See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Reactions with $NaHCO_3$ Producing $CO_2$ $NaHCO_3$ reacts with **carboxylic acids** (or compounds with acidic $-COOH$ groups) to produce $CO_2$: $$\text{R-COOH} + NaHCO_3 \rightarrow \text{R-COONa} + H_2O + CO_2 \uparrow$$ **Analysis of each option:** **(A) Oxalic acid** — Contains two $-COOH$ groups (dicarboxylic acid). Reacts with $NaHCO_3$ readily to produce $CO_2$. ✓ **(B) Salicylic acid** — Has both $-OH$ and $-COOH$ groups. The carboxylic acid group reacts with $NaHCO_3$ to produce $CO_2$. ✓ **(C) Phenol** — Contains only $-OH$ (hydroxyl) group, not $-COOH$. Phenols are too weakly acidic to react with the weak base $NaHCO_3$. ✗ **(D) Pyruvic acid** ($CH_3COCOOH$) — Contains a carboxylic acid group ($-COOH$). Reacts with $NaHCO_3$ to produce $CO_2$. ✓ **Key point:** Only compounds with carboxylic acid groups ($-COOH$) react with $NaHCO_3$ to produce $CO_2$. Phenols cannot do this as they are not strong enough acids.