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 + Br"_2 → "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 #"Br"_2# 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 + Br"_2#

Then

#"Br"_2 → "2Br·"#

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

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