Question

Write a stepwise mechanism that shows how a small amount of CH3CH3 could form during the bromination of CH4?

Answer 1

Here is the mechanism drawn out. I rectangled where the ethane gets produced.

Something to keep in mind here is to not get so caught up about the actual equiv. of molecules depicted here. These steps in the mechanism are not about keeping track of how many molecules are unreacted or reacted, but about tracking what kinds of key steps occur that describe the key points of the reaction.

INITIATION STEP

Tracing this mechanism, the initiation step homolytically cleaves the `"Br"-"Br"` bond, giving two bromine radicals.

PROPAGATION STEPS

Then, one of them reacts with methane to heterolytically cleave it and form one equiv. of a methyl radical and `"HBr"`.

Another equiv. of `"Br"_2` reacts with the methyl radical to form one equiv. of `"CH"_3"Br"` and one equiv. of a bromine radical. There are now at least two equiv. of bromine radical in the reaction vessel, which is the bare minimum for the first termination step.

Now, since at some point in the propagation steps, a methyl radical gets formed, if you have enough equiv. of each reactant, there is a chance that two methyl radicals will react in the termination steps, and at that point, you will form ethane.

This amount is small simply because in the actual propagation steps, it is depicted that the methyl radical that forms is supposed to react again with `"Br"_2`, and so if the amount of each reactant is such that a reaction goes partway to completion, you could get a few extra methyl radicals that haven't reacted with anything else yet. That is a good situation where they would react with each other, and the termination steps account for that possibility.

TERMINATION STEPS

The two equiv. of bromine radicals react to reform `"Br"_2`. Two equiv. of methyl radicals react to form ethane. Finally, another equiv. of `"CH"_3"Br"` forms from the reaction of a methyl radical with a bromine radical.