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17. Carbonyl compounds

Syllabus
9701–2028–2029
Section
17
Level
AS

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Topic 17.1

17.1 Aldehydes and ketones

Objectives in this topic

Oxidise primary alcohols to aldehydes and secondary alcohols to ketones

Acidified dichromate(VI) or manganate(VII) oxidises a primary alcohol to an aldehyde when the aldehyde is distilled off, while a secondary alcohol gives a ketone.

Distillation prevents a primary aldehyde from remaining in the oxidising mixture and being oxidised further to a carboxylic acid. A secondary alcohol has no hydrogen on the OH-bearing carbon for the same further oxidation pathway.

Propan-1-ol → propanal by controlled distillation; propan-2-ol → propanone under reflux. State reagent, heating and collection conditions.

Do not write a carboxylic acid as the isolated product when the syllabus specifies distillation of the aldehyde.

Carbonyl compounds are reduced to alcohols or add HCN to form hydroxynitriles

NaBH₄ or LiAlH₄ reduces an aldehyde to a primary alcohol and a ketone to a secondary alcohol. HCN adds across the C=O bond, with KCN acting as a catalyst, to form a hydroxynitrile.

Reduction supplies hydrogen/electron density to the carbonyl; cyanohydrin formation adds CN and OH to the former carbonyl carbon. The nitrile group adds one carbon to the skeleton.

Ethanal + NaBH₄ → ethanol; ethanal + HCN → CH₃CH(OH)CN. Propanone gives a tertiary alcohol on reduction and a substituted hydroxynitrile on addition.

Do not confuse reduction with oxidation, and do not forget the carbon-count increase in cyanohydrin formation.

HCN adds to a carbonyl by nucleophilic attack followed by proton transfer

The carbonyl carbon is electron-poor. CN⁻ attacks it with a lone pair, the C=O π electrons move to oxygen, and protonation gives the hydroxynitrile.

Curly arrows must begin at the cyanide lone pair and the C=O π bond, then show proton transfer. The carbonyl polarity explains why this is nucleophilic addition.

For ethanal, CN⁻ attacks CH₃CHO to give an alkoxide intermediate, which is protonated to CH₃CH(OH)CN. KCN helps generate CN⁻ without being consumed overall.

Do not draw electrophilic addition or start a curly arrow at the carbonyl carbon; electron pairs originate from a lone pair or bond.

2,4-DNPH detects aldehydes and ketones by forming an orange carbonyl derivative

2,4-dinitrophenylhydrazine reacts with aldehydes and ketones, which contain a C=O group, to form a yellow/orange precipitate of a 2,4-dinitrophenylhydrazone.

The test confirms a carbonyl compound but does not distinguish aldehyde from ketone. Use it alongside an oxidation test or spectroscopy when the structure is unknown.

An unknown gives an orange precipitate with 2,4-DNPH, showing it contains an aldehyde or ketone. Tollens’ reagent can then distinguish which one.

2,4-DNPH is not a general alcohol test and a positive result does not identify the exact carbon chain.

Fehling’s and Tollens’ tests distinguish aldehydes from ketones

Aldehydes are readily oxidised to carboxylic acids; ketones are not readily oxidised under these mild tests. Fehling’s gives a brick-red Cu₂O precipitate and Tollens’ gives a silver mirror for an aldehyde.

A positive oxidation test supports aldehyde identification; a negative result under valid conditions supports ketone, but controls, freshness and heating conditions matter.

An unknown is positive with 2,4-DNPH and gives a silver mirror. The combined evidence identifies an aldehyde rather than a ketone.

A negative Tollens’ test alone is not proof of a ketone if the reagent or conditions were wrong. Combine independent observations.

The iodoform test detects a CH₃CO– group in a carbonyl compound

An aldehyde or ketone containing CH₃CO– gives a yellow CHI₃ precipitate with alkaline iodine. The methyl group is oxidised and the carbonyl fragment becomes a carboxylate ion.

The test is positive for ethanal and methyl ketones, not for every aldehyde or ketone. Use the structural motif rather than memorising a list of names.

Propanone, CH₃COCH₃, gives the yellow precipitate and propanoate ion. Propanal, CH₃CH₂CHO, has no CH₃CO– group and is negative.

This is a different use of the iodoform test from identifying CH₃CH(OH)– alcohols: oxidation can make both routes converge on the same motif.

ConceptA-Level CAIE Chemistry AS