Question

What are \(X\) and \(Y\) respectively in the following reaction sequence? \((\mathrm{Me}=-\mathrm{CH_3})\) \[ Z \;(\text{Benzene}) \xrightarrow[\mathrm{AlCl_3}]{\mathrm{CH_3Br}} X \;(\text{Anisole}) \xrightarrow[\text{fusion}]{\mathrm{NaOH}} Y \]

• A. \textit{o}-Methylanisole ; Bromobenzene
• B. \textit{o}-Methylanisole ; Phenol
• C. \textit{p}-Methylanisole ; Bromobenzene
• D. \textit{p}-Methylanisole ; Phenol \bigskip

Hint:

The \(-\mathrm{OCH_3}\) group activates the benzene ring and directs electrophilic substitution mainly to the ortho and para positions. The para product is generally the major product due to lower steric hindrance.

Answer 1

The Correct Option is C

Step 1: X: Anisole + CH₃Cl/AlCl₃ → p-methylanisole (ortho/para, para major). Y: Bromobenzene + NaOH fusion then H⁺ → phenol.

Step 2: Write the final answer. \(\boxed{X=\textit{p}\text{-Methylanisole},\qquad Y=\text{Phenol}}\)