# How can alkyl halides be prepared from alkanes?

Feb 13, 2017

$\text{Free radical halogenation.........}$

#### Explanation:

And the reaction scheme is generally constructed as:

$i .$ $\text{Initiation: } {X}_{2} + h \nu \rightarrow 2 \dot{X}$

$i i .$ $\text{Propagation(i): } \dot{X} + {H}_{3} C R \rightarrow H X + {H}_{2} \dot{C} R$

$\text{Propagation(ii): } {H}_{2} \dot{C} R + {X}_{2} \rightarrow {H}_{2} X C R + \dot{X}$

$i i i .$ $\text{Termination: the coupling of two radical species,}$

${H}_{2} \dot{C} R + \dot{X} \rightarrow {H}_{2} C X R$

$2 {H}_{2} \dot{C} R \rightarrow {H}_{2} R C - C {H}_{2} R$

The presence of small quantities of say ethane in the product mix of halogenation of methane is good evidence for the radical mechanism proposed.

The propagation step is a cascade type of process, in that the reaction of a radical species generates another radical species that can continue the chain of the reaction.