In the following reaction,$C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2/HCl} X$ $\xrightarrow{Dil. NaOH} Y$. Here,$X$ and $Y$ are:

  • A
    $X = C_6H_5CH_2Cl, Y = \text{4-methylchalcone}$
  • B
    $X = C_6H_5CH_2Cl, Y = \text{chalcone derivative}$
  • C
    $X = C_6H_5CHO, Y = \text{4-methylchalcone}$
  • D
    $X = C_6H_5CHO, Y = \text{4-methylchalcone}$ (with correct structure)

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