How do alkenes react with bromine water?

1 Answer
Jul 26, 2016

They give the halohydrin, RCH(OH)CH_2Br, as the major product.

Explanation:

The reaction of RCH=CH_2 with bromine water, Br_2(aq), illustrates the mechanism of electrophilic addition.

The bromine molecule is polarizable, and is conceived to form a ""^(delta+)Br-Br^(delta+) intermediate, whose positive terminus acts as the electrophile. An RC^(+)HCH_2Br carbocation is formed in preference to a RCHBrC^(+)H_2 carbocation in that the 2""^@ is stabilized with respect to the alternative 1""^@ carbocation, as the secondary carbocation is more substituted.

Since the secondary carbocation is energetically more favoured, the reaction tends to follow this route. Once the first substitution is made however, the substituted olefin now reacts as an electrophile. The most likely nucleophile in bromine water is not Br^- but OH_2. Thus we could (finally!) write the rxn as:

RCH=CH_2 + Br_2(aq) rarr RCH(OH)CH_2Br + HBr

Thus bromine water would give a different product with an olefin, than it would with an olefin in an inert solvent.