# How do you write a stepwise mechanism that shows how a small amount of CH_3CH_3 could form during the bromination of #CH_4?

Sep 7, 2016

Bromination proceeds by a radical mechanism, in which $B r \cdot$ radicals are invoked. So let's see.

#### Explanation:

$\text{Initiation:}$ $B {r}_{2} + h \nu \rightarrow 2 B r \cdot$

$\text{Propagation:}$ $B r \cdot + C {H}_{4} \rightarrow \cdot C {H}_{3} + H - B r$

$B {r}_{2} + \cdot C {H}_{3} \rightarrow B r C {H}_{3} + B r \cdot$

$\text{Termination:}$ $B r \cdot + B r \cdot \rightarrow B {r}_{2}$, i.e. the coupling of 2 radicals could occur, or the coupling of hydrocarbyl radicals made in the propagation step:

${H}_{3} C \cdot + \cdot C {H}_{3} \rightarrow {H}_{3} C - C {H}_{3}$

And thus the coupling of 2 methyl radicals results in $C - C$ bond formation.

Overall, the bromination rxn is:

$C {H}_{4} + B {r}_{2} \rightarrow C {H}_{3} B r + H - B r$

The discovery of ethane in the reaction mixture is good evidence for the radical reaction as shown.