See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Caliene (C7H6) is a cross-conjugated bicyclic molecule consisting of a cyclopentadienyl ring connected through a methine bridge to a cyclopropenyl ring. To determine which resonance form contributes most to the resonance hybrid, we apply Hückel's rule of aromaticity. Step 1 – Identify the rings and electron counts: - The five-membered ring (cyclopentadienyl): if it bears a negative charge (gains one electron), it has 6 π-electrons (4n+2, n=1) → aromatic anion (cyclopentadienyl anion, Cp⁻). - The three-membered ring (cyclopropenyl): if it bears a positive charge (loses one electron), it has 2 π-electrons (4n+2, n=0) → aromatic cation (cyclopropenyl cation, Cy+). Step 2 – Evaluate aromaticity of each resonance form: - Option (a): neutral, no formal charges, no aromatic stabilization from charge delocalization — contributes but lacks extra aromatic stabilization. - Option (b): charges are localized on individual carbons (not delocalized over the rings), so no ring aromaticity is achieved — poor contributor. - Option (c): positive charge on five-membered ring and negative charge on three-membered ring. Five-membered ring with + has 4 π-electrons (antiaromatic), three-membered ring with - has 4 π-electrons (antiaromatic) — both rings are antiaromatic, very unstable contributor. - Option (d): negative charge delocalized over the five-membered ring (cyclopentadienyl anion, 6 π-electrons, aromatic) and positive charge delocalized over the three-membered ring (cyclopropenyl cation, 2 π-electrons, aromatic). Both rings are simultaneously aromatic, giving maximum stabilization. This structure also explains why Caliene is polar. Step 3 – Conclusion: Option (d) achieves double aromaticity — an aromatic cyclopentadienyl anion (6π) and an aromatic cyclopropenyl cation (2π) simultaneously — providing the greatest stabilization and thus contributing most to the resonance hybrid. This also accounts for the observed polarity of the molecule. Options (a), (b), and (c) either lack aromatic stabilization or introduce antiaromaticity, making them lesser contributors. Therefore, the correct answer is D.