See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Carbon acidity (pKa of C-H bonds) is determined by the stability of the conjugate base carbanion. Greater stabilization of the carbanion by electron-withdrawing groups (EWGs) through resonance and/or induction leads to stronger carbon acids (lower pKa). Step 1: Identify the acidic C-H position in each compound. - (a) Norbornene-2,3-dione: The C-H bonds are on the ring carbons adjacent to the two carbonyls, but the bridged bicyclic system and the alkene between the two carbonyls limits enolization; moreover, there is no active methylene between two EWGs. - (b) Norbornanedione: Similar bicyclic diketone; the C-H alpha to both carbonyls is at C1 (bridgehead), but in norbornane systems Bredt's rule restricts enolization at bridgehead positions, making these extremely poor carbon acids despite having two flanking carbonyls. - (c) CH2(CO2Et)2 (diethyl malonate): The methylene (-CH2-) is flanked by two ester groups (-CO2Et). The carbanion formed is stabilized by resonance with both ester carbonyl groups. pKa ~ 13. - (d) CH3COCH2COOC2H5 (ethyl acetoacetate): The methylene (-CH2-) is flanked by one ketone (CH3CO-) and one ester (-COOC2H5). The carbanion is stabilized by resonance delocalization into both the ketone carbonyl and the ester carbonyl. pKa ~ 11. Step 2: Compare (c) and (d). - In diethyl malonate (c), the carbanion is stabilized by two ester groups. Esters are moderate EWGs. - In ethyl acetoacetate (d), the carbanion is stabilized by one ketone and one ester. A ketone carbonyl is a better electron-withdrawing/resonance-stabilizing group than an ester carbonyl because the ester oxygen donates electron density back into the carbonyl (reducing its electrophilicity/stabilization ability), whereas the ketone methyl group does not do this. Thus the ketone provides stronger stabilization of the adjacent carbanion than the ester does. - Therefore, ethyl acetoacetate (d) has a lower pKa (~11) compared to diethyl malonate (c, pKa ~13), making (d) a stronger carbon acid. Step 3: Why (a) and (b) are weaker. - Options (a) and (b) are cyclic diketones where the carbons alpha to both carbonyls are constrained in a bicyclic framework. Bredt's rule prevents effective enolization at bridgehead positions, drastically reducing carbon acidity. Even for non-bridgehead alpha positions, the rigidity of the system does not provide the same stabilization as the freely rotating active methylene in (c) or (d). Step 4: Conclusion. - Ethyl acetoacetate (d) has the most acidic methylene proton among the four options due to dual resonance stabilization by a ketone and an ester group, with the ketone being a stronger stabilizer than a second ester. Therefore, the correct answer is D.