See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Acidity of a compound is determined by the stability of its conjugate base. The more stable the conjugate base, the stronger (more acidic) the acid. Step 1: Identify the acidic proton in cyclopentadiene. Cyclopentadiene has a sp3 CH2 group (C5) flanked by two double bonds. The proton on this sp3 carbon is the one that is lost upon deprotonation. Step 2: Analyze the conjugate base. When cyclopentadiene loses a proton from the sp3 carbon, it forms the cyclopentadienide anion (C5H5-). This carbanion has 6 pi electrons (4 from the two double bonds plus 2 from the lone pair on the carbanion carbon, all delocalized over the 5-carbon ring). Step 3: Apply Huckel's rule. The cyclopentadienide anion is a planar, cyclic, fully conjugated system with 6 pi electrons (4n+2 where n=1). This satisfies Huckel's rule for aromaticity, making the cyclopentadienide ion aromatic and exceptionally stable. Step 4: Compare with cyclopentane. Deprotonation of cyclopentane gives a simple carbanion with no resonance stabilization or aromatic stabilization, making it far less stable. Hence cyclopentane is a much weaker acid. Step 5: Evaluate why other options fail. - Option (a): Conjugated double bonds alone do not explain exceptional acidity; many conjugated dienes are not particularly acidic. - Option (b): Having both sp2 and sp3 carbons is a structural observation but does not by itself explain the high acidity. - Option (c): Cyclopentadiene is not entirely strain-free, and strain relief is not the primary reason for acidity; this option is irrelevant to acidity. - Option (d): Correctly identifies that the conjugate base (cyclopentadienide ion) is aromatic (6 pi electrons, Huckel's rule satisfied) and therefore highly stable, which is the fundamental reason for the high acidity of cyclopentadiene. Therefore, the correct answer is D.