See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the molecule: The structure shown is a cross-conjugated/extended system based on a cyclohexadiene ring bearing an exocyclic methylene (=CH2). The exocyclic methylene makes this portion resemble a ketone enol or a vinylogous system. In context, the exocyclic =CH2 is the enol/enolizable part derived from a carbonyl (the molecule is effectively a vinylogous carbonyl compound or the exocyclic methylene represents the carbonyl carbon equivalent in an enolization sense). Step 2 - Concept of vinylogous enolization: In a cross-conjugated or extended conjugated system, enolization does not have to occur at the carbon directly adjacent (alpha) to the carbonyl. Through the pi system, the gamma-hydrogen (the one at the gamma position relative to the carbonyl equivalent) can be removed because the resulting carbanion/enolate is stabilized by extended conjugation through the diene system. Step 3 - Assign positions: The exocyclic =CH2 carbon is the carbonyl carbon equivalent (C1). Moving through the conjugated diene system: the alpha carbon bears Ha, the beta carbon bears Hb, and the gamma carbon bears Hy. The gamma position is three carbons away from the carbonyl carbon through the conjugated pi system. Step 4 - Why gamma-H (Hy) is the one involved: Removal of Hy at the gamma position generates an extended enolate/dienolate that is fully conjugated through the diene system back to the exocyclic methylene. This is the most stabilized anion due to full delocalization across the conjugated framework (vinylogous enolization). The gamma-H is therefore the kinetically and thermodynamically preferred proton for enolization in such a cross-conjugated diene system. Step 5 - Why other options fail: - (a) alpha-H: Removal of Ha would give a simple localized enolate with less stabilization. - (b) beta-H: The beta carbon in a conjugated diene is a vinylic-type position; removal here would disrupt conjugation rather than extend it, and is not favorable for enolization. - (d) cannot be enolized: Incorrect, as the molecule clearly has an acidic gamma-H that can be removed to form a stable dienolate. Therefore, the correct answer is C.