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A2 · Practice questions · Primary and secondary amines

Find the lone pair.

Six original Cambridge-style questions on making amines and their basicity.

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

Three routes in, one lone pair out.

01
Recall
[3 marks]

Give the reagents and conditions for making propylamine, CH3CH2CH2NH2, from (a) 1-bromopropane, (b) propanenitrile, (c) propanamide.

  • (a) Excess NH3 in ethanol, heated in a sealed tube. ✓
  • (b) LiAlH4 in dry ether, or H2 with a nickel catalyst. ✓
  • (c) LiAlH4 in dry ether. ✓
02
Explain
[3 marks]

Explain why the order of base strength is ethylamine > ammonia > phenylamine.

  • Base strength depends on how available the N lone pair is to accept a proton. ✓
  • The ethyl group is electron-donating, so it increases the electron density on N in ethylamine: stronger than NH3. ✓
  • In phenylamine the lone pair overlaps with the ring π system and is partly delocalised, so it is less available: weaker than NH3. ✓
03
Calculation
[2 marks]

10.89 g of bromoethane (Mr 108.9) is heated with excess ammonia and 3.15 g of ethylamine (Mr 45.0) is obtained. Calculate the percentage yield.

  • n(bromoethane) = 10.89 ÷ 108.9 = 0.100 mol, so maximum ethylamine = 0.100 × 45.0 = 4.50 g. ✓
  • Yield = 3.15 ÷ 4.50 × 100 = 70.0%. ✓
04
Recall
[2 marks]

Write an equation for the formation of diethylamine from ethylamine and name the type of reaction.

  • CH3CH2NH2 + CH3CH2Br → (CH3CH2)2NH + HBr. ✓
  • Nucleophilic substitution: the N lone pair of ethylamine attacks the δ+ carbon. ✓
05
Calculation
[2 marks]

Calculate the volume of hydrogen at room conditions needed to reduce 0.0500 mol of ethanenitrile to ethylamine.

  • CH3CN + 2H2 → CH3CH2NH2, so n(H2) = 0.100 mol. ✓
  • Volume = 0.100 × 24.0 = 2.40 dm3. ✓
06
Explain
[2 marks]

Ethylamine is a gas that dissolves well in water, but ethylammonium chloride is a white solid. Explain what happens when HCl(aq) and then excess NaOH(aq) are added to ethylamine.

  • HCl: the N lone pair accepts H+, giving the ionic salt CH3CH2NH3+Cl−. ✓
  • NaOH: OH− removes H+ from CH3CH2NH3+, releasing ethylamine again (smell of fish). ✓

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.