32ppm and the peak at 𝛅24ppm are for C-C environments as there is no carbonyl group or nitrogen in the molecule.Β 

7. (a)

(b)