See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Solution: Basicity Order of Aniline Derivatives **Key Principle:** Basicity of anilines depends on electron density at nitrogen. Electron-donating groups (EDG) increase basicity; electron-withdrawing groups (EWG) decrease it. **Analysis of each compound:** 1. **p-$CH_3O$-$C_6H_4$-$NH_2$:** Methoxy ($-OCH_3$) is a strong EDG (ortho/para director) → **highest basicity** 2. **m-$CH_3O$-$C_6H_4$-$NH_2$:** Methoxy at meta position is less effective (meta doesn't significantly activate) → **lower basicity than (1)** 3. **o-$CH_3O$-$C_6H_4$-$NH_2$:** Methoxy at ortho position activates, BUT ortho-effect causes steric hindrance and hydrogen bonding between $-OCH_3$ and $-NH_2$, reducing basicity → **lowest basicity** 4. **$C_6H_5$-$NH_2$ (Aniline):** No substituent, baseline basicity → **intermediate** **Basicity Order (Decreasing):** $$\text{(1)} > \text{(4)} > \text{(2)} > \text{(3)}$$ **Why (1) > (4) > (2) > (3):** - (1): Para-methoxy provides maximum electron donation without steric issues - (4): Unsubstituted aniline (reference) - (2): Meta-methoxy weakly activating - (3): Ortho-methoxy suffers steric/intramolecular H-bonding effects **Answer: (A)**