Reaction with Ammonia

Halogenoalkanes react with ammonia by nucleophilic substitution to form primary amines, compounds containing the -NH₂ group.CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄BrBromoethane forms ethylamine (ethanamine). The by-product is ammonium bromide.Conditions: an excess of concentrated ammonia dissolved in ethanol, heated in a sealed tube. Ammonia is a gas, so the tube is sealed to stop it escaping; the reaction takes place under pressure.How it works: ammonia is a nucleophile because of the lone pair on the nitrogen atom. The lone pair attacks the δ+ carbon, and the C-Br bond breaks heterolytically, forming Br⁻. This gives the ion CH₃CH₂NH₃⁺. A second ammonia molecule then removes an H⁺ from it, forming the amine and NH₄⁺. This is why 2 mol of NH₃ are used per mole of halogenoalkane.Why an excess of ammonia? The amine product also has a lone pair on nitrogen, so it is a nucleophile too. It can attack more halogenoalkane, giving secondary and tertiary amines and eventually a quaternary ammonium salt. Using a large excess of ammonia makes it much more likely that the halogenoalkane meets an NH₃ molecule, which reduces this further substitution.Amines are named from the alkyl group: 1-chloropropane and ammonia give propylamine, CH₃CH₂CH₂NH₂.