Skip to main content

Nuclear magnetic resonance (NMR) spectroscopy: WACE Year 12 Chemistry

Syllabus dot point

“Interpret proton and carbon-13 NMR spectra to determine the number and types of chemical environments in a molecule”

8 min read

Quick answer

NMR spectroscopy maps the chemical environments of atoms. In proton (1^1H) NMR, the number of peaks tells you how many different hydrogen environments there are; the chemical shift tells you the type of environment; the peak area (integration) gives the ratio of hydrogens in each; and splitting (the n+1 rule) shows how many hydrogens are on neighbouring carbons. Carbon-13 NMR shows the number of carbon environments.

What this dot point is asking

Nuclear magnetic resonance (NMR) spectroscopy probes the environment of atomic nuclei (especially hydrogen and carbon-13) in a magnetic field. It reveals the carbon-hydrogen skeleton of a molecule, complementing the molar mass from mass spectrometry and the functional groups from infrared.

Chemical shift

Nuclei in different chemical environments resonate at slightly different frequencies, measured as chemical shift in parts per million (ppm) relative to a reference (TMS at 0 ppm).

Chemical environment

A set of equivalent atoms experiencing the same surrounding atoms and bonding. Atoms in different environments give separate NMR signals, while equivalent atoms give a single combined signal.

What proton NMR tells you

A 1^1H (proton) NMR spectrum carries four kinds of information:

  • Number of peaks (signals): equals the number of different hydrogen environments.
  • Chemical shift: the position of each peak indicates the type of environment (for example, hydrogens near an electronegative oxygen are shifted downfield to higher ppm).
  • Integration (peak area): the ratio of areas gives the ratio of the numbers of hydrogens in each environment.
  • Splitting (multiplicity): a peak splits into multiple lines according to the n+1 rule, where n is the number of hydrogens on the adjacent carbon. A single neighbour hydrogen gives a doublet; two give a triplet, and so on.
The n+1 rule

A hydrogen environment with n hydrogens on the neighbouring carbon atom is split into (n+1) peaks. So a CH3 next to a CH2 (n = 2) appears as a triplet, and the CH2 next to a CH3 (n = 3) appears as a quartet. This splitting reveals which groups are next to each other.

What carbon-13 NMR tells you

A 13^{13}C NMR spectrum is simpler: the number of peaks equals the number of different carbon environments, and the chemical shift indicates the type of carbon (for example a carbonyl carbon appears far downfield). It does not normally show splitting at this level, so it is a clean count of carbon environments.

Interpreting a proton NMR spectrum

Question

An organic compound, molar mass 46, shows three proton NMR signals with area ratio 3:2:1, and the signal of area 3 is a triplet while the signal of area 2 is a quartet. Deduce the structure.

Solution

Three signals mean three hydrogen environments. The area ratio 3:2:1 indicates 3, 2 and 1 hydrogens, totalling 6 hydrogens.

The triplet of area 3 is a CH3\text{CH}_3 group split by 2 neighbouring hydrogens (n+1 = 3). The quartet of area 2 is a CH2\text{CH}_2 group split by 3 neighbouring hydrogens (n+1 = 4). So CH3\text{CH}_3 and CH2\text{CH}_2 are adjacent. The single hydrogen (area 1) that is not coupled is consistent with an -OH.

This assembles to CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}, ethanol, which has molar mass 46. The data confirm ethanol.

Equivalent environments and symmetry

Counting environments correctly is the skill that decides most NMR questions, and it rests on symmetry. Hydrogens (or carbons) are in the same environment if they are related by the symmetry of the molecule, so swapping them gives an identical molecule. The three hydrogens of a CH3\text{CH}_3 group are always equivalent (free rotation makes them indistinguishable). The two CH3\text{CH}_3 groups in propanone are equivalent because the molecule is symmetric about the carbonyl, so they merge into one signal. By contrast, the two CH3\text{CH}_3 groups of butanone are in different environments, because one is next to the C=O and the other is not, so they give separate signals. Before assigning shifts or splitting, always work out the number of distinct environments from the structure, because that number must equal the number of signals.

Why this matters

NMR is the most powerful single technique for determining organic structure, because it shows both how many environments exist and how the atoms connect. Combined with mass spectrometry (molar mass) and infrared (functional groups), it lets a chemist assemble a complete structure, the integrated skill assessed in the analysis section of Unit 4.

Misapplying the n+1 rule or miscounting environments

The most common error is applying the n+1 rule to the hydrogens on the same carbon rather than the neighbouring carbon; splitting comes from hydrogens on adjacent carbons. Another is miscounting environments by treating equivalent hydrogens as different, or vice versa. Always identify each distinct environment, then count the neighbours for splitting.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks
A compound of molecular formula C3H6O2\text{C}_3\text{H}_6\text{O}_2 shows three proton NMR signals with integration ratio 3:2:13:2:1. The signal of area 3 is a triplet, the signal of area 2 is a quartet, and the signal of area 1 is a singlet far downfield (about 11.5 ppm11.5\ \text{ppm}). Deduce the structure and justify using each piece of evidence.
Show worked answer →

A 6 mark question rewards using the environments, splitting, integration and shift together.

Three signals mean three hydrogen environments; the 3:2:13:2:1 ratio means 3,23, 2 and 11 hydrogens (total 66, matching C3H6O2\text{C}_3\text{H}_6\text{O}_2).

The triplet of area 3 is a CH3\text{CH}_3 split by two neighbours (n+1=3n+1 = 3). The quartet of area 2 is a CH2\text{CH}_2 split by three neighbours (n+1=4n+1 = 4), so CH3\text{CH}_3 and CH2\text{CH}_2 are adjacent (CH3CH2−\text{CH}_3\text{CH}_2-). The singlet of area 1 far downfield at about 11.5 ppm11.5\ \text{ppm} is the very deshielded proton of a carboxylic acid -COOH\text{-COOH} (it has no neighbouring hydrogens, hence a singlet).

Assembling these: CH3CH2COOH\text{CH}_3\text{CH}_2\text{COOH}, propanoic acid, which fits C3H6O2\text{C}_3\text{H}_6\text{O}_2.

Markers reward the three environments from 3:2:13:2:1, the triplet/quartet adjacency for the ethyl group, the downfield singlet identifying -COOH\text{-COOH}, and the conclusion of propanoic acid.

WACE 2023-style5 marks
(a) State what the number of signals in a carbon-13 NMR spectrum indicates. (b) Propanone CH3COCH3\text{CH}_3\text{COCH}_3 shows only two carbon-13 signals; explain why. (c) Explain how proton NMR could distinguish propanone from propanal, isomers of C3H6O\text{C}_3\text{H}_6\text{O}.
Show worked answer →

A 5 mark answer needs the meaning of signal count, the symmetry argument, and the distinguishing feature.

(a) The number of signals in a carbon-13 NMR spectrum equals the number of different carbon (chemical) environments in the molecule.

(b) Propanone is symmetrical: the two CH3\text{CH}_3 groups are in identical environments (both attached to the central carbonyl carbon), so they give a single signal, and the carbonyl carbon gives a second signal. Hence only two carbon environments and two signals.

(c) In proton NMR, propanone CH3COCH3\text{CH}_3\text{COCH}_3 shows a single signal (one CH3\text{CH}_3 environment, a singlet, since the two methyls are equivalent and have no neighbouring hydrogens). Propanal CH3CH2CHO\text{CH}_3\text{CH}_2\text{CHO} shows three signals (a CH3\text{CH}_3 triplet, a CH2\text{CH}_2 multiplet, and a distinctive aldehyde -CHO\text{-CHO} proton far downfield around 9.5 ppm9.5\ \text{ppm}). The presence of three signals, including the downfield aldehyde proton, identifies propanal.

Markers reward signals equal to carbon environments, the symmetry giving two signals for propanone, and the single-versus-three-signal (with aldehyde proton) distinction.

Practise this

Sources & how we know this

ExamExplained