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#.