Question

Anisole is treated with HI under two different conditions.
$C+D\stackrel{HI\left(g\right)}{\leftarrow }{C}_6{H}_{5}OC{H}_3\stackrel{conc.HI}{\to }A+B$ The nature of A to D will be

• A. A and B are $C{H}_3$ I and $C_6{H}_{5}OH,$ while C and D are $C{H}_{3}OH$ and $C_6{H}_{5}I$
• B. A and B are $C{H}_{3}OH$ and $C_6{H}_{5}I$, while C and D are $C{H}_{3}I$ and $C_6{H}_{5}OH$
• C. Both A and B as well as both C and D are $C{H}_{3}I$ and $C_6{H}_{5}OH$
• D. A and B are $C{H}_{3}I$ and $C_6{H}_{5}OH$, while there is no reaction in the second case

Answer 1

The correct option is C

Both A and B as well as both C and D are $C{H}_{3}I$ and $C_6{H}_{5}OH$

Although in both cases products are $C{H}_{3}I$ and $C_6{H}_{5}OH$ ;the two reaction follow different mechanism.
$C_6{H}_5-O-C{H}_3\underset{S_{N^2}}{\overset{Hl\left(g\right)}{\to }}C{H}_{3}I+C_6{H}_{5}OH{C}_6{H}_5-O-C{H}_3\underset{S_{N^1}}{\overset{conc.Hl}{\to }}C{H}_{3}I+C_6{H}_{5}OH$
Remember that during $S_{N^1}$ reaction, $C{H}_3^{+}$ is formed because it is more stable than $C_6{H}_5^{+}.$