See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Resonance stabilization requires that the lone pair or charge can delocalize into an adjacent pi system (such as a C=O group) through orbital overlap. For this overlap to occur, the orbitals must be able to achieve the necessary geometry (ideally parallel p-orbitals). Step 1 - Analyze option (b): In cyclohexanone with a negative charge (carbanion) at the alpha carbon, the carbanion lone pair is adjacent to the C=O group. The p-orbital on the alpha carbon can overlap with the pi system of the carbonyl, giving resonance stabilization as an enolate. This is well-known resonance stabilization. Step 2 - Analyze option (a): The oxabicyclic compound has an oxygen with lone pairs adjacent to the carbonyl. The lone pairs on oxygen can delocalize into the C=O, providing resonance stabilization (similar to an ester or lactone-type resonance). Step 3 - Analyze option (d): The bicyclic system with a negative charge at the bridgehead adjacent to a double bond allows the carbanion to delocalize into the pi bond, providing resonance stabilization (allylic-type anion stabilization). Step 4 - Analyze option (c): This is the critical case. In the norbornane (bicyclo[2.2.1]) skeleton, the carbanion is located at the bridgehead carbon. Bredt's rule states that a bridgehead carbon in a small bicyclic system cannot bear a double bond (or effectively participate in pi overlap) because the geometry forces the bridgehead p-orbital to be perpendicular to the carbonyl pi system rather than parallel. The rigid bicyclic framework prevents the necessary orbital alignment for resonance delocalization between the bridgehead carbanion and the adjacent carbonyl group. Therefore, the carbanion at the bridgehead of norbornane adjacent to a ketone CANNOT be resonance stabilized - the p-orbital at the bridgehead is orthogonal to the C=O pi bond due to the rigid cage geometry (a violation of Bredt's rule situation). Why other options fail: Options (a), (b), and (d) all have geometries that allow proper orbital overlap for resonance delocalization. Option (c) is unique because the bicyclo[2.2.1] bridgehead geometry prevents the required orbital alignment. Therefore, the correct answer is C.