See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: NaHCO3 is a mild base (pKa of H2CO3 ≈ 6.35). It can deprotonate and react with (dissolve) acids that are stronger than carbonic acid, i.e., compounds with pKa < ~6.35. Step 1 – Evaluate each compound's acidity: (a) Cyclohexanol: aliphatic alcohol, pKa ≈ 16–17. Much weaker than carbonic acid. Does NOT react with NaHCO3. (b) Phenol: pKa ≈ 10. Weaker than carbonic acid. Does NOT react with NaHCO3 (phenol reacts with NaOH but not NaHCO3). (c) Benzoic acid: carboxylic acid, pKa ≈ 4.2. Stronger than carbonic acid. REACTS with NaHCO3, producing sodium benzoate + CO2 + H2O. ✓ (d) Benzenesulfonic acid: sulfonic acid, pKa ≈ –1 to 1. Very strong acid, much stronger than carbonic acid. REACTS with NaHCO3 readily. ✓ (e) Propyne (CH3–C≡CH): terminal alkyne, pKa ≈ 25. Extremely weak acid. Does NOT react with NaHCO3. Step 2 – Why other options fail: - (a) Cyclohexanol is far too weak an acid (pKa ~16–17) to be deprotonated by NaHCO3. - (b) Phenol (pKa ~10) is weaker than carbonic acid; NaHCO3 cannot neutralize it. - (e) Propyne (pKa ~25) is an extremely weak acid and does not react with NaHCO3. Step 3 – Conclusion: Only compounds (c) benzoic acid and (d) benzenesulfonic acid have pKa values lower than that of carbonic acid (~6.35), so only they react with (are soluble in) NaHCO3. Therefore, the correct answer is {"count": 2, "items": ["c", "d"]}.