How do you account for the formation of ethane during chlorination of methane?

Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps.

Step 1: Initiation:

The reaction begins with the homolytic cleavage of Cl – Cl bond as:

$\mathrm{Cl}-\mathrm{Cl} \stackrel{h \nu}{\longrightarrow} \dot{\mathrm{C}} 1+\dot{\mathrm{C}}$

Chlorine free radicals

Step 2: Propagation:

In the second step, chlorine free radicals attack methane molecules and break down the C–H bond to generate methyl radicals as:

$\mathrm{CH}_{4}+\dot{\mathrm{C}} \mathrm{l} \stackrel{\text { hv }}{\longrightarrow} \dot{\mathrm{C}} \mathrm{H}_{3}+\mathrm{H}-\mathrm{Cl}$

Methane

These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

$\begin{aligned} \dot{\mathrm{C}} \mathrm{H}_{3}+\mathrm{Cl}-\mathrm{Cl} \longrightarrow & \mathrm{CH}_{3}-\mathrm{Cl}+\dot{\mathrm{Cl}} \\ & \text { Methyl chloride } \end{aligned}$

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While $\mathrm{HCl}$ and $\mathrm{CH}_{3} \mathrm{Cl}$ are the major products formed, other higher halogenated compounds are also formed as:

$\mathrm{CH}_{3} \mathrm{Cl}+\dot{\mathrm{C}} \mathrm{l} \longrightarrow \dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{Cl}+\mathrm{HCl}$

$\dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{Cl}+\mathrm{Cl}-\mathrm{Cl} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\dot{\mathrm{Cl}}$

Step 3: Termination:

Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:

$\dot{\mathrm{C}} \mathrm{l}+\dot{\mathrm{C}} \mathrm{l} \longrightarrow \mathrm{Cl}-\mathrm{Cl}$

$\mathrm{H}_{3} \dot{\mathrm{C}}+\dot{\mathrm{C}} \mathrm{H}_{3} \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}$

(Ethane)

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.