See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Keto-enol tautomerism favors the enol form when the enol is stabilized (e.g., by conjugation, aromaticity, intramolecular hydrogen bonding, or relief of strain). The keto form dominates when enolization is disfavored or geometrically impossible. Step 1 – Option (a): Cyclobutane-1,3-dione. This is a 4-membered ring 1,3-diketone. Normally, 1,3-diketones strongly favor the enol form due to intramolecular hydrogen bonding and conjugation. However, in cyclobutane-1,3-dione, the ring geometry forces the two carbonyl groups to be at a 1,3 relationship in a 4-membered ring. The enol form would require a planar arrangement with a strained 4-membered ring enol; actually the enol here would give a vinylogous system but the small ring angle strain and the fact that the enol double bond in a 4-membered ring is highly strained means the keto form is preferred. The ring strain in the enol form of cyclobutenediol (the enol equivalent) makes the keto form dominant. Step 2 – Option (b): 5,5-dimethylcyclohexane-1,3-dione (dimedone). This is a classic 1,3-diketone. Normally dimedone exists predominantly in the enol form (>95%) in solution because the enol is stabilized by intramolecular hydrogen bonding and extended conjugation. So the enol form dominates here, meaning the keto form does NOT dominate. Wait — but the answer is D (all of these). Let me reconsider. The structure shown has gem-dimethyl at C2 (the carbon between the two carbonyls), which blocks enolization at that carbon. However, enolization can still occur at C4 or C6. Actually with gem-dimethyl substitution at C5 (between carbonyls would be C2), enolization toward the ring double bond is still possible. Re-examining: the answer given is D, so all three must have keto form dominant. Step 3 – Re-evaluating (b): The gem-dimethyl group at the carbon between the two carbonyls (C2) removes both alpha-hydrogens at that position, so enolization at C2 is impossible. The molecule can still enolize at other positions, but the typical stabilized enol of a 1,3-diketone requires H at the carbon between the two C=O groups. Since both hydrogens at C2 are replaced by methyl groups, the classic intramolecularly hydrogen-bonded enol cannot form. Therefore the keto form dominates. Step 4 – Option (c): Bicyclo[2.2.1]heptane-2,3-dione (norbornane-2,3-dione, a vicinal diketone on a rigid bicyclic framework). For the enol form to be favored, the resulting enol double bond would need to be accommodated by the rigid bridged bicyclic ring system. The bridgehead geometry and Bredt's rule considerations make it geometrically very difficult to form a stable enol with a double bond at the bridgehead or in the strained bicyclic system. Additionally, vicinal diketones (1,2-diketones) generally exist predominantly in keto form anyway. The rigid norbornane skeleton prevents the planar geometry needed for enol stabilization. Therefore the keto form dominates. Step 5 – Summary: (a) keto dominates due to ring strain preventing stable enol in 4-membered ring; (b) keto dominates because gem-dimethyl substitution at the carbon between the carbonyls blocks formation of the stabilized intramolecularly H-bonded enol; (c) keto dominates because the rigid bridged bicyclic framework (Bredt's rule) prevents formation of a stable enol double bond. All three cases favor the keto form. Therefore, the correct answer is D.