Considering the following reaction sequence,t | vedclass

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(C) $1$. The reaction of benzene with succinic anhydride in the presence of $AlCl_3$ (Friedel-Crafts acylation) gives $P$ ($4$-oxo$-4-$phenylbutanoic

Vedclass · Accepted Answer

$A, C$

Vedclass · Answer

(C) $1$. The reaction of benzene with succinic anhydride in the presence of $AlCl_3$ (Friedel-Crafts acylation) gives $P$ ($4$-oxo$-4-$phenylbutanoic acid),which is a carboxylic acid.$2$. Clemmensen reduction $(Zn/Hg, HCl)$ of $P$ gives $Q$ ($4$-phenylbutanoic acid),which is also a carboxylic acid. Thus,statement $A$ is correct.$3$. $Q$ reacts with $SOCl_2$ to form $R$ ($4$-phenylbutanoyl chloride). Intramolecular Friedel-Crafts acylation of $R$ gives $S$ ($1$-tetralone).$4$. $S$ ($1$-tetralone) is a ketone and does not contain a carbon-carbon double bond,so it does not decolorize bromine water. Statement $B$ is incorrect.$5$. Both $P$ (a ketone) and $S$ (a ketone) react with hydroxylamine $(NH_2OH)$ to form oximes. Statement $C$ is correct.$6$. Compound $R$ (an acid chloride) reacts with dialkylcadmium $(R'_2Cd)$ to give a ketone,not a tertiary alcohol. Statement $D$ is incorrect.$7$. Therefore,the correct statements are $A$ and $C$.

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