Question

Question #4d8fc

Answer 1

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

Explanation:

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`.