Question

Anisole is treated with HI under two different conditions. $\ce{ C + D [conc.HI] A + B }$ The nature of A to D will be

• A. A and B are $CH_3I$ and $C_6H_5OH$, while C and D are $CH_3OH$ and $C_6H_5I$
• B. A and B are $CH_3OH$ and $C_6H_5I$, while C and D are $CH_3I$ and $C_6H_5OH$
• C. Both A and B as well as both C and D are $CH_3I$ and $C_6H_5OH$
• D. A and B are $CH_3I$ and $C_6H_5OH$, while there

Answer 1

Answer: (C)

Both A and B as well as both C and D are $CH_3I$ and $C_6H_5OH$

Answer 2

Although in both cases products are $\ce{CH3I}$ and $\ce{C6H5OH}$; the two reactions follow different mechanism.
$\ce{ C_6H_5 - O - CH_3 ->[HI(g)][S_{N^2}] CH_3I + C_6H_5OH}$
$\ce{C_6H_5 - O - CH_3 ->[conc. HI][S_{N^1}] CH_3I + C_6H_5OH}$
Remember that during $S_{N^1}$ reaction, $CH_3^+$ is formed because it is more stable than $C_6H_5^+$ .