See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Electrophilic aromatic substitution (EAS) rate depends on how electron-rich the aromatic ring is. Electron-donating groups (EDGs) activate the ring toward EAS, while electron-withdrawing groups (EWGs) deactivate it. Step 2 - Analyze each compound's effect on the aromatic ring: (a) Tetralin: The saturated ring fused to the benzene ring is purely carbocyclic. There is no heteroatom, so the aromatic ring receives only a weak inductive/hyperconjugative effect from the alkyl carbons. Mild activation. (b) 1,2,3,4-Tetrahydroquinoline: The nitrogen (NH) is directly attached to the aromatic ring at position 1 (the nitrogen is part of the ring fused at the 1,2-bond of benzene). The nitrogen lone pair is in conjugation with the aromatic pi system, donating electron density directly into the aromatic ring via resonance. This is a powerful ortho/para-directing activating group — similar to an aniline-type nitrogen. Strong activation of the aromatic ring. (c) 1,2,3,4-Tetrahydroisoquinoline: The nitrogen is at the 2-position of the isoquinoline skeleton, meaning it is NOT directly bonded to the aromatic ring (it is one carbon removed from the ring). The nitrogen cannot donate its lone pair directly into the aromatic ring by resonance. Only weak inductive activation. (d) 3,4-Dihydroisoquinolin-1(2H)-one: The ring contains an amide carbonyl (C=O) fused to the benzene ring. The carbonyl is an EWG by resonance and induction, and the nitrogen lone pair is delocalized into the carbonyl rather than into the aromatic ring. This deactivates the aromatic ring toward EAS. Step 3 - Comparison: - (b) has an NH directly on the aromatic ring (vinylogous amine/aniline type), providing strong resonance donation into the ring — the fastest EAS rate. - (a) has only alkyl fusion — moderate activation. - (c) has nitrogen not directly conjugated with the ring — weaker activation than (b). - (d) has a deactivating amide carbonyl — slowest rate. Step 4 - Conclusion: Compound (b), 1,2,3,4-tetrahydroquinoline, has the most electron-rich aromatic ring due to direct resonance donation from the ring nitrogen (NH), making it undergo bromination (EAS) at the fastest rate. Therefore, the correct answer is B.