Everything you need to know about C-13 NMR spectroscopy

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 🙃.

• 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.

• sp and sp² hybridised carbons have a higher 𝛅 than sp³ carbon atoms.

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.

2. 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.

3. 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. 

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.

2. 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.

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