Q37 / 120 · VITEEE 2016 · Chemistry

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

  • A
    $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$, while $\mathrm{C}$ and $\mathrm{D}$ are $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}$
  • B
    $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}$, while $\mathrm{C}$ and $\mathrm{D}$ are $\mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
  • Correct answer
    C
    Both A and B as well as both C and D are $\mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
  • D
    $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{CH}_{3} \mathrm{I}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$, while there is no reaction in the second case.

Solution

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 $\mathrm{C}_{6} \mathrm{H}_{5}^{+}$

Hence, the correct answer is option: 'C'.

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