See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Resonance stabilization through cross-conjugation and aromaticity. When a fulvene-type molecule undergoes charge separation via resonance, the resulting ions can gain aromatic stabilization. The greatest resonance stabilization occurs when both fragments formed by polarization of the central double bond become aromatic. Step 1: Analyze each structure in terms of what happens when the exocyclic double bond polarizes (i.e., one ring gains electrons to become an anion, the other loses electrons to become a cation). Step 2 - Option (a): Benzene (6 pi, aromatic) connected to cyclopropane via exo double bond. If the double bond polarizes, cyclopropane ring fragment would need to become cyclopropenyl cation (2 pi, aromatic) or anion (4 pi, antiaromatic), and benzene side becomes cyclohexadienyl anion (not fully aromatic). Partial stabilization only. Step 3 - Option (b): Benzene ring (6 pi, aromatic) connected via exo double bond to cyclopentadienylidene. If the double bond polarizes: the cyclopentadienyl fragment gains electrons to become cyclopentadienyl anion (6 pi, aromatic), and the benzene ring loses electrons but remains benzenoid/aromatic (tropylium-like contribution is not applicable here, but benzene stays aromatic at 6 pi). This means BOTH rings simultaneously achieve or maintain aromatic character upon charge separation. Benzene (6 pi aromatic) + cyclopentadienyl anion (6 pi aromatic) = maximum resonance stabilization from both ends. Step 4 - Option (c): Two cyclopentadiene rings connected by exo double bond. Polarization gives cyclopentadienyl anion (6 pi, aromatic) on one side, but cyclopentadienyl cation (4 pi, antiaromatic) on the other. One side is stabilized, the other is destabilized. Net stabilization is less than (b). Step 5 - Option (d): Two cyclopropane rings connected by a double bond. Polarization could give cyclopropenyl cation (2 pi, aromatic) on one side and cyclopropenyl anion (4 pi, antiaromatic) on the other. Mixed stabilization/destabilization, minimal net benefit. Step 6: Comparing all options, option (b) — 6,5-fulvene (benzene connected to cyclopentadienylidene) — provides the greatest resonance stabilization because polarization of the central double bond allows BOTH the benzene ring (remains 6 pi aromatic) and the cyclopentadienyl fragment (becomes 6 pi aromatic anion) to simultaneously achieve aromatic stabilization, maximizing delocalization energy. Therefore, the correct answer is B.