The major product of the reaction,1-methyl-4- | vedclass

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Question

(B) The reaction is a Friedel-Crafts acylation of $1$-methyl-$4$-isopropylbenzene (p-cymene). In $p$-cymene,there are two alkyl groups: a methyl group

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(B) The reaction is a Friedel-Crafts acylation of $1$-methyl-$4$-isopropylbenzene (p-cymene). In $p$-cymene,there are two alkyl groups: a methyl group $(-CH_3)$ and an isopropyl group $(-CH(CH_3)_2)$. Both groups are ortho/para directing. The methyl group has $3 \alpha$-hydrogens,while the isopropyl group has only $1 \alpha$-hydrogen. The methyl group is a stronger activating group than the isopropyl group due to hyperconjugation. Therefore,the incoming acyl group $(-COCH_3)$ will be directed by the more activating methyl group. The position ortho to the methyl group is also ortho to the isopropyl group,but the steric hindrance from the bulky isopropyl group makes the position ortho to the methyl group (and meta to the isopropyl group) more favorable. Thus,the major product is $2$-acetyl-$1$-methyl-$4$-isopropylbenzene,which corresponds to option $B$.

The major product of the reaction,$1$-methyl-$4$-isopropylbenzene + $CH_3COCl \xrightarrow{AlCl_3}$ is:

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