Formation of Halohydrin from Alkenes

An alkene, such as propene, reacts with bromine in the presence of water to yield a halohydrin. Halohydrins contain a halogen and a hydroxyl group attached to adjacent carbons. When the halogen is bromine, it is called a bromohydrin, while a chlorohydrin has chlorine as the halogen.

Halohydrin formation commences when the π electrons of the alkene react with electrophilic bromine to form a bridged intermediate called a bromonium ion. Water, acting as a nucleophile, uses a lone pair of electrons to open the three-membered bromonium ion ring and forms a bond with the carbon in an S_N2 process. 

The deprotonation of the oxonium ion yields a hydronium ion and the neutral bromohydrin addition product.

The preferred opening of the cyclic bromonium ion intermediate by the regioselective attack of water on the more substituted carbon of the alkene can be explained based on two factors. Primarily, the electrostatic potential map of a bromonium ion shows that the more substituted carbon exhibits a greater carbocation character.

Additionally, the bond of the halogen with the more substituted carbon of the halonium ion is longer than that with the less substituted carbon.

This difference in bond lengths in the cyclic intermediate indicates that the ring-opening transition state can be attained more efficiently by the attack of the nucleophile at the more substituted carbon.

Since the mechanism involves a halonium ion, the stereochemistry of addition is anti. When 1-methylcyclohexene is treated with bromine, a pair of enantiomeric bromonium ions is obtained.

The anti addition of water gives trans-2-bromo-1-methylcyclohexanol as a racemic mixture.