See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: A substituent on benzene can direct incoming electrophiles and nucleophiles to specific positions. Ortho-para directors in electrophilic aromatic substitution (EAS) are typically electron-donating groups (EDGs) or certain groups that donate electrons via resonance. Ortho-para directors in nucleophilic aromatic substitution (NAS) are electron-withdrawing groups (EWGs) that stabilize the Meisenheimer complex at ortho and para positions. Step 2 - Analyzing the -NO (nitroso) group: The nitroso group (-NO) is an unusual substituent. In EAS, the nitroso group can act as an ortho-para director because the nitrogen lone pair can donate electron density into the ring via resonance (similar to how -NR2 or -OH direct), despite being an overall electron-withdrawing group by induction. The nitrogen of -NO has a lone pair that resonates with the ring, placing electron density at ortho and para positions, thus directing electrophiles ortho-para. Step 3 - In NAS, the nitroso group is an electron-withdrawing group (the N=O bond is polarized, making nitrogen electron-deficient, which withdraws electron density from the ring). EWGs stabilize the negative charge of the Meisenheimer complex when the nucleophile attacks at positions ortho or para to the EWG. Therefore, -NO also directs nucleophiles to ortho-para positions. Step 4 - Why other options fail: - -NO2 (option a): In EAS, -NO2 is a meta director (strong EWG, deactivates ring, directs meta). So it fails the EAS ortho-para criterion. - -SO3H (option c): In EAS, -SO3H is generally a meta director. It fails. - -SO2Me (option d): In EAS, -SO2Me is a meta director (strong EWG). It fails. Step 5 - The -NO (nitroso) group uniquely satisfies both conditions: ortho-para directing in EAS (due to lone pair resonance donation from nitrogen) and ortho-para directing in NAS (due to its electron-withdrawing character stabilizing the Meisenheimer complex at ortho/para positions). Therefore, the correct answer is B.