Reactions of aldehydes and ketonesAQA A-Level Chemistry: Revision notes

Both families contain the carbonyl group, C=O. In an aldehyde the carbonyl carbon is bonded to at least one hydrogen atom (RCHO); in a ketone it is bonded to two carbon groups (RCOR′).

The key chemical difference is that aldehydes are readily oxidised to carboxylic acids, whereas ketones are not easily oxidised. This difference is used in chemical tests. Both can be reduced to alcohols: aldehydes to primary alcohols and ketones to secondary alcohols.

An aldehyde is oxidised to a carboxylic acid, for example propanal to propanoic acid, using acidified potassium dichromate(VI) (orange to green). Ketones are not oxidised under these conditions.

Tollens' reagent is ammoniacal silver nitrate, which contains the complex ion [Ag(NH₃)₂]⁺. Warm the sample with the reagent in a water bath:

• Aldehyde: the Ag⁺ is reduced to silver metal and a silver mirror forms on the inside of the test tube.
• Ketone: no change, no silver mirror.

The aldehyde is the reducing agent: it is oxidised to the carboxylic acid (as its salt in alkaline solution).

Fehling's solution is an alkaline solution containing copper(II) ions, so it is blue. Warm the sample with the solution:

• Aldehyde: the copper(II) ions are reduced and the blue solution forms a red (brick-red) precipitate of copper(I) oxide, Cu₂O.
• Ketone: the solution stays blue.

Both Tollens' reagent and Fehling's solution therefore distinguish aldehydes from ketones. A good answer always states the observation for both compounds.

Sodium tetrahydridoborate(III), NaBH₄, in aqueous solution reduces carbonyl compounds at room temperature:

• aldehyde → primary alcohol
• ketone → secondary alcohol

NaBH₄ is the source of the hydride ion, H⁻, which is the nucleophile. In equations the reductant is written as [H], and each C=O needs 2[H]:

CH₃CHO + 2[H] → CH₃CH₂OH (ethanal to ethanol) CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃ (propanone to propan-2-ol)

NaBH₄ does not reduce C=C bonds, so it reduces only the carbonyl group.

The C=O bond is polar: oxygen is more electronegative than carbon, so the carbon is δ+ and the oxygen δ−. The mechanism for NaBH₄ reduction is nucleophilic addition:

1. The hydride ion, H⁻, uses its lone pair to attack the δ+ carbon (curly arrow from the lone pair to carbon).
2. The C=O π bond breaks, and the pair of electrons moves onto oxygen (curly arrow from the C=O bond to O), forming an alkoxide intermediate with a negative charge on oxygen.
3. The O⁻ is protonated by water to give the alcohol (and OH⁻).

The reaction is an addition because H and OH are added across the C=O bond, and it is nucleophilic because the attacking species is an electron-pair donor.

Use this to choose the right answer quickly:

• Aldehyde + oxidising agent → carboxylic acid; ketone: no reaction.
• Aldehyde + Tollens' → silver mirror; ketone: no change.
• Aldehyde + Fehling's → red precipitate; ketone: stays blue.
• Aldehyde + NaBH₄ → primary alcohol; ketone → secondary alcohol.

Example: butanal, CH₃CH₂CH₂CHO, gives a silver mirror and is reduced to butan-1-ol. Butanone, CH₃COCH₂CH₃, gives no silver mirror and is reduced to butan-2-ol.