Benzene + CH_3CH_2CH_2Cl xrightarrow{AlCl_3} | vedclass

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(A) $1$. The reaction of benzene with $n$-propyl chloride in the presence of $AlCl_3$ is a Friedel-Crafts alkylation. Due to the rearrangement of the

Vedclass · Accepted Answer

Isopropylbenzene and $CH_3COCH_3$

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(A) $1$. The reaction of benzene with $n$-propyl chloride in the presence of $AlCl_3$ is a Friedel-Crafts alkylation. Due to the rearrangement of the primary carbocation to a more stable secondary carbocation,the major product $M$ is isopropylbenzene (cumene).$2$. Cumene $(M)$ reacts with $O_2$ at high temperature to form cumene hydroperoxide.$3$. Upon treatment with $H_3O^+$,cumene hydroperoxide undergoes acid-catalyzed rearrangement to yield phenol and acetone $(CH_3COCH_3)$.$4$. Thus,$M$ is isopropylbenzene and $N$ is acetone $(CH_3COCH_3)$.

Column-$I$ (Reaction)	Column-$II$ (Product)
$(A)$ Dinitration of $4$-methylphenol	$(i)$ $2$-nitro-$4$-methylphenol + $2$-nitro-$5$-methylphenol
$(B)$ Mononitration of phenyl ethanoate	$(ii)$ $2,4$-dinitro-$5$-methylphenol
$(C)$ Dinitration of $3$-methylphenol	$(iii)$ $4$-nitrophenyl ethanoate
$(D)$ Mononitration of $3$-methylphenol	$(iv)$ $2,6$-dinitro-$4$-methylphenol

Benzene $+ CH_3CH_2CH_2Cl$ $\xrightarrow{AlCl_3} M$ $\xrightarrow{(i) O_2/\Delta , (ii) H_3O^+} N + \text{phenol}$. Here $M$ and $N$ are

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