How do alkenes react with bromine water?

Question

Madison Anderson · Accepted Answer

Bromine is the electrophile, and the nucleophile is the olefin.

#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 mechanism's initial step can be shown 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^+# Is that right?

Emilia Alvarez · Answer

undefined Alkenes react with bromine water through electrophilic addition, where the π bond of the alkene acts as a nucleophile attacking the electrophilic bromine atom, resulting in the formation of a bromonium ion intermediate. This intermediate is then attacked by water molecules, leading to the formation of a bromohydrin product.