Question

Write a stepwise mechanism that shows how a small amount of \[C{{H}_{3}}C{{H}_{3}}\]could form during the bromination of\[C{{H}_{4}}\]?

Answer 1

Hint: This reaction can be explained on the reaction followed by polymerization i.e. the concept by which a monomer single unit of any substance combines with itself or any other substances and forms a chain like compounds and form polymers.

Complete answer:
The mechanism is generally written in three steps:
1. Initiation step: In this step hemolytic cleavage occurs between $Br-Br$bond and we got the free radicals of bromine atoms and the reaction can be shown as:
$B{{r}_{2}}+C{{H}_{3}}\to BrC{{H}_{3}}+Br$
2. Propagation steps:
$Br+C{{H}_{4}}\to C{{H}_{3}}+H-Br$
$B{{r}_{2}}+C{{H}_{3}}\to BrC{{H}_{3}}+Br$
In this step one methane group is combined with bromine free radical and form one equivalent of a methyl radical and form free radical of$HBr$. Then another two equivalents of $B{{r}_{2}}$reacts with methyl radical and form one equivalent of $C{{H}_{3}}Br$and one equivalent of a bromine radical.
3. Termination steps:
$Br+Br\to B{{r}_{2}}$
This step includes the coupling of two bromine free radicals and the coupling of hydrocarbyl radicals occurs which forms in the propagation step and the reaction can be shown as:
$C{{H}_{3}}+C{{H}_{3}}\to C{{H}_{3}}-C{{H}_{3}}$
This indicates that coupling of 2 methyl radicals result in the formation of $C-C$bonds formation.
Thus in this way a small amount of \[C{{H}_{3}}C{{H}_{3}}\] could form during the bromination of\[C{{H}_{4}}\].

Note:
Bromination is generally kept under the type of halogenation reactions in which one of the hydrogen atom is replaced with the halogen atom where halogens are those compounds which are lack behind with only one electron in its outermost shell like fluorine, chlorine, bromine, iodine etc.