1
$\begingroup$

We have seen that $\ce{HBr}$ shows an anti-Markovnikov effect when it reacts with alkenes in the presence of peroxides. Why does only $\ce{HBr}$ shows this effect? How about the other hydrogen halides $\ce{HX}~(\ce{X} = \ce{F}, \ce{Cl}, \ce{I})$?

$\endgroup$
  • 2
    $\begingroup$ See here for detailed discussion or here for brief overview. $\endgroup$ – mythealias Aug 22 '13 at 14:41
4
$\begingroup$

Main reason is behind is the reaction mechanism involved. $\ce{HBr}$ or other halides usually react with alkenes through an electrophilic addition. Due to the weak $\ce{O-O}$ bond, peroxides act as a source of free radicals. Thus in the presence of peroxides, a free radical chain pathway is followed.

The other hydrogen halides are reluctant to this reaction, because the chain reaction can't proceed due to energy requirements for the generation of the corresponding halogen free radicals.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.