# Question #4d8fc

Jun 12, 2015

$\text{HCl}$ and $\text{HI}$ do not give anti-Markovnikov addition to alkenes because some of the steps in the chain reaction are endothermic.

#### Explanation:

Reaction with $\text{HBr}$

The mechanism for anti-Markovnikov addition of $\text{HBr}$ is:

Step 1. Abstraction of $\text{H}$ from $\text{HBr}$ (highly exothermic).

$\text{HO·" + "H-Br" → "HOH" + "·Br}$

Step 2. Addition of $\text{Br·}$ to the alkene (exothermic)

$\text{CH"_3"CH=CH"_2 + "·Br" → "CH"_3stackrel("·")("C")"H-CH"_2"Br}$

Step 3. The carbon radical abstracts another $\text{H}$ from $\text{HBr}$ (exothermic)

$\text{CH"_3stackrel("·")("C")"H-CH"_2"Br" + "H-Br" → "CH"_3"CH"_2"-CH"_2"Br" +"·Br}$

The reaction is favourable because all the steps are exothermic.

Reaction with $\text{HCl}$

The reaction with $\text{HCl}$ is unfavourable because the $\text{H-Cl}$ bond is much stronger than the $\text{H-Br}$ bond.

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

Reaction with $\text{HI}$

Step 1 is favourable, because the $\text{H-I}$ bond is relatively weak.

But Step 2, addition of $\text{I·}$ to the alkene, is unfavourable and endothermic because of the bulk of the iodine radical.

Instead, the iodine radicals combine to form ${\text{I}}_{2}$.