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Nucleophilic substitution reactions involving halogenalkanes

This page covers common nucleophilic substitution reactions of halogenalkanes using hydroxide ions (OH-), cynaide ions (CN-) and the neutral molecule ammonia (NH3) . If you are not familiar with the mechanism of nucleophilic substitution reactions I would suggest you review this before studying the page below.

Halogenalkanes and hydroxide ions

Alcohols are formed when halogenalkanes are warmed with aqueous solution of sodium hydroxide or potassium hydroxide. You are probably familiar with potassium and sodium hydroxide from any work you have done on acids and alkalis where the sodium hydroxide and potassium hydroxide would have been used as a base (a H+ acceptor). However the hydroxide ion (OH-) can also behave as a nucleophile. There are many similarities between nucleophiles and bases; for example they both possess lone pairs of electrons. Usually strong bases are also good nucleophiles and vice versa. However with sodium and potassium hydroxide we can alter the reaction conditions to "make it" behave as a base or a nucleophile.

In the reaction below sodium hydroxide is reacting with the halogenalkane bromoethane. In an aqueous solution of sodium hydroxide the hydroxide ion (OH-) behaves as a nucleophile, indeed this is a typical nucleophilic substitution reaction where the hydroxide ion uses one of its lone pairs of electrons to attack the partially charged carbon atom in the polar C-Br bond. The product of this reaction is the alcohol ethanol and the salt sodium bromide which of course will be in solution.

Mechanism and reaction conditions for the preparation of alcohols from halogenalkanes.  The reaction of bromoethane and sodium hydroxide is used as an example.

The main problem with the set-up above is that halogenalkanes are pretty much insoluble in water so the reaction is going to be very slow. You could add the alcohol ethanol to the sodium hydroxide to produce what is often called ethanolic sodium hydroxide solution. The halogenalkane would be soluble in this mixture but unfortunately this may lead to a different type of reaction called an elimination reaction; this elimination we would not produce the alcohol but instead lead to the formation of alkenes. A better solution to this problem would be to set-up a reflux reaction using the halogenalkane and aqueous sodium hydroxide, this would allow sufficient time for the reaction to take place and also help force it forwards to produce the product. A reflux set-up is shown below in the nitrile section of the page.

Halogenalkanes and cyanide ions

Nitriles contain the functional group R-CN. Nitriles are particularly useful in organic synthesis are they are one of the few ways in which it is possible to extend the carbon chain by 1 carbon atom. Nitriles are also reactive and are easily converted into other useful and reactive molecules such as amines, amides and carboxylic acids. The first two members of the nitriles homologous series are shown below:

3d models, displayed formula and molecular formula of ethanenitrile and propanenitrile. Nitriles can be made by reacting a halogenalkane with a solution of potassium or sodium cyanide in ethanol the reaction is carried out under reflux. For example bromoethane reacts with the cyanide ion (:CN-) to form propanenitrile. The cyanide ion uses its lone pair of electrons to attack the δ+ carbon atom in the C-Br bond. The mechanism for this reaction is shown below: Mechanism  and symbolic equation for the reaction of sodium cyanide with a halogen alkane, bromoethane.  The reflux apparatus diagram is also shown.

Halogenalkanes and ammonia

Ammonia reacts with halogenalkanes to produce amines. Amines are simply molecules of ammonia (NH3) where one or more of the hydrogen atoms on the ammonia molecule is replaced by an alkyl group.

It is possible to have four alkyl groups around the nitrogen atom from an ammonia molecule, these compounds with four alkyl groups are ionic solids and are called quaternary ammonium salts (see below for more details).

3d models, displayed formula and molecular formula of primary, secondary and tertiary amines.

Nucleophile or base?

Mechanism to show how ammonia can act as a nucleophile and as a base in its reactions with halogenalkanes or alkyl halides.

The reaction of ammonia with halogenalkanes needs to be carried out in sealed reaction vessels otherwise the ammonia gas will simply escape. Ammonia is a gas at room temperature and it is very soluble in water however when heated ammonia gas would be given off, so this reaction needs to be done in sealed vessels.
The mechanism for this nucleophilic substitution reaction is shown below; you should note that 2 moles of ammonia are used for every mole of the halogenalkane. One mole of the ammonia acts as a nucleophile and attacks the δ+ carbon atom in the halogenalkane while the other mole of ammonia acts as a base and abstracts a proton (H+) from the quaternary ammonium salt. The final product in this case is methylamine.

Mechanism to show how ammonia can act as a nucleophile and as a base in its reactions with halogenalkanes or alkyl halides.

However this is not the end of the story for this reaction. If you follow the mechanism you can clearly see that the ammonia simply swaps a hydrogen atom for a methyl group or if a different halogenalkane was used, for example bromoethane, then an ethyl group would replace a hydrogen on the ammonia to form ethylamine.

The problem is the product of the reaction, the primary amine is a stronger base and a better nucleophile than ammonia. So as the reaction proceeds its concentration will increase and it will take over from ammonia in the reaction mechanism. This will mean that one of the hydrogen atoms on the primary amine will be replaced by an alkyl group to form a secondary amine, this is outlined below:

Mechanism for the reaction of a halogenalkane and an amine, nucleophilic substitution.

I am sure you can see where this going! The product of the reaction above, the dimethylamine is a better nucleophile than the primary amine, methylamine. This means that as the concentration of the secondary amine, dimethylamine increases it will take over from the methylamine and form the tertiary amine trimethylamine. Even here the reaction will not stop! The trimethylamine will continue to react with the bromomethane and form the quaternary ammonium salt where all the hydrogen atoms from the original ammonia molecule have been replaced by -methyl groups.

This leads to a mixture of products which ultimately reduces the usefulness of this reaction. We can of course try to stop the reaction at the first step and only produce the methylamine, to do this we simply try to block out the methylamine by using a large excess of ammonia. This it is hoped by sheer weight of numbers the ammonia molecules will simply block the methylamine and limit the reaction to produce only the primary amine, methylamine.

Key Points

Practice questions

Check your understanding - Questions on Nucleophilic substitution reactions