See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of Base Strength **Step 1: Identify the lone pairs available for bonding** - **(A)** Imidazole: N has a lone pair; the other N is part of an aromatic $sp^2$ hybrid system - **(B)** Aniline: $NH_2$ lone pair on sp² carbon (aromatic) - **(C)** Cycloheptatriene with NH: lone pair on sp³ N (non-aromatic ring) - **(D)** Cyclohexane with NH: lone pair on sp³ N (saturated ring) **Step 2: Compare basicity factors** Base strength depends on lone pair availability and stability of the conjugate acid: | Factor | (A) | (B) | (C) | (D) | |--------|-----|-----|-----|-----| | **N hybridization** | $sp^2$ (one N) | $sp^2$ | $sp^3$ | $sp^3$ | | **Aromaticity loss** | Regained (imidazolium is aromatic) | Lost | Lost | N/A | | **Lone pair effect** | Enhanced by resonance | Delocalized to ring (weakens) | Localized | Localized | **Step 3: Why (A) is strongest** When imidazole is protonated: $$\text{Imidazole} + H^+ \rightarrow \text{Imidazolium}^+$$ The conjugate acid **imidazolium** is aromatic (6π electrons in 5-membered ring), making it highly stabilized. This stability makes the original imidazole the strongest base—it readily accepts a proton to form a stable aromatic cation. In contrast, aniline's lone pair (B) is delocalized into the aromatic ring, making it unavailable, while (C) and (D) lack this aromatic stabilization. **Answer: (A)** ✓