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(B) $1$. The starting material is $N-phenylbenzamide$. Friedel-Crafts alkylation with $Me_3C-Cl/AlCl_3$ occurs at the para-position of the aniline rin

Vedclass · Accepted Answer

$S$ is $4-tert-butyl-4'-(dimethylamino)azobenzene$

Vedclass · Answer

(B) $1$. The starting material is $N-phenylbenzamide$. Friedel-Crafts alkylation with $Me_3C-Cl/AlCl_3$ occurs at the para-position of the aniline ring because the amide group is ortho/para directing. Thus,$P$ is $N-(4-tert-butylphenyl)benzamide$.$2$. Acidic hydrolysis $(H_3O^+)$ of $P$ yields $4-tert-butylaniline$ $(Q)$.$3$. Diazotization of $Q$ with $NaNO_2/HCl$ at $0-5^{\circ}C$ gives $4-tert-butylbenzenediazonium$ chloride $(R)$.$4$. Coupling of $R$ with $N,N-dimethylaniline$ occurs in basic or weakly acidic medium. The reaction $R$ to $S$ is typically carried out at $pH \approx 4-5$ for coupling with amines,but the question asks for the correct statement. Option $B$ correctly identifies the structure of the azo dye $S$ as $4-tert-butyl-4'-(dimethylamino)azobenzene$.

Consider the following reaction sequence and identify the correct option:

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