The reaction of N-phenylbenzamide with NO_2^{ | vedclass

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(D) The starting material is $N$-phenylbenzamide,which contains two benzene rings: one attached to the nitrogen atom (aniline derivative) and one atta

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$N$-($4$-nitrophenyl)benzamide

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(D) The starting material is $N$-phenylbenzamide,which contains two benzene rings: one attached to the nitrogen atom (aniline derivative) and one attached to the carbonyl group (benzoyl derivative).The $-NH-CO-Ph$ group is an ortho/para-directing group because of the lone pair on the nitrogen atom,which can participate in resonance with the attached benzene ring.The $-CO-NH-Ph$ group is a meta-directing group because the carbonyl group is electron-withdrawing.In electrophilic aromatic substitution (nitration),the $-NH-CO-Ph$ group is a much stronger activating group than the $-CO-NH-Ph$ group is deactivating. Therefore,the nitration occurs on the ring attached to the nitrogen atom.Due to the steric hindrance of the bulky $-NH-CO-Ph$ group,the para-position is favored over the ortho-position. Thus,the major product is $N$-($4$-nitrophenyl)benzamide.

The reaction of $N$-phenylbenzamide with $NO_2^{\oplus}$ (nitronium ion) yields which product?

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