See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: $CO_2$ Liberation with $NaHCO_3$ **Key Principle:** Only compounds with $pK_a < 6.35$ (stronger acids than $HCO_3^-$) will liberate $CO_2$ when treated with aqueous $NaHCO_3$. The reaction mechanism: $$\text{HA} + NaHCO_3 \rightarrow NaA + H_2CO_3 \rightarrow NaA + H_2O + CO_2 \uparrow$$ This occurs only if the acid is strong enough to protonate $HCO_3^-$. **Analyzing each option:** | Compound | $pK_a$ | Stronger than $HCO_3^-$? | $CO_2$ Released? | |----------|--------|------------------------|------------------| | **(A) Benzoic acid** | ~4.2 | ✓ Yes | ✓ Yes | | **(B) Benzenesulphonic acid** | ~0.7 | ✓ Yes | ✓ Yes | | **(C) Salicylic acid** | ~3.0 | ✓ Yes | ✓ Yes | | **(D) Carbolic acid (Phenol)** | ~10 | ✗ No | ✗ **NO** | **Why phenol doesn't work:** Phenol is a very weak acid ($pK_a \approx 10$), much weaker than $HCO_3^-$ ($pK_a \approx 6.35$). Therefore, $HCO_3^-$ is not strong enough to deprotonate phenol, and no $CO_2$ is evolved. **Answer: (D) Carbolic acid (Phenol)**