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9476 / 2027
Organic Chemistry

Lesson 29 of 34

An amide nitrogen is electronically coupled to its carbonyl

Delocalisation suppresses ordinary basicity; hydrolysis and reduction break different parts of the group.

A-Level 9476 (2026-2027)

  • I can form amides from primary amines and acyl chlorides.Syllabus

    Syllabus 9476, 11.9(f). Form amides from primary amines and acyl chlorides.

  • I can explain amide neutrality.Syllabus

    Syllabus 9476, 11.9(g). Explain amide neutrality.

  • I can describe amide reactions using ethanamide.Syllabus

    Syllabus 9476, 11.9(h). Describe amide reactions using ethanamide.

An amide nitrogen is electronically coupled to its carbonyl

Delocalisation suppresses ordinary basicity; hydrolysis and reduction break different parts of the group.

A primary amine, RNH2, reacts with an acyl chloride, R'COCl, to form R'CONHR. The nitrogen keeps its own R group, and the acyl group comes from the chloride. For example, ethanoyl chloride and methylamine give N-methylethanamide, CH3CONHCH3. A second amine molecule, or another base, removes the HCl formed.

In an amide, the nitrogen lone pair spreads towards the C=O group. So it is much less available to accept H+ than the lone pair of an amine, and an amide is almost neutral in water. The C-N bond has some double-bond character. An amide can still be protonated in strong acid; it just does not act as a base in ordinary aqueous conditions.

Ethanamide as the reference
ConditionsEquationKey distinction
Aqueous acid and heatCH3CONH2 + H2O + H+ → CH3COOH + NH4+.The nitrogen product is protonated in acid.
Aqueous alkali and heatCH3CONH2 + OH- → CH3COO- + NH3.The organic product is a carboxylate and ammonia is released.
LiAlH4, dry ether, then work-upCH3CONH2 + 4[H] → CH3CH2NH2 + H2O.Reduction replaces C=O by CH2 while retaining C-N and the carbon skeleton.

When an N-substituted amide is hydrolysed, it releases the matching amine or its ammonium ion, not NH3 or NH4+. Reduction keeps the group on nitrogen too. For example, CH3CONHCH3 is reduced to CH3CH2NHCH3, a secondary amine. It does not lose its N-methyl group.

Check your understanding

Propanoyl chloride, CH3CH2COCl, is added to ethylamine at room temperature. Why are two moles of amine used for each mole of acyl chloride?

  1. The first mole is used up as a catalyst, and the second mole forms the amide product.
  2. The second mole reacts with the HCl formed, which would otherwise protonate the amine.
  3. Two amine molecules join the acyl carbon, giving a product with two ethyl groups on N.
Show the answer

The second mole reacts with the HCl formed, which would otherwise protonate the amine.

CH3CH2COCl + 2CH3CH2NH2 -> CH3CH2CONHCH2CH3 + CH3CH2NH3+Cl-. RNH3+ has no lone pair, so it could not act as a nucleophile. The product is N-ethylpropanamide.

Check your understanding

Ethylamine solution is alkaline, but a solution of ethanamide, CH3CONH2, is neutral. Which statement explains this?

  1. The C=O group donates electron density to the nitrogen, which then repels H+
  2. Ethanamide has no lone pair on its nitrogen atom, because the C=O group uses it
  3. Ethanamide is insoluble in water, so its solution contains no basic molecules
  4. The N lone pair is delocalised over C=O, so it is much less available to accept a proton
Show the answer

The N lone pair is delocalised over C=O, so it is much less available to accept a proton

The N lone pair overlaps with the C=O pi bond, so it is drawn towards the electronegative oxygen and is not free to bond to H+. In ethylamine the lone pair stays on N.