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

1 Answer
Ernest Z.
May 30, 2018

It's all a question of relative rates.

Explanation:

Bromine addition vs. allylic substitution

Consider the reaction

Bromination

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.

Alkane + Br2→bromonium ion

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

Alkane + Br⋅ → alkyl radical + HBr

At any one time, we have one equivalent of Br2 and only a small number of 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.

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

NBS + HBr → NHS + Br2

Then

Br2→2Br⋅

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

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