Multiple Nucleophilic Substitution Reactions of Halogenoalkanes and Ammonia

This page looks at further substitution in the nucleophilic substitution reaction between halogenoalkanes and ammonia following the formation of the primary amine. You could also think of this as a description of how primary, secondary and tertiary amines act as nucleophiles in a sequence of reactions with a halogenoalkane.

Only primary halogenoalkanes are considered on this page. Only one current A-level syllabus (AQA) is likely to ask about these reactions, and that only asks about primary halogenoalkanes in this context.

The lone pair on the nitrogen atom in an ammonia molecule is attracted towards the δ+ carbon in the halogenoalkane – in this example, bromoethane. It forms a bond with it – in the process expelling the bromine as a bromide ion.

A salt is formed – ethylammonium bromide.

There is then the possibility of another ammonia molecule removing a hydrogen from the positive ion to give a primary amine – ethylamine. A primary amine has the general formula RNH₂. This reaction is reversible, and you will only get significant amounts of the free amine if you use a large excess of ammonia.

At the end of the initial substitution there will be a certain amount of free primary amine formed – the CH₃CH₂NH₂ in the example above. There may not be very much of it, but there will be some. If that small amount reacts with something else, the reverse reaction can't happen any longer. The free amine will continue to be produced, and what was originally a reversible reaction becomes a one-way reaction – provided the free amine is removed by some other reaction as soon as it is formed.

Now compare the structures of ammonia and ethylamine:

You can think of the primary amine as a slightly modified ammonia molecule. It has a lone pair on the nitrogen atom and an even bigger δ- charge than in ammonia.

That means that the primary amine is going to be a better nucleophile than ammonia is. You can therefore get a reaction between it and a molecule of the halogenoalkane.

A secondary amine has the general formula R₂NH. It is like an ammonia molecule (NH₃) in which two of the hydrogens have been replaced by alkyl groups.

The lone pair on the nitrogen in the primary amine attacks the δ+ carbon exactly the same as the ammonia did. Bromine is lost as a bromide ion, and the immediate product is a salt called diethylammonium bromide – (CH₃CH₂)₂NH₂⁺ Br^-. This is essentially ammonium bromide in which two of the hydrogens attached to the nitrogen have been replaced by ethyl groups.

This then reacts with ammonia in a reversible reaction, exactly as we've seen before:

The organic product is a secondary amine – diethylamine. It is secondary because there are two alkyl groups attached to the nitrogen atom.

A tertiary amine has the general formula R₃N. It is like an ammonia molecule in which all three of the hydrogens have been replaced by alkyl groups.

The secondary amine still has an active lone pair of electrons on the nitrogen atom. That, in turn, can attack bromoethane if it happens to collide with it. The two step sequence is exactly as before:

In this first step, the bromine is again displaced as a bromide ion and you get a salt formed called triethylammonium bromide.

An ammonia molecule can then remove the hydrogen from the nitrogen in the reversible reaction:

The organic product of this reaction is the tertiary amine, triethylamine. It is tertiary because of the three alkyl groups attached to the nitrogen.

A quaternary ammonium salt is an ammonium salt (for example, NH₄⁺ Br^-) in which all the hydrogens have been replaced by an alkyl group – for example, (CH₃CH₂)₄N⁺ Br^-.

The tertiary amine still has an active lone pair on the nitrogen and, once again, that can attack the δ+ carbon in the bromoethane.

But this time there is nowhere else for the reaction to go. There is no longer a hydrogen atom on the nitrogen that an ammonia molecule could remove, and so the reaction finally comes to an end.

The product is a salt called tetraethylammonium bromide, (CH₃CH₂)₄N⁺ Br^-.

You will always get a mixture of all the above products unless you use a very large excess of bromoethane so that there is enough of it to carry the sequence all the way through to the quaternary ammonium salt.

You could favour formation of the primary amine (or its salt) by using a very large excess of ammonia. That way there is always a greater chance of a bromoethane molecule being hit by an ammonia molecule than any other possible nucleophile. Even so, there will always, by chance, be some collisions leading to the follow-on products.