Aldehydes and ketones are two families of organic compounds that contain the carbonyl group. Explain how the structure of an aldehyde differs from that of a ketone, name a chemical test that distinguishes between them, and state the result you would observe with each compound.

Eduqas A-Level Chemistry — 4.4 Aldehydes and ketones · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (5 marks)

An aldehyde has the carbonyl carbon bonded to at least one hydrogen, so the C=O group is at the end of the carbon chain. A ketone has the carbonyl carbon bonded to two carbon atoms, placing the C=O group in the middle of the chain.

A suitable test is Tollens’ reagent (ammoniacal silver nitrate). In the presence of an aldehyde a silver mirror is deposited on the test tube. With a ketone no silver mirror is formed.

(If Fehling’s or Benedict’s solution is used, the aldehyde gives a brick‑red/orange precipitate of Cu₂O, whereas the ketone gives no precipitate.)

Examiner tips

  • Use the exact wording ‘at the end of the chain’ for aldehydes and ‘in the middle of the chain’ for ketones. Include the name of the test and the specific visual change. Mention the negative result for the ketone clearly.
  • Keep the answer concise – 5 marks can be earned with two short sentences for structure, one for the test name, and two for the results.

Common mistakes

  • Confusing the position of the carbonyl group (placing it in the middle for an aldehyde). Using a test that is not a distinguishing one (e.g., bromine water) or omitting the result for the ketone. Not specifying the silver mirror or the colour of the precipitate.

Mark scheme (5 marks)

  1. In an aldehyde, the carbonyl group (C=O) is at the end of the carbon chain (bonded to at least one hydrogen)
  2. In a ketone, the carbonyl group (C=O) is in the middle of the carbon chain (bonded to two carbon atoms on either side)
  3. Names Tollens' reagent (silver mirror test / ammoniacal silver nitrate) OR Fehling's solution OR Benedict's solution as a suitable distinguishing test
  4. Correct positive result with the aldehyde — silver mirror forms (Tollens') OR brick-red/orange precipitate forms (Fehling's/Benedict's)
  5. Correct negative result with the ketone — no change / no precipitate / no silver mirror formed

Key terms in this question

aldehyde · ketone · carbonyl group

Related

More Aldehydes and ketones questions

▶ Try answering this question with AI marking (free) →