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:

\[\ce{\underset{Chlorine free radicals}{Cl - Cl ->[hv]\overset\bullet{\text{Cl}} + \overset\bullet{\text{C}}}}\]

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:

\[\ce{\underset{Methane}{CH4} + \overset\bullet{\text{Cl}} ->[hv] \overset\bullet{\text{C}}H3 + H - Cl}\]

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

\[\ce{\overset\bullet{\text{C}}H3 + Cl - Cl -> \underset{Methyl chloride}{CH3 - Cl +}\overset\bullet{\text{C}}l}\]

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and CH₃Cl are the major products formed, other higher halogenated compounds are also formed as:

\[\ce{CH3Cl + \overset\bullet{\text{C}}l -> \overset\bullet{\text{C}}H2Cl + HCl}\]

\[\ce{\overset\bullet{\text{C}}H2Cl + Cl - Cl -> CH2Cl2 + \overset\bullet{\text{C}}l}\]

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:

\[\ce{\overset\bullet{\text{C}}l + \overset\bullet{\text{C}}l -> Cl - Cl}\]

\[\ce{H3\overset\bullet{\text{C}} + \overset\bullet{\text{C}}H3 -> \underset{(Ethane)}{H3C - CH3}}\]

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