See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of Basic Strength of Anilines **Key Principle:** Basic strength of anilines depends on electron density at the $NH_2$ group. Electron-donating groups increase basicity; electron-withdrawing groups decrease it. ## Evaluating Option (1): $Ph-NH_2 > \text{(4-CH}_3\text{-aniline)} > \text{(4-NO}_2\text{-aniline)}$ **Step 1:** Compare aniline vs. 4-methylaniline - $CH_3$ is an electron-donating group (alkyl, +I effect) - 4-methylaniline should be MORE basic than aniline - **This ordering is WRONG** ✗ ## Evaluating Option (2): 4-methylaniline > aniline > 4-nitroaniline ✓ **Step 2:** Compare all three compounds - **4-methylaniline:** $CH_3$ donates electrons → increases $N$ electron density → **most basic** - **Aniline:** Reference baseline - **4-nitroaniline:** $NO_2$ is strongly electron-withdrawing (−M effect) → decreases $N$ electron density → **least basic** **Step 3:** Verify mechanism $NO_2$ withdraws electrons through resonance: $$\text{Lone pair on }N \text{ is delocalized into } NO_2 \text{ group}$$ This significantly reduces basicity. ## Why Option (1) is Incorrect Placing aniline above 4-methylaniline violates the electron-donating effect of the methyl group. The methyl group enhances basicity, so 4-methylaniline must be more basic than aniline. **Answer: Option (2)** correctly orders basic strength based on substituent effects.