Question #68c50

1 Answer
Milan S#
Dec 3, 2017

No HI won't give either Markonikoff's or Antimarkonikoff's product.

Explanation:

In the presence of peroxide free radicle addition takes place generally, For the radical reaction of HI the first propagation step {the breaking of C=C and C-H and C-I #sigma# bond formation} is endothermic #rarr# C-I sigma bond is weaker than C-I #pi#-bond.
Radical chain reactions are successful when propagation steps are exothermic. An endothermic propagation step is reversible and do not proceed. With HI, the first propagation step is endothermic, because H-I bond is weak. The formed Iodine free radicle i s oxidise to I2 readily so no reaction is observed.

If you want to refer mechanism of addition of HBr in presence of peroxide -link[
(https://socratic.org/questions/why-in-the-presence-of-peroxides-hcl-and-hi-do-not-give-anti-markovnikov-additio)

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