OxfordAQA A-Level Chemistry: Organic analysis (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 practises Organic analysis, section 3.3.6 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: test tube reactions and Required practical 5 (3.3.6.1), precise-mass work (3.3.6.2) and IR spectra (3.3.6.3). Section 3.3.6 is International AS content and is examined in Unit 2. Question 4 uses the acid–carbonate reaction from 3.3.9.1, an International A-level only statement.

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Data for these questions. Precise masses: C = 12.0000, H = 1.0078, O = 15.9949, N = 14.0031. IR ranges (cm⁻¹): C=O 1680–1750; C=C 1620–1680; C≡N 2220–2260; C–O 1000–1300; C–H 2850–3300; O–H in alcohols 3230–3550; O–H in acids 2500–3000; N–H 3300–3500.

Questions

1. Liquid paraffin is a mixture of alkanes. A technician suspects that a bottle of it is contaminated with alkenes. Describe a test that would show whether alkenes are present, giving the observation if they are. [2]

2. Two colourless liquids are 2-methylbutanal and pentan-2-one.

(a) Describe how Tollens’ reagent can be used to tell them apart. Give the result for each liquid. [3] (b) Explain why the tubes are warmed in a hot water bath rather than over a Bunsen flame. [1]

3. Separate samples of hexan-3-ol and hexan-3-one are heated gently with acidified potassium dichromate(VI). Give the observation for each, and explain why they differ. [3]

4. Butanedioic acid, HOOCCH₂CH₂COOH, is added to sodium carbonate solution. State what you would observe, give a test for the gas formed, and write an equation for the reaction. [3]

5. A high-resolution mass spectrum of compound J shows its molecular ion at m/z = 116.0834. J is C₆H₁₂O₂, C₇H₁₆O or C₅H₁₂N₂O. Show by calculation which is the molecular formula of J. [3]

6. Hexan-1-ol can be oxidised by acidified potassium dichromate(VI) to hexanal or to hexanoic acid.

(a) Explain how the method used decides which of these two products is obtained. [2] (b) Describe three ways in which the infrared spectrum of hexanoic acid differs from that of hexanal. Refer to bonds and wavenumber ranges. [3]

7. Butan-1-ol and 2-methylpropan-1-ol are isomers.

(a) Explain why absorptions above 1500 cm⁻¹ cannot be used to tell these two alcohols apart. [1] (b) Explain how infrared spectroscopy could still identify an unlabelled sample of one of them. [2]

8. A student heats 1-bromo-3-methylbutane with potassium cyanide to make 4-methylpentanenitrile, (CH₃)₂CHCH₂CH₂CN. After separating the organic layer, the student records its IR spectrum. It has a sharp absorption at 2250 cm⁻¹ and a broad absorption at 3340 cm⁻¹. What does this spectrum tell you about the product? Explain your answer. [4]

9. The concentration of methane in the atmosphere has risen. Explain, in terms of the bonds in the molecule, how this increase contributes to global warming. Name the bond in water vapour that acts in the same way. [4]

10. Compound W is a volatile liquid. Its high-resolution mass spectrum has a molecular ion at m/z = 70.0417.

(a) W is C₄H₆O, C₅H₁₀ or C₃H₆N₂. Use precise masses to show which formula is correct. [2] (b) W’s infrared spectrum has a strong absorption at 1692 cm⁻¹ and an absorption at 1645 cm⁻¹. There is no broad absorption above 3000 cm⁻¹. Identify the bond responsible for each absorption. [2] (c) W turns bromine water from orange to colourless, and gives a silver mirror when warmed with Tollens’ reagent. State what each result shows. [2] (d) Suggest a structure for W and name it. [2]

11. Compound X, C₇H₁₄O, gives a molecular ion at m/z = 114.1041. A student collects this evidence:

• Infrared: a broad band at 3365 cm⁻¹ and a band at 1642 cm⁻¹; nothing between 1680 and 1750 cm⁻¹.
• Bromine water: orange to colourless.
• Warm acidified potassium dichromate(VI): stays orange.

(a) Use the IR evidence to identify the functional groups present in X, and explain why X is not an aldehyde or a ketone. [3] (b) Explain what the two test tube results show. [3] (c) C₇H₁₄O allows only one double bond or one ring. Explain why X cannot be a cyclic alcohol. [1] (d) Suggest a structure for X. [2]

Answers

1. Shake a sample with bromine water [1]. If alkenes are present, the bromine water changes from orange to colourless [1]. [2] Examiner insight: An observation needs both colours; “goes clear” or “decolourises” alone loses this mark.

2. (a) Warm a sample of each liquid with Tollens’ reagent [1]. 2-methylbutanal: silver is deposited on the glass as a mirror [1]. Pentan-2-one: no change, no silver forms [1]. (b) The organic liquids are flammable, so a naked flame is a fire risk [1]. [4] Examiner insight: A distinguishing test needs the result for both compounds, the negative one included.

3. Hexan-3-ol: the solution changes from orange to green [1]. Hexan-3-one: the solution stays orange [1]. Hexan-3-ol is a secondary alcohol, so it is oxidised (to hexan-3-one) and Cr₂O₇²⁻ is reduced to Cr³⁺; hexan-3-one is a ketone and is not oxidised further [1]. [3] Examiner insight: Classify the alcohol in your explanation; an answer that only says “one reacts and one does not” does not explain the difference.

4. Effervescence (fizzing) [1]. The gas turns limewater milky, showing it is carbon dioxide [1]. HOOCCH₂CH₂COOH + Na₂CO₃ → NaOOCCH₂CH₂COONa + CO₂ + H₂O [1]. [3] Examiner insight: Both –COOH groups react, so one Na₂CO₃ is needed per acid molecule; balance the atoms before moving on.

5. C₆H₁₂O₂: 6(12.0000) + 12(1.0078) + 2(15.9949) = 116.0834 [1]. C₇H₁₆O: 7(12.0000) + 16(1.0078) + 15.9949 = 116.1197; C₅H₁₂N₂O: 5(12.0000) + 12(1.0078) + 2(14.0031) + 15.9949 = 116.0947 [1]. Only C₆H₁₂O₂ matches, so J is C₆H₁₂O₂ [1]. [3] Examiner insight: “Show by calculation” means every candidate’s total must appear; naming the right formula without the sums leaves the method unseen.

6. (a) For hexanal, distil the aldehyde out as soon as it forms, so it is not oxidised further [1]. For hexanoic acid, heat under reflux with excess oxidising agent, so the aldehyde is oxidised on to the acid [1]. (b) Only hexanoic acid shows the very broad acid O–H band across 2500–3000 cm⁻¹ [1]. Hexanoic acid shows C–O absorptions at 1000–1300 cm⁻¹, which hexanal lacks [1]. Their fingerprint regions (below about 1500 cm⁻¹) differ; C=O appears in both, so it does not distinguish them [1]. [5] Examiner insight: Quote the range with the bond each time; “the acid has an extra O–H peak” gives no range and does not show it is the acid O–H.

7. (a) Both contain the same bonds (O–H, C–H, C–O and C–C), so their absorptions above 1500 cm⁻¹ appear in the same ranges [1]. (b) Below about 1500 cm⁻¹ lies the fingerprint region, whose pattern is characteristic of one compound [1]. Compare the sample’s spectrum with reference spectra of both alcohols; the one that matches exactly identifies the sample [1]. [3] Examiner insight: Credit for fingerprinting depends on saying the spectrum is compared with known spectra; describing the region alone is not enough.

8. The absorption at 2250 cm⁻¹ lies in the C≡N range (2220–2260), so the nitrile has formed [1]. The broad absorption at 3340 cm⁻¹ lies in the O–H (alcohol) range (3230–3550) [1]. 4-methylpentanenitrile contains no O–H bond, so another substance must be present [1]. The O–H band comes from water left in the sample, or from 3-methylbutan-1-ol formed by substitution with hydroxide ions [1]. [4] Examiner insight: State that the expected product has no O–H bond before naming an impurity; the link between “unexpected band” and “impurity” carries the reasoning.

9. The Earth’s surface emits infrared radiation [1]. The C–H bonds in methane absorb some of this infrared radiation [1]. The energy is re-emitted in all directions, so some returns to the surface; more methane means more infrared absorbed and more warming [1]. In water vapour, the O–H bond absorbs infrared in the same way [1]. [4] Examiner insight: Name the bond and the radiation; “methane traps heat” without “C–H bonds absorb infrared” does not answer the “in terms of bonds” part.

10. (a) C₄H₆O: 4(12.0000) + 6(1.0078) + 15.9949 = 70.0417 [1]. C₅H₁₀ = 70.0780 and C₃H₆N₂ = 70.0530, so W is C₄H₆O [1]. (b) 1692 cm⁻¹: C=O (1680–1750) [1]. 1645 cm⁻¹: C=C (1620–1680) [1]. (c) Bromine water decolourised: W contains a C=C bond (it is unsaturated) [1]. Silver mirror: W contains an aldehyde group, –CHO [1]. (d) For example, structure CH₃CH=CHCHO [1], named but-2-enal [1]. Other named C₄H₆O structures with C=C and –CHO also fit, such as but-3-enal, CH₂=CHCH₂CHO. [8] Examiner insight: The structure must agree with every piece of evidence; but-3-en-2-one fits the IR and bromine results but would not give a silver mirror.

11. (a) The broad band at 3365 cm⁻¹ falls in the alcohol O–H range (3230–3550) [1]. The absorption at 1642 cm⁻¹ is C=C (1620–1680) [1]. There is no absorption at 1680–1750 cm⁻¹, so X has no C=O bond and cannot be an aldehyde or a ketone [1]. (b) Bromine water decolourised confirms a C=C bond [1]. Dichromate(VI) stays orange, so the alcohol is not primary or secondary [1]; X must be a tertiary alcohol, since these resist oxidation [1]. (c) The one double bond allowed by the formula is the C=C, so there is no ring [1]. (d) A tertiary alcohol with one C=C and seven carbons, for example 2-methylhex-5-en-2-ol, CH₂=CHCH₂CH₂C(OH)(CH₃)₂ [1]: the OH carbon carries three carbon groups and no hydrogen [1]. [9] Examiner insight: In a multi-clue question, tie each clue to a conclusion in turn; a correct structure written with no reasoning earns only the final step.

Where marks are usually lost

• Giving one colour only for a test, or writing “clear” for colourless.
• Giving the result for one compound in a “distinguish” question and not the other.
• Treating an orange-to-green dichromate(VI) result as proof of an alcohol when an aldehyde has not been ruled out.
• Picking a molecular formula from precise masses without the sum for every candidate.
• Calling the very broad 2500–3000 cm⁻¹ band an alcohol O–H.
• Describing the fingerprint region without comparing it with reference spectra.

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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 (first teaching 2019). Topic covered: Organic analysis, section 3.3.6, subsections 3.3.6.1, 3.3.6.2 and 3.3.6.3 (International AS), and Required practical 5.