Topic 6 of 12

Carbonyls, acids and amines

Carbonyl chemistry changes C=O; acid-base chemistry transfers protons.

A-Level 8873, revised syllabus (2026-2027)

Link oxidation level with acid-base behaviour

Carbonyl chemistry changes C=O; acid-base chemistry transfers protons.

Aldehydes and ketones share reduction but differ in ordinary oxidation
Compound and change	Reagent and conditions	Equation
Ethanal → ethanol	LiAlH₄, followed by aqueous work-up; or H₂ with Ni catalyst and heat.	CH₃CHO + 2[H] → CH₃CH₂OH
Propanone → propan-2-ol	LiAlH₄, followed by aqueous work-up; or H₂ with Ni catalyst and heat.	CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃
Ethanal → ethanoic acid	Acidified K₂Cr₂O₇ or acidified KMnO₄; heat.	CH₃CHO + [O] → CH₃COOH

In carbonyl reduction, one hydrogen is added to the carbonyl carbon and one to oxygen, converting C=O into CH-OH. An aldehyde therefore gives a primary alcohol; a ketone gives a secondary alcohol. LiAlH₄ is used in dry conditions, commonly dry ether, before the subsequent aqueous work-up; writing it as an aqueous reagent at the reduction stage is inappropriate. An aldehyde can be oxidised to a carboxylic acid without breaking its carbon skeleton. A ketone resists these usual aldehyde-oxidation conditions; do not invent a same-skeleton carboxylic acid product from it.

Ethanoic acid donates a proton from COOH. With an alkali it forms a carboxylate salt and water; with a carbonate it forms salt, water and carbon dioxide. The group changes from -COOH to -COO^-; the carbon skeleton is retained.

Acid and base reactions retain the organic skeleton
Reaction	What the equation explains
CH₃COOH + NaOH → CH₃COONa + H₂O	One mole of ethanoic acid neutralises one mole of hydroxide.
2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂	Two acidic protons are needed per carbonate ion; effervescence indicates gas evolution.
CH₃CH₂NH₂ + HCl → [CH₃CH₂NH₃]⁺Cl^-	Ethylamine accepts a proton at nitrogen, forming ethylammonium chloride. It is not converted into an amide by this acid-base step.

Worked example

Predict a carbonyl product without guessing the name

Reduce CH₃COCH₂CH₃ using H₂/Ni.

The starting carbonyl carbon is attached to CH₃ and CH₂CH₃, so it is a ketone.
Replace C=O by CH-OH while preserving both attached carbon groups: CH₃CH(OH)CH₂CH₃.
The product is butan-2-ol, a secondary alcohol. Both oxygen and all four carbons remain in the molecule.

Answer

Butan-2-ol forms; reduction adds hydrogen without extending or shortening the carbon chain.

Check your understandingWhere does the positive charge arise when ethylamine reacts with aqueous acid?Think it through, then reveal the answer

The nitrogen lone pair bonds to H⁺, giving CH₃CH₂NH₃⁺. The counter-ion, such as Cl^-, balances its charge. The added proton is on nitrogen, not on a carbon.