See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Resonance Energy Comparison **Key Principle:** Higher resonance energy indicates greater aromatic stabilization. We compare the aromatic character of each pair. ## Analysis of Each Option: **(A) Furan vs Cyclopentadienol:** - Furan is aromatic (5-membered ring with $O$ donating lone pair, 6 π-electrons) - Cyclopentadienol has disrupted aromaticity - Furan has significantly higher resonance energy ✗ **(B) Cyclopentadienyl anion vs Cyclopentadienyl cation:** - Anion: $C_5H_5^-$ has 6 π-electrons → aromatic (very stable) - Cation: $C_5H_5^+$ has 4 π-electrons → antiaromatic (unstable) - Anion has much higher resonance energy ✗ **(D) 1,4-Cyclohexanedione vs Phthalic anhydride:** - 1,4-Cyclohexanedione is non-aromatic - Phthalic anhydride is benzene-based (aromatic core) - Anhydride has higher resonance energy ✗ **(E) Pyrrole vs Furan:** - Both are aromatic 5-membered heterocycles - Similar resonance energies; both aromatic ✗ **✓ (C) Nitrobenzene vs Benzene with para-$NO_2$:** The question likely compares **benzene** (left, unlabeled) with **para-nitrobenzene** (right). - **Benzene:** $\Delta H_{res} \approx 150.4$ kJ/mol - **para-Nitrobenzene:** The electron-withdrawing $NO_2$ group destabilizes resonance structures through inductive withdrawal and resonance delocalization. This **reduces overall resonance energy** to ~$120$ kJ/mol. **Answer: (C)** has less resonance energy because the $NO_2$ group's electron-withdrawing nature weakens the aromatic stabilization compared to unsubstituted benzene.