Question

If light and radical initiators are present, radical addition of$\ce{HBr}$to alkenes takes place and outcompetes electrophilic addition of ice{HBr}to alkenes. Why isn't a similar thing same happening if we use ice{Br2}instead?

Answer 1

The radical addition of HBr to alkenes is due to the homolytic cleavage of H-Br bond by light, whereas Br₂ reacts with alkenes through an electrophilic addition mechanism forming a cyclic halonium ion without involving radical intermediates, thus not requiring radical initiators.

Step-by-step explanation:

The radical addition of HBr to alkenes in the presence of light and radical initiators occurs because the energy provided by light can homolytically cleave the H-Br bond to form bromide radicals, which add to alkenes to form radical intermediates. These intermediates then react with HBr to complete the addition reaction.

In contrast, with Br₂, a different process known as halogenation occurs where the alkene forms a cyclic halonium ion intermediate with Br₂, and then bromide ion can open this intermediate to give a dibromoalkane. This latter reaction does not involve radical intermediates and hence is not enhanced by the conditions favoring radical chemistry.

Additionally, the reaction with elemental bromine, Br₂, is an electrophilic addition that yields vicinal dibromides without necessarily forming radical intermediates or requiring radical initiators such as light. This explains why a similar radical process does not happen with Br₂ as it does with HBr under radical conditions.