See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Solubility in Sodium Bicarbonate **Key Principle:** Sodium bicarbonate ($NaHCO_3$) is a weak base that dissolves only **acidic compounds** (carboxylic acids and phenols with electron-withdrawing groups that increase acidity). ## Analysis of Each Compound: **(A) 2,4,6-Trinitrophenol** - Three $NO_2$ groups (strong electron-withdrawing) dramatically increase phenolic acidity - $pK_a \approx 0.4$ — highly acidic phenol - **Soluble** in $NaHCO_3$: $$C_6H_2(NO_2)_3OH + NaHCO_3 \rightarrow C_6H_2(NO_2)_3O^-Na^+ + H_2O + CO_2$$ **(B) Benzoic acid** - Carboxylic acid ($pK_a \approx 4.2$) - **Soluble** in $NaHCO_3$ (stronger acid than bicarbonate) **(C) O-Nitrophenol** - Only ONE $NO_2$ group provides limited electron-withdrawal - $pK_a \approx 7.2$ — weakly acidic - **NOT soluble** in $NaHCO_3$ (comparable acidity to $HCO_3^-$; no reaction occurs) **(D) Benzene sulphonic acid** - Strong sulfonic acid ($pK_a < 0$) - **Soluble** in $NaHCO_3$ **Answer: (C)** — O-Nitrophenol lacks sufficient electron-withdrawing character to be deprotonated by the weak base bicarbonate.