JEE Advanced levelSingle correct optionDifficulty 3 of 3g2157
Salicylic acid is treated with excess aqueous NaHCOX3. The first pKa (carboxylic acid) is 2.97 and the second (phenolic OH in the salicylate anion, which is intramolecularly hydrogen-bonded) is 13.6. Given pKa(HX2COX3)=6.4, the main product is
Adisodium 2-oxidobenzoate, with release of COX2
Bsodium 2-hydroxybenzoate (phenolic OH unchanged), with release of COX2
Csodium 2-hydroxybenzoate without release of COX2
Dno reaction
Answer: B
The carboxylic acid reacts: logK=6.4−2.97=3.4>0, giving HX2COX3 and thus COX2. The phenolic OH (pKa 13.6, raised by the intramolecular H-bond) would need logK=6.4−13.6=−7.2, so it stays protonated.