See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Enol stability is governed by (i) extended conjugation, (ii) intramolecular hydrogen bonding, and (iii) resonance stabilization (vinylogous or cross-conjugated systems). The most stable enol is the one that maximizes these stabilizing interactions. Step 2 - Structure of terric acid: Terric acid is 2,3-epoxy-5-methyl-1,4-cyclohexanedione. It has two ketone groups at C1 and C4 with an epoxide at C2-C3 and a methyl group at C5. The active methylene (CH between the two carbonyls at C5 bearing CH3) or the carbons alpha to each carbonyl can enolize. Step 3 - Evaluating option (c): In option (c), the enol OH is on the carbon adjacent to the upper carbonyl, and the C=C double bond forms between the two carbonyl-flanking carbons (i.e., between C5 bearing CH3 and the adjacent carbon). This enol places the OH between the two carbonyl groups, allowing the enol to be stabilized by: (a) conjugation with both flanking carbonyl groups simultaneously (vinylogous 1,3-dicarbonyl enol = very stable), and (b) intramolecular hydrogen bonding between the enol OH and one of the adjacent carbonyl oxygens. This is analogous to the highly stable enol of acetylacetone (2,4-pentanedione), where the enol content is ~80% due to chelation and conjugation. Step 4 - Why other options are less stable: - Option (a): The enol double bond is not positioned between the two carbonyl groups, so conjugation with both carbonyls simultaneously is not maximized; the OH is on a carbon that does not allow chelation with both carbonyls. - Option (b): The enol OH is on the carbon adjacent to the epoxide (lower position), and both carbonyls remain intact at the top. The C=C is not flanked by both carbonyls in a 1,3-relationship, reducing conjugation. No intramolecular H-bond chelation between both carbonyls and the OH. - Option (d): Similar to (b), the enol is at the lower position near the epoxide, the double bond is between carbons adjacent to the epoxide, and the arrangement does not allow simultaneous conjugation with both C=O groups or effective chelation. Step 5 - Conclusion: Option (c) represents the enol formed by enolization of the CH between the two carbonyl groups (or effectively placing the double bond between the two carbonyl carbons with OH on one of them), producing a fully conjugated, chelated enol stabilized by intramolecular hydrogen bonding and conjugation with both carbonyls - exactly analogous to the stable enol of a 1,3-diketone. Therefore, the correct answer is C.