Which position will be attacked most rapidly | vedclass

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(D) The molecule contains two benzene rings. One ring is attached to an electron-withdrawing acetyl group $(CH_3CO-)$,which is deactivating. The other

Vedclass · Accepted Answer

$D$

Vedclass · Answer

(D) The molecule contains two benzene rings. One ring is attached to an electron-withdrawing acetyl group $(CH_3CO-)$,which is deactivating. The other ring is attached to an electron-donating methoxy group $(OCH_3)$,which is strongly activating and ortho/para-directing due to its $+M$ effect.Nitration is an electrophilic aromatic substitution reaction,which occurs most rapidly on the most electron-rich ring.The methoxy group $(OCH_3)$ activates the ring it is attached to. The positions ortho to the methoxy group are $C$ and $D$. Since the methoxy group is a strong activator,the electrophile $(NO_2)^+$ will attack the ring containing the methoxy group.Between positions $C$ and $D$,both are ortho to the methoxy group. However,position $C$ is also meta to the other ring system,while position $D$ is para to the methoxy group. In electrophilic substitution,the para position is generally more favored due to less steric hindrance compared to the ortho position. Therefore,position $D$ is the most reactive site.

Which position will be attacked most rapidly by the nitronium ion $(NO_2)^+$ when the compound undergoes nitration with $HNO_3 / H_2SO_4$?

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