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

The chain mechanism is as follows, using the chlorination of methane as a typical example:

1. Initiation: Splitting or homolysis of a chlorine molecule to form two chlorine atoms, initiated by ultraviolet radiation or sunlight. A chlorine atom has an unpaired electron and acts as a free radical.

[Methane chlorination: initiation]

2. chain propagation (two steps): a hydrogen atom is pulled off from methane leaving a 1˚ methyl radical. The methyl radical then pulls a Cl· from Cl2.

[Methane chlorination: propagation]

This results in the desired product plus another chlorine radical. This radical will then go on to take part in another propagation reaction causing a chain reaction. If there is sufficient chlorine, other products such as CH2Cl2 may be formed.
3. chain termination: recombination of two free radicals:
[Methane chlorination: termination]
The last possibility in the termination step will result in an impurity in the final mixture; notably this results in an organic molecule with a longer carbon chain than the reactants.

The net reaction is:

[Methane chlorination overall reaction] or [CH_4 + Cl_2 \xrightarrow{uv\ light} CH_3Cl + HCl]

In the case of methane or ethane, all the hydrogen atoms are equivalent and thus have an equal chance of being replaced. This leads to what is known as a statistical product distribution. For propane and higher alkanes, the hydrogen atoms which form part of CH2 (or CH) groups are preferentially replaced.

The reactivity of the different halogens varies considerably. The relative rates are: fluorine (108) > chlorine (1) > bromine (7 × 10−11) > iodine (2 × 10−22). Hence the reaction of alkanes with fluorine is difficult to control, that with chlorine is moderate to fast, that with bromine is slow and requires high levels of UV irradiation while the reaction with iodine is practically non-existent and thermodynamically unfavorable.