OCR A Level Chemistry A: Analytical techniques (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 4.2.4 Analytical techniques in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, testing every outcome from (a) to (h) in Module 4. H432 is the full A level, and the longer problems, Questions 8 to 10, combine elemental analysis, mass spectra and IR spectra.

Revise first with the Analytical techniques study guide and the Analytical techniques revision notes. See the OCR A Level Chemistry hub for related topics and keep score on the printable H432 checklist. The free diagnostics give a wider check.

Data for all questions. Relative atomic masses: carbon 12.0, hydrogen 1.0, oxygen 16.0. Use these IR absorptions:

Bond	Wavenumber / cm⁻¹
O–H (alcohol)	3200–3600, broad
O–H (carboxylic acid)	2500–3300, very broad
C–H	2850–3100
C=O	1630–1820
C=C	1620–1680
C–O	1000–1300

Questions

1. Infrared spectroscopy.

(a) State what happens to a covalent bond when it absorbs infrared radiation of a matching frequency. [1] (b) Explain why almost every organic compound absorbs near 3000 cm⁻¹, and why this is not used to identify a functional group. [2]

2. Carbon dioxide, water vapour and methane in the atmosphere all absorb infrared radiation.

(a) For each gas, name the bond responsible. [2] (b) Explain how this absorption is linked to global warming. [2] (c) Describe one resulting change in energy use. [1]

3. IR absorption is used in breathalysers and roadside monitors.

(a) Explain how an IR breathalyser measures the concentration of ethanol in a breath sample. [2] (b) Suggest why such an instrument measures absorption at more than one wavenumber. [1] (c) Name two gases from car emissions that can be monitored by IR absorption. [1]

4. A colourless liquid is cyclopentanol, cyclopentanone or cyclopentanecarboxylic acid. Its IR spectrum has absorptions at 2880–2970 cm⁻¹ and a strong absorption at 1745 cm⁻¹. There is no broad absorption anywhere between 2500 and 3600 cm⁻¹.

(a) Identify the liquid. Explain your answer. [2] (b) Describe two absorptions that together would identify cyclopentanecarboxylic acid. [2]

5. A student oxidises 2-methylbutan-1-ol using acidified potassium dichromate(VI). Explain what the product’s IR spectrum would show to confirm oxidation, and how it would tell 2-methylbutanal from 2-methylbutanoic acid. [3]

6. Predict the IR spectrum of pent-3-en-2-one, CH₃COCH=CHCH₃. Give two expected absorptions with ranges, and one you would not expect. [3]

7. A student writes: “The molecular ion peak is always the last peak on the right of a mass spectrum, and its m/z value is the relative molecular mass.” Evaluate this statement. [2]

8. Hydrocarbon H contains 83.7% carbon and 16.3% hydrogen by mass. Its mass spectrum has peaks at m/z 86 (with a very small peak at 87), 71, 43 (the tallest peak) and 15. There are no peaks at m/z 29 or 57.

(a) Calculate the empirical formula of H. [2] (b) Deduce the molecular formula of H. [1] (c) Identify the ions responsible for the peaks at m/z 71 and 43. [2] (d) Write an equation showing how the ion at m/z 43 forms from the molecular ion. [1] (e) Use the missing peaks to suggest the structure and name of H. Explain your answer. [2]

9. Alcohol D has the molecular formula C₅H₁₂O. Its IR spectrum shows a broad absorption at 3350 cm⁻¹, absorptions at 2870–2960 cm⁻¹ and 1040 cm⁻¹, and nothing between 1630 and 1820 cm⁻¹. Its mass spectrum has peaks at m/z 88 (very small), 73, 57 (the tallest), 31 and 15, with no peaks at 29 or 43.

(a) Explain how the IR spectrum shows that D is an alcohol and not a carbonyl compound. [1] (b) Identify the ions at m/z 31 and 57, and the radical lost when each forms from the molecular ion. [2] (c) Suggest the structure and name of D. Explain how the missing peaks help. [2] (d) Refluxing D with an excess of acidified potassium dichromate(VI) gives E. Name E and give the m/z of its molecular ion peak. [2] (e) Describe how the IR spectrum of E differs from that of D. [1]

10. Compound K contains 55.8% carbon, 7.0% hydrogen and 37.2% oxygen by mass. Its mass spectrum has peaks at m/z 86 (with a very small peak at 87), 71, 45 and 41. In its IR spectrum there is a very broad band across 2500–3300 cm⁻¹, a strong absorption at 1700 cm⁻¹ and a medium one at 1650 cm⁻¹.

(a) Calculate the empirical formula of K. [3] (b) Deduce the molecular formula of K. [1] (c) Deduce the functional groups in K from the IR spectrum. [3] (d) Identify the ions at m/z 45 and 41. [2] (e) Use the peak at m/z 71 to explain why K is not but-3-enoic acid, CH₂=CHCH₂COOH, and suggest two structures for K that fit all the data. [2]

Answers

1. (a) The bond absorbs the energy and vibrates (stretches or bends) more [1]. (b) C–H bonds absorb at 2850–3100 cm⁻¹ [1]. Almost all organic compounds contain C–H bonds, so the peak does not tell one functional group from another [1]. Examiner insight: “The bond breaks” gains nothing; IR makes the bond vibrate more.

2. (a) C=O in CO₂ and O–H in H₂O [1]; C–H in CH₄ [1]. (b) The bonds absorb IR given out by the Earth’s surface and vibrate more [1]. The energy is re-emitted in all directions, some back towards the surface, so less escapes to space and the lower atmosphere warms [1]. (c) Policies to generate more energy from renewable sources such as wind or solar instead of fossil fuels [1]. Examiner insight: “Trap heat” alone misses the bond absorption being tested.

3. (a) IR passes through the breath and bonds in ethanol, such as C–H and C–O, absorb at their own wavenumbers [1]. The greater the absorption, the greater the ethanol concentration [1]. (b) Other substances in breath may absorb at one wavenumber, so a second reading helps tell ethanol apart [1]. (c) Carbon monoxide, CO, and nitrogen monoxide, NO [1]. Examiner insight: In (a), link the size of the absorption to concentration.

4. (a) Cyclopentanone [1]. The strong absorption at 1745 cm⁻¹ is C=O, and the lack of any broad O–H absorption rules out the alcohol and the acid [1]. (b) A strong C=O absorption in 1630–1820 cm⁻¹ [1] together with a very broad O–H absorption across 2500–3300 cm⁻¹ [1]. Examiner insight: Say what is absent; the missing O–H rules out two options.

5. A C=O absorption appears at 1630–1820 cm⁻¹ in either product, showing oxidation [1]. For 2-methylbutanal, the broad alcohol O–H at 3200–3600 cm⁻¹ has gone and no O–H absorption remains [1]. For 2-methylbutanoic acid, a very broad O–H absorption across 2500–3300 cm⁻¹ is present with the C=O [1]. Examiner insight: “The O–H peak changes” is too vague; give range and shape for each.

6. A strong C=O absorption at 1630–1820 cm⁻¹ [1]. A C=C absorption at 1620–1680 cm⁻¹ [1]. No broad O–H absorption in 2500–3600 cm⁻¹ [1]. Examiner insight: A C–H peak is true but not specific, so choose other bonds.

7. It is not always the last peak: a very small M+1 peak, one unit higher, comes from molecules that contain a carbon-13 atom [1]. The molecular ion peak is the tall peak at highest m/z apart from M+1, and its m/z equals Mr because the ions are singly charged [1]. Examiner insight: “Evaluate” needs a judgement on each part of the statement.

8. (a) C: 83.7 ÷ 12.0 = 6.98; H: 16.3 ÷ 1.0 = 16.3 [1]. Ratio 1 : 2.34; multiply by 3 to give 3 : 7, so the empirical formula is C₃H₇ [1]. (b) C₃H₇ has mass 43.0; 86 ÷ 43.0 = 2, so the molecular formula is C₆H₁₄ [1]. (c) m/z 71 is C₅H₁₁⁺, formed by loss of •CH₃ [1]. m/z 43 is C₃H₇⁺ [1]. (d) C₆H₁₄⁺ → C₃H₇⁺ + •C₃H₇ [1] (e) No peak at 29 (C₂H₅⁺) or at 57 (loss of •C₂H₅) suggests H has no ethyl group [1]. H is 2,3-dimethylbutane, (CH₃)₂CHCH(CH₃)₂; breaking the central C–C bond gives two C₃H₇ pieces, which fits the tall peak at 43 [1]. Examiner insight: Rounding 2.34 to 2 gives CH₂, which is wrong; multiply by 3.

9. (a) Broad O–H absorption at 3350 cm⁻¹ with no C=O absorption in 1630–1820 cm⁻¹ [1]. (b) m/z 31 is CH₂OH⁺, formed by loss of •C₄H₉ [1]. m/z 57 is C₄H₉⁺, formed by loss of •CH₂OH [1]. (c) CH₂OH⁺ shows a primary alcohol, and no peaks at 29 or 43 suggest no ethyl or propyl chain, so the C₄H₉ group is (CH₃)₃C– [1]. D is 2,2-dimethylpropan-1-ol, (CH₃)₃CCH₂OH [1]. (d) E is 2,2-dimethylpropanoic acid [1]; molecular ion peak at m/z 102 [1]. (e) E shows a strong C=O absorption at about 1700 cm⁻¹, and its O–H becomes a very broad absorption across 2500–3300 cm⁻¹ [1]. Examiner insight: Show fragment ions with a + charge and lost pieces as neutral radicals.

10. (a) C: 55.8 ÷ 12.0 = 4.65; H: 7.0 ÷ 1.0 = 7.0; O: 37.2 ÷ 16.0 = 2.325 [1]. Divide by 2.325: C 2.00, H 3.01, O 1.00 [1]. Empirical formula C₂H₃O [1]. (b) C₂H₃O has mass 43.0; 86 ÷ 43.0 = 2, so C₄H₆O₂ [1]. (c) Strong C=O at 1700 cm⁻¹ [1] with a very broad O–H at 2500–3300 cm⁻¹ shows a carboxylic acid group [1]. The absorption at 1650 cm⁻¹ shows a C=C bond [1]. (d) m/z 45 is COOH⁺ [1]. m/z 41 is C₃H₅⁺, formed by loss of •COOH (86 − 45 = 41) [1]. (e) 86 − 71 = 15, so a CH₃ group is lost; but-3-enoic acid has no CH₃ group [1]. K is but-2-enoic acid, CH₃CH=CHCOOH, or 2-methylpropenoic acid, CH₂=C(CH₃)COOH; both fit all the data given [1]. Examiner insight: If two isomers fit, name both with a reason rather than guessing.

Where marks are usually lost

• Using the C–H absorption near 3000 cm⁻¹ as evidence for a functional group.
• Quoting an O–H absorption without saying whether it is the alcohol range or the very broad acid range.
• Saying greenhouse gases “trap heat” without naming the C=O, O–H or C–H bond that absorbs IR.
• Not linking breathalyser absorption to ethanol concentration.
• Taking the M+1 peak as the molecular ion.
• Missing the positive charge on fragment ions.
• Rounding a ratio of 2.33 or 2.34 instead of multiplying by 3.
• Ignoring missing peaks when choosing between isomers.

Next steps

• Recall tables and a quick self-test: Analytical techniques revision notes
• Full explanations and worked examples: Analytical techniques study guide
• All H432 topics: OCR chemistry hub
• Track your progress: printable checklist
• Whole-course check: 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 4.2.4 Analytical techniques, learning outcomes (a) to (h). Published by OCR.