See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of Option B **The reaction shown:** $$\text{O}_2\text{N-C}_6\text{H}_3\text{(-NO}_2)_2\text{-OH} + \text{NaHCO}_3 \rightarrow \text{O}_2\text{N-C}_6\text{H}_3\text{(-NO}_2)_2\text{-ONa} + \text{CO}_2 + \text{H}_2\text{O}$$ **Why this does NOT occur in practice:** 1. **Electron-withdrawing groups (EWG) deactivate phenols:** The two $\text{NO}_2$ groups are strongly electron-withdrawing, which significantly decreases the acidity of the phenolic $-\text{OH}$ group by destabilizing the phenoxide anion. 2. **$\text{NaHCO}_3$ is too weak a base:** Sodium bicarbonate is a mild base ($\text{pK}_a$ of $\text{HCO}_3^- \approx 10.3$). It can deprotonate phenols with **lowered $\text{pK}_a$** values (like polynitrophenols, $\text{pK}_a \sim 4$), but only weakly. 3. **The critical issue:** With **three** electron-withdrawing $\text{NO}_2$ groups on the benzene ring, the phenolic $\text{pK}_a$ is still too high for $\text{NaHCO}_3$ to effectively deprotonate it. Stronger bases like $\text{Na}_2\text{CO}_3$ or $\text{NaOH}$ are required. **In contrast:** Options A, C, and D all occur readily under their specified conditions (A: deprotonation by alkyl anion; C: enolate formation and alkylation; D: single $\text{NO}_2$ makes phenol acidic enough for $\text{NaHCO}_3$).