The addition of HI in the presence of peroxide does not follow anti-markonikov's rule because:
A
HI bond is too strong to be broken Homolytically
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B
iodine atom is not reactive enough to add on to a double bond
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C
iodine combines with H to give back HI
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D
HI a reducing agent
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Solution
The correct option is B iodine atom is not reactive enough to add on to a double bond Because Iodine free radical formed as H-I bond is weaker but iodine free radicals readily combine with each other to form iodine molecules rather than to attack double bond.