See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Deprotonation to Form Carbanion **Key Principle:** Carbanion stability depends on stabilization of the negative charge by adjacent electron-withdrawing groups (EWGs) and resonance effects. ## Analysis of Each Option: **(A) Diethyl ester of malonic acid:** - Two ester groups ($CO_2R$) are EWGs - Deprotonation occurs at the **central $CH_2$** between two carbonyls - The resulting carbanion is stabilized by resonance with **both** carbonyl groups simultaneously - Forms a stable resonance-stabilized carbanion: $\ce{RO2C-\overset{-}{C}H-CO2R} \leftrightarrow$ resonance forms **(B) Monomethyl ester of malonic acid:** - Only one ester group and one aldehyde - Less stabilization than (A) due to only one carbonyl EWG for resonance **(C) Acetylacetone (2,4-pentanedione):** $$\ce{CH3-C(=O)-CH2-C(=O)-CH3}$$ - Two **ketone** groups (more reactive carbonyl electrophiles than esters) - Central $CH_2$ can be deprotonated - Carbanion stabilized by resonance with **both** adjacent ketone groups - **Most acidic proton** ($pK_a \approx 9$) due to strong stabilization of enolate - Ketones are better EWGs than esters for this purpose **(D) Unsaturated aldehyde:** - Single carbonyl; weaker stabilization than compounds with two carbonyls ## Answer: **Option (C)** The acetylacetone compound has the most readily deprotonatable proton because the resulting carbanion is stabilized by resonance delocalization into **two ketone groups**, making it the most stable carbanion among the options.