See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Identify acidic groups in the molecule. The compound contains: (i) a sulfonic acid group (-SO3H), (ii) a carboxylic acid group (-CO2H), (iii) a phenolic -OH group on the ring, (iv) an alcoholic -CH-OH group, and (v) a terminal alkyne -C≡CH. Step 2 - Understand reactivity with NaHCO3. NaHCO3 is a mild base (pKa of H2CO3 ~ 6.35 for first dissociation). It reacts with acids stronger than carbonic acid. The groups that react with NaHCO3 are: sulfonic acids (very strong acid) and carboxylic acids (pKa ~4-5). Phenols (pKa ~10), alcohols (pKa ~16), and terminal alkynes (pKa ~25) do NOT react with NaHCO3. Step 3 - First reaction with NaHCO3 (superscript 14 indicates labeled carbon). The most acidic group reacts first. Sulfonic acid (pKa ~ -1) is stronger than carboxylic acid (pKa ~4-5). However, both -SO3H and -CO2H react with NaHCO3 to release CO2 gas. Wait - sulfonic acids react with NaHCO3 but do NOT release CO2; they simply form sodium sulfonate salt. Carboxylic acids react with NaHCO3 to release CO2 gas (A). So in the first step: -CO2H + NaHCO3 → CO2 (gas A) + H2O + sodium carboxylate salt. The labeled 14 on NaHCO3 means the CO2 produced is 14CO2, but molecular mass of CO2 = 44. Wait, but the superscript 14 means carbon-14, so molecular mass of 14CO2 = 14 + 16 + 16 = 46. Gas A = 14CO2, molecular mass = 46. Step 4 - Second reaction with NaHCO3. After the -CO2H has reacted, product (B) still contains -SO3H group. Sulfonic acid is a strong acid (pKa ~ -1) and reacts with NaHCO3 to release CO2 gas (C): -SO3H + NaHCO3 → sodium sulfonate + CO2 (gas C) + H2O. Gas C = CO2 (regular, unlabeled from NaHCO3), molecular mass = 44. Step 5 - Sum of molecular masses: A + C = 46 + 44 = 90. Step 6 - Check other options: (a) 88 would be 44+44, ignoring the 14C label; (c) 92 would be 46+46, incorrectly labeling both; (d) 40 is too small. Option (b) 90 = 46 + 44 correctly accounts for 14CO2 from the carboxylic acid reacting first with labeled NaH14CO3, and regular CO2 from sulfonic acid reacting with regular NaHCO3 in the second step. Therefore, the correct answer is b.