# How does propane react with bromine under radical conditions?

Feb 28, 2017

It is conceived to be a radical process......

#### Explanation:

Halogenation of alkanes is a radical process, that tends to be indiscriminate (i.e. dihalogenated products are possible in that radical mechanisms can be quite unselective):

$\text{H"_3"CCH"_2"CH"_3 +"Br"_2 stackrel(hnu)rarr "H"_3"CCHBrCH"_3 +"HBr}$

The mechanism of halogenation is in your text, and the starting point is the homolysis of the bromine molecule by UV light:

$B {r}_{2} + h \nu \rightarrow 2 \dot{B} r$

The $\dot{B} r$ radical, a 7 electron species, is very reactive inasmuch as when it reacts it generates another radical species to continue the chain of reaction:

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

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

And the $\dot{B} r$ radical can continue the chain of reaction.......The reaction can terminate by the coupling of 2 radicals:

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

The presence of such $C - C$ coupled products is good evidence for the intermediacy of these radicals. Most of the time, however, radical halogenation of alkanes will lead to a mess (whereas radical halogenation of benzyl groups is a bit more discriminating in that the stabilization of a benzyl radical allows selective formation of $P h C {H}_{2} B r$).