AQA A-Level Chemistry: Aldehydes and ketones (7405) – 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 Aldehydes and ketones of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards. The whole section is A-level only. It is examined on A-level Paper 2, and Paper 3 can test any content. Questions 7, 10 and 11 also draw on alcohols, optical isomerism and organic analysis.

Learn the ideas first in the aldehydes and ketones study guide and the aldehydes and ketones revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. For a quick check across the whole course, try the free 10-minute diagnostics. Use Ar values C = 12.0, H = 1.0, N = 14.0, O = 16.0, Cu = 63.5, Ag = 107.9.

Questions

1. Explain why the carbon atom of a carbonyl group is attacked by nucleophiles. [2]

2. 2-methylpropanal is warmed with Tollens’ reagent.

(a) State what is observed. [1] (b) Name the organic product, and state the role of 2-methylpropanal in this reaction. [2] (c) Explain why butanone gives no reaction with Tollens’ reagent. [1]

3. Write an overall equation, using [H], for each reduction by NaBH₄ in aqueous solution. Name each organic product.

(a) pentan-3-one [2] (b) 3-methylbutanal [2]

4. Outline the mechanism for the reduction of propanal by NaBH₄ in aqueous solution. Show the nucleophile as H⁻. [4]

5. Write an overall equation, using HCN, for the formation of a hydroxynitrile from each compound. Name each product.

(a) propanal [2] (b) pentan-3-one [2]

6. Potassium cyanide is used, followed by dilute acid, to make hydroxynitriles.

(a) State two hazards of using KCN in this reaction. [2] (b) Give one reason why KCN is used rather than hydrogen cyanide. [1]

7. Butanone reacts with KCN followed by dilute acid.

(a) Outline the mechanism for this reaction. [4] (b) The product is a mixture that has no effect on plane-polarised light. Explain why. [3]

8. A student warms 2.16 g of butanal (Mr 72.0) with an excess of a mild oxidising agent.

(a) Calculate the maximum mass of silver formed if the reagent is Tollens’. [2] (b) Calculate the maximum mass of copper(I) oxide formed if the reagent is Fehling’s solution. Assume one mole of Cu₂O forms per mole of butanal. [2]

9. 4.40 g of ethanal is converted into 2-hydroxypropanenitrile, CH₃CH(OH)CN, using KCN followed by dilute acid. The student obtains 5.33 g of product. Calculate the percentage yield. [4]

10. X, Y and Z are isomers with the formula C₄H₈O. Each contains a C=O group and no C=C bond.

• X and Y each give a red precipitate when warmed with Fehling’s solution. Z does not.
• Reduction of X with NaBH₄ gives an alcohol with an unbranched chain.

(a) Identify X, Y and Z by name. [3] (b) Give the name of the product when Z is reduced by NaBH₄, and state whether this product is formed as a racemic mixture. Explain your answer. [3] (c) The IR spectrum of the reduction product of Y shows a broad absorption between 3230 and 3550 cm⁻¹ and no absorption between 1680 and 1750 cm⁻¹. Using these ranges, state what each feature shows. [2]

11. Describe a two-step synthesis of 2-hydroxybutanenitrile from propan-1-ol. For each step give the reagent(s) and conditions, the organic product, and an equation (use [O] for the oxidising agent and HCN for the overall second step). [5]

12. Methanal, propanone and pentan-2-one each react with KCN followed by dilute acid. State which of these gives a mixture of enantiomers, and explain why the other two do not. [3]

Answers

1. Oxygen is more electronegative than carbon, so the C=O bond is polar [1]; the carbon is δ+ (electron deficient), so an electron-pair donor is attracted to it [1] [2] Examiner insight: “The carbon is positive” without linking it to the electronegativity of oxygen usually earns only one of the two marks.

2. (a) A silver mirror forms [1] (b) 2-methylpropanoic acid (as its carboxylate in the alkaline reagent) [1]; it is a reducing agent (it is oxidised) [1] (c) Ketones are not easily oxidised: the C=O carbon has no hydrogen atom to lose [1] Examiner insight: “Silver precipitate” or “grey solid” is usually accepted in place of a mirror, but “red” or “white” is not.

3. (a) CH₃CH₂COCH₂CH₃ + 2[H] → CH₃CH₂CH(OH)CH₂CH₃ [1]; pentan-3-ol [1] (b) (CH₃)₂CHCH₂CHO + 2[H] → (CH₃)₂CHCH₂CH₂OH [1]; 3-methylbutan-1-ol [1] Examiner insight: An equation with a single [H] is unbalanced and loses the equation mark even when the product formula is right.

4. Cδ+=Oδ− shown on CH₃CH₂CHO and an arrow from the lone pair on H⁻ to the carbonyl carbon [1]; arrow from the C=O bond to the oxygen [1]; intermediate CH₃CH₂CH₂–O⁻ drawn with the negative charge on O [1]; arrow from a lone pair on O⁻ to H of H₂O (or to H⁺), giving propan-1-ol [1] [4] Examiner insight: The first arrow must end on the carbonyl carbon; an arrow from H⁻ to the oxygen shows the wrong atom being attacked and the later arrows then make no sense.

5. (a) CH₃CH₂CHO + HCN → CH₃CH₂CH(OH)CN [1]; 2-hydroxybutanenitrile [1] (b) CH₃CH₂COCH₂CH₃ + HCN → CH₃CH₂C(OH)(CN)CH₂CH₃ [1]; 2-ethyl-2-hydroxybutanenitrile [1] Examiner insight: The nitrile carbon counts in the chain, so (a) is a butanenitrile, not a propanenitrile; a wrong chain length loses the name mark.

6. (a) KCN is highly toxic [1]; with acid it releases HCN, which is highly toxic and volatile [1] (b) HCN is a weak acid and supplies few CN⁻ ions (or HCN is volatile and hard to handle safely) [1] Examiner insight: “Toxic” twice is one point; the second mark needs the separate idea of HCN being released when acid is added.

7. (a) Cδ+=Oδ− on butanone and an arrow from the lone pair on the carbon of :CN⁻ to the carbonyl carbon [1]; arrow from the C=O bond to O [1]; intermediate CH₃C(CN)(O⁻)CH₂CH₃ [1]; arrow from a lone pair on O⁻ to H⁺, giving 2-hydroxy-2-methylbutanenitrile [1] (b) The C=O group is planar (trigonal planar carbon) [1]; CN⁻ attacks from either side with equal probability [1]; the two enantiomers form in equal amounts, a racemic mixture, whose opposite rotations cancel [1] [7] Examiner insight: In (b) “the product is chiral” is not an explanation; the marks come from the planar group, equal attack from both sides and equal amounts of each enantiomer.

8. (a) n(butanal) = 2.16 ÷ 72.0 = 0.0300 mol, so n(Ag) = 2 × 0.0300 = 0.0600 mol [1]; mass = 0.0600 × 107.9 = 6.47 g [1] (b) Mr(Cu₂O) = 2(63.5) + 16.0 = 143.0, n(Cu₂O) = 0.0300 mol [1]; mass = 0.0300 × 143.0 = 4.29 g [1] Examiner insight: Using a 1 : 1 ratio for silver gives 3.24 g; the method mark depends on the 1 : 2 ratio, so show it.

9. n(ethanal) = 4.40 ÷ 44.0 = 0.100 mol [1]; Mr(C₃H₅NO) = 71.0 [1]; theoretical mass = 0.100 × 71.0 = 7.10 g [1]; yield = 5.33 ÷ 7.10 × 100 = 75.1 % [1] [4] Examiner insight: An error in Mr is normally carried forward if the rest of the method is right, so write the Mr down rather than going straight to the answer.

10. (a) X is butanal [1]; Y is 2-methylpropanal [1]; Z is butanone [1] (b) Butan-2-ol [1]; yes, it is racemic [1]: C2 of butan-2-ol carries four different groups (H, OH, CH₃, C₂H₅) and H⁻ attacks the planar C=O from either side equally [1] (c) Broad 3230–3550 cm⁻¹: O–H of an alcohol is present [1]; nothing at 1680–1750 cm⁻¹: no C=O, so the reduction is complete [1] [8] Examiner insight: For (a), names must match the evidence: the Fehling’s result separates the aldehydes from the ketone, and the unbranched product separates X from Y.

11. Step 1: acidified potassium dichromate(VI) [1]; heat gently with the alcohol in excess and distil the aldehyde off as it forms [1]; CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O, giving propanal [1]. Step 2: KCN followed by dilute acid [1]; CH₃CH₂CHO + HCN → CH₃CH₂CH(OH)CN, giving 2-hydroxybutanenitrile [1] [5] Examiner insight: “Reflux” in step 1 gives propanoic acid, not propanal, so the condition mark is lost and the second step no longer works.

12. Pentan-2-one gives a mixture of enantiomers [1]; methanal has two H atoms on the C=O carbon [1] and propanone has two CH₃ groups, so in each case the product carbon does not have four different groups and is not chiral [1] [3] Examiner insight: Name the identical groups in each non-chiral case (two H atoms; two CH₃ groups); “it is symmetrical” alone does not show why no chiral centre forms.

Where marks are usually lost

• Drawing the cyanide arrow from nitrogen instead of the lone pair on carbon.
• Omitting the arrow from the C=O bond to oxygen.
• Leaving the alkoxide O⁻ unprotonated at the end of a mechanism.
• Writing one [H] in reduction equations.
• Using a 1 : 1 aldehyde-to-silver ratio.
• Miscounting carbons when naming hydroxynitriles.
• Explaining optical inactivity without “planar” and “equal attack from both sides”.
• Saying symmetrical ketones give enantiomers.
• Refluxing a primary alcohol when the aldehyde is the target.

Next steps

• Aldehydes and ketones revision notes
• Aldehydes and ketones study guide
• Alcohols practice questions
• Organic analysis practice questions
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.8 Aldehydes and ketones (A-level only).