Give the major product of the following react | vedclass

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(C) The reaction is a Friedel-Crafts alkylation of $N$-phenylbenzamide (benzanilide) with $CH_3Cl$ in the presence of $AlCl_3$. In $N$-phenylbenzamide

Vedclass · Accepted Answer

$4-$methyl-$N$-phenylbenzamide

Vedclass · Answer

(C) The reaction is a Friedel-Crafts alkylation of $N$-phenylbenzamide (benzanilide) with $CH_3Cl$ in the presence of $AlCl_3$. In $N$-phenylbenzamide,the $-NHCOPh$ group is attached to the benzene ring. The $-NHCOPh$ group is a deactivating group because the lone pair on the nitrogen atom is involved in resonance with the carbonyl group $(-C=O)$,making it less available for resonance with the benzene ring. However,the $-NHCOPh$ group is still ortho/para-directing due to the presence of the lone pair on the nitrogen atom. Between the ortho and para positions,the para position is sterically less hindered. Therefore,the major product is the para-substituted isomer,which is $4$-methyl-$N$-phenylbenzamide. Thus,the correct option is $(C)$.

Give the major product of the following reaction:

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