Question:
What are \(X\) and \(Y\) respectively in the following reaction sequence? \((\mathrm{Me}=-\mathrm{CH_3})\)
\[
Z
\;(\text{Benzene})
\xrightarrow[\mathrm{AlCl_3}]{\mathrm{CH_3Br}}
X
\;(\text{Anisole})
\xrightarrow[\text{fusion}]{\mathrm{NaOH}}
Y
\]
What are \(X\) and \(Y\) respectively in the following reaction sequence? \((\mathrm{Me}=-\mathrm{CH_3})\)
\[
Z
\;(\text{Benzene})
\xrightarrow[\mathrm{AlCl_3}]{\mathrm{CH_3Br}}
X
\;(\text{Anisole})
\xrightarrow[\text{fusion}]{\mathrm{NaOH}}
Y
\]
Show 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.
Updated On: Jul 9, 2026
- \textit{o}-Methylanisole ; Bromobenzene
- \textit{o}-Methylanisole ; Phenol
- \textit{p}-Methylanisole ; Bromobenzene
- \textit{p}-Methylanisole ; Phenol \bigskip
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Formation of \(X\).
Anisole undergoes Friedel--Crafts alkylation with methyl chloride in the presence of anhydrous \(\mathrm{AlCl_3}\).
The methoxy group is an ortho/para-directing group, and the para product predominates due to less steric hindrance.
Hence,
\[
X=\textit{p}\text{-Methylanisole}
\]
Step 2: Formation of \(Y\). Bromobenzene on fusion with NaOH followed by acidification gives phenol. Thus, \[ Y=\mathrm{C_6H_5OH} \] (Phenol).
Step 3: Final conclusion. Therefore, \[ \boxed{X=\textit{p}\text{-Methylanisole},\qquad Y=\text{Phenol}} \] Hence, the correct option is \(\boxed{(D)}\).
Step 2: Formation of \(Y\). Bromobenzene on fusion with NaOH followed by acidification gives phenol. Thus, \[ Y=\mathrm{C_6H_5OH} \] (Phenol).
Step 3: Final conclusion. Therefore, \[ \boxed{X=\textit{p}\text{-Methylanisole},\qquad Y=\text{Phenol}} \] Hence, the correct option is \(\boxed{(D)}\).
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