See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic Aromatic Substitution (EAS) on fused ring systems is directed by the activating/deactivating and ortho/para-directing effects of substituents, as well as the hyperconjugative and inductive activation from the fused alkyl ring. Step 1: Identify the structure. The compound is an indane (benzene fused with cyclopentane) derivative with gem-dimethyl groups on the cyclopentane ring. The benzene ring has positions labeled a (top, para to the ring fusion bottom), b (bottom, at the ring fusion junction area), and c (left side, ortho/meta positions relative to the fusion). Step 2: Determine the directing effects. In indane-type systems, the fused cyclopentane ring acts as an alkyl substituent attached at two positions of the benzene ring (C1 and C3 of indane, i.e., the two carbons where the rings are fused). The CH2 groups of the cyclopentane ring attached to the benzene ring are electron-donating by hyperconjugation and induction, making them ortho/para directors. Step 3: Identify the activated positions. The two ring-junction carbons (where cyclopentane meets benzene) are both ortho/para directors. The positions ortho and para to both ring junctions are activated. In indane, the C5 position (which corresponds to position 'b' in this labeling — the bottom position of the benzene ring, between the two fused positions) is activated by being ortho to one ring-junction carbon and para to the other, making it doubly activated. Step 4: Position 'b' (bottom of benzene ring) is ortho to one fusion carbon and para to the other fusion carbon simultaneously, giving it the greatest electron density and making it the most reactive position toward electrophilic attack. Step 5: Why other options fail: - Position 'a' (top of benzene ring) is meta to both fusion carbons, so it is deactivated relative to b and c. - Position 'c' (side of benzene ring) is ortho to one fusion carbon but meta to the other, making it less activated than position b. - Since positions are not equivalent and b is most activated, 'all positions identical' is incorrect. Therefore, the correct answer is B.