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

Answer

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:
$\text{Cl}-\text{Cl}\xrightarrow{\ \ \ \text{hv}\ \ \ \ \ }\dot{\text{Cl}}+\dot{\text{C}}\\ \ \text{Chlorine free radicals}$
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:
$\text{CH}_4+\dot{\text{Cl}}\xrightarrow{\ \ \ \ \text{hv}\ \ \ \ }\dot{\text{C}}\text{H}_3+\text{H}-\text{Cl}\\\text{Methane}$
These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.
$\dot{\text{C}}\text{H}_3+\text{Cl}-\text{Cl}\xrightarrow{\ \ \ \ }\text{CH}_3-\text{Cl}+\dot{\text{Cl}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methyl chloride}$
Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and $\mathrm{CH}_3 \mathrm{Cl}$ are the major products formed, other higher halogenated compounds are also formed as:
$\text{CH}_3\text{Cl}+\dot{\text{Cl}}\xrightarrow{\ \ \ \ }\dot{\text{C}}\text{H}_2\text{Cl}+\text{HCl}$
$\dot{\text{C}}\text{H}_2\text{Cl}+\text{Cl}-\text{Cl}\xrightarrow{\ \ \ \ \ }\text{CH}_2\text{Cl}_2+\dot{\text{Cl}}$
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:
$\dot{\text{Cl}}+\dot{\text{Cl}}\xrightarrow{\ \ \ \ }\text{Cl}-\text{Cl}$
$\text{H}_3\dot{\text{C}}+\dot{\text{CH}_3}\xrightarrow{\ \ \ \ \ }\text{H}_3\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Ethane)}$
Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

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