Reactions of alkanes and free radical substitutionEdexcel International A Level Chemistry: Revision notes

Alkanes are saturated hydrocarbons and are fairly unreactive because C–C and C–H bonds are strong and almost non-polar. They do react in two ways.

Combustion in plenty of oxygen gives carbon dioxide and water: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. With limited oxygen, incomplete combustion gives carbon monoxide or carbon.

With halogens in ultraviolet light, a hydrogen atom is replaced by a halogen atom. This is a substitution reaction: $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$. Bromine reacts in the same way but more slowly.

A free radical is a species with an unpaired electron, shown by a single dot, e.g. Cl• and •CH₃. Radicals are very reactive because they have an unpaired electron.

Radicals form by homolytic fission: a covalent bond breaks so each atom takes one electron from the shared pair. This is shown with curly half-arrows (a single-headed arrow, one for each electron).

In initiation, ultraviolet light supplies enough energy to break the weakest bond, the halogen–halogen bond, by homolytic fission:

$Cl_2 \rightarrow 2Cl\bullet$

This step creates radicals from a molecule with no radicals. C–H bonds in the alkane are stronger and are not broken here.

In propagation, a radical reacts with a molecule to make a new radical, so the chain continues. For methane:

$CH_4 + Cl\bullet \rightarrow \bullet CH_3 + HCl$

$\bullet CH_3 + Cl_2 \rightarrow CH_3Cl + Cl\bullet$

The chlorine radical is regenerated, so one initiation event can lead to thousands of cycles: a chain reaction. Every propagation step has one radical on each side of the arrow.

In termination, two radicals collide and combine to form a molecule, so no radical is left. Examples:

$Cl\bullet + Cl\bullet \rightarrow Cl_2$

$Cl\bullet + \bullet CH_3 \rightarrow CH_3Cl$

$\bullet CH_3 + \bullet CH_3 \rightarrow C_2H_6$

The last equation shows how a longer-chain alkane forms as a trace by-product.

Once chloromethane has formed, its C–H bonds can still be attacked, giving $CH_2Cl_2$, $CHCl_3$ and $CCl_4$. Longer alkanes also give several isomers because any hydrogen can be replaced.

The product is therefore a mixture that must be separated, and the yield of any one compound is low. Free-radical substitution is of limited use in synthesis for this reason. Using a large excess of alkane reduces further substitution.