See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The acidity of substituted phenols depends on whether the substituent is electron-donating (EDG) or electron-withdrawing (EWG). EWGs stabilize the phenoxide anion (conjugate base) by delocalizing negative charge, increasing acidity (lower pKa). EDGs destabilize the phenoxide anion, decreasing acidity (higher pKa). Step 1: Identify the substituents in each option. (a) Cl (para) – weakly electron-withdrawing by induction (though pi-donor), net effect increases acidity slightly relative to phenol. (b) NO2 (para) – strong electron-withdrawing group (both inductive and resonance), greatly increases acidity (very low pKa ~7.15). (c) CH3 (para) – electron-donating group (hyperconjugation and induction), destabilizes phenoxide anion, decreases acidity (highest pKa ~10.26). (d) CN (para) – electron-withdrawing group (inductive and resonance), increases acidity (lower pKa ~7.95). Step 2: Rank by pKa (least acidic = highest pKa). - 4-nitrophenol: pKa ≈ 7.15 (most acidic) - 4-cyanophenol: pKa ≈ 7.95 - 4-chlorophenol: pKa ≈ 9.38 - 4-methylphenol: pKa ≈ 10.26 (least acidic) Step 3: The methyl group in option (c) is an electron-donating group. It increases electron density on the ring and on the oxygen, making it harder to lose a proton. This destabilizes the phenoxide anion relative to phenol itself, resulting in the highest pKa among the four options. Why other options fail: - (a) Cl is electron-withdrawing (net), so it increases acidity compared to (c). - (b) NO2 is a powerful EWG, making 4-nitrophenol the most acidic of the four. - (d) CN is an EWG, increasing acidity compared to (c). Therefore, the correct answer is C.