See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

To solve this problem, we apply the principles of electrophilic aromatic substitution (EAS): electron-donating groups (EDG) activate the ring and direct ortho/para; electron-withdrawing groups (EWG) deactivate the ring and direct meta. The more activated ring is preferentially attacked. Compound 1: PhC(=O)CH2Ph (deoxybenzoin). - Ring A is directly attached to the carbonyl (C=O), which is a strong EWG via resonance and induction. Ring A is deactivated (meta director). - Ring B is attached via CH2, which insulates it from the carbonyl. Ring B behaves essentially as a plain benzene ring (mildly activated relative to A). - Therefore Ring B is more reactive and substitution goes ortho/para. - Answer: B, ortho/para. Compound 2: A-N(CH3)-SO2-B (N-methyl sulfonamide). - Ring B is directly attached to SO2, a strong EWG. Ring B is strongly deactivated. - Ring A is attached to the nitrogen of the sulfonamide. The nitrogen lone pair, although partially withdrawn by SO2, still donates electron density into Ring A via resonance (N is EDG toward Ring A). Ring A is more activated. - Therefore Ring A is preferentially attacked, ortho/para direction. - Answer: A, ortho/para. Compound 3: A-C(=O)-O-B (phenyl benzoate). - Ring A is attached to the carbonyl carbon of the ester (acyl side). The C=O is EWG, deactivating Ring A. - Ring B is attached to the oxygen of the ester (phenoxy side). Oxygen donates electrons into Ring B via resonance (lone pair on O into ring), activating Ring B as an EDG, ortho/para director. Despite the partial withdrawal by the carbonyl, the net effect on Ring B through oxygen is activation relative to Ring A. - Therefore Ring B is preferentially attacked, ortho/para. - Answer: B, ortho/para. Compound 4: Benzodiazepinedione with Ring A and Ring B both fused to the lactam ring. - Both Ring A and Ring B are fused to a seven-membered ring containing two amide (lactam) carbonyls and two N-methyl groups. - Each ring is flanked by an amide nitrogen (EDG, donates into ring) and an amide carbonyl (EWG, withdraws from ring). The two effects partially cancel. - Due to the symmetry of the molecule (both rings experience similar electronic environments) and the competing EDG/EWG effects, all positions on both rings are approximately equally reactive. Ring A is attacked preferentially (or equivalently) — but importantly the net activation from the nitrogen lone pairs means both rings react at all positions. - The answer given is Ring A, all positions. - Answer: A, all sites. Compound 5: A-C(=O)-CH2-N(CH3)-C(=O)-B. - Ring A is attached to a ketone carbonyl, which is EWG. Ring A is deactivated. - Ring B is attached to an amide carbonyl C(=O)-N. The amide nitrogen donates into the carbonyl, but the carbonyl still withdraws from Ring B. However, relative to Ring A (plain aryl ketone, strong deactivation), Ring B attached to an amide C=O is also deactivated. The amide carbonyl is EWG toward Ring B (meta director), but we must compare both rings. - Actually Ring B is directly bonded to the amide carbonyl carbon. In amides, C=O is still EWG toward the attached ring → Ring B is meta-directed. Ring A (aryl ketone) is also meta-directed but more strongly deactivated. - Comparing: the amide C=O attached to Ring B is less deactivating than the ketone C=O attached to Ring A (amide resonance reduces electrophilicity of carbonyl, making it somewhat less EWG toward the ring). So Ring B is relatively more reactive than Ring A. - Ring B is attacked preferentially, meta orientation. - Answer: B, meta. Compound 6: Fluorene/indanone-type fused system with Ring A and Ring B connected through a five-membered ring bearing C(CH3)2 and C=O. - Ring A is fused to the five-membered ring containing the C=O (ketone). The carbonyl is EWG toward Ring A. - Ring B is fused to the five-membered ring on the C(CH3)2 side (no carbonyl directly on Ring B side). The gem-dimethyl quaternary carbon is slightly EDG (alkyl groups are weakly electron-donating). Ring B experiences no direct EWG effect. - Ring B is more electron-rich and thus more reactive toward electrophiles, with ortho/para direction relative to the ring junction. - Answer: B, ortho/para. Therefore, the correct answer is {"1": {"reactivity": "B", "substitution": "ortho/para"}, "2": {"reactivity": "A", "substitution": "ortho/para"}, "3": {"reactivity": "B", "substitution": "ortho/para"}, "4": {"reactivity": "A", "substitution": "all position"}, "5": {"reactivity": "B", "substitution": "meta"}, "6": {"reactivity": "B", "substitution": "ortho/para"}}.