# How do alkenes decolourise bromine water?

Apr 28, 2016

By electrophilic addition to give the halohydrin, $R C \left(O H\right) H - C {H}_{2} B r$.

#### Explanation:

Olefins are electron rich species and react with electrophiles or potential electrophiles:

i.e. $R C H = C {H}_{2} + B {r}_{2} \rightarrow R C H B r - C {H}_{2} B r$.

The brown colour of the bromine would dissipate (i.e. go to colourless!).

But (of course) there is an added sting in the tail of this question. It asked for the reaction of the olefin with bromine water, i.e. $B {r}_{2} \left(a q\right)$ not $B {r}_{2}$ per se.

So looking at intermediate steps:

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

So in the first addition, the olefin reacts as a nucleophile, as an electron rich species. The intermediate carbocation reacts of course as an electrophile. Because in bromine water $B {r}_{2} \left(a q\right)$, by far the most concentrated nucleophile is the WATER molecule, this reaction would give $R C \left(O H\right) H - C {H}_{2} B r$, the halohydrin, as the major product.

This is at a 1st/2nd year level rather than A levels.