OCR A Level Chemistry A: Spectroscopy (H432) – 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 covers 6.3.2 Spectroscopy in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Questions target each outcome, 6.3.2(a) to (e). Questions 9 and 10 combine elemental analysis, mass spectra, IR and NMR.

Background reading: the study guide and revision notes. Also useful: the OCR A Level Chemistry hub, the H432 checklist and the free diagnostics.

Data for all questions:

Carbon	δ / ppm	Proton	δ / ppm
alkyl C–C	5–55	alkyl C–H	0.7–1.8
C–Cl	30–70	H–C–C=O	2.0–2.7
C–N	30–65	H–C–N	2.3–3.0
C–O	50–90	H–C–Cl	3.0–4.5
alkene C=C	115–150	H–C–O	3.3–4.5
aromatic C	110–160	aromatic H	6.5–8.2
acid, ester C=O	160–185	carboxylic acid O–H	10.0–12.5
aldehyde, ketone C=O	190–220	aldehyde H	9.4–10.0
C≡N	110–125	alcohol O–H, amine N–H	0.5–5.0

IR: alcohol O–H 3200–3600 cm⁻¹ (broad); acid O–H 2500–3300 cm⁻¹ (very broad); N–H 3300–3500 cm⁻¹; C=O 1630–1820 cm⁻¹. Use Ar values of 12.0 for carbon, 1.0 for hydrogen, 14.0 for nitrogen and 16.0 for oxygen.

Questions

1. State the chemical shift of the tetramethylsilane (TMS) signal, and give two reasons why TMS is the NMR standard. [3]

2. A student dissolves a sample in CHCl₃ for proton NMR. Explain why this is a poor choice and state what to use instead. [2]

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

(a) propane-1,2,3-triol, HOCH₂CH(OH)CH₂OH [1] (b) 4-methylphenol, CH₃C₆H₄OH [1] (c) 3-methylbut-1-ene, CH₂=CHCH(CH₃)₂ [1] (d) hexane-2,5-dione, CH₃COCH₂CH₂COCH₃ [1]

4. Predict the high resolution proton NMR spectrum of butanoyl chloride, CH₃CH₂CH₂COCl. Give the number of signals, their relative areas and the splitting of each, with reasons. [4]

5. Several isomers have the formula C₄H₈.

(a) Isomer A gives two carbon-13 lines, at δ 126 and δ 17. Identify A, and explain why it cannot be cyclobutane. [2] (b) Isomer B gives three carbon-13 lines, at δ 142, 112 and 24. Identify B, and explain why it cannot be methylcyclopropane. [2] (c) Predict the proton NMR spectrum of B. [2]

6. Compound C, C₅H₁₀O₂, is either methyl butanoate, CH₃CH₂CH₂COOCH₃, or propyl ethanoate, CH₃COOCH₂CH₂CH₃. Its proton spectrum has a 3H singlet at δ 3.7 and a 2H triplet at δ 2.3.

(a) Explain which compound C is. [3] (b) State the expected shift range of the other compound’s 3H singlet. [1]

7. Compound D, C₅H₁₀O₂, has IR absorptions at 3450 cm⁻¹ (broad) and 1710 cm⁻¹ (strong). Its proton spectrum has three singlets: δ 1.38 (6H), 2.22 (3H) and 3.7 (1H, broad, absent after shaking with D₂O).

(a) Deduce the structure of D, explaining the evidence. [4] (b) State the number of lines in the carbon-13 spectrum of D. [1]

8. Predict the NMR spectra of 2-chloropropanoic acid, CH₃CHClCOOH.

(a) the carbon-13 spectrum: number of lines and a shift range for each [2] (b) the proton spectrum: relative areas, shift ranges and splitting [4]

9. Compound E contains 65.8% C, 15.1% H and 19.1% N by mass. Mass spectrum: m/z 73 (small), 58 and 44 (tallest). IR: two sharp absorptions near 3300 and 3370 cm⁻¹, none at 1630–1820 cm⁻¹. Carbon-13: δ 49, 33, 23 and 10. Proton: δ 0.90 (3H, triplet), 1.05 (3H, doublet), 1.2 (2H, broad singlet, absent after shaking with D₂O), 1.35 (2H, quintet), 2.80 (1H, sextet).

(a) Show that the molecular formula of E is C₄H₁₁N. [3] (b) Deduce the structure of E from the IR and NMR data. [5] (c) Suggest the formula of the ion responsible for the peak at m/z 44. [1]

10. Compound F contains 72.0% C, 6.7% H and 21.3% O by mass. Mass spectrum: m/z 150, 105 (tallest) and 77. IR: strong absorption at 1720 cm⁻¹, no broad absorption at 2500–3600 cm⁻¹. Carbon-13: seven lines, δ 166.6, 61.0, 14.4 and four at δ 128–133. Proton: δ 1.39 (3H, triplet), 4.37 (2H, quartet), 7.4–8.1 (5H).

(a) Determine the molecular formula of F. [3] (b) Deduce the structure of F from the IR and NMR data. [4] (c) Identify the ion at m/z 105. [1]

11. The proton spectrum of compound G, C₇H₁₄O₂, is:

δ / ppm	Splitting	Step height / mm
0.92	doublet	36
1.52	quartet	12
1.68	nine peaks	6
2.04	singlet	18
4.09	triplet	12

(a) Give the proton ratio and show it fits the formula. [1] (b) Identify the groups responsible for the signals at δ 2.04 and δ 4.09. [2] (c) Explain the nine-peak signal and give the structure of G. [2]

Answers

1. δ = 0 [1]. Its 12 equivalent H give one sharp signal [1]. The signal is away from sample signals, or it is inert, or volatile and easily removed [1]. [3] Examiner insight: “It has 12 hydrogens” is not enough; say they are equivalent and give one signal.

2. The H in CHCl₃ gives a large solvent signal that swamps the sample’s signals [1]. Use a deuterated solvent such as CDCl₃, which has no ¹H [1]. [2] Examiner insight: Name a deuterated solvent; “use a different solvent” is too vague.

3. (a) 2: the two CH₂OH carbons, and the CH carbon [1]. (b) 5: CH₃, ring C–CH₃, two pairs of ring C–H, ring C–OH [1]. (c) 4: =CH₂, =CH, CH, two equivalent CH₃ [1]. (d) 3: C=O, CH₂, CH₃, each pair equivalent by symmetry [1]. Examiner insight: Sketch the molecule and mark its symmetry first; parts (b) and (d) are easy to overcount.

4. Three signals, relative areas 3 : 2 : 2 [1]. CH₃: triplet, 2 adjacent H [1]. Middle CH₂: sextet, 3 + 2 = 5 adjacent H [1]. CH₂ next to COCl: triplet, 2 adjacent H [1]. [4] Examiner insight: Give the number of adjacent H with each pattern to show the n + 1 rule being applied.

5. (a) A is but-2-ene, CH₃CH=CHCH₃ (either geometric isomer); δ 126 is alkene C [1]. Cyclobutane would give one line and no alkene carbon [1]. (b) B is 2-methylpropene, (CH₃)₂C=CH₂ [1]. Methylcyclopropane has no C=C, so it would give no line at δ 115–150 [1]. (c) Two signals, ratio 6 : 2 (3 : 1) [1]. Both are singlets, because the only carbon next to each set of protons is the central =C<, which carries no hydrogen [1]. Examiner insight: To rule out an isomer, quote the evidence it fails to match.

6. (a) The 3H singlet is a CH₃ with no adjacent H [1]. δ 3.7 is in the H–C–O range, so the CH₃ is on oxygen [1]. The δ 2.3 triplet is a CH₂ next to C=O, not on oxygen: C is methyl butanoate [1]. (b) δ 2.0–2.7 (CH₃ next to C=O) [1]. Examiner insight: Both isomers have a 3H singlet, so the shift decides what the CH₃ is bonded to.

7. (a) IR: alcohol O–H and C=O; with no aldehyde proton, a ketone [1]. The δ 3.7 signal disappears with D₂O, so it is the O–H [1]. δ 2.22 (3H, singlet) is CH₃ on C=O [1]. δ 1.38 (6H, singlet) comes from a pair of equivalent methyl groups on a carbon that carries no hydrogen, so D is CH₃COC(OH)(CH₃)₂, 3-hydroxy-3-methylbutan-2-one [1]. (b) 4 [1]. Examiner insight: In a “deduce” question, tie every signal to a group; a bare structure shows no reasoning.

8. (a) 3 lines [1]: C=O at δ 160–185, C–Cl at δ 30–70, CH₃ at δ 5–55 [1]. (b) Ratio 3 : 1 : 1 [1]. CH₃: δ 0.7–1.8, doublet (1 adjacent H) [1]. CH: δ 3.0–4.5, quartet (3 adjacent H) [1]. COOH: δ 10.0–12.5, singlet [1]. Examiner insight: Each proton environment needs shift, area and splitting; leaving one out loses credit.

9. (a) C: 65.8 ÷ 12.0 = 5.483; H: 15.1 ÷ 1.0 = 15.1; N: 19.1 ÷ 14.0 = 1.364 [1]. Dividing by 1.364 gives 4.02 : 11.07 : 1, so the empirical formula is C₄H₁₁N [1]. Its mass is 73.0, which matches the molecular ion at m/z 73, so the molecular formula is C₄H₁₁N [1]. (b) IR: N–H and no C=O, so an amine [1]. Four carbon environments, with δ 49 a C–N carbon [1]. The 2H signal removed by D₂O is NH₂: a primary amine [1]. Triplet (3H) + quintet (2H) are CH₃CH₂ next to a CH; the doublet (3H) is CH₃ on that CH, whose sextet shows 3 + 2 = 5 neighbours [1]. E is CH₃CH(NH₂)CH₂CH₃, butan-2-amine [1]. (c) CH₃CHNH₂⁺ (C₂H₆N⁺), from loss of the ethyl group [1]. Examiner insight: A “show that” answer needs the division and ratio, not the given formula repeated.

10. (a) C: 72.0 ÷ 12.0 = 6.0; H: 6.7 ÷ 1.0 = 6.7; O: 21.3 ÷ 16.0 = 1.331 [1]. Ratio 4.51 : 5.03 : 1, doubled gives C₉H₁₀O₂ [1]. Mass 150.0 matches m/z 150, so the molecular formula is C₉H₁₀O₂ [1]. (b) C=O with no O–H, and a line at δ 166.6: an ester [1]. Four aromatic lines and 5H at δ 7.4–8.1: a C₆H₅ group [1]. The 3H triplet and 2H quartet at δ 4.37: OCH₂CH₃ [1]. F is C₆H₅COOCH₂CH₃, ethyl benzoate [1]. (c) C₆H₅CO⁺ [1]. Examiner insight: Do not round a ratio such as 4.51 to 5; multiply through by 2 when a ratio ends in about .5.

11. (a) Dividing by 6 mm gives 6 : 2 : 1 : 3 : 2, total 14, matching the 14 H in C₇H₁₄O₂ [1]. (b) δ 2.04 (3H, singlet): CH₃ next to C=O with no adjacent H [1]. δ 4.09 (2H, triplet): CH₂ on oxygen next to a CH₂ [1]. (c) Nine peaks means 8 adjacent H: a CH between two CH₃ (6H) and a CH₂ (2H) [1]. G is CH₃COOCH₂CH₂CH(CH₃)₂, 3-methylbutyl ethanoate [1]. Examiner insight: Use the trace’s step heights, not peak heights, and simplify the ratio before assigning groups.

Where marks are usually lost

• Counting adjacent protons on one side only, so a sextet is called a triplet.
• Splitting a signal by an O–H or N–H proton, or by protons in the same equivalent group.
• Assigning a C=O line at δ 160–185 to a ketone.
• Saying D₂O makes a signal shift or split, rather than disappear.
• Rounding an empirical ratio of 4.5 up or down instead of doubling.
• Not checking that integration totals match the formula.

Next steps

• Spectroscopy revision notes
• Spectroscopy study guide
• Analytical techniques practice for IR and mass spectra
• Printable H432 checklist
• OCR A Level Chemistry hub
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Source document: OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.3.2 Spectroscopy, learning outcomes (a) to (e). Published by OCR.