# How do alkenes react with bromine water?

Jul 26, 2016

They give the halohydrin, $R C H \left(O H\right) C {H}_{2} B r$, as the major product.

#### Explanation:

The reaction of $R C H = C {H}_{2}$ with bromine water, $B {r}_{2} \left(a q\right)$, 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 $R {C}^{+} H C {H}_{2} B r$ carbocation is formed in preference to a $R C H B r {C}^{+} {H}_{2}$ 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 $B {r}^{-}$ but $O {H}_{2}$. Thus we could (finally!) write the rxn as:

$R C H = C {H}_{2} + B {r}_{2} \left(a q\right) \rightarrow R C H \left(O H\right) C {H}_{2} B r + H B r$

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