OxfordAQA A-Level Chemistry: Aldehydes and ketones (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.

These questions cover section 3.3.8 (subsection 3.3.8.1), Aldehydes and ketones, of the OxfordAQA International AS and A-level Chemistry (9620) specification (Version 5.3; International AS and A-level exams May/June 2020 onwards). This content is International A-level only. Questions on it belong to Unit 4 (Organic 2 and Physical 2), with Unit 5 (practical and synoptic) also able to test it. Some questions also draw on 3.3.5.1 and 3.3.7.1.

Revise first with the study guide and revision notes. See also the course hub, the printable checklist and the free 10-minute diagnostics. Ar values: C 12.0, O 16.0, H 1.0, N 14.0, K 39.1, Cu 63.5, Ag 107.9.

Questions

1. Two isomers, 3-ethylpentanal and 3-ethylpentan-2-one, are in unlabelled bottles. Describe how Tollens’ reagent could be used to identify them. Give the result for each compound. [3]

2. 3,3-dimethylbutanal, (CH₃)₃CCH₂CHO, is readily oxidised.

(a) Write an equation, using [O], for its oxidation, and name the organic product. [2] (b) Explain why ketones are not readily oxidised. [1]

3. Give an overall equation, using [H], for reducing each compound with aqueous NaBH₄, and name the alcohol formed.

(a) 2-methylhexan-3-one, CH₃CH₂CH₂COCH(CH₃)₂ [2] (b) 5-methylhexanal, (CH₃)₂CHCH₂CH₂CH₂CHO [2]

4. Aqueous NaBH₄ reduces 5-methylhexan-2-one, CH₃COCH₂CH₂CH(CH₃)₂. Outline the mechanism, representing the nucleophile as H⁻, and name the product. [5]

5. For each compound, give the overall equation (with HCN) for making its hydroxynitrile, and name the hydroxynitrile.

(a) 3,3-dimethylbutanal [2] (b) nonan-5-one, CH₃(CH₂)₃CO(CH₂)₃CH₃ [2]

6. 3-ethylpentan-2-one, CH₃COCH(C₂H₅)₂, is treated with KCN and then dilute acid.

(a) Outline the mechanism. [4] (b) The product has exactly one chiral centre. Identify it, and explain why carbon 3 of the product is not a chiral centre. [2]

7. A student writes this plan for making a hydroxynitrile: “Dissolve potassium cyanide in water in a beaker on the open bench, add the ketone, then pour in all the acid at once.” Identify two hazards of this reaction and suggest a control measure for each. [4]

8. 0.600 g of a saturated, non-cyclic aldehyde X (general formula CₙH₂ₙO) is warmed with excess Fehling’s solution. 0.858 g of copper(I) oxide, Cu₂O, forms. One mole of X gives one mole of Cu₂O.

(a) Calculate the relative molecular mass of X. [3] (b) Deduce the molecular formula of X. [1] (c) In X, the carbon atom next to the –CHO group carries no hydrogen atoms. Name X. [1] (d) State whether X gives a mixture of enantiomers with KCN followed by dilute acid. [1]

9. A student reacts 5.70 g of 4,4-dimethylpentan-2-one (C₇H₁₄O) with 2.93 g of potassium cyanide, then adds excess dilute acid. The ketone and KCN react in a 1 : 1 ratio. 4.76 g of 2-hydroxy-2,4,4-trimethylpentanenitrile (C₈H₁₅NO) is obtained.

(a) Show that KCN is the limiting reagent. [2] (b) Calculate the percentage yield. [3]

10. A pure sample of one enantiomer of heptan-3-ol, CH₃CH₂CH(OH)CH₂CH₂CH₂CH₃, rotates the plane of plane-polarised light. It is oxidised with acidified potassium dichromate(VI) to ketone K. K is then reduced with NaBH₄ in aqueous solution.

(a) Name K and write an equation, using [O], for its formation. [2] (b) Write an equation, using [H], for the reduction of K. [1] (c) The heptan-3-ol recovered from K does not rotate plane-polarised light. Explain this. [3] (d) A second portion of K is treated with KCN, then dilute acid. Give the product’s name and say whether the sample is optically active. [2]

11. This question links the silver test, amounts and chirality.

(a) Write a half-equation for the reduction of the silver complex in Tollens’ reagent. [1] (b) Calculate the maximum mass of silver deposited when 0.500 g of 3,3-dimethylbutanal (C₆H₁₂O) is warmed with excess Tollens’ reagent. [2] (c) Explain why reducing any aldehyde with NaBH₄ never creates a chiral centre at the old carbonyl carbon, whereas KCN followed by dilute acid can. [2]

Answers

1. Warm a sample of each with Tollens’ reagent in a hot water bath [1]; 3-ethylpentanal: a silver mirror forms [1]; 3-ethylpentan-2-one: no silver mirror, the mixture stays colourless [1] [3] Examiner insight: Give an observation for both compounds; “no silver mirror” for the ketone is stronger than “nothing happens”.

2. (a) (CH₃)₃CCH₂CHO + [O] → (CH₃)₃CCH₂COOH [1]; 3,3-dimethylbutanoic acid [1] (b) The C=O carbon of a ketone has no hydrogen atom, so oxidation would need a C–C bond to break [1] Examiner insight: Keep the carbon skeleton unchanged in the acid; drawing the acid with one carbon fewer is a common slip and the name then no longer matches the equation.

3. (a) CH₃CH₂CH₂COCH(CH₃)₂ + 2[H] → CH₃CH₂CH₂CH(OH)CH(CH₃)₂ [1]; 2-methylhexan-3-ol [1] (b) (CH₃)₂CHCH₂CH₂CH₂CHO + 2[H] → (CH₃)₂CHCH₂CH₂CH₂CH₂OH [1]; 5-methylhexan-1-ol [1] Examiner insight: Check that each equation balances; a single [H] leaves the equation two hydrogen atoms short, and an unbalanced equation is unlikely to gain credit even with the right product.

4. Cδ+=Oδ− shown, with an arrow starting at the H⁻ lone pair and ending on the carbonyl carbon [1]; arrow from the double bond on to O [1]; alkoxide intermediate CH₃CH(O⁻)CH₂CH₂CH(CH₃)₂ with the charge on oxygen [1]; arrow from an O⁻ lone pair to a hydrogen of water (or to H⁺) [1]; product 5-methylhexan-2-ol [1] [5] Examiner insight: Start each arrow at a lone pair or a bond and end it at an atom; an arrow starting from the minus sign of H⁻, or ending between atoms, is ambiguous.

5. (a) (CH₃)₃CCH₂CHO + HCN → (CH₃)₃CCH₂CH(OH)CN [1]; 2-hydroxy-4,4-dimethylpentanenitrile [1] (b) CH₃(CH₂)₃CO(CH₂)₃CH₃ + HCN → CH₃(CH₂)₃C(OH)(CN)(CH₂)₃CH₃ [1]; 2-butyl-2-hydroxyhexanenitrile [1] Examiner insight: Number from the nitrile carbon, which is C1; in (a) that puts both methyl groups on C4, and answers that number from the other end give a different, wrong name.

6. (a) Cδ+=Oδ− shown, with an arrow from the carbon-end lone pair of :CN⁻ to the δ+ carbon [1]; arrow from the double bond on to O [1]; intermediate CH₃C(O⁻)(CN)CH(C₂H₅)₂ [1]; arrow from an O⁻ lone pair to H⁺ [1] (b) Carbon 2, which carries CH₃, OH, CN and CH(C₂H₅)₂ [1]; carbon 3 carries two identical ethyl groups [1] [6] Examiner insight: The arrow must leave the carbon end of the cyanide ion; starting it at nitrogen implies a C–N bond to the carbonyl carbon, which is the wrong product.

7. Potassium cyanide is highly toxic by contact or if swallowed [1]; wear gloves and eye protection [1]. Adding acid forms hydrogen cyanide, a highly toxic, volatile gas [1]; work in a fume cupboard and add dilute acid slowly (not all at once) [1] [4] Examiner insight: Each control must match the hazard it answers; “wear gloves” does nothing against HCN gas, so pair the fume cupboard with the gas and gloves with the solid.

8. (a) Mr(Cu₂O) = 127.0 + 16.0 = 143.0, so n(Cu₂O) = 0.858 ÷ 143.0 = 0.00600 mol [1]; n(X) = 0.00600 mol [1]; Mr(X) = 0.600 ÷ 0.00600 = 100.0 [1] (b) 14n + 16 = 100, so n = 6: C₆H₁₂O [1] (c) 2,2-dimethylbutanal, CH₃CH₂C(CH₃)₂CHO [1] (d) Yes: the new carbon carries H, OH, CN and C(CH₃)₂C₂H₅, four different groups, so a racemate forms [1] Examiner insight: Use the 1 : 1 ratio stated for Fehling’s here; carrying over the 1 : 2 ratio from the silver test doubles the Mr to 200, which gives no whole-number value of n.

9. (a) n(ketone) = 5.70 ÷ 114.0 = 0.0500 mol [1]; n(KCN) = 2.93 ÷ 65.1 = 0.0450 mol, which is less than 0.0500 mol in a 1 : 1 reaction, so KCN is limiting [1] (b) Mr(C₈H₁₅NO) = 96.0 + 15.0 + 14.0 + 16.0 = 141.0 [1]; maximum mass = 0.0450 × 141.0 = 6.35 g [1]; yield = 4.76 ÷ 6.35 × 100 = 75.0 % [1] Examiner insight: Base the yield on the limiting reagent; using the ketone gives 7.05 g as the maximum and a yield of 67.5 %, which does not answer the question as set.

10. (a) Heptan-3-one [1]; CH₃CH₂CH(OH)CH₂CH₂CH₂CH₃ + [O] → CH₃CH₂COCH₂CH₂CH₂CH₃ + H₂O [1] (b) CH₃CH₂COCH₂CH₂CH₂CH₃ + 2[H] → CH₃CH₂CH(OH)CH₂CH₂CH₂CH₃ [1] (c) The C=O carbon in K is trigonal planar, so the chirality of the original alcohol is lost [1]; H⁻ attacks either face of the plane with equal probability [1]; equal amounts of the two enantiomers form (a racemate), and their equal and opposite rotations cancel [1] (d) 2-ethyl-2-hydroxyhexanenitrile, CH₃CH₂C(OH)(CN)CH₂CH₂CH₂CH₃ [1]; not optically active, because it forms as a racemate [1] [8] Examiner insight: In (c), the planar intermediate, attack from either face and equal amounts are three separate points; an answer that says only “a racemic mixture forms” covers the conclusion but not the reasons.

11. (a) [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃ [1] (b) n(aldehyde) = 0.500 ÷ 100.0 = 0.00500 mol, so n(Ag) = 2 × 0.00500 = 0.0100 mol [1]; mass = 0.0100 × 107.9 = 1.08 g [1] (c) Reduction turns –CHO into –CH₂OH, so the old carbonyl carbon carries two hydrogen atoms [1]; with KCN it carries H, OH, CN and the R group, which can be four different groups [1] [5] Examiner insight: In (b), write the 1 : 2 ratio in your working; a final answer of 0.54 g shows the ratio was missed, and clear working lets the examiner see which steps were right.

Where marks are usually lost

• Using a 1 : 1 aldehyde-to-silver ratio, or 1 : 2 for Fehling’s.
• Cyanide arrows that start at N; the attacking lone pair is on C.
• BH₄⁻ drawn as the nucleophile instead of H⁻.
• No arrow from the double bond on to oxygen.
• O⁻ left unprotonated at the end of a mechanism.
• Writing [H] rather than 2[H] in reduction equations.
• Basing a yield on the reagent in excess instead of the limiting reagent.
• Explaining a racemate without the planar carbonyl group and equal attack from either face.
• Pairing a hazard with a control that does not address it.

Next steps

• Aldehydes and ketones revision notes
• Aldehydes and ketones study guide
• Optical isomerism practice questions
• OxfordAQA A-Level Chemistry 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: section 3.3.8 Aldehydes and ketones (3.3.8.1), International A2 content.