The use of carbon-13 and proton (high-resolution) NMR spectroscopy. The number of peaks and chemical shift indicating different environments. Integration giving the ratio of hydrogen atoms. Spin-spin splitting interpreted with the n+1 rule, and the use of TMS as a reference and deuterated solvents.

What this dot point is asking

AQA wants you to interpret carbon-13 and high-resolution proton NMR spectra: relate the number of peaks to environments, use chemical shift to identify groups, use integration for hydrogen ratios, apply the n+1 rule to splitting, and explain the use of TMS and deuterated solvents.

What NMR measures

Nuclei such as carbon-13 and hydrogen-1 behave like tiny magnets and absorb radio-frequency energy in a strong magnetic field. The exact frequency depends on the electronic environment of the nucleus, measured as a chemical shift ($\delta$) in parts per million (ppm) relative to TMS.

Carbon-13 NMR

The number of peaks equals the number of different carbon environments in the molecule. The chemical shift of each peak indicates the type of carbon (e.g. $C=O$ around $\delta = 190-220$, aromatic carbons around $\delta = 110-160$, alkyl carbons around $\delta = 5-50$).

Proton (high-resolution) NMR

Three features are read together:

1. Number of peaks = number of different hydrogen environments.
2. Chemical shift identifies the type of proton (e.g. $-CH_3$ around $\delta = 0.7-1.2$, $-O-CH_2-$ around $\delta = 3.3-4.3$, aldehyde $-CHO$ around $\delta = 9-10$).
3. Integration (peak area) gives the ratio of hydrogen atoms in each environment.

Spin-spin splitting and the n+1 rule

Adjacent non-equivalent hydrogens split a peak into a multiplet.

The $OH$ (and $NH$) protons can be identified by adding $D_2O$: they exchange with deuterium and the peak disappears, so running the spectrum before and after a $D_2O$ shake confirms which signal is the labile proton. These protons usually appear as a single broad peak and are not split by, and do not split, their neighbours, because they exchange too fast for the coupling to be seen.

A structured way to solve an NMR problem is to read the three pieces of information in turn and cross-check them against the molecular formula: the number of signals gives the number of different environments, the integration gives the ratio of hydrogens in each, the chemical shift suggests the type of group, and the splitting pattern (via the n+1 rule) tells you how many hydrogens are on the neighbouring carbons. Combining proton NMR with carbon-13 NMR, infrared and mass spectrometry lets chemists assign a full structure with confidence, which is why NMR is the central tool of modern structure determination.

Try this

Q1. How many peaks would carbon-13 NMR of propanone, $CH_3COCH_3$, show? [1 mark]

• Cue. Two (the two $CH_3$ carbons are equivalent; the $C=O$ is different).

Q2. A proton signal is a quartet. How many hydrogens are on the adjacent carbon(s)? [1 mark]

• Cue. Three (n + 1 = 4, so n = 3).