Whiteboard Chemistry with Joe White

Aldehydes & Ketones

The carbonyl group and why it is attacked by nucleophiles, oxidation and reduction, the two nucleophilic-addition mechanisms (NaBH4 and HCN) in full, and the tests that identify a carbonyl compound.

AQA 7404/7405 Paper 2 A-level only
Hero motif · to come Nucleophilic addition to C=O
Where this sits

You met aldehydes and ketones as the oxidation products of alcohols at AS, and their tests in organic analysis. Here they get their own reactions, all driven by one feature: the polar C=O double bond. Because the carbon is electron-poor (δ+), carbonyl compounds are attacked by nucleophiles — and two nucleophilic-addition mechanisms are the heart of the topic.

The carbonyl group: oxidation & reduction

Aldehydes (R–CHO) and ketones (R–CO–R′) both contain the carbonyl group, C=O. Oxygen is far more electronegative than carbon, so the double bond is polar: the carbon is δ+ and the oxygen δ−. That δ+ carbon is the target for nucleophiles.

Oxidation. An aldehyde is readily oxidised to a carboxylic acid (by warming with acidified potassium dichromate(VI), orange → green). A ketone is not oxidised this way — the basis of the tests that tell them apart.

Reduction. Both are reduced by NaBH4 (sodium tetrahydridoborate) to alcohols: an aldehyde gives a primary alcohol, a ketone a secondary alcohol. Using [H] for the reductant:

Figure F1 · illustration to come The polar C=O and its reactions Oxidation, reduction, and nucleophilic addition.
Everything follows from the δ+ carbon — that is where nucleophiles attack.
🧪 Exam-style questions
Q1 [2 marks]

A student reduces 2-methylbutanal with NaBH4 but adds too little, so the reduction is incomplete. Give a chemical test, with the observation, that confirms an aldehyde is still present.

Show answer

Add Tollens’ reagent (or Fehling’s solution) and warm. 1 mark

A silver mirror forms (Tollens’) / a brick-red precipitate forms (Fehling’s) — showing an aldehyde remains. 1 mark

Source: AQA A-Level Chemistry past papers.

Nucleophilic addition

Both key reactions add a nucleophile across the C=O by the same three moves: the nucleophile attacks the δ+ carbon, the C=O double bond breaks onto the oxygen (forming a negative alkoxide), and the oxygen is then protonated. Learn one mechanism and you have both.

Reduction by NaBH4

NaBH4 delivers a hydride ion, H, which acts as the nucleophile:

The NaBH4 mechanism, arrow by arrow
  • Arrow 1 — from the H (a B–H bond of BH4) to the δ+ carbon of C=O.
  • Arrow 2 — from the C=O double bond to the oxygen, forming a negatively charged alkoxide intermediate.
  • Arrow 3 — a lone pair on the alkoxide O takes an H+ (from water / dilute acid), giving the alcohol.

NaBH4 reduces the C=O but not a C=C double bond: H is a nucleophile, and only the C=O is polar with a δ+ carbon to attack. A C=C is non-polar and electron-rich, so it does not attract the nucleophile.

Figure F2 · mechanism to come NaBH4 reduction of a carbonyl H → δ+ C; C=O → O; protonate → alcohol.
Nucleophilic addition of hydride: attack, break, protonate.

Addition of HCN (using KCN)

The cyanide ion, CN, is the nucleophile. Addition across the C=O gives a hydroxynitrile (a 2-hydroxynitrile), which is one carbon longer than the carbonyl — a useful way to extend a carbon chain.

The HCN / KCN mechanism, arrow by arrow
  • Arrow 1 — from a lone pair on the C of CN to the δ+ carbon of C=O.
  • Arrow 2 — from the C=O double bond to the oxygen, forming the alkoxide.
  • Arrow 3 — the alkoxide O takes an H+ (from the dilute acid / HCN), giving the hydroxynitrile.
Figure F3 · mechanism to come HCN addition to a carbonyl CN → δ+ C; C=O → O; protonate → hydroxynitrile.
A chiral hydroxynitrile forms as a racemate — the planar C=O is attacked from both sides equally.
Precision points
  • KCN is very toxic (it releases toxic HCN). It is used instead of HCN because HCN is a volatile, extremely toxic gas and a weak acid, so KCN provides a higher, controlled concentration of the CN nucleophile.
  • Every curly arrow must start from a lone pair or a bond and finish on an atom or bond — the nucleophile’s lone pair to the δ+ carbon, the C=O to the oxygen.
🧪 Exam-style questions
Q2 [5 marks]

Show the first step of the mechanism of the reaction between NaBH4 and 2-methylbutanal (include two curly arrows), and explain why NaBH4 reduces 2-methylbutanal but has no reaction with 2-methylbut-1-ene.

Show answer

Arrow from the H (hydride) to the δ+ carbon of the C=O. 1 mark

Arrow from the C=O double bond to the oxygen (forming the alkoxide intermediate). 1 mark

The C=O bond is polar, with a δ+ carbon. 1 mark

H is a nucleophile, so it is attracted to that δ+ carbon. 1 mark

The C=C bond in the alkene is non-polar / electron-rich, so it does not attract the nucleophile — no reaction. 1 mark

Q3 [4 marks]

Outline the mechanism for the reaction of propanone with KCN followed by dilute acid.

Show answer

Arrow from a lone pair on the C of CN to the δ+ carbon of the C=O. 1 mark

Arrow from the C=O double bond to the oxygen. 1 mark

Correct alkoxide intermediate with the negative charge on oxygen. 1 mark

Arrow from the O lone pair to H+ (dilute acid) → 2-hydroxy-2-methylpropanenitrile. 1 mark

Q4 [2 marks]

Propanone reacts with the weak acid HCN, but the hydroxynitrile is usually made using KCN followed by dilute acid instead. State the hazard of KCN, and suggest a reason (other than safety) why KCN is used instead of HCN.

Show answer

Hazard: KCN is (very) toxic / poisonous (it can release toxic HCN). 1 mark

HCN is only a weak acid, so it provides a low concentration of the CN nucleophile; KCN gives a higher concentration of CN, so the reaction is faster. 1 mark

Q5 [2 marks]

Ethanal reacts with KCN then dilute acid to form 2-hydroxypropanenitrile, a mixture of equal amounts of two isomers. Name the mechanism, and name this type of mixture.

Show answer

Mechanism: nucleophilic addition. 1 mark

Type of mixture: a racemic mixture (racemate). 1 mark

The planar C=O is attacked from both sides equally, giving the two enantiomers in equal amounts.

Source: AQA A-Level Chemistry past papers.

Tests for carbonyl compounds

Two questions: is it a carbonyl at all, and is it an aldehyde or a ketone? Three reagents answer them.

ReagentAldehydeKetone
2,4-DNP (Brady’s reagent)orange pptorange ppt
Tollens’ reagentsilver mirrorno change
Fehling’s solutionbrick-red pptno change
Acidified K2Cr2O7orange → greenno change

2,4-DNP (Brady’s reagent) gives an orange precipitate with any carbonyl — it confirms a C=O. To identify which carbonyl, the precipitate is purified by recrystallisation and its melting point is measured and compared with known values. Tollens’ and Fehling’s then separate aldehydes from ketones: only the easily-oxidised aldehyde gives a positive result.

Figure F4 · illustration to come The carbonyl tests 2,4-DNP for any carbonyl; Tollens’/Fehling’s for aldehydes.
2,4-DNP finds the carbonyl; Tollens’ and Fehling’s find the aldehyde.
🧪 Exam-style questions
Q6 [2 marks]

Identify a reagent that would give a positive result with an aldehyde but not a ketone, and state the observation with the aldehyde.

Show answer

Tollens’ reagent (or Fehling’s solution). 1 mark

Silver mirror forms (Tollens’) / brick-red precipitate forms (Fehling’s). 1 mark

Q7 [3 marks]

A ketone is reacted with 2,4-DNP to form a crystalline solid, which is then identified by its melting point. Describe how the crystalline solid is separated and purified.

Show answer

Separate the solid by filtration (under reduced pressure). 1 mark

Recrystallise: dissolve in the minimum volume of hot solvent, then cool so the pure product crystallises; filter off the crystals. 1 mark

Wash the crystals with a little cold solvent and dry them. 1 mark

Source: AQA A-Level Chemistry past papers.

Identify & react

Bring it together. First find out what you have, then use its reactions.

The method
  • Is it a carbonyl? 2,4-DNP gives an orange precipitate for any aldehyde or ketone.
  • Aldehyde or ketone? Tollens’ (silver mirror) or Fehling’s (brick-red) — positive for aldehydes only.
  • Its reactions: reduce with NaBH4 (→ alcohol), or add HCN/KCN (→ hydroxynitrile, extending the chain).
Precision points
  • Give the reagent and the observation for every test — “Tollens’ → silver mirror”, not just “Tollens’”.
  • In the mechanisms, the nucleophile’s arrow starts from its lone pair and the C=O arrow finishes on the oxygen; show the negative alkoxide intermediate.
  • A chiral hydroxynitrile forms as a racemate — the planar C=O is attacked from both sides equally.
🧪 Capstone question
Q8 [3 marks]

Two unlabelled liquids are propanal and propanone. Describe how you would use chemical tests to identify which is which, giving the reagents and observations.

Show answer

Both give an orange precipitate with 2,4-DNP, confirming both are carbonyls. 1 mark

Add Tollens’ reagent (or Fehling’s) and warm. 1 mark

Propanal (the aldehyde) gives a silver mirror (or brick-red precipitate); propanone (the ketone) gives no change. 1 mark

Source: AQA A-Level Chemistry past papers.

3.3.8 Aldehydes & ketones — Quick-reference summary
  • Carbonyl C=O is polar (Cδ+=Oδ−), so it undergoes nucleophilic addition.
  • Oxidation: aldehyde → carboxylic acid (acidified K2Cr2O7, orange→green); ketone not oxidised.
  • Reduction (NaBH4): nucleophilic addition of hydride (H) — aldehyde → 1° alcohol; ketone → 2° alcohol.
  • HCN / KCN: nucleophilic addition of CN gives a hydroxynitrile (extends the chain by one carbon); a chiral product forms as a racemate (planar C=O attacked from both sides).
  • Tests: 2,4-DNP gives an orange precipitate with any carbonyl (mp identifies which); Tollens’ (silver mirror) and Fehling’s (brick-red) are positive for aldehydes only.

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