How many marks are NMR questions worth in AQA Chemistry?

It varies with the question. A short 'deduce and explain' may credit each reasoned step, while a longer structure-deduction question is often marked by levels of response, where the quality of your explanation determines the band. In both cases the reasoning is marked, not just the final structure — 'triplet, so two hydrogens on the next carbon' is the creditable step, not the bare label 'triplet'.

Do you have to memorise NMR chemical shift values?

No. AQA gives you the chemical shift tables (Table B for proton NMR and Table C for carbon-13) in the Data Booklet in the exam. You are not tested on recall of the numbers — you are tested on using them to identify environments. Practising with the tables open is exactly the right way to revise.

What is the difference between carbon-13 and proton NMR questions?

Carbon-13 NMR is the lighter version: you count the number of carbon environments (equivalent carbons share a peak) and place each peak in a shift region from Table C. Proton NMR carries more information per peak — integration for the number of hydrogens and splitting for the neighbours — so proton questions are where the full three-step method earns the most marks.

How to Answer AQA NMR Long-Answer Questions (2026)

Q: How do you work out a structure from an NMR spectrum?

Take one peak at a time and read all three pieces of information. Integration (relative peak area) tells you how many hydrogens are in that environment. Splitting tells you how many hydrogens are on the neighbouring carbon, using the n+1 rule — the number of lines minus one. Chemical shift, read from Table B of the Data Booklet, tells you what the environment is. Write each peak up as one fragment, then join the fragments so they fit the molecular formula.

Where NMR marks are won and lost

The hard part of an NMR question is rarely reading a single peak — it is pulling every peak, its splitting and the molecular formula together into one structure, and then writing that reasoning down. Examiner reports find that, in spectroscopy questions, only the strongest answers integrate all the spectral and formula data into a structure, and that weaker answers frequently reach the right structure without enough explanation to access the top marks. The lesson is to state the deduction, not just the label — write “triplet, so two hydrogens on the neighbouring carbon” rather than just “triplet”.

These questions also follow a fixed shape. They usually give you the molecular or empirical formula, often an infrared or ¹³C clue alongside the ¹H peaks — and they want the same three things read off each peak. Learn to work one peak at a time and the whole question becomes a routine.

The method: one peak at a time

Every proton (¹H) NMR peak carries three separate pieces of information. Take them in this order and squeeze all three out of each peak before moving on:

The method, per peak

Integration → how many H. The relative peak area (the integration trace, or the numbers given) tells you how many hydrogens are in that environment.
Splitting → how many H on the neighbour. Number of lines minus one, by the n+1 rule. A triplet means 2 neighbouring H, a quartet means 3.
Shift → the environment. Read δ against Table B of the Data Booklet — is this H near an O, a halogen, a C=O, or just an alkyl chain?

Write each peak up as a single fragment, then join the fragments so the hydrogen count matches the molecular formula.

The shift table is given to you in the exam — you are not tested on recalling the numbers, only on using them. That is why the right way to revise is with the table open, which is exactly how the analyser further down works.

Why the reasoning matters, not just the answer

A common misconception is that getting the structure right guarantees the marks. It doesn’t. The longer “deduce the structure” questions are often marked by levels of response — the band you land in depends on how completely you justify the structure, not only on whether it is correct. Examiner reports describe the top level as reserved for answers that integrate all the data; a right structure with thin explanation is capped below it.

So the safest way to reach the top band is to make your reasoning visible for every peak — write the deduction as a chain, not as a private thought process with only the final answer shown.

A worked example, start to finish

This is a real AQA past-paper question — a six-mark “deduce the structure” task that combines three techniques. Work it yourself first, then read the stage-by-stage analysis, a top-band answer, and how the levels-of-response marks are awarded.

AQA past paper [6 marks]

This question is about compound X with the empirical formula C₂H₄O. Figure 2 shows the infrared spectrum of X. Figure 3 shows the ¹³C NMR spectrum of X. The ¹H NMR spectrum of X shows four peaks with different chemical shift values. Table 3 gives data for these peaks.

Figure 2

Figure 3

Table 3

Proton (¹H) NMR data for compound X
Chemical shift δ / ppm	3.9	3.7	2.1	1.2
Splitting pattern	quartet	singlet	singlet	doublet
Integration value	1	1	3	3

Show how information from Figure 2, Figure 3 and Table 3 can be used to deduce the structure of compound X.

Source: AQA A-level Chemistry, Paper 2 (7405/2), June 2022, Question 6.

Notice the command: “show how… can be used to deduce”. That phrasing signals a levels-of-response question — the marks are for the chain of reasoning across all three techniques, not just the final structure. The mark scheme groups the work into three stages: infrared, then ¹H NMR, then ¹³C NMR.

Stage 1 — infrared: the functional groups

Two absorptions, two clues:

Broad absorption at 3400 cm⁻¹ → an O–H of an alcohol.
Strong peak at 1720 cm⁻¹ → a C=O.

So X contains both a hydroxyl group and a carbonyl.

Stage 2 — ¹H NMR: the hydrogens and the formula

Each peak, read in full

δ 2.1, singlet, 3 H → a CH₃ with no H on the neighbour; the shift (2.1–2.6) puts it next to C=O → CH₃–CO–.
δ 1.2, doublet, 3 H → a CH₃ next to a single H (a CH) → CH₃–CH–.
δ 3.9, quartet, 1 H → one H on a carbon bonded to O; quartet, so next to a CH₃ → –CH–O.
δ 3.7, singlet, 1 H → the non-coupling O–H proton, matching the IR alcohol.
Integration totals 1 + 1 + 3 + 3 = 8 H. The empirical formula C₂H₄O has 4 H, so the molecular formula is twice it: C₄H₈O₂.

Stage 3 — ¹³C NMR: confirm the carbons

Four peaks means four carbon environments — consistent with C₄H₈O₂:

δ 210 → C=O of an aldehyde or ketone (here a ketone).
δ 75 → a carbon singly bonded to O (the C–OH).
δ 25 and 20 → two different alkyl carbons (the two CH₃ groups).

Putting it together

The three techniques have handed you four fragments and a formula. Now connect them one bond at a time:

Assembling the fragments

List the pieces from the ¹H NMR: CH₃–CO– (from δ 2.1), CH₃–CH– (from δ 1.2), –CH–O (from δ 3.9) and –O–H (from δ 3.7).
The doublet CH₃ and the quartet CH are neighbours, so join them: CH₃–CH–. That CH is the one on a C–O, and it carries the O–H — so it is a CH₃–CH(OH)– unit (confirmed by the δ 75 C–OH in the ¹³C).
That CH still has one bond spare. The only fragment left is the methyl ketone, so attach it there: CH₃–CO–CH(OH)–CH₃.
Check against everything: C₄H₈O₂ (8 H, 4 C), four carbon environments in the ¹³C, and both the C=O and O–H the IR demanded. All accounted for.

So compound X is CH₃COCH(OH)CH₃ — 3-hydroxybutan-2-one.

How to present it — a top-band answer

Because this is levels-marked, the examiner is looking for a coherent chain that moves through all three stages. This is what a Level 3 answer reads like:

🎯 Model answer — deducing the structure of X

Infrared — the broad peak at 3400 cm⁻¹ shows an O–H (alcohol); the peak at 1720 cm⁻¹ shows a C=O.

¹H NMR — take each peak in turn and draw the fragment it gives:

δ 2.1 (singlet, 3 H) → a CH₃ with no neighbouring H, so next to the C=O:

δ 1.2 (doublet, 3 H) → a CH₃ next to a single H (a CH):

δ 3.9 (quartet, 1 H) → that CH — one H on a carbon bonded to O:

δ 3.7 (singlet, 1 H) → the hydroxyl O–H:

The integrations total 1 + 1 + 3 + 3 = 8 H, so the molecular formula is C₄H₈O₂.

¹³C NMR — four peaks = four carbon environments: δ 210 (C=O), δ 75 (C–OH), and δ 25 and 20 (the two CH₃ carbons).

Structure — the fragments join into 3-hydroxybutan-2-one, CH₃COCH(OH)CH₃:

How the marks are awarded

The biggest misconception is that the right structure earns full marks automatically. It doesn’t. This question is marked by levels of response, so the band depends on how completely and coherently you cover the three stages:

The levels-of-response bands

Level 3 (5–6): all three stages covered, generally correct, with a logical progression from the IR clue through the two NMR spectra to the structure.
Level 2 (3–4): all three stages touched but incomplete or with slips, or two stages done fully.
Level 1 (1–2): isolated correct statements, not tied into a line of reasoning.

That is why a correct structure with thin working is capped low — the marks live in the chain of reasoning, so write every stage down.

The mistakes that cost the marks

Precision points — the marks that get dropped

Reading splitting the wrong way: lines = neighbours + 1, so a triplet means 2 H next door, not 3.
Counting environments wrong — equivalent H share one peak, so symmetry lowers the count.
Letting an –OH or –NH split its neighbour — it usually doesn’t couple, and often shows as a broad singlet.
Ignoring the molecular formula — your integration H-total must add up to it.
Treating the IR and NMR clues in isolation, or not accounting for every signal — examiner reports flag both as reasons answers stall below the top level.

Practise until it’s automatic — the NMR analyser

Reading about the method is not the same as owning it. This is the same interactive I built for my A-Level NMR revision notes, dropped in here so you can drill it now. Read a spectrum works exactly like the exam: take each peak in turn, fix its integration, splitting and Data-Booklet environment, collect the exam-answer lines, then sketch the structure before revealing it. Predict a spectrum runs the drill in reverse. The ¹³C drill does the lighter carbon version — count the environments, place the peaks.

First revise the topic, then drill it

If the analyser is exposing gaps — you’re unsure what a shift means, or how splitting arises — go back to the full notes before you keep drilling. My A-Level revision page covers chemical shift and Table B, integration, the n+1 splitting rule, the ¹³C version and more worked spectra, all with the interactive built in.

Revise the topic first

AQA A-Level NMR spectroscopy — full revision notes · chemical shift, integration, splitting, ¹³C, and the shift tables. Free, no sign-up, works on your phone.

Common questions

How do you work out a structure from an NMR spectrum?

One peak at a time. Integration gives the number of hydrogens in that environment, splitting gives the number of neighbours (lines − 1), and the chemical shift — read from Table B of the Data Booklet — gives the environment. Write each peak as a fragment, then join the fragments to fit the molecular formula.

How many marks are NMR questions worth?

It varies with the question — a short “deduce and explain” may credit each reasoned step, while a longer structure-deduction is often marked by levels of response, where the quality of your explanation sets the band. In both cases the reasoning is what’s marked, not just the final structure, so always show it.

Do I have to memorise the chemical shift values?

No — AQA gives you Table B (proton) and Table C (carbon-13) in the Data Booklet. You’re tested on using them, not recalling them, so revise with the table open.

What’s the difference between ¹³C and ¹H NMR questions?

Carbon-13 is lighter: count the carbon environments (equivalent carbons share a peak) and place each in a Table C region. Proton NMR carries more per peak — integration and splitting — so it’s where the full method earns the most.