Question

Photochemical iodination of alkane is carried out in the presence of :

• A. $HI$
• B. $I_2$
• C. $HI{O}_3$
• D. None of the above

Answer 1

The correct option is C $HI{O}_3$

Rate of photochemical iodination of alkane is very slow and is a reversible process.
Here, the product $HI$ formed in the forward reaction is a strong reducing agent which will convert the formed alkyl iodide back to alkane. To encounter this backward reaction, a strong oxidising agent like $HI{O}_3/HN{O}_3$ is used. These reagent will oxidise the formed $HI$ to $I_2$. Thus the product HI is removed from the reversible reaction and this will enhance the forward reaction by Le Chatelier's principle.
$C{H}_4+I_2\rightleftharpoons C{H}_{3}I+HIHI{O}_3+5HI\to 3I_2+3H_{2}O$
Thus, option (c) is correct.