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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions 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. All of this content is A-level only. Organic chemistry is examined on A-level Paper 2, and Paper 3 can test any content. Later questions add mass spectra and IR data.

Learn the content in the study guide and the revision notes. For IR and mass spectrometry questions, see the Organic analysis practice set. The course hub and the printable checklist list the whole course, and the free diagnostics show where your gaps are.

Questions

Where a question needs chemical shift ranges, use these approximate values, written for this practice set. For your own revision, work from the AQA Chemistry Data Booklet itself.

¹H environment	δ / ppm	¹³C environment	δ / ppm
Alkyl CH₃, CH₂, CH	0.7–1.7	Alkyl C–C	5–50
H–C next to C=O	2.0–2.7	C–O	50–90
H–C next to O	3.3–4.3	C=O in acids and esters	160–185
Aldehyde –CHO	9.0–10.0	C=O in aldehydes and ketones	190–220
Acid –COOH	10.0–12.0

For infrared, take O–H (alcohols) as a broad absorption at 3230–3550 cm⁻¹, O–H (acids) as a very broad absorption at 2500–3000 cm⁻¹, and C=O at 1680–1750 cm⁻¹.

1. Tetramethylsilane (TMS) is added to samples before NMR spectra are recorded. Give three reasons why TMS is a suitable standard. [3]

2. A student plans to dissolve a sample in trichloromethane, CHCl₃, to record its ¹H NMR spectrum. Explain why CDCl₃ or CCl₄ should be used instead. [2]

3. State the number of peaks in the ¹³C NMR spectrum of each compound.

(a) propan-1-ol, CH₃CH₂CH₂OH [1] (b) propan-2-ol, CH₃CH(OH)CH₃ [1] (c) 2,2-dimethylbutane, (CH₃)₃CCH₂CH₃ [1] (d) cyclohexane, C₆H₁₂ [1]

4. Compound P has the molecular formula C₇H₁₄O₂. Its ¹H NMR spectrum has three peaks:

δ / ppm	Splitting	Integration trace height / mm
4.1	quartet	6
1.3	triplet	9
1.2	singlet	27

(a) Use the integration data to find the number of H atoms in each environment. [2] (b) Explain the splitting of each peak. [3] (c) Use the chemical shift of the quartet to suggest a structure for P. [1]

5. 1,1-dichloroethane and 1,2-dichloroethane are isomers. Predict the ¹H NMR spectrum of each, giving the number of peaks, their relative integration and their splitting, and explain how the spectra distinguish the two compounds. [4]

6. Three ketones have the formula C₅H₁₀O: pentan-2-one, pentan-3-one and 3-methylbutan-2-one. State the number of peaks in the ¹³C NMR spectrum of each, and so explain whether ¹³C NMR can distinguish all three. [3]

7. Predict the ¹H NMR spectrum of ethanal, CH₃CHO. For each peak, give its approximate chemical shift range, its splitting and its relative integration. [3]

8. Compound Q has the formula C₄H₁₀O.

• Infrared: broad absorption at 3360 cm⁻¹.
• ¹³C NMR: two peaks, at δ 69 and δ 31.
• ¹H NMR: two singlets, with integration ratio 9 : 1.

(a) Identify the functional group shown by the infrared spectrum. [1] (b) Deduce the structure of Q, explaining how the NMR data support it. [4] (c) Explain why the ¹³C spectrum rules out butan-1-ol. [1]

9. Compound R gives a molecular ion at m/z = 102. Its molecular formula is C₅H₁₀O₂. Its infrared spectrum has a strong absorption at 1740 cm⁻¹ and no absorption from O–H. Its ¹H NMR spectrum is:

δ / ppm	Splitting	Relative integration
4.1	quartet	2
2.3	quartet	2
1.3	triplet	3
1.1	triplet	3

(a) Deduce the structure of R. Explain how each piece of data supports your answer. [6] (b) State the number of peaks you would expect in the ¹³C NMR spectrum of R. [1]

10. Compound S is an isomer of R in question 9. It also gives a molecular ion at m/z = 102.

• Infrared: a very broad absorption across 2500–3000 cm⁻¹ and a strong absorption at 1700 cm⁻¹.
• ¹³C NMR: three peaks, at δ 185, 39 and 27.
• ¹H NMR: a singlet at δ 11.9 (1H) and a singlet at δ 1.2 (9H).

(a) Use all the data to deduce the structure of S. Explain your reasoning. [6] (b) Pentanoic acid, CH₃CH₂CH₂CH₂COOH, is another isomer. Give two differences between its NMR spectra and those of S. [2]

11. Explain why ¹³C NMR spectra are simpler than ¹H NMR spectra. [2]

Answers

1. All 12 H atoms are equivalent, so it gives a single strong peak [1]. Its peak is at lower δ than almost all organic compounds, so it does not overlap the sample’s peaks [1]. It is inert, so it does not react with the sample [1]. (Being volatile, so easily removed, can replace any one of these.) [3] Examiner insight: “It gives a peak at zero” is a definition, not a reason; give a property and why it helps.

2. CHCl₃ contains ¹H, which would give a large peak that swamps or overlaps the sample’s peaks [1]. CDCl₃ contains deuterium instead, which gives no signal in a ¹H spectrum, and CCl₄ contains no H at all [1]. [2] Examiner insight: Say that the solvent gives no signal; calling it “inert” or “a good solvent” does not answer the question.

3. (a) 3 [1] (b) 2 (the two CH₃ carbons are equivalent) [1] (c) 4 (three equivalent CH₃, the quaternary C, the CH₂ and the end CH₃) [1] (d) 1 (all six carbons are equivalent) [1] Examiner insight: Mark the symmetry on a displayed formula before counting environments.

4. (a) Divide by the smallest height: 6 : 9 : 27 = 2 : 3 : 9 [1]. 2 + 3 + 9 = 14, which equals the 14 H in C₇H₁₄O₂, so the peaks are 2H (δ 4.1), 3H (δ 1.3) and 9H (δ 1.2) [1]. (b) The 2H quartet is next to a carbon with 3 H, a CH₃ [1]. The 3H triplet is next to a carbon with 2 H, a CH₂, so these two form an ethyl group [1]. The 9H singlet is three equivalent CH₃ groups on a carbon with no H, a (CH₃)₃C– group [1]. (c) δ 4.1 means the CH₂ is bonded to O, so P is the ester (CH₃)₃CCOOCH₂CH₃ (ethyl 2,2-dimethylpropanoate) [1]. [6] Examiner insight: Show the ratio and the check against the formula; a bare list of H numbers gives no evidence that the integration was used.

5. 1,1-dichloroethane, CH₃CHCl₂: two peaks, ratio 3 : 1 [1]; the CH₃ is a doublet (next to 1 H) and the CH is a quartet (next to 3 H) [1]. 1,2-dichloroethane, ClCH₂CH₂Cl: one peak [1], a singlet, because all four H are equivalent and equivalent protons do not split each other [1]. So the number of peaks (and the splitting) distinguishes them. [4] Examiner insight: Applying n+1 blindly to ClCH₂CH₂Cl gives a triplet, which is wrong; state the equivalence explicitly.

6. Pentan-2-one: 5 peaks [1]. Pentan-3-one: 3 peaks [1]. 3-methylbutan-2-one: 4 peaks, so all three give different numbers of peaks and ¹³C NMR can distinguish them [1]. [3] Examiner insight: The conclusion rests on three different counts, so one wrong count breaks it.

7. Two peaks: the –CHO H at δ 9.0–10.0, a quartet (adjacent carbon has 3 H), integration 1 [1]; the CH₃ at δ 2.0–2.7 (next to C=O), a doublet (adjacent carbon has 1 H), integration 3 [1]. Ratio 1 : 3 with both peaks identified [1]. [3] Examiner insight: Give shift, splitting and integration for each peak; leaving one out loses credit.

8. (a) Alcohol O–H (broad, 3230–3550 cm⁻¹) [1]. (b) Two ¹³C peaks for four carbons means three carbons are equivalent [1]. δ 69 is a C–O carbon and δ 31 is the three equivalent CH₃ carbons [1]. The 9H singlet is three equivalent CH₃ groups on a carbon with no H, and the 1H singlet is the O–H [1]. So Q is 2-methylpropan-2-ol, (CH₃)₃COH [1]. (c) Butan-1-ol has four different carbon environments, so it would give 4 peaks, not 2 [1]. [6] Examiner insight: Link each conclusion to a named piece of data; “the NMR fits” without saying which peak shows what does not show the reasoning asked for.

9. (a) The IR absorption at 1740 cm⁻¹ is C=O, and with no O–H the compound is an ester, not an acid [1]. There are two quartet (2H) + triplet (3H) pairs, so R contains two ethyl groups [1]. The CH₂ at δ 4.1 is bonded to O, so one ethyl group is the O–CH₂CH₃ of an ester [1]. The CH₂ at δ 2.3 is next to C=O, so the other ethyl group is attached to the C=O [1]. Structure: CH₃CH₂COOCH₂CH₃, ethyl propanoate [1]. Check: C₅H₁₀O₂, Mr = 5(12.0) + 10(1.0) + 2(16.0) = 102.0, matching m/z = 102 [1]. (b) 5 peaks (every carbon is in a different environment) [1]. [7] Examiner insight: Both ethyl groups give the same splitting, so only the chemical shifts place them; reasoning that never uses δ 4.1 and δ 2.3 cannot fix which side of the ester each one is on.

10. (a) The very broad IR absorption at 2500–3000 cm⁻¹ with C=O at 1700 cm⁻¹ shows a carboxylic acid [1]. The ¹H singlet at δ 11.9 is the –COOH proton [1]. The 9H singlet is three equivalent CH₃ groups on a carbon with no H: a (CH₃)₃C– group [1]. ¹³C: δ 185 is the acid C=O, δ 27 the three equivalent CH₃ carbons and δ 39 the central carbon, giving three environments [1]. C₅H₁₀O₂ has Mr = 102.0, matching the molecular ion [1]. S is (CH₃)₃CCOOH, 2,2-dimethylpropanoic acid [1]. (b) Pentanoic acid gives 5 ¹³C peaks, not 3 [1]. Its ¹H spectrum has 5 peaks, not 2 (also accept: no 9H singlet) [1]. [8] Examiner insight: In a combined-data question, use every data set; an answer built on the NMR alone leaves the IR and mass spectrum evidence unused.

11. In ¹³C spectra each carbon environment gives a single line, with no splitting to interpret [1]. The δ range for ¹³C is much wider than for ¹H, so peaks rarely overlap [1]. [2] Examiner insight: Two separate ideas are needed; rewording one idea does not add a second.

Where marks are usually lost

• Counting atoms instead of environments, especially in (CH₃)₃C– groups and symmetrical molecules.
• Using n+1 on the protons in the group itself instead of those on adjacent carbons.
• Splitting a CH₃ next to C=O or O, which has no H neighbours and is a singlet.
• Forgetting that equivalent protons do not split each other.
• Giving an integration ratio without scaling it to the H count of the molecular formula.
• Explaining TMS only as “the zero point” and missing the single-peak, inert and volatile points.
• Ignoring the IR or mass spectrum in a combined question.

Next steps

• Revision notes
• Study guide
• AQA A-Level Chemistry course hub
• Printable checklist
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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 questions cover section 3.3.15 Nuclear magnetic resonance spectroscopy (A-level only).