Question #68c50

1 Answer
Dec 3, 2017

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


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[