See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The acidity of C-H bonds is determined by the stability of the carbanion formed upon deprotonation, which is enhanced by resonance delocalization into adjacent electron-withdrawing groups (carbonyl groups). Step 1 – Identify the environment of each labeled carbon: - Carbon 1: The methyl (CH3) carbon of the ketone portion (the terminal CH3 of the acetyl group). It is alpha to ONE carbonyl (the ketone C=O). Its conjugate base (carbanion) is stabilized by resonance into one carbonyl. - Carbon 2: The alpha carbon between BOTH the ketone carbonyl and the ester carbonyl (the active methylene, CH2). It is alpha to TWO carbonyls simultaneously. Its conjugate base is stabilized by resonance into both carbonyls, making it much more acidic. This is the classic beta-ketoester active methylene with pKa ~11. - Carbon 3: The CH2 of the ethyl group on the ester oxygen (-O-CH2-CH3). This carbon is on the oxygen side of the ester; it is NOT alpha to a carbonyl. It is simply an alkyl C-H with no adjacent stabilization. Very weakly acidic. Step 2 – Rank by acidity (most acidic = lowest pKa): - Carbon 2: flanked by two carbonyls → highest acidity (pKa ~11) - Carbon 1: alpha to one carbonyl (ketone) → moderate acidity (pKa ~20) - Carbon 3: no adjacent carbonyl, just an ether-like alkyl C-H → least acidic (pKa ~50) So the order from most to least acidic is: 2 > 1 > 3. Step 3 – Why other options fail: - (b) 1 > 2 > 3: Incorrect because C2 (between two carbonyls) is far more acidic than C1 (alpha to only one carbonyl). - (c) 2 > 3 > 1: Incorrect because C3 (ethyl ester O-CH2) has no adjacent carbonyl stabilization and is much less acidic than C1. - (d) 3 > 2 > 1: Incorrect for the same reasons; C3 is the least acidic, not the most. Therefore, the correct answer is A.