See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The starting material is 1-*C-cyclopenta-2,4-diene (cyclopentadiene labeled with C-14 at C1, the sp3 carbon bearing two hydrogens). KH is a strong base that removes a proton (-H2 means loss of H2 in the sense of deprotonation releasing H2 gas or simply deprotonation). KH deprotonates cyclopentadiene at C1 (the sp3 CH2) to form the cyclopentadienyl anion (A), which is aromatic (6 pi electrons counting the lone pair). The C-14 label (*C) is at C1. Step 2 - Resonance delocalization in cyclopentadienyl anion: The cyclopentadienyl anion is fully delocalized and aromatic. The negative charge (and the labeled carbon C-14 at C1) participates in resonance across all five carbons. The anion has five equivalent resonance structures where the carbanion character is delocalized over C1, C2, C3, C4, and C5. Step 3 - Protonation with H2O: When A (cyclopentadienyl anion) is treated with H2O, protonation occurs. The proton can be delivered to any of the carbon positions that bear negative charge in the resonance structures. Due to the symmetry of the delocalized anion, protonation can occur at C1 (regenerating starting material with * at C1), or at C2/C5 (equivalent positions adjacent to C1), or at C3/C4 (equivalent positions). However, the major product consideration involves which protonation gives a stable diene product. Step 4 - Determining major products: When the cyclopentadienyl anion (with *C at C1) is protonated at C1, we get back the original cyclopentadiene with * at C1 (option a). When protonation occurs at C2 or C5 (the carbons adjacent to *C), the *C ends up as one of the sp2 carbons of the diene system, giving option (b) (* at C5 position) or option (c) (* at C2/bottom position). Since C2 and C5 are equivalent by the symmetry of the molecule (both adjacent to *C1), protonation at these positions gives two products that correspond to options (b) and (c). These represent the positions where protonation from water on the delocalized anion places the double bond system such that *C is now an internal sp2 carbon. Step 5 - Why (b) and (c) are major and not (a): In the cyclopentadienyl anion, the charge is delocalized. Kinetic protonation by water (a weak proton donor) tends to occur at the more nucleophilic carbons. Statistical and electronic factors in the symmetric cyclopentadienyl anion mean protonation at C2 and C5 (adjacent to *C) gives products (b) and (c) as the major products. Option (a) would require re-protonation at C1 (the original sp3 carbon), which is less favored because the resonance contributors place more electron density at the other positions, and the conjugated diene products from protonation at C2/C5 are thermodynamically stable. Both (b) and (c) are formed in equal amounts due to symmetry. Step 6 - Why option (a) fails as major product: Protonation back at C1 regenerates starting material but is not the dominant pathway under these conditions. Therefore, the correct answer is D.