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A2 · Practice questions · Amides

Split the C–N bond.

Six original Cambridge-style questions on amides.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Acyl chloride in, acid or alkali out.

01
Recall
[2 marks]

Write equations for the reactions of ethanoyl chloride with (a) ammonia and (b) ethylamine. Name the organic products.

  • (a) CH3COCl + 2NH3 → CH3CONH2 + NH4Cl; ethanamide. ✓
  • (b) CH3COCl + 2CH3CH2NH2 → CH3CONHCH2CH3 + CH3CH2NH3+Cl−; N-ethylethanamide. ✓
02
Recall
[3 marks]

N-ethylpropanamide, CH3CH2CONHCH2CH3, is heated under reflux with (a) HCl(aq) and (b) NaOH(aq). Give the organic products in each case.

  • (a) Propanoic acid, CH3CH2COOH, and ethylammonium chloride, CH3CH2NH3+Cl−. ✓
  • (b) Sodium propanoate, CH3CH2COO−Na+, and ethylamine, CH3CH2NH2. ✓
  • The C–N bond is broken in both; only the ionisation state of the products differs. ✓
03
Explain
[3 marks]

Explain why an aqueous solution of ethylamine is alkaline but a solution of ethanamide is neutral.

  • In ethylamine the N lone pair is available (enhanced by the electron-donating ethyl group) and accepts H+ from water, releasing OH−. ✓
  • In ethanamide the N lone pair overlaps with the C=O π bond and is delocalised towards oxygen. ✓
  • So it is not available to accept a proton, and the solution stays neutral. ✓
04
Calculation
[2 marks]

7.85 g of ethanoyl chloride (Mr 78.5) reacts with excess ammonia, giving 4.72 g of ethanamide (Mr 59.0). Calculate the percentage yield.

  • n(CH3COCl) = 7.85 ÷ 78.5 = 0.100 mol, so maximum ethanamide = 0.100 × 59.0 = 5.90 g. ✓
  • Yield = 4.72 ÷ 5.90 × 100 = 80.0%. ✓
05
Analysis
[2 marks]

A white solid X, C2H5NO, gives no reaction with dilute acid at room temperature. Warming X with NaOH(aq) gives a gas that turns damp red litmus blue. Identify X and the gas.

  • X is ethanamide, CH3CONH2 (neutral, so no salt forms with dilute acid). ✓
  • Alkaline hydrolysis releases ammonia, NH3, which is alkaline. ✓
06
Recall
[2 marks]

Give the reagent and the product when propanamide is reduced, and write an equation using [H].

  • LiAlH4 in dry ether; product propylamine, CH3CH2CH2NH2. ✓
  • CH3CH2CONH2 + 4[H] → CH3CH2CH2NH2 + H2O. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.