See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Acidity of Cyclopentadiene **Step 1: Identify the $sp^3$ carbon in cyclopentadiene** Cyclopentadiene has a methylene group ($-CH_2-$) flanked by two $sp^2$ carbons of the double bonds. The hydrogens on this $sp^3$ carbon are unusually acidic. **Step 2: Understand conjugate base stabilization** When a **proton** ($H^+$) is removed from the $sp^3$ carbon, the resulting anion (cyclopentadienyl anion, $C_5H_5^-$) forms a fully conjugated, aromatic system with 6π electrons (4n+2, n=1), making it exceptionally stable. **Step 3: Why other options are incorrect** - **(1) Hydride ion ($H^-$)**: Would require C–H bond cleavage and loss of a hydride; not favorable - **(2) Hydrogen molecule ($H_2$)**: Requires two separate C–H bond cleavages and H–H bond formation; mechanistically implausible - **(4) Hydrogen atom ($H \cdot$)**: Would form a radical; cyclopentadiene acts as a Brønsted acid, not a radical donor **Step 4: Confirmation** Cyclopentadiene is famously acidic ($pK_a \approx 16$) because removal of a **proton** generates the aromatic cyclopentadienyl anion—a classic example of Brønsted acidity enhanced by resonance stabilization. **Answer: (3) Proton** — because it's removed as $H^+$ in a Brønsted acid–base mechanism.