See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Aromaticity and acidity of cyclopentadiene. Step 1: Identify the structure. The molecule is cyclopenta-2,4-diene (cyclopentadiene). It has a five-membered ring with two double bonds (at C2-C3 and C4-C5) and one sp3 carbon at C1 bearing two hydrogens. The arrow points to one of the sp3 C-H hydrogens at C1. Step 2: Consider what happens when the indicated H is removed as H+. Removal of H+ (a proton) from C1 leaves behind a carbanion (C1 bears a lone pair/negative charge). This carbanion at C1 gives the cyclopentadienyl anion, which has 6 pi electrons (2 from the lone pair/carbanion + 4 from the two double bonds) delocalized over all 5 carbons in the ring. Step 3: Apply Huckel's rule. The cyclopentadienyl anion has 6 pi electrons (4n+2 where n=1), making it aromatic. Aromaticity provides exceptional stabilization to the anion, making the loss of H+ very favorable. The pKa of cyclopentadiene is approximately 16, which is remarkably acidic for a hydrocarbon. Step 4: Evaluate other options. Removal as H- (hydride, option b) would give a carbocation, which is a high-energy antiaromatic (4 pi electrons) cyclopentadienyl cation - very unfavorable. Removal as H radical (option c) gives a radical species, which is not particularly stabilized by aromaticity. Option d (H^-2) is chemically nonsensical. Step 5: Conclusion. The thermodynamic driving force of aromaticity in the resulting cyclopentadienyl anion makes removal of the hydrogen as H+ (a proton) the easiest and most favorable pathway. Therefore, the correct answer is A.