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AS · Practice questions · Aldehydes and ketones

Climb the ladder.

Six original Cambridge-style questions on making and reducing aldehydes and ketones.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Primary to aldehyde, secondary to ketone, NaBH4 back down.

01
Recall
[3 marks]

Name each compound and state whether it is an aldehyde or a ketone: (a) CH3CH2CH2CHO, (b) CH3COCH2CH2CH3, (c) CH3CH2COCH2CH3.

  • (a) butanal, an aldehyde (C=O at the chain end). ✓
  • (b) pentan-2-one, a ketone. ✓
  • (c) pentan-3-one, a ketone. ✓
02
Explain
[3 marks]

Explain why propanone (Mr 58) boils higher than butane (Mr 58) but lower than propan-1-ol (Mr 60).

  • All three have similar id-id forces, as they have similar numbers of electrons. ✓
  • Propanone also has permanent dipole-dipole forces from the polar C=O, so it boils above butane. ✓
  • Propan-1-ol forms hydrogen bonds (O–H), which are stronger still; propanone has no H on O, so it cannot. ✓
03
Recall
[3 marks]

Describe how ethanal is prepared from ethanol. Give the reagents, the apparatus and an equation using [O].

  • Warm ethanol with acidified potassium dichromate(VI), adding the oxidant slowly. ✓
  • Use distillation apparatus so ethanal (b.p. 21 °C) distils off as it forms and is not oxidised to ethanoic acid. ✓
  • CH3CH2OH + [O] → CH3CHO + H2O. ✓
04
Calculation
[2 marks]

1.16 g of propanone (Mr 58.0) is reduced with excess NaBH4. Calculate the maximum mass of propan-2-ol (Mr 60.0) formed.

  • n(propanone) = 1.16 ÷ 58.0 = 0.0200 mol; the ratio is 1 : 1. ✓
  • Mass = 0.0200 × 60.0 = 1.20 g. ✓
05
Analysis
[3 marks]

Compound P, C4H8O, is reduced by NaBH4 to a secondary alcohol. Identify P and the alcohol, and write the equation using [H].

  • Reduction to a secondary alcohol means P is a ketone: butanone, CH3COCH2CH3. ✓
  • The alcohol is butan-2-ol, CH3CH(OH)CH2CH3. ✓
  • CH3COCH2CH3 + 2[H] → CH3CH(OH)CH2CH3. ✓
06
Analysis
[2 marks]

Pent-3-en-2-one, CH3CH=CHCOCH3, is treated with NaBH4. Give the structure of the product and explain why the C=C is unchanged.

  • Product: pent-3-en-2-ol, CH3CH=CHCH(OH)CH3. ✓
  • NaBH4 supplies a nucleophile (H−) that attacks the δ+ carbon of the polar C=O; the non-polar C=C repels nucleophiles, so it is not reduced. ✓

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