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(D) Nucleophilic aromatic substitution reactions are facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para p

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m-Nitrochlorobenzene

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(D) Nucleophilic aromatic substitution reactions are facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at the ortho and para positions,as these groups stabilize the intermediate carbanion (Meisenheimer complex) through resonance.At the meta position,the electron-withdrawing group cannot stabilize the negative charge of the intermediate through resonance,making it much less effective at accelerating the reaction compared to ortho or para positions.Therefore,$m$-nitrochlorobenzene (option $D$) will have the lowest rate of nucleophilic aromatic substitution among the given compounds.

The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with $OH^-$ is

Similar Questions

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