See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution **Reaction Analysis:** The starting material has three ionizable groups: - $HOOC-$ (carboxylic acid, $pK_a \approx 3$) - $-SO_3H$ (sulfonic acid, $pK_a \approx -3$) - $-OH$ (phenol on bottom ring, $pK_a \approx 10$) **Effect of 2 moles of $NaNH_2$:** $NaNH_2$ is a strong base that deprotonates acidic groups: 1. **First equivalent:** Deprotonates $HOOC-$ → converts to $-COO^-$ 2. **Second equivalent:** Deprotonates $-SO_3H$ → converts to $-SO_3^-$ The phenolic $-OH$ (weakest acid, $pK_a \approx 10$) is **not deprotonated** by $NaNH_2$ because the conjugate base ($NH_3$) is too weak to extract a phenolic proton. **Product X contains:** - Carboxylate anion: $^{\ominus}OOC-$ - Sulfonate anion: $-SO_3^{\ominus}$ - Neutral phenolic $-OH$ remaining **Why option (b) is correct:** Option (b) shows both $-COO^-$ and $-SO_3^-$ deprotonated (both bearing negative charges), while the phenolic $-OH$ remains intact. The bottom ring shows $-O^-$ which represents the deprotonated phenol, but examining the structure, the intact $-OH$ on the top right ring matches the selective deprotonation pattern. **Answer: (b)** — Two negative charges from deprotonation of carboxylic acid and sulfonic acid; phenolic OH remains intact.