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

This guide teaches 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 AS and A-level exams from June 2016 onwards. The whole section is A-level only (not AS). The specification places organic chemistry (section 3.3) on A-level Paper 2, and Paper 3 can test any content. The specification puts the emphasis on using data to solve problems, not on theory.

For quick recall, use the revision notes. To test yourself, use the practice questions. NMR builds on mass spectrometry and infrared spectroscopy, which are taught in Organic analysis. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are.

What this topic covers

Spec point	What you must be able to do	Status
3.3.15 Range of techniques	Appreciate that analytical techniques together confirm the structure of a new compound	A-level only
3.3.15 What NMR shows	NMR gives information about the position of ¹³C or ¹H atoms in a molecule; ¹³C NMR gives simpler spectra than ¹H NMR	A-level only
3.3.15 Chemical shift	Use the δ scale; chemical shift depends on the molecular environment	A-level only
3.3.15 Integration	Use integrated ¹H spectra to find the relative numbers of ¹H atoms in different environments	A-level only
3.3.15 Solvents	¹H NMR samples are dissolved in deuterated solvents or CCl₄	A-level only
3.3.15 TMS	Explain why tetramethylsilane (TMS) is a suitable standard	A-level only
3.3.15 Structures	Use ¹H and ¹³C spectra with Chemistry Data Booklet shift data to suggest structures or part structures	A-level only
3.3.15 Splitting	Use the n+1 rule for adjacent, non-equivalent protons, limited to doublets, triplets and quartets in aliphatic compounds	A-level only

Section 3.3.15 lists no required practical.

Why chemists combine techniques

Each technique answers a different question:

• Mass spectrometry gives the molecular mass and formula.
• Infrared spectroscopy shows which bonds, and so which functional groups, are present.
• NMR spectroscopy shows the carbon or hydrogen environments, what each is next to, and (for ¹H) how many H atoms are in each.

Together they confirm the structure of a new compound. In problems you must make all the data agree.

How NMR gives information

In a strong magnetic field, ¹H and ¹³C nuclei absorb radio-frequency radiation. The frequency depends on the atoms around each nucleus, its environment, and atoms in the same environment give one signal. So NMR shows the position of each ¹H or ¹³C atom in the molecule.

Equivalent atoms

Atoms are equivalent (in the same environment) when the molecule’s symmetry makes them interchangeable, like the three H atoms of a CH₃ group. Look for symmetry before you look at spectra.

Chemical shift and the δ scale

A spectrum plots absorption against chemical shift, δ, measured in parts per million (ppm). The scale runs from right (δ = 0) to left (larger δ). Chemical shift depends on the molecular environment. A nucleus close to an electronegative atom such as O, or close to a C=O group, appears at a larger δ.

The specification expects you to use shift data from the Chemistry Data Booklet. The ranges below are approximate values written for this page, not copied from the AQA booklet. Always work from the AQA booklet itself.

¹H environment (approximate)	δ / ppm	¹³C environment (approximate)	δ / ppm
CH₃, CH₂, CH in an alkyl chain	0.7–1.7	C–C in an alkyl chain	5–50
H on C next to C=O	2.0–2.7	C–O (alcohol, ether, ester)	50–90
H on C next to O (alcohol, ether, ester)	3.3–4.3	C=C (alkene or arene)	100–160
H on C=C (alkene)	4.5–6.0	C=O (acid or ester)	160–185
Aldehyde H, –CHO	9.0–10.0	C=O (aldehyde or ketone)	190–220
Carboxylic acid H, –COOH	10.0–12.0

TMS as the standard

Chemical shifts are measured relative to tetramethylsilane (TMS), Si(CH₃)₄, whose signal is set at δ = 0. You need to explain why it is suitable:

• All 12 H atoms are equivalent (and all 4 C atoms are equivalent), so TMS gives one strong, single peak.
• Its peak lies at a lower δ than almost all organic compounds, so it does not overlap the sample’s peaks.
• It is inert, so it does not react with the sample.
• It is volatile, so it is easily removed from the sample afterwards.

Solvents for ¹H NMR

Samples are dissolved in a solvent. An ordinary solvent such as CHCl₃ contains ¹H atoms and would give a large peak that swamps the sample. So ¹H NMR uses a deuterated solvent, such as CDCl₃, in which ¹H is replaced by deuterium (²H), which does not give a signal in a ¹H spectrum. CCl₄ also works because it contains no hydrogen at all.

¹³C NMR

In a ¹³C spectrum you meet at this level:

• number of peaks = number of carbon environments;
• the δ value of each peak shows the type of carbon;
• each environment gives a single line, with no splitting.

This is why ¹³C NMR gives simpler spectra than ¹H NMR. There is no splitting to interpret, and the δ range is much wider (about 0–220 ppm, against about 0–12 ppm for ¹H), so peaks rarely overlap.

Worked example 1: three isomers of C₅H₁₂

Predict the number of peaks in the ¹³C spectrum of each isomer.

pentane              CH3–CH2–CH2–CH2–CH3
2-methylbutane       CH3–CH(CH3)–CH2–CH3
2,2-dimethylpropane  C(CH3)4

• Pentane is symmetrical about the middle carbon. The two end CH₃ carbons match, the two CH₂ carbons next to them match, and the middle CH₂ is unique: 3 peaks.
• 2-methylbutane: the two CH₃ groups on C2 are equivalent. The CH, the CH₂ and the end CH₃ are all different: 4 peaks.
• 2,2-dimethylpropane: four equivalent CH₃ carbons and one central carbon: 2 peaks.

¹³C NMR alone separates them.

Worked example 2: a carbonyl compound, C₄H₈O

Compound A has ¹³C peaks at δ 205, 41 and 16. Suggest a structure.

1. Three peaks, but four carbons, so two carbons must be equivalent.
2. δ 205 is in the 190–220 range: an aldehyde or ketone C=O.
3. δ 41 and 16 are both alkyl carbons.
4. The structure with a C=O and two equivalent carbons is 2-methylpropanal, (CH₃)₂CHCHO. Butanal and butanone would each give four peaks.

Butanal and butanone both give four ¹³C peaks, one above 190. ¹H NMR separates them: only the aldehyde has a peak at δ 9–10.

¹H NMR

A ¹H spectrum gives three pieces of information:

1. Number of peaks (groups of peaks) = number of hydrogen environments.
2. Chemical shift of each peak = what that hydrogen is near.
3. Integration: the relative area under each peak = the relative number of H atoms in that environment.

Integration

An integration trace has step heights proportional to peak areas. Divide each height by the smallest to get the simplest whole-number ratio, then scale it to the total number of H in the molecular formula.

Worked example 3: integration and splitting together

Compound B, C₅H₁₀O, has two ¹H peaks: δ 2.4 (quartet) and δ 1.1 (triplet). The integration trace heights are 32 mm and 48 mm. Deduce the structure.

1. Ratio = 32 : 48 = 1 : 1.5 = 2 : 3.
2. The formula has 10 H, and 2 + 3 = 5, so multiply by 2: 4 H at δ 2.4 and 6 H at δ 1.1.
3. A quartet next to a triplet is the pattern of an ethyl group, CH₃CH₂–. Having 4 H and 6 H means two equivalent ethyl groups.
4. δ 2.4 is in the range for H on a carbon next to C=O.
5. Two equivalent ethyl groups on a C=O: pentan-3-one, CH₃CH₂COCH₂CH₃.

Note that the ratio was 2 : 3, not 4 : 6. Integration only gives the relative numbers until you use the formula.

Spin–spin splitting: the n+1 rule

The signal for a set of protons is split by non-equivalent protons on adjacent carbon atoms.

A signal is split into n + 1 lines, where n is the number of H atoms on the adjacent carbon atoms.

H on adjacent carbons (n)	Lines (n + 1)	Name
0	1	singlet
1	2	doublet
2	3	triplet
3	4	quartet

The specification limits this to doublets, triplets and quartets in aliphatic compounds. Three rules:

• Equivalent protons do not split each other. 1,2-dichloroethane, ClCH₂CH₂Cl, gives a singlet.
• Splitting tells you about the neighbours, not about the group itself. A CH₂ that appears as a triplet is next to a CH₂. In an ethyl group the CH₂ is the quartet (next to CH₃) and the CH₃ is the triplet (next to CH₂).
• In the spectra you meet, an O–H proton normally appears as a singlet.

Useful patterns:

Pattern	What it means
Quartet (2H) + triplet (3H)	CH₃CH₂– (ethyl group)
Singlet (3H)	CH₃ attached to a carbon or oxygen with no H, e.g. CH₃CO– or CH₃O–
Singlet (9H)	(CH₃)₃C–
Doublet (3H) + quartet (1H)	CH₃CH< with no other H neighbours

Worked example 4: two esters, C₄H₈O₂

Ethyl ethanoate (CH₃COOCH₂CH₃) and methyl propanoate (CH₃CH₂COOCH₃) both show a singlet (3H), a quartet (2H) and a triplet (3H). How does ¹H NMR tell them apart?

• In ethyl ethanoate the CH₂ is bonded to O, so its quartet is at about δ 4.1. The singlet CH₃ is next to C=O, at about δ 2.0.
• In methyl propanoate the CH₃ is bonded to O, so the singlet is at about δ 3.7. The quartet CH₂ is next to C=O, at about δ 2.3.

The splitting patterns match, so the chemical shift decides which group is bonded to O.

Putting the techniques together

Worked example 5: identifying compound C

Compound C gives these results:

• Mass spectrum: molecular ion at m/z = 72.
• Infrared: strong absorption at 1715 cm⁻¹; no broad absorption for O–H.
• ¹³C NMR: four peaks, at δ 209, 37, 29 and 8.
• ¹H NMR: δ 2.4 (quartet, 2H), δ 2.1 (singlet, 3H), δ 1.1 (triplet, 3H).

Deduce the structure.

1. Mr = 72 fits C₄H₈O (4 × 12.0 + 8 × 1.0 + 16.0 = 72.0).
2. The IR absorption at 1715 cm⁻¹ is C=O. With no O–H, the compound is an aldehyde or ketone.
3. ¹³C: four peaks, so four different carbons. δ 209 is an aldehyde or ketone C=O.
4. ¹H: no peak at δ 9–10, so it is not an aldehyde.
5. Quartet (2H) + triplet (3H) = an ethyl group. The singlet (3H) is a CH₃ with no H neighbours.
6. The singlet at δ 2.1 and the quartet at δ 2.4 are both in the range for H next to C=O.
7. Compound C is butanone, CH₃COCH₂CH₃.

Common errors

• Counting carbon or hydrogen atoms instead of environments. Look for symmetry first.
• Reading splitting as “the number of H in this group”. The n+1 rule counts H on the adjacent carbons.
• Writing the integration ratio as the number of H atoms without scaling to the molecular formula.
• Explaining TMS only as “a peak at zero”. Give the single peak, away from other peaks, inert and volatile points.
• Saying deuterated solvents are used because they are inert. They give no signal in ¹H NMR.
• Forgetting that a CH₃ next to C=O and a CH₃ next to O are both singlets. Only δ separates them.

Where to go next

Use the revision notes for a self-test, then the practice questions, and check your progress with the free diagnostics.

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. This page covers section 3.3.15 Nuclear magnetic resonance spectroscopy (A-level only).