Highest enol content will be shown by:

Solution

Chemistry ma’am · 4 steps

Ask which compound becomes a benzene ring on enolising: aromaticity pulls it almost fully to the enol.

1. What drives enolisation. The keto form is normally favoured because C=O is stronger than C=C. The enol wins only when it gains extra stability from conjugation or, best of all, from aromaticity.
2. Look for active methylene. A CH₂ sitting between two C=O groups has very acidic hydrogens and enolises easily. Count such groups in each ring and see what the fully enolised form looks like.

   Compound	CH₂ between two C=O	Enol form	Enol content
   (a) cyclohexane-1,3-dione	one (C-2)	conjugated enone, not aromatic	high
   (b) cyclohexane-1,3,5-trione	all three	benzene-1,3,5-triol (phloroglucinol), aromatic	highest, almost 100%
   (c) cyclohexane-1,4-dione	none	isolated enol, no conjugation	low
   (d) cyclohexane-1,2,3-trione	none	conjugated enol, but direct enolisation does not give an aromatic ring	moderate

3. Why (b) wins. In (b) each CH₂ gives one H to a neighbouring oxygen. The ring then has three C=C bonds in a six-membered ring, a complete aromatic sextet. That stability far outweighs losing three C=O bonds.
4. So. Cyclohexane-1,3,5-trione, option (b), shows the highest enol content because its enol is the aromatic phenol phloroglucinol.

The idea to keep. For enol content, check two things: a CH₂ flanked by two C=O groups, and whether the enol becomes aromatic. An aromatic enol beats every other factor.

Answer from the NTA final answer key. Solution by PrepWiser.