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

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Chlorination of methane is a free radical reaction which occurs by the following mechanism.

Step I: Chain Initiation step The reaction is initiated by homolytic cleavage of Chlorine molecule in the presence of light or heat. The Cl-Cl bond is weaker than the C-C and C-H bond and hence is easiest to break. 

Cl-Cl ------> Cl. + Cl. 

  Chlorine free radicals

Step II: Chain Propagation step Chlorine free radicals attacks the methane molecules and takes the reaction in the forward direction by breaking the C-H bond to generate methyl free radicals with the formation of HCl. 

(a) CH4 + Cl.-------->  CH3 + H-Cl 

The methyl radicals attack the second molecule of chlorine to form CH3-Cl with the liberation of another Cl by homolysis of Cl2 molecule 

(b) CH3 + Cl-Cl ------> CH3-Cl + Cl. 

The chlorine and methyl free radicals repeats steps (a) and (b) respectively and thereby setup a chain of reactions. 

The propagation steps (a) and (b) are those which directly gives principal products, but many other propagation steps are possible.

Two such steps shown here explains how more halogenated products are formed 

 CH3Cl + Cl. ------>  CH.2Cl + HCl 

CH. 2Cl + Cl-Cl -------> CH2Cl2 + Cl

Step III: Termination 

The reaction stops after some time due to consumption of reactants and / or due to the following side reactions : 

(a) Cl. + Cl. --------> Cl-Cl 

(b) CH.3 + CH.3 ---------> H3C-CH3 

(c) CH.3 + Cl. H3C-Cl 

Though in (c) CH3-Cl , the one of the product is formed but free radicals are consumed and the chain is terminated.

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