Idea involved: keto–enol tautomerism

A carbonyl compound shows enol form when an $\alpha$-hydrogen shifts and the carbonyl oxygen gets protonated: $\text{keto} \rightleftharpoons \text{enol}$

The enol content is maximum when the enol form is especially stabilized.

Main factors stabilizing the enol form

Enol content increases when the enol is stabilized by:

1. Conjugation
2. Intramolecular hydrogen bonding
3. Presence of two electron-withdrawing groups on the carbon bearing the enolizable hydrogen

In particular, $1,3$-dicarbonyl compounds often have very high enol content because the enol is stabilized by both conjugation and a six-membered intramolecular H-bond.

Examine each option

Option A: $CH_3COCH_3$ (acetone)

This is a simple ketone.

• It can form an enol.
• But the enol is not strongly stabilized.
• So enol content is low.

Option B: $CH_3COCH_2COCH_3$ (acetylacetone)

This is a $1,3$-diketone. Its enol form is: $CH_3COCH=C(OH)CH_3$ This enol is highly stabilized by:

• Conjugation over the $C=C-C=O$ system
• Strong intramolecular hydrogen bonding between $OH$ and the other carbonyl oxygen

Hence this compound has very high enol content.

Option C: $CH_3COCH_2COOC_2H_5$ (a $\beta$-keto ester)

This is also a $1,3$-dicarbonyl compound. It can form a stabilized enol and shows appreciable enol content. However, compared with a $1,3$-diketone, the ester group stabilizes the enol less effectively than a ketone. So its enol content is less than option B.

Option D: $CH_3COCH_2CONH_2$ (a $\beta$-keto amide)

This is also a $1,3$-dicarbonyl-type system, but the amide carbonyl is strongly resonance-stabilized: $-CONH_2$ Because of amide resonance, it participates less effectively in enol stabilization than a ketone. Hence enol content is lower than in the diketone.

Comparison

General trend among these: $\beta\text{-diketone} > \beta\text{-keto ester} > \beta\text{-keto amide} > \text{simple ketone}$

Therefore, the compound with maximum enol content is: $CH_3COCH_2COCH_3$

Final answer

Option B is correct.