Draw structures for all monoenol forms of the following molecule. Which would you expect to be the most stable? Explain.

The given molecule is cyclohexane-1,3-dione.

The monoenol forms of the given molecule are

Enol II and III are equivalent.

Enol IV and V are equivalent.

From these figures you can say that enol II and III are the most stable monoenol forms of the given compound as the carbonyl group takes part in conjugation with the double bond of the ring.

Cyclohexane-1,3-dione has two carbonyl groups. Using the Hydrogen atoms at the alpha carbons, these carbonyl groups participate in enolisation. After enolisation, it forms enol II and enol IV. In enol II and III, the carbonyl group is in conjugation with the double bond. But in enol IV and V, the carbonyl group is not conjugated. As you know that, conjugated systems are more stable than the nonconjugated systems. Hence, enol II and III are the most stable.

Enol II and III are most stable monoenols of the given molecule.