JEE Chemistry 2025 Question with Solution

Given: We must identify which compounds produce $CO_2$ with aqueous $NaHCO_3$ solution.

Find: The correct set of compounds.

Aqueous sodium bicarbonate, $NaHCO_3$, reacts with compounds that are sufficiently acidic. Carboxylic acids react with $NaHCO_3$ to liberate $CO_2$. Ordinary phenols are weaker acids than carbonic acid and usually do not react, but strongly electron-withdrawing groups can increase phenolic acidity enough for reaction.

From the given compounds:

• A. Benzoic acid is a carboxylic acid, so it reacts with $NaHCO_3$ and produces $CO_2$.
• B. Phenol is a weak acid and does not react with $NaHCO_3$.
• C. Picric acid (2,4,6-trinitrophenol) is a strongly acidic phenol because of three nitro groups, so it reacts with $NaHCO_3$ and produces $CO_2$.
• D. Cyclohexanecarboxylic acid is a carboxylic acid, so it reacts with $NaHCO_3$ and produces $CO_2$.
• E. 4-Methoxyphenol is a weak phenol and does not react with $NaHCO_3$.

Therefore, the compounds that produce $CO_2$ are A, C, and D only. The correct option is C.

Given: Compounds A to E are tested with aqueous $NaHCO_3$.

Find: Which ones evolve $CO_2$.

Only acids stronger than carbonic acid react with sodium bicarbonate. Thus, the key comparison is acidity.

• Carboxylic acids are sufficiently acidic and give $CO_2$.
• Phenol is not acidic enough and does not give $CO_2$.
• Strongly electron-withdrawing substituents such as nitro groups can make a phenol acidic enough to react.

Applying this to each compound:

1. A = Benzoic acid → carboxylic acid → reacts.
2. B = Phenol → weak phenol → does not react.
3. C = 2,4,6-Trinitrophenol → three $NO_2$ groups strongly increase acidity → reacts.
4. D = Cyclohexanecarboxylic acid → carboxylic acid → reacts.
5. E = p-Methoxyphenol → methoxy group is electron-donating, so acidity decreases → does not react.

Hence the compounds producing $CO_2$ are A, C, and D only. Therefore, the correct option is C.