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Anisole is treated with HI under two different conditions.
$\mathrm{C}+\mathrm{D} \stackrel{\mathrm{HI}(\mathrm{g})}{\longleftarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OCH}_{3} \stackrel{\text { conc.HI }}{\longrightarrow} \mathrm{A}+\mathrm{B}$
The nature of A to D will be
Options:
$\mathrm{C}+\mathrm{D} \stackrel{\mathrm{HI}(\mathrm{g})}{\longleftarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OCH}_{3} \stackrel{\text { conc.HI }}{\longrightarrow} \mathrm{A}+\mathrm{B}$
The nature of A to D will be
Solution:
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Verified Answer
The correct answer is:
Both A and B as well as both C and D are $\mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
Although in both cases products are $\mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} ;$ the two reactions follow different mechanism.
$$
\begin{array}{l}
\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{O}-\mathrm{CH}_{3} \frac{\mathrm{HI}(\mathrm{g})}{\mathrm{S}_{\mathrm{N}} 2} \\
\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{O}-\mathrm{CH}_{3} \frac{\mathrm{conc} . \mathrm{HI}}{\mathrm{S}_{\mathrm{N}} 1} \mathrm{CH}_{3} \mathrm{I}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\
\mathrm{CH}_{3} \mathrm{I}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}
\end{array}
$$
Remember that during $\mathrm{S}_{\mathrm{N}^{1}}$ reaction, $\mathrm{CH}_{3}^{+}$ is formed because it is more stable than $\mathrm{C}_{6} \mathrm{H}_{5}^{+}$
$$
\begin{array}{l}
\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{O}-\mathrm{CH}_{3} \frac{\mathrm{HI}(\mathrm{g})}{\mathrm{S}_{\mathrm{N}} 2} \\
\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{O}-\mathrm{CH}_{3} \frac{\mathrm{conc} . \mathrm{HI}}{\mathrm{S}_{\mathrm{N}} 1} \mathrm{CH}_{3} \mathrm{I}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\
\mathrm{CH}_{3} \mathrm{I}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}
\end{array}
$$
Remember that during $\mathrm{S}_{\mathrm{N}^{1}}$ reaction, $\mathrm{CH}_{3}^{+}$ is formed because it is more stable than $\mathrm{C}_{6} \mathrm{H}_{5}^{+}$
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