#RCH=CH_2(g) + Br_2(aq) rarr RCH(OH)-CH_2Br + Br^-#
The bromine molecule is polarizable and can form a transient #""^(delta-)Br-Br^(delta+)# dipole. The #Br^(delta+)# site can react with an electron rich species such as an olefin, #RCH=CH_2#.
This first step of the meachanism can be represented as:
#RCH=CH_2 + Br_2 rarr RC^(+)(H)-CH_2Br + Br^-#
This carbocation intermediate could THEN react with the #Br^-# nucleophile, and indeed it would if the reaction were carried out in dry petroleum ether. However, because it was performed in bromine water, by far the most abundant nucleophile is the water molecule, which would lose a proton to give the halohydrin:
#RC^+(H)-CH_2Br + OH_2 rarr RC(OH)(H)-CH_2Br +H^+#
Capisce?
