JEE Main 2025 · 7 April · morning · Single correct · +4 / −1
Which of the following compounds is least likely to give effervescence of CO₂ in presence of aq. NaHCO₃?
- (a) Ph-N((+))H₃ Cl((-))
- (b)
- (c)
- (d)
Answer and solution
Answer (d)
Solution
Chemistry ma’am · 3 steps
Only an acid stronger than carbonic acid can push CO2 out of bicarbonate, so compare each pKa with 6.4.
- The test. An acid HA gives HA + HCO₃⁻ → A⁻ + H₂CO₃, and H₂CO₃ then breaks into CO₂ + H₂O. This goes forward only if HA is a stronger acid than H₂CO₃, whose pKₐ is about 6.4.
Compound Approx. pKₐ Stronger than H₂CO₃? CO₂ with NaHCO₃ (a) anilinium chloride, PhNH₃⁺ 4.6 yes effervescence (b) 4-nitrobenzoic acid 3.4 yes effervescence (c) picric acid (2,4,6-trinitrophenol) 0.4 yes, very strong effervescence (d) 3-nitrophenol 8.4 no no effervescence - Why m-nitrophenol fails. From the meta position, -NO₂ can pull electrons only by its inductive effect, not by resonance. So 3-nitrophenol stays a weak acid (pKₐ ≈ 8.4), weaker than carbonic acid. In picric acid, three -NO₂ groups at the ortho and para positions spread the phenoxide charge by resonance, which makes it far stronger.
- So. 3-nitrophenol is the only compound that cannot release CO₂ from NaHCO₃. The answer is (d).
The idea to keep. The bicarbonate test is a strength check against H₂CO₃ (pKₐ ≈ 6.4). Carboxylic acids pass. Plain phenols fail unless ortho or para nitro groups boost them, as in picric acid.
Answer from the NTA final answer key. Solution by PrepWiser.