A Level Chemistry: NMR Spectroscopy — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: NMR Spectroscopy revision notes

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Section A

1. State why TMS is used as a reference standard, giving two properties. [3]

2. State the n+1 rule and say precisely what n counts. [2]

3. Explain why a deuterated solvent such as CDCl₃ is used. [2]

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Section B

4. Predict the number of peaks in the ¹³C NMR spectrum of:

(a) propan-2-ol, CH₃CH(OH)CH₃ [1]

(b) ethyl ethanoate, CH₃COOCH₂CH₃ [1]

(c) 1,4-dimethylbenzene [1]

(d) Explain your answer to (c). [2]

5. A compound of molecular formula C₃H₆O₂ gives the following ¹H NMR data.

δ / ppm	Integration	Splitting
1.2	3	triplet
2.4	2	quartet
11.2	1	broad singlet

(a) Identify the functional group indicated by the peak at 11.2 ppm. [1]

(b) Explain what the triplet at 1.2 ppm indicates about its neighbours. [2]

(c) Deduce the structure of the compound, showing your reasoning. [3]

(d) Describe a simple test to confirm the peak at 11.2 ppm, and state what would be observed. [2]

6. Ethanol gives three peaks in its ¹H NMR spectrum.

(a) State the splitting pattern and integration for each. [3]

(b) Explain why the OH peak is not split. [2]

7. Propan-1-ol (CH₃CH₂CH₂OH) and propan-2-ol ((CH₃)₂CHOH) are structural isomers with the same molecular formula.

(a) Predict the number of ¹³C NMR peaks for propan-1-ol. [1] (b) Explain why propan-1-ol and propan-2-ol give different numbers of peaks despite sharing a molecular formula. [2]

8. An amine, R–NH₂, shows a broad ¹H NMR peak. Describe a test using D₂O to confirm this peak arises from N–H protons, and state the expected observation. [2]

9. A ¹H NMR spectrum shows peak areas in the ratio 6:4:2. Explain why this does not necessarily mean the molecule has exactly 6, 4 and 2 of each type of proton. [2]

10. Shaking a sample with D₂O genuinely exchanges each labile H for D in the molecules present (e.g. R–OH + D₂O ⇌ R–OD + HOD) — it isn’t an illusion confined to the printed spectrum. Does this amount to a permanent change to the compound’s carbon-skeleton structure? Explain. [2]

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Answers

1. Any two properties [1] [1] plus the reason [1]: it is inert and does not react with the sample; it is volatile, so it is easily removed; it is non-toxic; it gives a single sharp peak because all twelve protons are equivalent, well away from most other signals — so it is a convenient zero point.

2. A signal is split into n + 1 lines [1], where n is the number of hydrogen atoms on the adjacent carbon atoms — not on the same carbon [1].

3. An ordinary solvent containing hydrogen would produce its own large proton peaks [1], obscuring those of the sample; deuterium is not detected in ¹H NMR [1].

4. (a) 2 [1]. (b) 4 [1]. (c) 3 [1].

(d) The molecule is symmetrical [1], so the two methyl carbons are equivalent, and the ring carbons fall into only two distinct environments — those bearing methyl groups and those bearing hydrogen — giving three environments in total [1].

5. (a) Carboxylic acid (–COOH) [1].

(b) It has two hydrogen atoms on the adjacent carbon [1], since 2 + 1 = 3 lines — indicating a neighbouring CH₂ group [1].

(c) Integration 3 : 2 : 1 gives CH₃, CH₂ and COOH [1]. The triplet–quartet pair indicates an adjacent CH₃CH₂ unit [1]. The structure is CH₃CH₂COOH — propanoic acid [1].

(d) Add a few drops of D₂O and re-run the spectrum [1]; the peak at 11.2 ppm disappears, because the labile hydrogen is exchanged for deuterium, which is not detected [1].

6. (a) CH₃ — triplet, 3H [1]; CH₂ — quartet, 2H [1]; OH — singlet, 1H [1].

(b) The OH proton exchanges rapidly with the solvent [1], so it does not couple with neighbouring protons and appears as a single unsplit peak [1].

7. (a) 3 peaks — the terminal CH₃, the central CH₂, and the CH₂ attached to OH are all in different environments [1]. (b) In propan-2-ol the two CH₃ groups are equivalent by symmetry, both attached identically to the central CHOH carbon, giving only 2 distinct environments; propan-1-ol has no such symmetry, so all three carbons are distinct [1] — this is exactly why ¹³C NMR can distinguish structural isomers that share a molecular formula [1].

8. Add a few drops of D₂O and re-run the spectrum [1]; the broad N–H peak disappears, since the N–H proton exchanges for deuterium, which is not detected in ¹H NMR [1].

9. Peak area gives only a ratio, not an absolute count [1]; a 6:4:2 ratio is equally consistent with the simplest 3:2:1 ratio (or any other multiple) — the actual numbers must be fixed using the molecular formula, often from mass spectrometry [1].

10. No, not to the carbon skeleton — the exchange R–OH + D₂O ⇌ R–OD + HOD is a real, reversible chemical equilibrium that does swap H for D throughout the sample, not merely something that vanishes once the spectrum is recorded [1]; but it only changes which isotope of hydrogen is attached at labile O–H/N–H positions, leaving every C–H bond and the carbon connectivity completely unchanged, and the exchange itself can run in reverse if the sample is later exposed to plenty of ordinary water [1].

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Where marks are usually lost

• Counting all carbons rather than distinct carbon environments.
• Applying the n+1 rule to hydrogens on the same carbon.
• Treating integration as an absolute count rather than a ratio.
• Expecting OH to be split.
• Forgetting that ¹³C NMR shows no splitting at this level.
• Assuming isomers with the same molecular formula must give the same number of ¹³C peaks — symmetry, not formula, determines the number of distinct environments.
• Forgetting D₂O exchange identifies both O–H and N–H protons, not only O–H.
• Treating a peak-area ratio as an absolute proton count rather than a scalable ratio.
• Believing D₂O exchange alters the molecule’s carbon-skeleton structure, rather than being a real but reversible swap of isotope at labile O–H/N–H positions only.