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Multiple Free Radical Substitutions in the Reaction of Methane and Bromine

Warning! We are just about to muddy the water quite considerably! Don't go on until you are sure that you understand the mechanism for the production of bromomethane – and are confident that you could write it in an exam. If you aren't sure about it, go back to that reaction and look at it again.

>It would be worth checking your syllabus and past exam papers to see if you need to know about these further substitution reactions.

The Facts

When a mixture of methane and bromine is exposed to ultraviolet light – typically sunlight – a substitution reaction occurs and the organic product is bromomethane.

CH_4 + \text{Br}_2 \longrightarrow CH_3\text{Br} + H\text{Br}

However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms. That means that you could get any of bromomethane, dibromomethane, tribromomethane or tetrabromomethane.

\begin{aligned} CH_4 + \text{Br}_2 &\longrightarrow CH_3\text{Br} + H\text{Br} \\ \\ CH_3\text{Br} + \text{Br}_2 &\longrightarrow CH_2\text{Br}_2 + H\text{Br} \\ \\ CH_2\text{Br}_2 + \text{Br}_2 &\longrightarrow CH\text{Br}_3 + H\text{Br} \\ \\ CH\text{Br}_3 + \text{Br}_2 &\longrightarrow C\text{Br}_4 + H\text{Br} \\ \end{aligned}

You might think that you could control which product you got by the proportions of methane and bromine you used, but it isn't as simple as that. If you use enough bromine you will eventually get CBr4, but any other proportions will always lead to a mixture of products.

The Mechanisms

The formation of multiple substitution products like di-, tri- and tetrabromomethane can be explained in just the same sort of way as the formation of the original bromomethane. You just have to look at the likely collisions as the reaction progresses.

Making Dibromomethane

You will remember that the over-all equation for the first stage of the reaction is

CH_4 + \text{Br}_2 \longrightarrow CH_3\text{Br} + H\text{Br}

As the reaction proceeds, the methane is getting used up and bromomethane is taking its place. That means that the argument about what a bromine radical is likely to hit changes during the course of the reaction. As time goes by there is an increasing chance of it hitting a bromomethane molecule rather than a methane molecule.

When that happens, the bromine radical can take a hydrogen from the bromomethane just as well as it could from a methane. In this new case:

CH_3\text{Br} + \text{Br}{\bullet} \longrightarrow {\bullet}CH_2\text{Br} + H\text{Br}

Notice: The dot representing the electron was placed against the carbon which is the atom with the unpaired electron. It would be potentially confusing to leave it next to the bromine.

The bromomethyl radical formed can then interact with a bromine molecule in a new propagation step

{\bullet}CH_2\text{Br} + \text{Br}_2 \longrightarrow CH_2\text{Br}_2 + \text{Br}{\bullet}

and so dibromomethane is formed and a bromine radical regenerated.

These propagation steps continue until the chain is terminated by any two radicals colliding and combining together.

Making Tri- and Tetrabromomethane

Obviously, as time goes on, there is an increasing chance of the dibromomethane being hit by a bromine radical – producing these propagation steps giving tribromomethane:

\begin{aligned} CH_2\text{Br}_2 + \text{Br}{\bullet} &\longrightarrow {\bullet}CH\text{Br}_2 + H\text{Br} \\ \\ {\bullet}CH\text{Br}_2 + \text{Br}_2 &\longrightarrow CH\text{Br}_3 + \text{Br}{\bullet} \end{aligned}

Care! Don't just skip lightly over these equations. Look carefully at each one so that you understand what is happening, and can relate it to what has gone before. Talk through the equations with yourself.

For example: "A bromine radical hits the dibromomethane molecule and steals a hydrogen. That leaves a new radical (I don't know what it's called, but that doesn't really matter, as long as I can work out its formula if I have to!), which then bumps into a bromine molecule – etc, etc."

Doing this helps you to focus properly on the equations. If you just read them quickly, you'll have forgotten all about them again in 15 seconds!

As the amount of tribromomethane builds up, then you will get these steps giving tetrabromomethane:

\begin{aligned} CH\text{Br}_3 + \text{Br}{\bullet} &\longrightarrow {\bullet}C\text{Br}_3 + H\text{Br} \\ \\ {\bullet}C\text{Br}_3 + \text{Br}_2 &\longrightarrow C\text{Br}_4 + \text{Br}{\bullet} \end{aligned}

This is why you will always get a mixture of products whatever the reaction proportions of methane and bromine you use. The whole process is simply governed by chance. Having produced some bromomethane there is no way that you can prevent it from being hit by bromine radicals, and similarly for dibromomethane and tribromomethane.

Trying to Produce Mainly One Product

If you wanted tetrabromomethane, you could of course get it by using a large excess of bromine, so that eventually all the hydrogens would be replaced.

If you wanted mainly bromomethane, you could favour this by using a huge excess of methane so that the chances were always greater of a bromine radical hitting a methane rather than anything else – but even so, you would still get some mixture of products.

There is no obvious way of getting mainly dibromomethane or tribromomethane.