Part 1: Carbonyl reactions and identification

Connect the position of C=O to oxidation, reduction and a defensible sequence of qualitative tests.

Start with what is attached to the carbonyl carbon

A carbonyl group contains C=O. In an aldehyde the carbonyl carbon is bonded to at least one H; the general formula is RCHO, with methanal as the case where both substituents are H. In a ketone it is bonded to two carbon groups, RCOR′. The aldehyde carbon is part of the parent chain and is numbered carbon 1.

Oxygen is more electronegative than carbon, so the C=O bond is polar: C is δ+ and O is δ−. The carbonyl carbon is susceptible to nucleophiles. Keep this local polarity separate from the formal charges that appear after electrons move in a mechanism.

Propanal, CH₃CH₂CHO, and propanone, CH₃COCH₃, have the same molecular formula C₃H₆O but different connectivity. Identify the groups attached to C=O before predicting chemistry; merely counting oxygen atoms cannot distinguish them.

Why aldehydes and ketones give different oxidation tests

Acidified dichromate(VI) oxidises an aldehyde to a carboxylic acid on warming. The orange reagent turns green as chromium(VI) is reduced to chromium(III). RCHO + [O] → RCOOH adds one oxygen in this shorthand representation. Do not change the carbon skeleton.

Ketones resist oxidation under the usual test conditions. Oxidising them further would generally involve breaking a carbon–carbon bond, so they do not show the aldehyde reaction under those conditions. This does not mean a ketone can never be oxidised by any possible reagent.

An orange-to-green change alone does not prove an aldehyde because primary and secondary alcohols can also reduce dichromate. Combine evidence: first establish that a carbonyl is present, then use an appropriate aldehyde–ketone distinction.

Worked transformation: retain the skeleton and add two H

NaBH₄ reduces aldehydes to primary alcohols and ketones to secondary alcohols. The carbonyl C receives H and the oxygen becomes OH after protonation; the C=O π bond is replaced by two single bonds. Use 2[H] in the overall equation, not [H].

For butan-2-one, CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃. The product is butan-2-ol because OH remains on the carbon that originally carried C=O. Reduction does not move the oxygen to the end of the chain.

In the OCR toolkit NaBH₄ does not reduce a carboxylic acid group under the usual carbonyl-reduction conditions. In a molecule containing both a ketone and COOH, change the ketone to a secondary alcohol while retaining COOH. Check each functional group separately before accepting the product.

Two tests answer two different questions

2,4-DNPH, also called Brady’s reagent, produces a yellow/orange precipitate with an aldehyde or ketone. It establishes a reactive carbonyl group in this test scheme; it does not by itself distinguish the two classes. The observation is formation of a precipitate, not simply an orange solution.

A purified 2,4-DNPH derivative can be identified by measuring its melting point and comparing with reference data. Recrystallise the derivative, dry it and compare the measured range with reliable values. An impure or wet derivative gives a less useful melting-point comparison. The specification does not require recall of the derivative’s structure or its formation equation.

Tollens’ reagent gives a silver mirror, or a grey/silver deposit, with an aldehyde on suitable gentle warming. Silver(I) is reduced to silver while the aldehyde is oxidised. An ordinary ketone gives no silver mirror in the standard comparison. Some additional functional arrangements can reduce Tollens’ reagent, so unfamiliar structures require interpretation rather than an absolute claim that every positive result must be an aldehyde.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources