AS Chemistry: Aldehydes and Ketones — Revision Notes

Condensed for the final weeks. For the full explanation, use the Carbonyl Compounds study guide, and test yourself with the practice questions.

The carbonyl group

C=O is polar — oxygen is far more electronegative, so the carbon carries a δ+ charge and is attacked by nucleophiles.

	Aldehyde	Ketone
Structure	RCHO — carbonyl at the end of the chain	RCOR’ — carbonyl within the chain
Suffix	-al	-one
Oxidised further?	Yes → carboxylic acid	No

That last row explains every distinguishing test in the topic.

Preparation

PRIMARY alcohol   + [O], DISTIL    ->  aldehyde
SECONDARY alcohol + [O], DISTIL    ->  ketone

The syllabus specifies distillation for both preparations. Distillation is critical for the aldehyde route — without it, the aldehyde stays in contact with the oxidising mixture and is oxidised further to a carboxylic acid. For the ketone route, distillation isn’t strictly needed to prevent over-oxidation (a ketone can’t be oxidised further under these conditions either way) — but the syllabus specifies it regardless, so use distillation for both, not reflux.

Reduction

Reducing agent: NaBH₄ (or LiAlH₄ in dry ether).

aldehyde + 2[H]  ->  PRIMARY alcohol
ketone   + 2[H]  ->  SECONDARY alcohol

Nucleophilic addition with HCN

Conditions: HCN with a trace of KCN (to supply CN⁻), around pH 8.

Step 1  CN- lone pair attacks the delta+ carbonyl carbon
        The C=O pi bond breaks, both electrons going to the oxygen
Step 2  The resulting O- is protonated by H+ (or HCN)

Product:  a HYDROXYNITRILE   R-C(OH)(CN)-R'

The product has one more carbon than the starting carbonyl — a chain-lengthening step worth remembering for synthesis routes.

Curly arrows: from the CN⁻ lone pair to the carbon; from the C=O π bond to the oxygen.

If the carbonyl carbon has two different groups, the product is chiral and forms as a racemic mixture (“racemic mixture” and “optically active” are A Level terms — at AS it’s enough to say equal amounts of both enantiomers form), because the planar carbonyl can be attacked equally from either face.

The identifying tests

Test	Reagent	Aldehyde	Ketone
2,4-DNPH (Brady’s)	2,4-dinitrophenylhydrazine	Orange precipitate	Orange precipitate
Tollens’	Ammoniacal silver nitrate	Silver mirror	No change
Fehling’s	Alkaline copper(II) complex	Brick-red precipitate	No change

2,4-DNPH identifies a carbonyl but does not distinguish the two — both give an orange precipitate. Use Tollens’ or Fehling’s to tell them apart. Candidates routinely get this the wrong way round.

The melting point of the purified 2,4-DNPH derivative identifies the specific carbonyl compound.

Worked example. Distinguish propanal (CH₃CH₂CHO) from propanone (CH₃COCH₃), and explain why the iodoform test would not be the best choice here.

Warm each with Tollens' reagent:
propanal (aldehyde) -> silver mirror forms
propanone (ketone)  -> no reaction

Iodoform is a poor choice: propanone DOES contain a CH3CO- group and
would give a positive result, but so might other compounds sharing that
fragment -- the aldehyde/ketone (Tollens'/Fehling's) test targets exactly
the distinction asked for, while iodoform targets a different feature.

The iodoform test

Warming with alkaline I₂(aq) gives a yellow precipitate of CHI₃ with any compound containing a CH₃CO– group — that is, a methyl group directly bonded to the carbonyl carbon — in either an aldehyde or a ketone. Both ethanal (CH₃CHO) and propanone (CH₃COCH₃) test positive, since both have a methyl group directly on the carbonyl carbon.

This is a different test from Tollens’/Fehling’s: those two distinguish aldehyde from ketone by oxidation; the iodoform test identifies a specific structural fragment, regardless of which type of carbonyl it belongs to. Don’t assume “ketone” automatically means a positive iodoform result — check the actual structure for a CH₃CO– group.

Exam traps

• Claiming 2,4-DNPH distinguishes aldehydes from ketones — it does not.
• Forgetting that HCN addition adds a carbon.
• Curly arrows starting from an atom rather than a lone pair or bond.
• Omitting the racemic outcome when a chiral centre is created.
• Confusing Tollens’ (silver mirror) with Fehling’s (brick-red).
• Using [O] for reduction — reduction is [H].
• Assuming the iodoform test distinguishes aldehydes from ketones — it identifies a CH₃CO– group, in either type, not the aldehyde/ketone distinction.
• Assuming a compound is a “ketone” therefore gives a negative iodoform result — check the actual structure for a CH₃CO– fragment; propanone (a ketone) is positive.

Self-test

1. Why is the carbonyl carbon attacked by nucleophiles?
2. What does a positive 2,4-DNPH test tell you, and what does it not?
3. Give the reagent and conditions for converting a ketone to a secondary alcohol.
4. Why is the hydroxynitrile from ethanal a racemic mixture?
5. Name the two tests that distinguish an aldehyde from a ketone and their positive results.
6. What does a positive iodoform test tell you, and what does a negative result NOT rule out?
7. Explain why propanone gives a positive iodoform test even though it is a ketone.

Answers: 1. Oxygen is much more electronegative than carbon, so the C=O bond is polar and the carbon carries a partial positive charge. 2. That a carbonyl group is present (aldehyde or ketone); it does not distinguish between them. 3. NaBH₄ (or LiAlH₄ in dry ether), giving the secondary alcohol. 4. The carbonyl is planar, so CN⁻ attacks with equal probability from either face, producing equal amounts of both enantiomers. 5. Tollens’ — silver mirror with an aldehyde; Fehling’s — brick-red precipitate with an aldehyde. Neither reacts with a ketone. 6. That the compound contains a CH₃CO– group; a negative result does not rule out the compound being an aldehyde or a ketone, only that this specific fragment is absent. 7. Its carbonyl carbon has a CH₃ group directly attached (CH₃–CO–CH₃), which is exactly the structural feature the iodoform test detects, regardless of whether the compound is an aldehyde or a ketone.