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8873 / 2027
Polymers and Organic Chemistry

Lesson 6 of 12

Carbonyls, acids and amines

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

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

  • I can read and name the required organic families and formula representations.Syllabus

    Syllabus 8873, 9(a). Read and name the required organic families and formula representations.

  • I can use organic reaction terminology.Syllabus

    Syllabus 8873, 9(b). Use organic reaction terminology.

  • I can apply the specified organic reactions, conditions and analogous product predictions.Syllabus

    Syllabus 8873, 9(g). Apply the specified organic reactions, conditions and analogous product predictions.

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 changeReagent and conditionsEquation
Ethanal → ethanolLiAlH4, followed by aqueous work-up; or H2 with Ni catalyst and heat.CH3CHO + 2[H] → CH3CH2OH
Propanone → propan-2-olLiAlH4, followed by aqueous work-up; or H2 with Ni catalyst and heat.CH3COCH3 + 2[H] → CH3CH(OH)CH3
Ethanal → ethanoic acidAcidified K2Cr2O7 or acidified KMnO4; heat.CH3CHO + [O] → CH3COOH

In carbonyl reduction, one hydrogen adds to the carbonyl carbon and one to the oxygen, so C=O becomes CH-OH. An aldehyde gives a primary alcohol, and a ketone gives a secondary alcohol. LiAlH4 is used in dry conditions, usually dry ether, and water is added afterwards. Do not write it as an aqueous reagent. An aldehyde can be oxidised to a carboxylic acid with the same carbon chain. A ketone is not oxidised by these usual conditions, so do not give it a carboxylic acid product.

Ethanoic acid gives a proton from its COOH group. With an alkali it forms a carboxylate salt and water. With a carbonate it forms a salt, water and carbon dioxide. The group changes from -COOH to -COO-, and the carbon chain stays the same.

Acid and base reactions retain the organic skeleton
ReactionWhat the equation explains
CH3COOH + NaOH → CH3COONa + H2OOne mole of ethanoic acid neutralises one mole of hydroxide.
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2Two 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 examplePredict a carbonyl product without guessing the name

Reduce CH3COCH2CH3 using H2/Ni.

Try it first. Then show the worked stepsHide the worked steps
  1. The starting carbonyl carbon is attached to CH3 and CH2CH3, so it is a ketone.
  2. Replace C=O by CH-OH while preserving both attached carbon groups: CH3CH(OH)CH2CH3.
  3. 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 understanding

Ethylamine reacts with aqueous acid. Where is the positive charge in the product?

  1. On the carbon next to nitrogen, which takes the H+ from the acid.
  2. On the chloride ion, which takes the extra electrons from nitrogen.
  3. On nitrogen, as its lone pair bonds to H+ to give CH3CH2NH3+.
Show the answer

On nitrogen, as its lone pair bonds to H+ to give CH3CH2NH3+.

The lone pair on N accepts the proton. The counter-ion, such as Cl-, balances the charge.