See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

**Part A – Strongest Bronsted Acid:** Concept: A Bronsted acid donates a proton; acidity of an N–H bond increases when the resulting conjugate base (nitrogen anion) is better stabilized by resonance or inductive effects. Compound I (piperidine): Simple secondary amine, N–H with no adjacent electron-withdrawing groups. Conjugate base is poorly stabilized. Very weakly acidic (pKa ~35). Compound II (2-piperidinone / delta-lactam): N–H flanked by one carbonyl. The lone pair on N can be delocalized into one C=O (amide resonance), stabilizing the conjugate base to some extent. More acidic than I (pKa ~25 range). Compound III (delta-valerolactone): Contains no N–H; the acidic proton would have to come from a C–H. Not relevant for N–H acidity comparison; it is not a Bronsted acid in the N–H sense. Compound IV (glutarimide): N–H flanked by TWO carbonyls. The conjugate base (nitrogen anion) is stabilized by resonance delocalization into both carbonyl groups simultaneously, providing far greater stabilization than in compound II. This makes glutarimide significantly more acidic (pKa ~10 range, similar to other imides like succinimide pKa ~9.6). Reasoning: The two flanking carbonyls in IV provide double resonance stabilization of the N-anion, making IV the strongest Bronsted acid among N–H containing compounds. III lacks N–H entirely and is not competing here. Why other options fail: I has no EWG stabilization; II has only one carbonyl; III has no N–H; IV has two carbonyls giving maximum N–H acidity. Answer A: (d) IV **Part B – Strongest Lewis Base:** Concept: A Lewis base donates an electron pair. For nitrogen-containing compounds, basicity depends on the availability of the nitrogen lone pair. Electron-withdrawing groups (carbonyls) reduce lone pair availability via resonance and induction. Compound I (piperidine): The nitrogen lone pair is fully available (not delocalized into any carbonyl). This is a standard aliphatic amine, highly basic (pKa of conjugate acid ~11). Compound II (2-piperidinone): One carbonyl withdraws electron density from N via amide resonance, reducing lone pair availability. Less basic than I. Compound III (delta-valerolactone): No nitrogen; the oxygen in the lactone ring is the only potential Lewis base, but ester oxygens are very weak bases. Not competitive with amine nitrogen. Compound IV (glutarimide): Two carbonyls withdraw electron density from N extensively. The lone pair is heavily delocalized; nitrogen is essentially non-basic as a Lewis base. Reasoning: Compound I has no electron-withdrawing groups on nitrogen; its lone pair is maximally available for donation, making it the strongest Lewis base. Why other options fail: II, III, and IV all have reduced electron density at the basic site due to one or two carbonyls or absence of nitrogen altogether. Answer B: (a) I Therefore, the correct answer is {"A": "d", "B": "a"}.