Chlorination of alkanesAQA A-Level Chemistry: Revision notes

Alkanes react with chlorine in ultraviolet (UV) light. A hydrogen atom is substituted by a chlorine atom. For methane:

$CH_4 + Cl_2 \to CH_3Cl + HCl$

The reaction goes by a free-radical substitution mechanism. A radical is a species with an unpaired electron, shown with a dot, for example Cl• and •CH₃. Radicals are very reactive.

The mechanism has three stages: initiation, propagation and termination.

In initiation, UV light provides the energy to break the Cl–Cl bond. The bond breaks by homolytic fission, so each chlorine atom takes one electron and two chlorine radicals form:

$Cl_2 \to 2Cl\bullet$

The Cl–Cl bond is broken because it is weaker than the C–H and C–C bonds in the alkane.

In the propagation steps a radical reacts with a molecule to form a new molecule and a new radical. The chain reaction continues because a radical is regenerated:

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

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

The chlorine radical formed in the second step goes on to react with another methane molecule, so one chlorine radical can cause many reactions. Adding the two steps gives the overall equation.

In termination, two radicals collide and form a stable molecule. No radicals are left to continue the chain:

$Cl\bullet + Cl\bullet \to Cl_2$

$\bullet CH_3 + Cl\bullet \to CH_3Cl$

$\bullet CH_3 + \bullet CH_3 \to C_2H_6$

The last equation shows how ethane is formed as a by-product.

Chloromethane still has C–H bonds, so further substitution can give $CH_2Cl_2$, $CHCl_3$ and $CCl_4$. This multiple substitution produces a mixture. It is reduced by using a large excess of methane, so that Cl• is more likely to collide with CH₄ than with a chlorinated product.