See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Acidity of phenols is determined by the stability of the phenoxide ion formed after loss of a proton. Electron-withdrawing groups (EWG) stabilize the phenoxide ion (increase acidity), while electron-donating groups (EDG) destabilize the phenoxide ion (decrease acidity). Step 1: Analyze option (d). - First compound: 4-methylphenol (para-cresol). The methyl group (CH3) at the para position is an electron-donating group (EDG) via hyperconjugation and inductive effect. This increases electron density on the oxygen, destabilizing the phenoxide ion, making para-cresol LESS acidic than phenol. pKa of para-cresol ≈ 10.26. - Second compound: Phenol (no substituent). pKa ≈ 9.99 (or ~10.0). - Since para-cresol (first) is less acidic than phenol (second), the second compound IS more acidic than the first. This matches the condition. Step 2: Check why other options fail. - Option (a): BrCH2NO2 vs CH3CH3. BrCH2NO2 has both Br and NO2 as EWGs, making it significantly more acidic than ethane. So the first compound is MORE acidic than the second, not less. The second is NOT more acidic than the first. - Option (b): CH3COCH2CN vs CH3COCH3. CH3COCH2CN has both a carbonyl and a cyano group activating the alpha hydrogens, making it more acidic than acetone. So the first compound is MORE acidic than the second. The second is NOT more acidic than the first. - Option (c): 4-acetylphenol vs phenol. The acetyl group (CH3C=O) at the para position is an EWG, which stabilizes the phenoxide ion via resonance, making 4-acetylphenol MORE acidic than phenol. So the first compound is MORE acidic than the second. The second is NOT more acidic than the first. Step 3: Conclusion. Only in option (d) is the second compound (phenol) more acidic than the first compound (4-methylphenol / para-cresol), because the electron-donating methyl group reduces the acidity of para-cresol relative to phenol. Therefore, the correct answer is D.