Animated Solution for Chemistry - Organic Chemistry: Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are

To determine which compounds will liberate carbon dioxide gas when treated with sodium bicarbonate (${\text{NaHCO}}_3$), we must first understand the underlying principle of the reaction. Sodium bicarbonate is a weak base. It will only undergo an acid-base reaction to release ${\text{CO}}_2$ effervescence if it reacts with an acid that is strictly stronger than carbonic acid (${\text{H}}_2{\text{CO}}_3$).

Carbonic acid has a $\text{p}{K}_a$ of approximately $6.4$. Therefore, any organic compound with a $\text{p}{K}_a$ lower than $6.4$ will successfully drive the equilibrium forward, liberating ${\text{CO}}_2$ gas. Let's evaluate the acidic strength of the three given compounds.

Compound B is benzoic acid. Carboxylic acids are inherently strong organic acids due to the highly resonance-stabilized carboxylate anion formed after deprotonation. The $\text{p}{K}_a$ of benzoic acid is approximately $4.2$, which is significantly lower than that of carbonic acid.

Because it is a stronger acid, benzoic acid will readily react with sodium bicarbonate to form sodium benzoate, water, and carbon dioxide gas. Thus, Compound B gives a positive test.

Compound C is 2,4,6-trinitrophenol, commonly known as picric acid. While simple phenol is a weak acid ($\text{p}{K}_a\approx 10$) and does not react with ${\text{NaHCO}}_3$, picric acid is a dramatic exception.

It possesses three nitro ($-{\text{NO}}_2$) groups located at the ortho and para positions. These nitro groups are powerful electron-withdrawing groups (EWG) that exert strong $-R$ (resonance) and $-I$ (inductive) effects. They intensely withdraw electron density from the phenoxide oxygen, dispersing the negative charge across the entire ring and onto the oxygen atoms of the nitro groups. This extreme stabilization makes picric acid highly acidic ($\text{p}{K}_a\approx 0.38$), even stronger than many carboxylic acids! Consequently, it will definitely liberate ${\text{CO}}_2$.

Compound A is 2,4,6-triaminophenol. Here, the phenol ring is substituted with three amino ($-{\text{NH}}_2$) groups. Unlike nitro groups, amino groups are strong electron-donating groups (EDG) due to their $+R$ effect.

They push electron density into the aromatic ring, which severely destabilizes the negative charge on the phenoxide ion formed after deprotonation. This destabilization makes 2,4,6-triaminophenol an even weaker acid than normal phenol. Since it is much weaker than carbonic acid, it will not react with sodium bicarbonate.

Since only benzoic acid (B) and picric acid (C) are stronger acids than carbonic acid, they are the only compounds that will liberate carbon dioxide gas. Therefore, the correct option is (c).