Why in the presence of peroxides HCl and HI do not give anti-markovnikov addition to alkenes?

##"HCl"## and ##"HI"## do not give anti-Markovnikov addition to alkenes because some of the steps in the chain reaction are endothermic.

Reaction with ##"HBr"##

The mechanism for anti-Markovnikov addition of ##"HBr"## is:

Step 1. Abstraction of ##"H"## from ##"HBr"## (highly exothermic).

##"HO·" + "H-Br" → "HOH" + "·Br"##

Step 2. Addition of ##"Br·"## to the alkene (exothermic)

##"CH"_3"CH=CH"_2 + "·Br" → "CH"_3stackrel("·")("C")"H-CH"_2"Br"##

Step 3. The carbon radical abstracts another ##"H"## from ##"HBr"## (exothermic)

##"CH"_3stackrel("·")("C")"H-CH"_2"Br" + "H-Br" → "CH"_3"CH"_2"-CH"_2"Br" +"·Br"##

The reaction is favourable because all the steps are exothermic.

Reaction with ##"HCl"##

The reaction with ##"HCl"## is unfavourable because the ##"H-Cl"## bond is much stronger than the ##"H-Br"## bond.

Step 1 is endothermic, and the reaction becomes too slow to be useful.

Reaction with ##"HI"##

Step 1 is favourable, because the ##"H-I"## bond is relatively weak.

But Step 2, addition of ##"I·"## to the alkene, is unfavourable and endothermic because of the bulk of the iodine radical.

Instead, the iodine radicals combine to form ##"I"_2##.