AQA A-Level Chemistry 7405 · 3.3.8 Aldehydes and ketones

Part 1: Carbonyl structure, tests and reduction

All 2 parts available · worked answers and exam guidance included. Reviewed 2 October 2026.

Connect the polar carbonyl group to aldehyde oxidation, identification tests and reduction with sodium borohydride.

Locate the carbonyl carbon

An aldehyde has a carbonyl carbon bonded to at least one hydrogen: RCHO, with methanal HCHO as the simplest case. A ketone has that carbon bonded to two carbon-containing groups: RCOR′. The functional groups are –CHO and >C=O respectively. Not every compound with C=O is an aldehyde or ketone: carboxylic acids, esters and amides also contain carbonyl groups.

Use –al for an aldehyde and –one with a position number when needed for a ketone. CH₃CH₂CHO is propanal; CH₃COCH₂CH₃ is butan-2-one. The carbonyl carbon counts as part of the chain. The C=O bond is polarised towards oxygen because oxygen is more electronegative, leaving carbon δ⁺ and oxygen δ⁻. A nucleophile attacks the electron-deficient carbon.

Explain physical properties with the right interactions

Simple aldehydes and ketones have London forces and permanent dipole–dipole attractions. They lack an O–H or N–H donor, so pure samples cannot hydrogen-bond to themselves in the usual model. Their oxygen lone pairs can accept hydrogen bonds from water, explaining appreciable water solubility of smaller members. Solubility generally decreases as the non-polar hydrocarbon part grows.

Compare similar-sized molecules when explaining boiling points. Hydrogen bonding usually makes the corresponding alcohol boil higher than a carbonyl compound; molecular size and shape still matter. Do not state that a carbonyl oxygen cannot hydrogen-bond merely because the molecule cannot donate an H bond.

Aldehydes oxidise under mild conditions

Heating an aldehyde with acidified potassium dichromate(VI) gives a carboxylic acid, with orange Cr₂O₇²⁻ becoming green Cr³⁺. Use sulfuric acid to acidify the reagent. Ordinary ketones do not react with this mild test; harsher conditions can break carbon–carbon bonds, so “ketones can never be oxidised” is too broad.

Aldehyde oxidation retains the number of carbon atoms. The shorthand [O] represents an oxidising equivalent, not a bottle of atomic oxygen. Under alkaline aldehyde-test conditions, the organic oxidation product is the carboxylate rather than predominantly free carboxylic acid.

CH₃CH₂CHO + [O] → CH₃CH₂COOH
3CH₃CHO + Cr₂O₇²⁻ + 8H⁺ → 3CH₃COOH + 2Cr³⁺ + 4H₂O

State the reagent and a visible observation

Tollens’ reagent contains [Ag(NH₃)₂]⁺; silver(I) gains electrons to form silver metal. In Fehling’s solution copper(II) is reduced to copper(I) oxide. Fehling’s is not a universal test for every aromatic aldehyde, and other reducing substances can also respond to aldehyde tests: draw conclusions within the supplied candidate set.

Use fresh separate portions. A dichromate change cannot by itself distinguish an aldehyde from an oxidisable alcohol. Do not replace a requested observation with “oxidation happens”. Tollens’ reagent must be freshly prepared and promptly disposed of through the laboratory’s procedure, never stored or allowed to dry.

CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O + 3H₂O
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag + 4NH₃ + 2H₂O
Aldehyde versus a typical simple ketone
Reagent and conditionsAldehyde resultKetone result
Tollens’ reagent, warm gentlySilver mirror or grey/silver metal depositNo visible change
Fehling’s solution, heat gentlyBrick-red Cu₂O precipitate for typical aliphatic aldehydesNo visible change
K₂Cr₂O₇ with dilute H₂SO₄, warmOrange solution turns greenNo visible change in this mild test

NaBH₄ supplies the hydride nucleophile

Aqueous sodium borohydride, NaBH₄, reduces aldehydes to primary alcohols and ketones to secondary alcohols under mild conditions; aqueous ethanol can be used where needed to dissolve the organic reactant. The exam mechanism represents the transferred hydride as H⁻ with its electron pair, not H⁺ or a hydrogen radical.

First draw a full-headed curly arrow from the hydride electron pair to carbonyl carbon and another from the C=O π bond to oxygen. The tetrahedral intermediate has an O⁻ group and a new C–H bond. Then O⁻ accepts a proton from water or another appropriate proton source, giving the alcohol. Proton transfer from water also requires an arrow from its O–H bond back to that water oxygen.

June 2023 Paper 2 Q03.3 assessed the hydride arrow, C=O electron movement, intermediate and mechanism name. A correct final alcohol does not replace a requested mechanism.

CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH
CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃

Diagram placeholder

Hydride addition to butan-2-one to add

Labels to include:

  • Cδ⁺=Oδ⁻ of CH₃COCH₂CH₃
  • H⁻ with lone pair
  • Arrow H lone pair → carbonyl C
  • Arrow C=O π bond → O
  • CH₃C(H)(O⁻)CH₂CH₃ intermediate
  • Arrow O⁻ lone pair → H of water; water O–H bond → water O
  • Butan-2-ol product and OH⁻ after water protonation

The carbonyl carbon has three bonds before attack and four single bonds afterwards. Retain every carbon and the negative charge on intermediate oxygen. The two newly added hydrogens end up on carbon and oxygen.

Quick checks

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

Q1. Name CH₃CH₂COCH₂CH₃ and identify its functional group.Show answer

Pentan-3-one; it is a ketone. The carbonyl carbon is attached to two ethyl groups.

Q2. Why can propanone dissolve in water even though pure propanone does not hydrogen-bond to itself?Show answer

Its oxygen lone pairs accept hydrogen bonds from water O–H groups. Propanone has no suitable O–H/N–H donor for hydrogen bonding between its own molecules.

Q3. How would Tollens’ reagent distinguish propanal from propanone?Show answer

Warm separate samples with fresh Tollens’ reagent. Propanal gives silver metal as a mirror or deposit; propanone gives no visible change under the test conditions.

Q4. What is the organic product when ethanal reacts with Fehling’s solution?Show answer

Ethanoate, CH₃COO⁻, because the test mixture is alkaline. Cu₂O forms as a brick-red precipitate.

Q5. Describe the first two electron-pair movements in reduction by NaBH₄.Show answer

The hydride electron pair moves to carbonyl carbon; the C=O π electrons move to oxygen. This makes a C–H bond and an O⁻ tetrahedral intermediate, which is then protonated.

Sources

Sources and examiner guidance (reviewed 2 October 2026)

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