See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The energy of non-bonding (lone pair) electrons on nitrogen depends on how much the lone pair is stabilized (lower energy) or destabilized (higher energy). Lone pairs that are delocalized into pi systems (conjugation) or are held in lower-energy orbitals are at lower energy. Lone pairs in higher-energy orbitals or those not stabilized by resonance are at higher energy. In terms of basicity/nucleophilicity: more stabilized lone pairs = lower energy = less basic; less stabilized lone pairs = higher energy = more basic. Step 2 - Analyze Nitrogen C (amide nitrogen, lactam): Nitrogen C is directly bonded to a carbonyl group (C=O) in a lactam. Its lone pair is delocalized into the carbonyl pi system via resonance (N-C=O <-> N+=C-O-). This resonance donation into the electron-withdrawing carbonyl significantly stabilizes (lowers the energy of) the lone pair. This makes the lone pair on C the lowest energy (least available, least basic) among the three. Step 3 - Analyze Nitrogen B (bridgehead/vinylogous amine adjacent to aromatic ring): Nitrogen B is at a bridgehead position with an N-methyl group and is adjacent to the aromatic benzene ring. Its lone pair can participate in conjugation with the aromatic pi system (enamine/amine conjugation into the ring). This partial delocalization into the aromatic system stabilizes the lone pair, but less so than the full amide resonance of nitrogen C. Therefore, nitrogen B has a lone pair of intermediate energy: higher energy than C but lower energy than A. Step 4 - Analyze Nitrogen A (sp3, saturated, non-conjugated amine): Nitrogen A is a fully saturated sp3 nitrogen in a pyrrolidine-type ring with two methyl groups and no adjacent pi system. Its lone pair cannot be delocalized by resonance into any pi system. It resides in a pure sp3 orbital and is the most available (highest energy, most basic) lone pair of the three. Step 5 - Ranking in order of increasing energy: C (amide, most stabilized by resonance with C=O, lowest energy) < B (adjacent to aromatic ring, partially conjugated, intermediate energy) < A (sp3 saturated amine, no conjugation, highest energy). Step 6 - Why other orders fail: If A were lower than B, that would imply sp3 amine lone pairs are more stabilized than those conjugated into an aromatic ring, which is incorrect. If B were lower than C, that would imply aromatic conjugation stabilizes a lone pair more than amide resonance with a carbonyl, which is also incorrect since amide resonance is stronger due to the highly electron-withdrawing carbonyl. Therefore, the correct answer is C<B<A.