Primary amines

AS Level Chemistry · Nitrogen compounds. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

A primary amine, RNH2, is made by heating a halogenoalkane with an excess of ammonia dissolved in ethanol. The mixture is heated under pressure in a sealed tube, because ammonia is a gas and would escape.

Ammonia acts as a nucleophile. The lone pair on nitrogen attacks the δ+ carbon and the halide ion leaves, so this is nucleophilic substitution. A second ammonia molecule then acts as a base and removes H+, forming an ammonium ion.

The amine formed also has a lone pair on nitrogen, so it can attack another halogenoalkane molecule. This further substitution gives unwanted products such as diethylamine. A large excess of ammonia makes this less likely.

Rules to remember

Common mistake

Students write HBr as the second product. HBr is an acid and ammonia is a base, so they react at once: the other product is ammonium bromide, NH4Br, and the equation needs 2NH3.

Worked example

6.15 g of 1-bromopropane (Mr 123) is heated with excess ammonia in ethanol. 1.77 g of propylamine (Mr 59.0) is obtained. What is the percentage yield?

  1. Moles of 1-bromopropane = 6.15 ÷ 123 = 0.0500 mol.
  2. 1 mol of halogenoalkane gives 1 mol of amine, so theoretical mass = 0.0500 × 59.0 = 2.95 g.
  3. Percentage yield = 1.77 ÷ 2.95 × 100 = 60.0%.

Answer: 60.0%

Practice questions

Try each one, then open the answer.

1. Why is a large excess of ammonia used when preparing ethylamine from bromoethane?

  1. A
    to make the reaction exothermic
  2. B
    to reduce further substitution, in which the ethylamine formed reacts with more bromoethane
  3. C
    to neutralise the ethylamine
  4. D
    because ammonia is the catalyst
Show answer

Answer: B. Ethylamine is itself a nucleophile and can attack another bromoethane molecule, giving secondary and tertiary amines. Excess ammonia makes it more likely that bromoethane meets NH3 rather than the amine.

2. Why is the reaction of a halogenoalkane with ammonia in ethanol carried out in a sealed tube?

  1. A
    to keep out oxygen, which would oxidise the amine
  2. B
    ammonia is volatile, so a sealed tube keeps it in the heated mixture under pressure
  3. C
    to stop the ethanol reacting with the halogenoalkane
  4. D
    to prevent light from starting a free-radical reaction
Show answer

Answer: B. Ammonia would escape on heating in an open vessel. Sealing the tube keeps a high concentration of NH3 in contact with the halogenoalkane at the raised temperature and pressure.

3. Bromoethane is heated with ammonia in ethanol, but the ammonia is not in excess. Which additional product is likely to form?

  1. A
    ethene
  2. B
    ethanol
  3. C
    ethanenitrile
  4. D
    diethylamine, (CH3CH2)2NH
Show answer

Answer: D. Ethylamine also has a lone pair on nitrogen, so it can attack more bromoethane, giving diethylamine and further products. A large excess of ammonia makes this less likely.

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