Aldehydes and ketones as carbonyl compounds. Oxidation of aldehydes to carboxylic acids and the use of Tollens' and Fehling's reagents to distinguish them from ketones. Reduction with NaBH4 to alcohols. Nucleophilic addition of HCN to form hydroxynitriles and the production of a racemic mixture.

What this dot point is asking

AQA wants you to compare the reactions of aldehydes and ketones, oxidise aldehydes and use Tollens' and Fehling's tests to distinguish them from ketones, reduce carbonyls with $NaBH_4$, and give the nucleophilic addition mechanism of HCN including why it gives a racemate.

The carbonyl group

The $C=O$ bond is polar because oxygen is more electronegative, giving the carbon a $\delta+$ charge. This makes the carbon vulnerable to attack by nucleophiles, which is the basis of both the reduction by hydride and the addition of cyanide. In an aldehyde the carbonyl carbon carries at least one hydrogen ($\text{RCHO}$), so the molecule can be oxidised; in a ketone the carbonyl carbon sits between two carbon groups ($\text{RCOR}'$), so there is no hydrogen to remove and oxidation does not occur under mild conditions. This single structural difference explains why the distinguishing tests work, and the carbonyl carbon is sp2 hybridised and trigonal planar, which is why nucleophilic addition that creates a chiral centre gives a racemate.

Oxidation and distinguishing tests

Aldehydes are easily oxidised to carboxylic acids; ketones are not.

• Tollens' reagent (ammoniacal silver nitrate): aldehyde gives a silver mirror; ketone, no change.
• Fehling's solution: aldehyde turns it from blue to a brick-red precipitate of $Cu_2O$; ketone, no change.

Reduction with $NaBH_4$

Sodium tetrahydridoborate(III), $NaBH_4$, provides the hydride ion $H^-$ as the nucleophile. The mechanism is nucleophilic addition: the hydride attacks the $\delta+$ carbonyl carbon, breaking the $C=O$ pi bond to give an alkoxide, which is then protonated to the alcohol.

• Aldehyde to primary alcohol: $CH_3CHO + 2[H] \rightarrow CH_3CH_2OH$
• Ketone to secondary alcohol: $CH_3COCH_3 + 2[H] \rightarrow CH_3CH(OH)CH_3$

This is the exact reverse of the oxidation of alcohols by acidified dichromate, so the two reactions together let you move freely between alcohols and carbonyl compounds in a synthesis. $NaBH_4$ is selective: it reduces the polar $C=O$ of an aldehyde or ketone but leaves a non-polar $C=C$ double bond untouched, which is useful when a molecule contains both.

Nucleophilic addition of HCN

Because the carbonyl carbon is planar, $CN^-$ attacks from either side equally, creating a new chiral centre and producing a racemic mixture (optically inactive). The hydroxynitrile product is valuable in synthesis because it contains both an $-OH$ group and a $-CN$ group, and the nitrile can be hydrolysed to a carboxylic acid, giving a 2-hydroxy acid one carbon longer than the original carbonyl compound.

Try this

Q1. Name the product of reducing propanone with $NaBH_4$. [1 mark]

• Cue. Propan-2-ol.

Q2. Why does HCN addition to butanone give a racemic mixture? [2 marks]

• Cue. The planar carbonyl is attacked equally from both faces, giving equal amounts of two enantiomers.