See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Basicity Comparison of Aniline Derivatives **Step 1: Identify the structures** - Left compound: o-toluidine (2-methylaniline) — $NH_2$ group ortho to $CH_3$ - Right compound: p-methylaniline (4-methylaniline) — $NH_2$ group para to $CH_3$ **Step 2: Analyze electron-donating effects** The methyl group ($CH_3$) is an electron-donating group via the inductive effect. It increases electron density on the benzene ring, which increases electron density on the nitrogen atom. **Step 3: Compare basicity** A more electron-rich nitrogen lone pair → stronger base (higher $K_b$, higher $pK_a$ of conjugate acid). Since both isomers have $CH_3$ attached to the benzene ring, **both are stronger bases than unsubstituted aniline** ($C_6H_5NH_2$). **Step 4: Evaluate statements** - Statement (A): "weaker base than [2-methylaniline]" — **FALSE** (p-isomer is actually slightly weaker due to resonance effects, but both are stronger than plain aniline) - Statement (B): "stronger base than [4-methylaniline]" — **FALSE** (same reasoning; o-isomer is slightly stronger but both are stronger than the reference) **Conclusion:** Neither statement correctly compares the two methylanilines to aniline itself. Both compounds are **stronger bases than aniline**, making both (A) and (B) incorrect. **Answer: (D) None is correct**