AQA A-Level Chemistry: Nuclear magnetic resonance spectroscopy (7405) – Revision Notes

Condensed notes for the final weeks. For full explanations and worked examples, use the NMR study guide.

These notes cover section 3.3.15 Nuclear magnetic resonance spectroscopy of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for exams from June 2016 onwards. The whole section is A-level only. Organic chemistry is examined on A-level Paper 2, and Paper 3 can test any content. Test yourself with the practice questions, see the course hub and the printable checklist, and find your gaps with the free diagnostics. Mass spectrometry and IR are in the Organic analysis revision notes.

3.3.15 at a glance (all A-level only)

• Scientists use a range of analytical techniques together to confirm the structure of a new compound: mass spectrometry, infrared and NMR.
• NMR gives information about the position of ¹³C or ¹H atoms in a molecule.
• ¹³C NMR gives simpler spectra than ¹H NMR.
• Chemical shift is recorded on the δ scale and depends on the molecular environment.
• Integrated ¹H spectra give the relative numbers of ¹H atoms in each environment.
• ¹H samples are dissolved in deuterated solvents or CCl₄.
• TMS is the standard.
• Splitting follows the n+1 rule, limited to doublets, triplets and quartets in aliphatic compounds.
• No required practical is listed for this section.

Key definitions

Term	Meaning
Environment	The set of atoms around a nucleus. Atoms in the same environment give one signal
Equivalent atoms	Atoms made interchangeable by the molecule’s symmetry
Chemical shift, δ	Position of a signal in ppm, measured relative to TMS at δ = 0
Integration	Relative area under a ¹H peak, proportional to the number of H in that environment
Spin–spin splitting	Splitting of a ¹H signal by non-equivalent H on adjacent carbon atoms
TMS	Tetramethylsilane, Si(CH₃)₄, the standard

The δ scale

Larger δ means the nucleus is near an electronegative atom (such as O) or a C=O group. The specification expects you to read shifts from the Chemistry Data Booklet. The approximate ranges below are written for these notes and are not copied from the AQA booklet. Always work from the AQA booklet itself.

¹H (approximate)	δ / ppm
Alkyl CH₃, CH₂, CH	0.7–1.7
H–C next to C=O	2.0–2.7
H–C next to O	3.3–4.3
Alkene H	4.5–6.0
Aldehyde –CHO	9.0–10.0
Acid –COOH	10.0–12.0

¹³C (approximate)	δ / ppm
Alkyl C–C	5–50
C–O	50–90
C=C	100–160
C=O in acids and esters	160–185
C=O in aldehydes and ketones	190–220

TMS: four reasons it is suitable

1. 12 equivalent H atoms (and 4 equivalent C atoms), so it gives one strong single peak.
2. Its peak is at lower δ than almost all organic protons, so it does not overlap the sample’s peaks.
3. It is inert and does not react with the sample.
4. It is volatile, so it is easily removed afterwards.

Solvents

• CDCl₃ and other deuterated solvents: deuterium (²H) gives no signal in a ¹H spectrum, so the solvent does not swamp the sample’s peaks.
• CCl₄: contains no H atoms at all.
• CHCl₃ would be unsuitable: its ¹H would give a large peak.

¹³C spectra

• Number of peaks = number of carbon environments.
• Each peak’s δ tells you the type of carbon.
• No splitting, wide δ range, little overlap, so the spectrum is simpler than a ¹H spectrum.
• Don’t try to read numbers of carbons from ¹³C peak heights.

Worked reminder. Methyl ethanoate, CH₃COOCH₃, has three carbon environments: the C=O (160–185), the O–CH₃ (50–90) and the CH₃ on the C=O (5–50). That gives 3 peaks.

¹H spectra: method in steps

1. Count the peaks (groups of lines): that is the number of H environments.
2. Convert integration heights to the simplest whole-number ratio, then scale to the H count in the molecular formula.
3. Read each δ from the data to say what each set of H is near.
4. Use the n+1 rule to find how many H are on adjacent carbons.
5. Assemble fragments, then check the structure gives every peak, ratio and splitting.

Worked reminder. A compound C₄H₁₀O gives two ¹H peaks with integration heights 15 mm and 10 mm. Ratio = 15 : 10 = 3 : 2. The formula has 10 H and 3 + 2 = 5, so scale by 2: 6 H and 4 H. Two equivalent ethyl groups are likely, which fits ethoxyethane, CH₃CH₂OCH₂CH₃.

Combining techniques

Technique	What it gives you
Mass spectrometry	Molecular ion, so Mr and (with precise masses) the molecular formula
Infrared	Bonds present, e.g. C=O, O–H, so the functional group
¹³C NMR	Number and type of carbon environments
¹H NMR	Number of H environments, their ratio, their neighbours

Use them in that order: formula, functional group, carbon skeleton, then the H detail.

The n+1 rule

A signal is split into n + 1 lines by n H atoms on adjacent carbon atoms.

n	Pattern	You can say
0	singlet	no H on neighbouring carbons (e.g. CH₃ next to C=O or O)
1	doublet	next to a CH
2	triplet	next to a CH₂
3	quartet	next to a CH₃

• Equivalent H do not split each other: ClCH₂CH₂Cl gives one singlet.
• An O–H proton usually shows as a singlet in the spectra you meet.
• Quartet (2H) + triplet (3H) = ethyl group, CH₃CH₂–.
• Singlet (9H) = (CH₃)₃C–.

Worked reminder. Chloroethane, CH₃CH₂Cl: the CH₂ (2H) has 3 neighbouring H, so it is a quartet. The CH₃ (3H) has 2 neighbouring H, so it is a triplet. Ratio 2 : 3.

Must-know distinctions

• ¹³C vs ¹H: ¹³C counts carbon environments and shows no splitting. ¹H counts hydrogen environments and gives integration and splitting.
• Peak position vs splitting: δ tells you what the H is near. Splitting tells you how many H are on the next carbons.
• Ratio vs actual number: integration gives a ratio. You need the molecular formula for actual numbers.
• CH₃ next to C=O vs CH₃ next to O: both are singlets. δ separates them (about 2.0–2.7 vs 3.3–4.3).
• Aldehyde vs ketone: both show C=O near 190–220 in ¹³C. Only an aldehyde has a ¹H peak at δ 9–10.
• Carboxylic acid vs ester: only the acid has a ¹H peak at δ 10–12.

Quick self-test

1. How many peaks are in the ¹³C spectrum of propanone, CH₃COCH₃?
2. How many peaks are in the ¹H spectrum of propanone, and what is their splitting?
3. How many peaks are in the ¹³C spectrum of 2-methylpropan-2-ol, (CH₃)₃COH?
4. Integration heights are 18 mm, 12 mm and 6 mm. Give the ratio.
5. A ¹H signal is a quartet. How many H atoms are on the adjacent carbon atoms?
6. Give two reasons why TMS is used as the standard.
7. Why is CDCl₃ used instead of CHCl₃ as the solvent?
8. A ¹H peak appears at δ 9.7. What does it suggest?
9. A ¹³C peak appears at δ 205. What type of carbon is it?
10. Predict the ¹H spectrum of methoxymethane, CH₃OCH₃.
11. Why does ClCH₂CH₂Cl give a singlet, not a triplet?
12. State one way in which ¹³C spectra are simpler than ¹H spectra.

Answers

1. 2 (the two equivalent CH₃ carbons, and the C=O).
2. 1 peak, a singlet: all 6 H are equivalent, and the neighbouring carbon (C=O) has no H.
3. 2 (three equivalent CH₃ carbons, and the C–O carbon).
4. Divide by 6: 3 : 2 : 1.
5. 3.
6. Any two: a single strong peak from 12 equivalent H; peak away from (lower δ than) other peaks; inert; volatile, so easily removed.
7. CDCl₃ gives no signal in the ¹H spectrum. CHCl₃ would give a large peak from its H.
8. An aldehyde H, –CHO (δ 9.0–10.0).
9. A C=O in an aldehyde or ketone (190–220).
10. One singlet: all 6 H are equivalent, and the adjacent atom is O, which carries no H.
11. All four H are equivalent, and equivalent protons do not split each other.
12. Either: no splitting, so one line per environment; or the wider δ range means peaks rarely overlap.

Where marks are usually lost

• Counting atoms instead of environments, especially missing symmetry in molecules such as (CH₃)₃C– groups or symmetrical ketones.
• Applying n+1 to the H in the group itself rather than the H on adjacent carbons.
• Splitting a CH₃ next to a C=O or O. It has no H neighbours, so it is a singlet.
• Leaving an integration ratio unscaled when the molecular formula is given.
• Saying TMS “has a peak at zero” without the reason: one strong peak, away from other peaks.
• Saying deuterated solvents are chosen because they are unreactive, rather than because they give no ¹H signal.
• Proposing a structure that fits the splitting but not the δ values (for example, mixing up which ester group is bonded to O).
• Forgetting to check the final structure against every peak, ratio and splitting pattern.

Official syllabus

AQA AS and A-level Chemistry (7404/7405) specification, version 1.2, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. These notes cover section 3.3.15 Nuclear magnetic resonance spectroscopy (A-level only).