Interpreting NMR Spectra and Integration Revision Notes | A Level AQA Chemistry

1. Why NMR Spectroscopy Unlocks Molecular Structure

Imagine you have a colourless liquid and you need to identify it without any labels. Mass spectrometry gives you the molecular mass. Infrared spectroscopy hints at functional groups. But neither tells you how the atoms are connected or how many of each type of proton the molecule contains. That is exactly what nuclear magnetic resonance (NMR) spectroscopy does.

NMR works because certain atomic nuclei — particularly ¹H (protons) and ¹³C (carbon-13) — behave like tiny magnets when placed in a strong external magnetic field. They absorb radiofrequency radiation at frequencies that depend on their electronic environment. The result is a spectrum where each peak position tells you what kind of environment a nucleus sits in, and in ¹H NMR, the peak area tells you how many protons share that environment.

At A Level, you will use two complementary techniques: ¹H NMR (proton NMR) and ¹³C NMR (carbon-13 NMR). Together, they let you build up a picture of a molecule's structure from spectral evidence alone — a skill that appears consistently in AQA Paper 2.

2. Chemical Shift: Reading the Horizontal Axis

The horizontal axis of any NMR spectrum is the chemical shift scale, measured in parts per million (ppm) and given the symbol δ. The reference point — δ = 0 ppm — is set by tetramethylsilane (TMS), the compound Si(CH₃)₄. TMS is chosen because its twelve equivalent protons resonate at a lower frequency than almost any proton in an organic molecule, so it sits conveniently at the far right of the spectrum without overlapping real signals.

Deshielding is the key concept here. Electronegative atoms (O, N, halogens) pull electron density away from nearby protons, reducing the shielding effect of those electrons. Less shielding means the proton 'feels' more of the external magnetic field and resonates at a higher δ value — further to the left on the spectrum. A proton next to a carbonyl group (C=O) is more deshielded than one on a simple alkyl chain, so it appears at a higher chemical shift.

Your Chemistry Data Booklet contains a table of typical ¹H chemical shift ranges and a separate table for ¹³C shifts. In the exam, you are expected to use these tables — you do not need to memorise exact values, but you do need to know the approximate regions and what they indicate.

Notice the pattern: the more electronegative the neighbouring group, the higher the δ value. Aldehyde protons sit at δ 9–10 ppm because they are adjacent to a C=O group and directly bonded to the carbonyl carbon. Carboxylic acid O–H protons appear even further downfield (δ 10–12 ppm) because the O–H is part of a highly polar system.

3. ¹³C NMR: Counting Carbon Environments

¹³C NMR gives you a peak for each distinct carbon environment in the molecule. The number of peaks tells you the number of chemically non-equivalent carbon atoms. For example, ethanol (CH₃CH₂OH) has two carbon environments — the methyl carbon and the carbon bearing the OH group — so its ¹³C spectrum shows two peaks.

Unlike ¹H NMR, the peaks in a ¹³C spectrum are not integrated to give ratios. Each peak confirms the presence of a carbon in a particular environment. The chemical shift scale for ¹³C runs from 0 to approximately 220 ppm, with carbonyl carbons (C=O) appearing at the highest shifts (δ 160–220 ppm) and alkyl carbons at the lowest (δ 5–40 ppm).

A useful exam strategy: count the number of ¹³C peaks and compare it to the number of carbons in the molecular formula. If the molecule has a plane of symmetry, some carbons will be equivalent and you will see fewer peaks than the total carbon count. For instance, propan-2-ol has three carbons but only two ¹³C peaks, because the two CH₃ groups are in identical environments.

4. ¹H NMR Integration: Counting Protons in Each Environment

Here is where ¹H NMR becomes particularly powerful. The area under each peak — called the integration — is directly proportional to the number of protons in that chemical environment. This is the core principle behind integration, and it is one of the three specification requirements for this topic.

In practice, integration is displayed as a stepped line drawn over the spectrum. Each step rises by an amount proportional to the number of protons contributing to that peak. You read the relative heights of the steps, not absolute values. If one step is twice the height of another, there are twice as many protons in that environment.

Consider ethanol, CH₃CH₂OH. Its ¹H NMR shows three peaks: one for the CH₃ group, one for the CH₂ group, and one for the OH proton. The integration ratio is 3: 2: 1, matching the proton count in each environment. This ratio is what you report — not the absolute number of protons, unless you know the molecular formula and can scale accordingly.

Worked example: Using Integration to Determine Proton Ratios

A ¹H NMR spectrum of a compound with molecular formula C₄H₈O₂ shows three peaks. The integration step heights are measured as 6 mm, 2 mm, and 2 mm. Determine the ratio of protons in each environment and suggest a possible structure.

1. Step heights: 6 mm, 2 mm, 2 mm — These heights are proportional to the number of protons in each environment. We do not need to know the absolute scale — only the ratios matter.
2. Divide each height by the smallest value (2 mm): 6/2 = 3, 2/2 = 1, 2/2 = 1. Ratio = 3: 1: 1 — This tells us the three environments contain protons in the ratio 3: 1: 1.
3. Molecular formula C₄H₈O₂ contains 8 protons. Ratio parts: 3 + 1 + 1 = 5 parts. 8 ÷ 5 is not a whole number, so the ratio must be scaled. Try multiplying by 2: 6: 2: 2, giving 6 + 2 + 2 = 10 protons — too many. Check: the ratio 3: 1: 1 with 8 H does not scale to whole numbers directly, so consider whether the OH proton may be absent or broadened. Alternatively, try ratio 3: 1: 1 scaled to give 6H + 1H + 1H = 8H — this works if the 3-part group represents 6H (two equivalent CH₃ groups), the 1-part groups represent 1H each. — Scaling: if the '3' unit = 6H, then each unit = 2H. So the three environments contain 6H, 2H, and 2H — but that gives 10H total, not 8H. Re-examine: if each unit = 1H, then 3H + 1H + 1H = 5H, not 8H. The most consistent interpretation with 8H is a ratio of 3: 1: 1 where the smallest unit represents 2H, giving 6H: 2H: 2H — but that sums to 10H. A cleaner fit: the ratio is actually 3: 1: 1 representing 3H, 1H, 1H with the remaining 3H accounted for by a broad or absent OH signal. This is a common exam scenario.
4. If the compound is methyl propanoate (CH₃CH₂COOCH₃): environments are CH₃CH₂– (3H, δ ~1.1 ppm), –CH₂– (2H, δ ~2.3 ppm), –OCH₃ (3H, δ ~3.7 ppm). Integration ratio = 3: 2: 3. — Methyl propanoate gives a 3: 2: 3 ratio (8H total), consistent with C₄H₈O₂. The OCH₃ and CH₃ groups both give 3H signals but at very different δ values, allowing them to be distinguished.
5. Proposed structure: methyl propanoate, CH₃CH₂COOCH₃ — Three environments, integration 3: 2: 3, molecular formula C₄H₈O₂ — all consistent. In the exam, always state the ratio, the environments, and the δ values used to support your answer.

Answer: Integration ratio 3: 2: 3 (total 8H), consistent with methyl propanoate (CH₃CH₂COOCH₃). Three distinct proton environments: CH₃ (δ ~1.1 ppm), CH₂ (δ ~2.3 ppm), OCH₃ (δ ~3.7 ppm).

The worked example above shows the full reasoning chain examiners expect. Notice that integration alone does not identify a structure — you must combine the ratio with the chemical shift values to assign each environment. The two pieces of information work together.

5. Combining Chemical Shift and Integration to Suggest Structures

In an AQA exam question, you will typically be given a ¹H NMR spectrum (or a table of peaks with their δ values and integration data) and asked to suggest a structure or part-structure. The method is always the same: work through the evidence systematically rather than guessing.

• Step 1 — Count the peaks: each peak represents a distinct proton environment.
• Step 2 — Read the chemical shifts: use the Data Booklet table to assign each peak to a likely environment (e.g. δ 9–10 ppm → aldehyde C–H).
• Step 3 — Read the integration ratios: determine the relative number of protons in each environment.
• Step 4 — Cross-check with the molecular formula: scale the ratio to match the total number of protons in the formula.
• Step 5 — Assemble the part-structures: combine the environments into a consistent structural proposal.

This systematic approach is especially important when the question uses the command word 'suggest'. There may be more than one valid answer, but your response must be consistent with all the spectral data provided. Examiners award marks for correct reasoning, not for guessing a name.

The ethyl ethanoate example above illustrates how integration and chemical shift work together. Without integration, you could not distinguish the two CH₃ groups (one at δ 1.25 ppm, one at δ 2.05 ppm) from a single CH₃ group — both would appear as single peaks. The integration confirms each accounts for 3H, and the different δ values confirm they are in different environments.

6. Worked Exam-Style Structure Determination

The following worked example mirrors the style of AQA Paper 2 questions. Work through each step before revealing the answer — this is the most effective way to build the reasoning skill the examiner is testing.

Worked example: Structure Determination from ¹H and ¹³C NMR Data

A compound has molecular formula C₃H₆O₂. Its ¹³C NMR shows three peaks at δ 21 ppm, δ 52 ppm, and δ 171 ppm. Its ¹H NMR shows two peaks: δ 2.0 ppm (integration 3H) and δ 3.7 ppm (integration 3H). Suggest a structure for this compound, justifying your answer.

1. δ 21 ppm → alkyl carbon (C–C or C–H environment, e.g. CH₃ adjacent to C=O)\nδ 52 ppm → carbon bonded to oxygen (C–O, typical of O–CH₃ in an ester)\nδ 171 ppm → carbonyl carbon C=O (ester or carboxylic acid range: δ 160–185 ppm) — Three distinct carbon environments match three carbons in C₃H₆O₂. The peak at δ 171 ppm strongly suggests an ester carbonyl.
2. δ 2.0 ppm → CH₃ group adjacent to a carbonyl C=O (typical δ ~2.0–2.1 ppm for CH₃CO–)\nδ 3.7 ppm → CH₃ adjacent to oxygen (O–CH₃ in ester: typical δ 3.3–3.8 ppm) — The δ 3.7 ppm signal is characteristic of an O–CH₃ group in an ester (methoxy group). The δ 2.0 ppm signal is consistent with a CH₃ group directly attached to a carbonyl carbon.
3. δ 2.0 ppm: 3H → one CH₃ group (CH₃CO–)\nδ 3.7 ppm: 3H → one CH₃ group (O–CH₃)\nTotal: 6H accounted for. C₃H₆O₂ has 6H — consistent. — Both peaks integrate for 3H each, giving a 1:1 ratio. With only two proton environments and six protons total, both must be CH₃ groups in different electronic environments.
4. Part-structures identified:\n– CH₃CO– (δ 2.0 ppm, 3H; ¹³C δ 21 ppm)\n– O–CH₃ (δ 3.7 ppm, 3H; ¹³C δ 52 ppm)\n– C=O (ester carbonyl; ¹³C δ 171 ppm)\nCombining: CH₃–C(=O)–O–CH₃ — This gives methyl ethanoate (CH₃COOCH₃), molecular formula C₃H₆O₂. All three ¹³C environments and both ¹H environments are accounted for.
5. Methyl ethanoate CH₃COOCH₃:\n¹³C: CH₃CO– (δ ~21 ppm) ✓, O–CH₃ (δ ~52 ppm) ✓, C=O ester (δ ~171 ppm) ✓\n¹H: CH₃CO– (3H, δ ~2.0 ppm) ✓, O–CH₃ (3H, δ ~3.7 ppm) ✓\nMolecular formula: C₃H₆O₂ ✓, total 6H ✓ — All spectral data are fully consistent with methyl ethanoate. The two CH₃ groups are in clearly different environments — one adjacent to the carbonyl (δ 2.0 ppm) and one bonded to oxygen (δ 3.7 ppm) — so they appear as two distinct ¹H peaks with equal integration.
6. Proposed structure: methyl ethanoate, CH₃COOCH₃ — The ¹H NMR δ 2.0 ppm (3H) matches CH₃CO–; δ 3.7 ppm (3H) matches O–CH₃. The ¹³C peaks at δ 21 ppm (CH₃CO–), δ 52 ppm (O–CH₃), and δ 171 ppm (ester C=O) are all consistent. Integration ratio 3:3 accounts for all 6 protons in C₃H₆O₂.

Answer: Methyl ethanoate, CH₃COOCH₃. Justified by: ¹H NMR δ 2.0 ppm (3H, CH₃CO–), δ 3.7 ppm (3H, O–CH₃); ¹³C NMR δ 21 ppm (CH₃CO– alkyl C), δ 52 ppm (O–CH₃ carbon), δ 171 ppm (ester C=O). Integration ratio 3:3 accounts for all 6 protons in C₃H₆O₂.

7. Exam Technique: Writing Mark-Winning NMR Answers

NMR questions on AQA papers reward structured, evidence-based reasoning. The most common reason students lose marks is not getting the chemistry wrong — it is failing to link their structural proposal back to the spectral data. Every claim you make about a structure must be supported by a specific peak, its δ value, and (for ¹H NMR) its integration.

When the command word is 'suggest', you are expected to apply your knowledge to unfamiliar data. There may be more than one acceptable answer, but your answer must be internally consistent with all the evidence given. Partial credit is available for correct part-structures even if the full structure is wrong — so always write down what you can deduce with confidence.

For 'explain' questions about NMR, the mark scheme typically requires you to state the environment, the δ value, and the reason for that shift (usually referencing deshielding by an electronegative group or the C=O system). A one-word answer such as 'aldehyde' will not score full marks — you need to say why the aldehyde proton appears at that δ value.

Integration tells you the relative number of protons in each environment. Once the molecular formula is known, the spectrum should be interpreted by matching both chemical shift and integration ratio to the candidate structure.