Halohydrin Formation Reaction - Reaction Mechanism

Reaction Mechanism

In the first step, the pi bond of the alkene attacks the halogen. For the case where bromine is the halogen, a three-membered C, C, Br ring, known as a bromonium ion, is formed.

The addition of water, which provides the hydroxide anion, gives the desired halohydrin with high anti stereospecificity as a consequence of the S_N2 opening of the bromonium ion by hydroxide or water. Markovnikov's rule applies: the positive charge resides on the more-substituted carbon, hence the hydroxide anion adds here.

However, in the case of bromine water reacting with cyclohexene, the solubility of bromine in water is about 0.21 moles per litre, and the concentration of HOBr formed in the equilibrating reaction between bromine and water is about 1.15 x 10−3 moles per litre. So it is thousands of times more likely that Br₂ will be the species that attacks the C=C bond.

A couple of other factors have some bearing too:

1. Br₂ is much more soluble in the non-polar alkene than either water or HOBr.
2. When the Br+ ion is attacked by the C=C bond electrons, the Br- is right on the spot to add to the other side of the bond. In this manner, steric issues are minimized.

So while there can be no doubt that some halohydrin is formed when a halogen-water mixture is mixed with an alkene, the main product will be the dihalo compound. For example, when bromine water reacts with cyclohexene, the main product is 1,2-dibromocyclohexane.