Identification of Compound Failing to Liberate Carbon Dioxide with Sodium Bicarbonate

To determine which compound does not liberate $\text{CO}_2$ upon reaction with an aqueous solution of sodium bicarbonate ($\text{NaHCO}_3$), we consider the acid-base equilibrium involved:

$\text{HA} + \text{NaHCO}_3 \rightleftharpoons \text{NaA} + \text{H}_2\text{CO}_3 \rightleftharpoons \text{NaA} + \text{H}_2\text{O} + \text{CO}_2\uparrow$

An acid ($\text{HA}$) can decompose sodium bicarbonate to liberate carbon dioxide ($\text{CO}_2$) only if it is a stronger acid than carbonic acid ($\text{H}_2\text{CO}_3$), which has a $\text{p}K_a$ of approximately $6.35$. In other words, the acid must have a $\text{p}K_a < 6.35$.

Let us examine the $\text{p}K_a$ values of the given compounds:

1. Benzenesulphonic acid ($\text{C}_6\text{H}_5\text{SO}_3\text{H}$): It is a strong mineral-like acid with $\text{p}K_a \approx -2.8$. Since $\text{p}K_a < 6.35$, it readily reacts with $\text{NaHCO}_3$ to liberate $\text{CO}_2$.
2. Salicylic acid ($2\text{-hydroxybenzoic acid}$): The ortho-hydroxy group stabilizes the conjugate base through intramolecular hydrogen bonding (the ortho-effect), giving it a $\text{p}K_a \approx 2.98$. Since $\text{p}K_a < 6.35$, it liberates $\text{CO}_2$.
3. Benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$): A carboxylic acid with $\text{p}K_a \approx 4.20$. Since $\text{p}K_a < 6.35$, it decomposes $\text{NaHCO}_3$ to liberate $\text{CO}_2$.
4. Carbolic acid (Phenol, $\text{C}_6\text{H}_5\text{OH}$): A very weak acid with $\text{p}K_a \approx 9.95$. Because its $\text{p}K_a$ is substantially greater than that of carbonic acid ($\text{p}K_a \approx 6.35$), phenol is a weaker acid than $\text{H}_2\text{CO}_3$. The acid-base equilibrium lies heavily to the left, meaning it fails to decompose $\text{NaHCO}_3$ and therefore does NOT liberate $\text{CO}_2$ gas.

Thus, carbolic acid (phenol) does not liberate $\text{CO}_2$.

Correct Option: (D)