JEE Main 2026 · 28 January · morning · Single correct · +4 / −1
Consider the following reaction sequence
Compound (x); [76.6% C, 6.38% H,; vapour density 47] →((i) CO₂, NaOH, 120°C,; high pressure) Compound (y) (Major Product)Compound (y) develops characteristic colour with neutral FeCl₃ solution.
Identify the INCORRECT statement from the following for the above sequence.
- (a) Compound y will dissolve in NaHCO₃ and evolve a gas.
- (b) Compound x is more acidic than compound y .
- (c) Both compounds x and y will dissolve in NaOH .
- (d) Both compounds x and y will burn with sooty flame.
Answer and solution
Answer (b) Compound x is more acidic than compound y .
Solution
Chemistry ma’am · 3 steps
Vapour density 47 means M = 94, so x is phenol and the Kolbe reaction turns it into salicylic acid.
- Find compound x. The molar mass is 2 × 47 = 94 g mol⁻¹. Carbon is 0.766 × 94 ≈ 72 (6 C), hydrogen is 0.0638 × 94 ≈ 6 (6 H), and the remaining 16 is one O. So x is C₆H₆O, which is phenol, C₆H₅OH.
- Kolbe reaction. Phenol with NaOH and CO₂ at 120°C and high pressure, then acid, gives salicylic acid (2-hydroxybenzoic acid). So y is salicylic acid. It still has a phenolic -OH, so it gives a violet colour with neutral FeCl₃.
Statement Check True or false (a) y gives a gas with NaHCO₃ Its -COOH group is stronger than carbonic acid, so CO₂ bubbles off True (b) x is more acidic than y Phenol has pKₐ ≈ 10, salicylic acid has pKₐ ≈ 3 False (c) both dissolve in NaOH Both are acidic enough to form sodium salts with a strong base True (d) both burn with a sooty flame Both are aromatic, with a high carbon content True - So. Salicylic acid is far more acidic than phenol, so statement (b) is the incorrect one.
The idea to keep. Phenol reacts with NaOH but not with NaHCO3. A -COOH group makes a compound a much stronger acid, strong enough to release CO₂ from bicarbonate.
Answer from the NTA final answer key. Solution by PrepWiser.