Question

Given below are two reactions :
$\left(i\right)C_6{H}_6+ICl\stackrel{Anhyd.AlC{l}_3}{\to }\text{iodobenzene}$
$\left(ii\right)C_6{H}_6+I_2\stackrel{Anhyd.AlC{l}_3}{\to }\text{iodobenzene}$

• A. Reaction (i) is more favored for preparation of iodobenzene as compared to reaction (ii)
• B. Reaction (ii) is more favored for preparation of iodobenzene as compared to reaction (i)
• C. Both are equally favored for preparation of iodobenzene.
• D. Chlorobenzene is formed in reaction (i) instead of iodobenzene.

Answer 1

The correct option is A Reaction (i) is more favored for preparation of iodobenzene as compared to reaction (ii)

Since iodine is less electronegative than chlorine, $AlC{l}_3$ reacts with $ICl$ to form iodonium ion $\left(I^{+}\right)$ as an electrophile. Hence iodobenzene is formed in reaction (i).

Iodination of benzene is done with $I-Cl+AlC{l}_3$ because if we take
$I_2+AlC{l}_3$ then $HI$ will be formed as byproduct. $HI$ being a strong reducing agent will reduce the iodobenzene making the reaction reversible.
So to prevent the reversibility of reaction we use $I-Cl$ with lewis acid.
So, reaction (i) is more favored for preparation of iodobenzene as compared to reaction (ii)