The addition of HI in the presence of a peroxide does not lead to anti-Markovnikov behaviour because

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The iodine free radicals fromed readily combine with each other to give an $I_{2}$ molecule

The HI bond is too strong to be broken homolytically

HI is a reducing agent

I combines with H to give back HI

answer is A.

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Detailed Solution

In the presence of organic peroxides, bromides will take part in radical based reactions. Fission of the peroxide O-O linkage will induce a Br radical which behaves differently to Bromide, by adding to the less substituted side of the alkene (anti - markovnikov). HI and HCl do not do this for energetic reasons.

The addition of Cl and I radicals to the alkene in an anti-markovnikov fashion is an endothermic reaction and therefore unfavorable. Without the presence of peroxide, all hydrogen halides will add according to the markovnikov rule.

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