Question

What are the steps of radical halogenation of alkanes?

Answer 1

`"i. Initiation. ii. Propagation. iii. Termination"`

Explanation:

`"Initiation:"`

`Cl_2 + hnu rarr 2Cl*`

The `hnu` represents UV light, the radical halide can abstract an hydrogen to make the carbon radical:

`Cl* + H_3C-CH_3 rarr *CH_2-CH_3 + H-Cl`

So we have made a ethyl radical, which serves to propagate the reaction by subsequent reaction with UNREACTED halogen. This serves to propagate the reaction:

`"Propagation:"`

`H_3C-CH_2* + Cl-Cl rarr ClCH_2-CH_3 + Cl*`

And thus by making a `C-Cl` bond, another radical species is generated, which thus serves to propagate the reaction. Thus the radical halogenations tend to take off.

`"Termination:"`

The reaction is terminated when two radicals couple to close down the chain mechanism:

`H_3C-CH_2* + *Cl rarr H_3C-CH_2-Cl`

OR `Cl* + *Cl rarr Cl-Cl`

OR THE COUPLING OF TWO ETHYL RADICALS:

`H_3C-CH_2* + *H_2C-CH_3 rarr H_3C-CH_2CH_2CH_3`

The appearance of small quantities of butane in the product mix is excellent evidence for the intermediacy of ethyl radicals in the reaction mechanism.

The net reaction is:

`C_2H_6 + Cl_2 rarr H_3C-CH_2Cl + HCl`