# What is the halogenation of alkanes?

Apr 5, 2016

A way to introduce functionality into an otherwise unreactive alkyl chain.

#### Explanation:

$B {r}_{2} + h \nu \rightarrow 2 B r \cdot$

$B r \cdot$ is a neutral bromine radical. Radical species are quite reactive in that in the act of reaction they generate another radical, that continue the chain of reactivity. UV light is known to produce radicals from bromine molecules. This is the propagation step.

$B r \cdot + C {H}_{4} \rightarrow H - B r + \cdot C {H}_{3}$

The methyl radical is a very reactive species, and can react with bromine to give the product:

${H}_{3} C \cdot + B {r}_{2} \rightarrow {H}_{3} C - B r + \cdot B r$

The termination step involves the coupling of two radicals:

${H}_{3} C \cdot + \cdot B r \rightarrow {H}_{3} C - B r$

Or two methyl radicals can couple:

${H}_{3} C \cdot + \cdot C {H}_{3} \rightarrow {H}_{3} C - C {H}_{3}$

The observation of ethane where a $C - C$ has been formed is good evidence to support the radical mechanism.

Overall the rxn is:

$C {H}_{4} + B {r}_{2} \rightarrow C {H}_{3} B r + H B r$