See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of Acidity Order **Identify each compound:** - **I:** Cyclohexane-1,4-dione (two ketones) - **II:** Cyclohexanone (one ketone) - **III:** Tetrahydropyran-4-one (ketone with ether oxygen) - **IV:** Cyclohexanone carboxylic acid methyl ester (ketone + ester) **Acidity ranking principles:** Acidity increases with: (1) electron-withdrawing groups, (2) resonance stabilization of conjugate base, (3) inductive effects **Step-by-step comparison:** 1. **II vs I:** Compound I has two ketone groups creating stronger electron-withdrawal and better conjugation stabilization of the deprotonated form → **I > II** 2. **III vs II:** The ether oxygen in III is electron-withdrawing (inductive effect) compared to CH₂ in II, stabilizing the anion better → **III > II** but weakly 3. **IV vs others:** The ester group ($-COOCH_3$) is highly electron-withdrawing and the conjugate base is resonance-stabilized. This makes α-hydrogens extremely acidic → **IV >> all others** 4. **Overall order:** $$\text{II} < \text{III} < \text{I} < \text{IV}$$ **Answer: (D) II < III < I < IV** ✓ The ester functional group in IV makes it by far the most acidic, while the two ketones in I make it more acidic than the single ketone in II, and III (with its ether activation) falls between them.