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AS · Practice questions · Nitriles and hydroxynitriles

Add a carbon, then convert.

Six original Cambridge-style questions on nitriles and hydroxynitriles.

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

KCN in ethanol, HCN with KCN, hydrolyse or reduce.

01
Recall
[3 marks]

Bromoethane is converted into propanoic acid in two steps. Give the reagents and conditions for each step and name the intermediate.

  • Step 1: KCN in ethanol, heat under reflux. ✓
  • Intermediate: propanenitrile, CH3CH2CN. ✓
  • Step 2: dilute HCl (or H2SO4), heat under reflux, giving CH3CH2COOH. ✓
02
Explain
[4 marks]

Describe the mechanism of the reaction between propanone and HCN in the presence of KCN.

  • The C=O bond is polar; the carbonyl carbon is δ+. ✓
  • CN− attacks the δ+ carbon with its lone pair; the π bond electrons move onto O, forming an O− intermediate. ✓
  • The O− removes H+ from HCN, forming (CH3)2C(OH)CN. ✓
  • CN− is regenerated, so it is a catalyst: nucleophilic addition. ✓
03
Calculation
[2 marks]

10.9 g of bromoethane (Mr 108.9) is converted into propanoic acid (Mr 74.0). The mass of acid obtained is 5.18 g. Calculate the overall percentage yield.

  • n(bromoethane) = 10.9 ÷ 108.9 = 0.100 mol, so the maximum is 0.100 mol × 74.0 = 7.40 g. ✓
  • Yield = 5.18 ÷ 7.40 × 100 = 70.0%. ✓
04
Calculation
[2 marks]

5.50 g of propanenitrile (Mr 55.0) is completely reduced. Calculate the mass of propylamine (Mr 59.0) formed.

  • n(propanenitrile) = 5.50 ÷ 55.0 = 0.100 mol; 1 : 1 ratio. ✓
  • Mass = 0.100 × 59.0 = 5.90 g. ✓
05
Explain
[2 marks]

Explain why the product of ethanal and HCN does not rotate plane-polarised light, and why the product of propanone and HCN does not rotate it either.

  • CH3CH(OH)CN has a chiral carbon, but the planar C=O is attacked equally from both faces, giving a racemic mixture whose rotations cancel. ✓
  • (CH3)2C(OH)CN has two identical CH3 groups on that carbon, so it has no chiral centre at all. ✓
06
Analysis
[3 marks]

Suggest a two-step route from ethanal to 2-hydroxypropanoic acid, CH3CH(OH)COOH, with reagents, conditions and the intermediate.

  • Step 1: HCN with a trace of KCN (about 10 to 20 °C). ✓
  • Intermediate: 2-hydroxypropanenitrile, CH3CH(OH)CN. ✓
  • Step 2: heat under reflux with dilute acid to hydrolyse –CN to –COOH. ✓

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