OxfordAQA A-Level Chemistry: Nuclear magnetic resonance spectroscopy (9620) – 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.

This set tests section 3.3.15, Nuclear magnetic resonance spectroscopy (3.3.15.1), of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Every outcome here is International A-level only. Unit 4 examines it, and Unit 5 (practical and synoptic) may also include it.

Related pages: the study guide, the revision notes, the Organic analysis practice set, the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

Use these approximate shift ranges, written for this set. In the exam, use the Chemistry data booklet.

¹H environment	δ / ppm	¹³C environment	δ / ppm
Alkyl H	0.7–1.7	Alkyl C	5–45
H on C next to C=O	2.0–3.0	C next to C=O	20–55
H on C bonded to Cl or Br	3.0–4.5	C bonded to Cl or Br	30–75
H on C bonded to O	3.1–4.3	C bonded to O	50–90
Alcohol O–H (variable)	1.0–5.0	C=O, acid or ester	160–185
Acid COOH	10.0–12.0	C=O, aldehyde or ketone	190–220

For infrared, take alcohol O–H as a broad absorption at about 3230–3550 cm⁻¹ and C=O at about 1680–1750 cm⁻¹.

1. Before a spectrum is run, a little TMS is mixed into the sample.

(a) Give the molecular formula of TMS. [1] (b) Explain two properties that make TMS suitable as the standard. [2]

2. A chemist has three possible solvents for a ¹H NMR spectrum: CDCl₃, CCl₄ and ethanol, C₂H₅OH. Explain why the first two are suitable and ethanol is not. [3]

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

(a) pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃ [1] (b) 2,3-dimethylbutane, (CH₃)₂CHCH(CH₃)₂ [1] (c) methyl 2-methylpropanoate, (CH₃)₂CHCOOCH₃ [1] (d) 1,3-dichloropropane, ClCH₂CH₂CH₂Cl [1]

4. Compound J, C₄H₇ClO₂, is an ester. Its ¹H NMR spectrum is:

δ / ppm	Splitting	Integration trace height / mm
4.2	quartet	16
4.1	singlet	16
1.3	triplet	24

(a) Work out how many H atoms each signal represents. [2] (b) Deduce the structure of J, using the splitting and the shifts. [3]

5. Predict the ¹H NMR spectrum of methoxyethane, CH₃OCH₂CH₃. For each signal give an approximate chemical shift range, the splitting and the relative integration. [3]

6.

(a) Explain why chemists use several analytical techniques, rather than one, to confirm the structure of a new compound. [1] (b) Suggest two reasons why a ¹³C NMR spectrum is easier to interpret than a ¹H NMR spectrum. [2]

7. A student writes: “The CH₃ signal in a spectrum is a doublet, so the CH₃ group must be next to a carbon with 2 hydrogen atoms.” Correct the statement and explain the rule. [2]

8. Compound K has a molecular ion at m/z = 70 and the formula C₅H₁₀. It does not decolourise bromine water. Its ¹³C spectrum has one line at δ 26, and its ¹H spectrum has one singlet. Identify K and explain why pent-1-ene is ruled out by the NMR data. [3]

9. Compound L has the formula C₆H₁₂O₂ and the following data.

• Infrared: broad absorption at 3450 cm⁻¹ and strong absorption at 1710 cm⁻¹.
• ¹³C NMR: five lines, at δ 210, 70, 54, 32 and 29.
• ¹H NMR: four singlets.

δ / ppm	Integration trace height / mm
3.8	3
2.6	6
2.2	9
1.2	18

(a) Convert the trace heights into numbers of H atoms. [2] (b) Deduce the structure of L. Explain how each set of data supports your answer. [6]

10. 1-chlorobutan-2-one, ClCH₂COCH₂CH₃, and 4-chlorobutan-2-one, CH₃COCH₂CH₂Cl, are isomers.

(a) Predict the ¹H NMR spectrum of 1-chlorobutan-2-one (shift, splitting, integration). [3] (b) Do the same for 4-chlorobutan-2-one. [3] (c) State the feature of the spectra that most clearly distinguishes the two isomers. [1]

11. Compound M, C₅H₁₁Cl, gives these spectra.

• ¹³C NMR: four lines, at δ 71, 39, 32 and 9.
• ¹H NMR:

δ / ppm	Splitting	Integration trace height / mm
1.7	quartet	10
1.6	singlet	30
1.0	triplet	15

(a) Work out the H count for each signal. [2] (b) Deduce the structure of M, explaining your reasoning. [4] (c) Explain why the ¹³C spectrum rules out 1-chloro-2,2-dimethylpropane, (CH₃)₃CCH₂Cl. [1]

Answers

1. (a) Si(CH₃)₄, or SiC₄H₁₂ [1] (b) All 12 H are equivalent, so it gives a single intense signal [1]. Its signal lies upfield of (lower δ than) nearly every organic signal, so the two do not clash (also accept: inert, so it does not react with the sample; or volatile, so it is easily removed) [1]. [3] Examiner insight: Give each property with its consequence; a bare “it is inert” may not earn the mark.

2. CDCl₃ contains deuterium instead of ¹H, and deuterium gives no signal in a ¹H spectrum [1]. CCl₄ contains no hydrogen at all [1]. Ethanol contains ¹H and is present in large excess, so its signals would swamp the sample’s [1]. [3] Examiner insight: Calling the solvents “unreactive” misses the point: they give no ¹H signal.

3. (a) 3 (the two ethyl groups are equivalent) [1] (b) 2 (four equivalent CH₃ and two equivalent CH) [1] (c) 4 (C=O, O–CH₃, CH, and the two equivalent CH₃ on the CH) [1] (d) 2 (two equivalent CH₂Cl carbons and the central CH₂) [1] [4] Examiner insight: Mark equivalent carbons on a displayed formula first; a near-miss count earns nothing.

4. (a) 16 : 16 : 24 = 2 : 2 : 3 [1]. 2 + 2 + 3 = 7, which equals the 7 H in C₄H₇ClO₂, so the signals are 2H (δ 4.2), 2H (δ 4.1) and 3H (δ 1.3) [1]. (b) The 2H quartet and 3H triplet are an ethyl group, CH₃CH₂– [1]. The quartet is at δ 4.2, so this CH₂ is bonded to the ester O: –OCH₂CH₃ [1]. The 2H singlet at δ 4.1 is a CH₂ with no H neighbours, bonded to Cl and next to C=O, so J is ClCH₂COOCH₂CH₃ (ethyl chloroethanoate) [1]. [5] Examiner insight: Show that the scaled ratio adds up to the formula’s H count, so the method is seen.

5. CH₃–O: δ 3.1–4.3, singlet (the O has no H), integration 3 [1]. O–CH₂: δ 3.1–4.3, quartet (next to CH₃), integration 2 [1]. CH₃ of the ethyl group: δ 0.7–1.7, triplet (next to CH₂), integration 3 [1]. [3] Examiner insight: A signal described without its δ range, its pattern or its H count is incomplete, and that signal’s mark can be lost.

6. (a) Each technique gives different information, and only together do they fix one structure [1]. (b) ¹³C signals are not split, so each environment gives one line [1]. The ¹³C δ range is much wider than the ¹H range, so lines rarely overlap [1]. [3] Examiner insight: “It is easier to read” repeats the question; name the feature that makes it simpler.

7. A doublet means the adjacent carbon carries 1 H, not 2 [1]. A signal is split into n + 1 peaks by n non-equivalent H on the adjacent carbons, so 2 neighbouring H would give a triplet [1]. [2] Examiner insight: State the rule in terms of H on adjacent carbons; rules worded around the group’s own H are marked as wrong.

8. K is a saturated compound with all 5 C and all 10 H equivalent, so it is cyclopentane [1]. Pent-1-ene has five different carbon environments, so it would give 5 ¹³C lines, not 1 [1]. It would also give several ¹H signals, including H on C=C at δ 4.5–6.0, not a single singlet [1]. [3] Examiner insight: The question asks for NMR evidence; the bromine water result alone does not answer it.

9. (a) 3 : 6 : 9 : 18 = 1 : 2 : 3 : 6 [1]. 1 + 2 + 3 + 6 = 12, matching the 12 H of C₆H₁₂O₂, so the signals are 1H (δ 3.8), 2H (δ 2.6), 3H (δ 2.2) and 6H (δ 1.2) [1]. (b) The broad IR band at 3450 cm⁻¹ is an alcohol O–H, and 1710 cm⁻¹ is C=O [1]. The ¹³C line at δ 210 shows a ketone (or aldehyde) C=O, and there is no ¹H signal at δ 9–10, so it is a ketone [1]. The ¹³C line at δ 70 is a carbon bonded to O, the C–OH [1]. The 6H singlet means a pair of identical methyls on a carbon bearing no H; the 1H singlet at δ 3.8 is the O–H; so the fragment is (CH₃)₂C(OH)– [1]. The 3H singlet at δ 2.2 is CH₃CO–, and the 2H singlet at δ 2.6 is a CH₂ next to C=O with no H neighbours [1]. L is CH₃COCH₂C(CH₃)₂OH, 4-hydroxy-4-methylpentan-2-one, which has five carbon environments, matching the five ¹³C lines [1]. [8] Examiner insight: Name the signal behind each deduction; “the NMR shows” links no evidence.

10. (a) ClCH₂: δ 3.0–4.5, singlet (neighbour is C=O), 2H [1]. CH₂ of the ethyl group: δ 2.0–3.0, quartet, 2H [1]. CH₃: δ 0.7–1.7, triplet, 3H [1]. (b) CH₃CO: δ 2.0–3.0, singlet, 3H [1]. COCH₂: δ 2.0–3.0, triplet, 2H [1]. CH₂Cl: δ 3.0–4.5, triplet, 2H [1]. (c) Splitting: only 1-chlorobutan-2-one shows a quartet (from its ethyl group); 4-chlorobutan-2-one shows two triplets instead [1]. [7] Examiner insight: Both isomers give three signals, so “the number of peaks” cannot earn part (c).

11. (a) 10 : 30 : 15 = 2 : 6 : 3 [1]. 2 + 6 + 3 = 11, matching the 11 H in C₅H₁₁Cl, so 2H, 6H and 3H [1]. (b) The 2H quartet and 3H triplet are an ethyl group, CH₃CH₂– [1]. The 6H singlet is two equivalent CH₃ groups on a carbon that has no H [1]. The ¹³C line at δ 71 is the carbon bonded to Cl, and it carries no H, because no ¹H signal is in the 3.0–4.5 range [1]. M is (CH₃)₂CClCH₂CH₃, 2-chloro-2-methylbutane, with four carbon environments [1]. (c) 1-chloro-2,2-dimethylpropane has only three carbon environments, so it would give 3 lines, not 4 [1]. [7] Examiner insight: A missing signal is evidence too: no ¹H signal at 3.0–4.5 points to a tertiary chloride.

Where marks are usually lost

• Miscounting environments because equivalent CH₃ or CH₂ groups were not spotted.
• Applying n+1 to the group’s own H instead of the H on adjacent carbons.
• Giving a CH₂ next to C=O or O a splitting when its other neighbour carries no H.
• Stopping at the simplest integration ratio when the formula requires scaling.
• Using the NMR alone in a combined question and ignoring the infrared or molecular formula.
• Giving a structure without checking it against the molecular formula.

Next steps

• NMR revision notes
• NMR study guide
• OxfordAQA A-Level Chemistry course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. This set covers section 3.3.15 Nuclear magnetic resonance spectroscopy (3.3.15.1), International A-level only.