See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

**Analysis of Enol Form Stability:** Enol forms exist when a compound has an alpha-hydrogen adjacent to a carbonyl group, allowing keto-enol tautomerism. The enol is stabilized by resonance and hydrogen bonding. **Examining each compound:** **(A) 1,4-Benzoquinone ($p$-quinone):** - Has two carbonyl groups in a conjugated aromatic ring - No alpha-hydrogens present (carbons adjacent to $C=O$ are part of the aromatic ring) - Cannot form enol tautomer ✗ **(B) 1,4-Benzimidazoquinone (4-quinone imine):** - Similar structure to (A) but with $NH$ instead of one $C=O$ - Still part of aromatic system with no alpha-hydrogens available - No alpha-H to generate enol form ✗ **(C) $CH_3-C(=O)-CH_2-COOC_2H_5$:** - The methyl ketone has alpha-hydrogens on the adjacent $CH_2$ group - These hydrogens can tautomerize to form enol: $CH_3-C(=OH)-CH=COOC_2H_5$ - Can exist in enol form ✓ **Conclusion:** Both (A) and (B) lack alpha-hydrogens due to their aromatic conjugated structures, making enol formation impossible. Compound (C) possesses alpha-hydrogens and readily exists in enol form. **Answer: (D) Both (A) and (B)** — neither can exist in enol form because they lack alpha-hydrogens necessary for keto-enol tautomerism.