See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Analysis of Basicity Order **Identifying the compounds:** - (i) Pyridine (aromatic, N in ring) - (ii) Pyrrole (aromatic, N-H in ring) - (iii) Imidazole (aromatic, two N atoms with one N-H) **Key factors determining basicity:** 1. **Pyrrole (ii):** The lone pair on N is part of the aromatic sextet (6π electrons). Donating this lone pair disrupts aromaticity, making protonation highly unfavorable. **Least basic.** 2. **Pyridine (i):** The lone pair is NOT part of the aromatic system (it's in an $sp^2$ orbital perpendicular to the ring). Protonation maintains aromaticity of the pyridinium cation. **Moderately basic** ($pK_b \approx 8.75$). 3. **Imidazole (iii):** Has two nitrogens. When protonated on one N, the positive charge is delocalized between both nitrogens through resonance: $$\text{Im-H}^+ \leftrightarrow \text{Im}^+ \text{ (charge distributed)}$$ This resonance stabilization makes the conjugate acid much more stable. **Most basic** ($pK_b \approx 7.05$). **Basicity order:** $$\text{(iii)} > \text{(i)} > \text{(ii)}$$ **Answer: (C)** ✓