See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Most Acidic Compound Analysis **Step 1: Identify the acidic functional groups** All four compounds contain a $-COOH$ (carboxylic acid) group, which is the acidic moiety. The acidity depends on how easily the $H^+$ is released, determined by the stability of the resulting carboxylate anion. **Step 2: Recognize the role of electron-withdrawing/donating groups** - **Electron-withdrawing groups (EWG):** Stabilize the negative charge on the carboxylate anion ($COO^-$) by pulling electron density away → **increases acidity** - **Electron-donating groups (EDG):** Destabilize the anion by pushing electron density toward it → **decreases acidity** **Step 3: Analyze each option** - **(A)** $-OCH_3$ (ortho): EDG via resonance → decreases acidity - **(B)** $-OH$ (ortho): EDG via resonance → decreases acidity - **(C)** Two $-OH$ groups (ortho and meta positions): These are EWGs through the **inductive effect** (oxygen withdraws electrons through the $\sigma$-bond), especially when adjacent to electron-rich regions - **(D)** $-OCH_3$ and $-H_3CO$ groups: Both are EDGs → decreases acidity **Step 4: Why option C is most acidic** In compound **(C)**, the hydroxyl groups act as **inductive electron-withdrawing groups** that stabilize the carboxylate anion more effectively than the resonance-donating effects in other compounds. Multiple OH groups create cumulative electron withdrawal, making the carboxylic acid significantly more acidic. **Answer: (C)**