Aldehydes and Ketones: Structure, Oxidation, and Reduction

What is the structural difference between an aldehyde and a ketone?

Both contain the carbonyl group (C=O), but their position differs. In an aldehyde, the carbonyl carbon is at the end of the carbon chain and is bonded to at least one hydrogen. In a ketone, the carbonyl carbon is in the middle of the chain, bonded to two other carbon atoms

Why is the C=O bond polar, and what effect does this have on the chemistry of carbonyl compounds?

Oxygen is significantly more electronegative than carbon, so the bonding electrons are drawn towards the oxygen, making the oxygen δ− and the carbon δ+. This creates a permanent dipole across the C=O bond. The δ+ carbon is electron-deficient and is therefore susceptible to attack by nucleophiles — this drives the nucleophilic addition reactions characteristic of aldehydes and ketones.

Why do small carbonyl compounds have higher boiling points than alkanes of similar mass, but lower boiling points than equivalent alcohols?

Small carbonyls are polar (due to the C=O dipole), so they experience permanent dipole–dipole interactions in addition to London forces — giving higher boiling points than non-polar alkanes of similar mass. However, carbonyls cannot form hydrogen bonds between their own molecules (there is no O–H or N–H group), so they have lower boiling points than alcohols of similar mass, which can hydrogen bond intermolecularly.

What happens when an aldehyde is oxidised, and what reagent is used?

Aldehydes are oxidised to carboxylic acids by acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄). The orange dichromate solution turns green as Cr₂O₇²⁻ is reduced to Cr³⁺. The general equation using [O] to represent the oxidising agent is:
RCHO + [O] → RCOOH
Example: CH₃CHO + [O] → CH₃COOH (ethanal → ethanoic acid)

What is the reagent for reducing carbonyl compounds, and what are the products?

(sodium borohydride, NaBH₄) in aqueous solution is used. It acts as a source of hydride ions (H⁻), which are the nucleophile. The products are:

• Aldehyde → primary alcohol (e.g. ethanal → ethanol)

• Ketone → secondary alcohol (e.g. propanone → propan-2-ol)

The general equations are:
RCHO + 2[H] → RCH₂OH
RCOR’ + 2[H] → RCHOHR’

Why is NaBH₄ described as providing a hydride ion (H⁻) nucleophile?

NaBH₄ contains BH₄⁻, which delivers :H⁻ (a hydride ion). H⁻ acts as a nucleophile, attacking the δ+ carbon of the carbonyl group. This is why the reaction is classified as nucleophilic addition — the H⁻ adds across the C=O bond.

Describe the mechanism for the nucleophilic addition of H⁻ (from NaBH₄) to an aldehyde such as ethanal.

The mechanism has two steps:

1. The H⁻ ion (nucleophile) donates its pair of electrons to the δ+ carbon of the C=O bond. A curly arrow goes from H⁻ to the carbon, and simultaneously a curly arrow goes from the C=O π bond to the oxygen (which becomes O⁻). This forms an alkoxide intermediate: CH₃CH(H)O⁻.

2. The O⁻ is then protonated — it accepts H⁺ from water (or the solvent): CH₃CH(H)O⁻ + H₂O → CH₃CH₂OH + OH⁻. The product is the primary alcohol ethanol.

Key points in the mechanism: show the lone pair on H⁻, both curly arrows in step 1, the negative charge on oxygen in the intermediate, and the protonation step.

Image of the mechanism for the nucleophilic addition of H⁻ (from NaBH₄) to an aldehyde such as ethanal.