# Why do halogens do not add to double bond in allylic halogenation?

May 30, 2018

#### Answer:

It's all a question of relative rates.

#### Explanation:

Bromine addition vs. allylic substitution

Consider the reaction

Why is the product almost exclusively the 1,2-dibromide?

Relative rates

The rate-determining step in an addition reaction is the attack of a bromine molecule to form a cyclic bromonium ion.

$\text{Alkane + Br"_2 → "bromonium ion}$

The rate-determining step in a substitution reaction is the attack of a bromine atom on an alkane.

$\text{Alkane + Br· → alkyl radical + HBr}$

At any one time, we have one equivalent of ${\text{Br}}_{2}$ and only a small number of $\text{Br·}$ atoms, so addition predominates.

If we could decrease the concentration of bromine molecules, the rate of addition would decrease, and the relative rate of substitution ould increase.

Allylic bromination

We can use "NBS" (N"-bromosuccinimide") as a radical source.

$\text{NBS}$ is usually contaminated with a little $\text{HBr}$, and the reaction between $\text{NBS}$ and $\text{HBr}$ delivers the bromine molecules that form the bromine radicals.

${\text{NBS + HBr → NHS + Br}}_{2}$

Then

$\text{Br"_2 → "2Br·}$

Thus, the concentration of bromine stays low and allows the free-radical substitution to out-compete the alkene addition.