In the following conversion,
$C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) MeMgBr} X$ $\xrightarrow[H_3O^+]{NaOH/I_2} Y$
the major products $X$ and $Y$,respectively are

  • A
    $I$
    Option A
  • B
    $II$
    Option B
  • C
    $III$
    Option C
  • D
    $IV$
    Option D

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