Question

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

Answer 1

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:
$\underset{Chlorinefreeradicals}{Cl-Cl\stackrel{hv}{\to }\stackrel{.}{C}l+\stackrel{.}{C}}$

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:
$\underset{Methane}{C{H}_4}+\stackrel{.}{C}l\stackrel{hv}{\to }\stackrel{.}{C}{H}_3+H-Cl$
These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.
$\stackrel{.}{C}{H}_3+Cl-Cl\to \underset{Methylchloride}{C{H}_3-Cl+\stackrel{.}{C}l}$
Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and $C{H}_{3}Cl$ are the major products formed, other higher halogenated compounds are also formed as:
$C{H}_{3}Cl+\stackrel{.}{C}l\to \stackrel{.}{C}{H}_{2}Cl+HCl$
$\stackrel{.}{C}{H}_{2}Cl+Cl-Cl\to C{H}_{2}C{l}_2+\stackrel{.}{Cl}$

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:
$\stackrel{.}{Cl}+\stackrel{.}{Cl}\to Cl-Cl$
$\stackrel{.}{C}{H}_3+\stackrel{.}{C}{H}_3\to \underset{\left(Ethane\right)}{H_{3}C-C{H}_3}$
Hence, by this process, ethane is obtained as a by-product of the chlorination of methane.