Which of the following compounds will show the maximum 'enol' content?

The phenomenon we are examining is keto-enol tautomerism, which is the equilibrium

$$\text{keto form}\; \rightleftharpoons \;\text{enol form}$$

The percentage of the enol form present at equilibrium depends mainly on two factors.

1. Acidity of the $$\alpha$$-hydrogen: The more acidic this hydrogen is, the more easily the enolate ion is formed, and the greater is the conversion to the enol after protonation.

2. Stabilization of the enol formed: If the enol can be stabilized by conjugation and / or by intramolecular hydrogen bonding (chelation), its concentration rises.

We now analyse each option one by one, keeping both points in mind.

Option D: $$\text{CH}_3\text{COCH}_3$$ (acetone)

This molecule has only one carbonyl group. The $$\alpha$$-hydrogens (those on the carbon between the carbonyl carbon and the methyl group) have a pKa of about $$19$$. Such hydrogens are not very acidic, so very little enolate—and therefore very little enol—is produced. Moreover, any enol that does form has no special stabilization beyond ordinary C=C-O conjugation. Hence the enol content is extremely small, usually < 0.1 %.

Option A: $$\text{CH}_3\text{COCH}_2\text{CONH}_2$$

This is a $$1,3$$-dicarbonyl system in which the second carbonyl is an amide. The central $$\alpha$$-hydrogen lies between a ketone carbonyl and an amide carbonyl. Because two electron-withdrawing groups flank it, its pKa drops to roughly $$13$$, so deprotonation is easier than in acetone.

However, when the enol form is drawn

$$\text{CH}_3\text{CO-CH}=\text{CONH}_2 \quad\text{with}\quad \text{OH on the left carbonyl}$$

the amide carbonyl is a poor hydrogen-bond acceptor; its oxygen is less basic because it is involved in resonance with the nitrogen (-C=O↔-C-O--N+). As a result, the expected six-membered intramolecular O-H···O hydrogen bond is not very strong. Conjugation is still present, but the reduced hydrogen-bond energy lowers stabilization. Hence the enol content is higher than acetone’s but still moderate.

Option C: $$\text{CH}_3\text{COCH}_2\text{COOC}_2\text{H}_5$$ (a $$\beta$$-keto ester)

Here the central hydrogen is between a ketone carbonyl and an ester carbonyl. Its pKa is about $$11$$, making it even more acidic than in Option A, so enolate formation is easier.

The enol obtained can be represented as

$$\text{CH}_3\text{C(OH)=CH-COOC}_2\text{H}_5$$

and this enol can indeed form a six-membered ring via intramolecular hydrogen bonding between the newly formed -OH and the ester carbonyl oxygen. Conjugation extends over the C=C-C=O system. Nevertheless, an ester carbonyl is less basic than a ketone carbonyl because the -OR group donates electron density by resonance: $$-C(=O)OR \leftrightarrow -C(-O^{-})=O-R^{+}$$. Consequently the hydrogen bond is weaker than it is with a ketone carbonyl, so the net stabilization—though appreciable—is not maximal.

Option B: $$\text{CH}_3\text{COCH}_2\text{COCH}_3$$ (acetylacetone)

This is a true $$\beta$$-diketone. Both flanking groups are ketone carbonyls, each strongly electron withdrawing. Hence the pKa of the central hydrogen is around $$9$$, the smallest among all the given molecules, so the enolate forms most readily.

The corresponding enol is

$$\text{CH}_3\text{C(OH)=CH-COCH}_3$$

In this structure we have two powerful stabilizing effects simultaneously:

(i) Conjugation: The C=C is directly conjugated with a carbonyl group, giving the usual enol resonance

$$\text{C(OH)=C-C=O} \;\;\rightleftharpoons\;\; \mathrm{C(O}^{-}\mathrm{)-C}^+\mathrm{-C=O}$$

(ii) Intramolecular hydrogen bonding (chelation): The enolic -OH can donate a hydrogen to the oxygen of the second ketone carbonyl, creating a six-membered chelate ring. A ketone carbonyl oxygen is more basic than an ester or amide oxygen, so this hydrogen bond is very strong. The extra stabilization may be depicted as

$$\text{C(OH)…O=C}$$

The combined resonance and hydrogen-bond energies make the enol of a $$\beta$$-diketone so stable that, in solvents like benzene, the equilibrium mixture is as high as $$70$$-$$80\,\%$$ enol.

We now compare the options quantitatively in the order of the two criteria discussed:

$$\text{Enol content} \propto \dfrac{\text{ease of enolate formation}}{\text{pK}_a} \times \text{stabilization of the enol}$$

Option B: lowest pKa (≈9) & strongest chelation ⇒ highest enol content.

Option C: slightly higher pKa (≈11) & somewhat weaker chelation ⇒ less than B.

Option A: pKa (≈13) & poor chelation ⇒ still lower.

Option D: pKa (≈19) & no special stabilization ⇒ the least.

Putting all observations together, the $$\beta$$-diketone $$\text{CH}_3\text{COCH}_2\text{COCH}_3$$ in Option B possesses the maximum enol content.

Hence, the correct answer is Option B.