JEE Chemistry 2025 Question with Solution

Given: We must identify which compound is least likely to give effervescence of $\mathrm{CO_2}$ with aqueous $\mathrm{NaHCO_3}$.

Find: The correct option among the four given structures.

Sodium bicarbonate gives effervescence with compounds that are sufficiently acidic to liberate $\mathrm{CO_2}$.

Carboxylic acids react readily with $\mathrm{NaHCO_3}$, whereas ordinary phenols generally do not. A more strongly acidic phenol such as trinitrophenol can react, but a simple nitrophenol is much less likely to do so.

From the given structures:

1. Option $1$ is a highly acidic nitro-substituted phenol.
2. Option $2$ contains a carboxylic acid group and will react with $\mathrm{NaHCO_3}$.
3. Option $3$ is an anilinium chloride salt and is acidic in nature.
4. Option $4$ is nitrophenol, which is least likely to liberate $\mathrm{CO_2}$ with $\mathrm{NaHCO_3}$.

Therefore, the correct option is D.

Given: Effervescence with aqueous $\mathrm{NaHCO_3}$ is used to test whether a compound is acidic enough to produce $\mathrm{CO_2}$.

Find: Which given compound is least likely to show this behavior.

Principle:

So we compare the acidic character of the four compounds shown.

• A compound with a carboxylic acid group reacts readily.
• A strongly electron-withdrawing substituted phenol becomes more acidic.
• A simple nitrophenol is still much weaker than a carboxylic acid and is least likely to give brisk effervescence with $\mathrm{NaHCO_3}$.

The solution states that the correct option is D, corresponding to option $4$.

Hence, the compound in option $4$ is least likely to give effervescence of $\mathrm{CO_2}$ in presence of aqueous $\mathrm{NaHCO_3}$.