See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The basic strength of aromatic amines depends on the availability of the lone pair on nitrogen for protonation. Electron-donating groups increase basicity; resonance delocalization into the aromatic ring decreases basicity. Steric effects and inductive effects of alkyl groups also play a role. Step 1: Identify the compounds. - A: Aniline (Ph-NH2) — lone pair on N delocalized into the phenyl ring, reducing basicity. - B: N-methylaniline (Ph-NH-Me) — one methyl group donates electron density inductively, partially counteracting ring delocalization; more basic than aniline. - C: N,N-dimethylaniline (Ph-N(Me)2) — two methyl groups donate electron density inductively; despite steric hindrance, the inductive effect of two methyls makes the nitrogen more electron-rich than in B; more basic than B in aqueous solution (pKa ~ 5.1 vs ~ 4.85 for B vs ~ 4.6 for A). - D: o-toluidine (2-methylaniline) — methyl group at ortho position donates electrons inductively, but the ortho methyl creates steric hindrance that inhibits solvation of the conjugate acid (protonated amine), reducing effective basicity. The ortho effect makes it less basic than aniline itself in many comparisons, though the methyl is electron-donating. Step 2: Rank the compounds. - C (Ph-N(Me)2): Two methyl groups provide the greatest inductive donation to nitrogen; most basic among the aromatic amines listed. pKa of conjugate acid ~ 5.1. - B (Ph-NH-Me): One methyl group, moderate inductive donation. pKa ~ 4.85. - A (Ph-NH2): No alkyl substituent on N, baseline aniline basicity. pKa ~ 4.6. - D (o-toluidine): The ortho methyl group causes steric inhibition of resonance (slightly) but more importantly causes steric hindrance to solvation of the ammonium ion, making it less basic than aniline. pKa ~ 4.44. Step 3: Order is C > B > A > D. Step 4: Why other options fail. - (a) A > B > C > D: Incorrect; alkyl groups on N increase basicity, so B and C should be more basic than A. - (b) B > A > C > D: Incorrect; C has two methyl groups and should be more basic than B. - (d) C > B > D > A: Incorrect; D (o-toluidine) is less basic than A (aniline) due to the ortho steric effect on solvation, not more basic. Therefore, the correct answer is C.