In 1-phenylbutane-1,3-dione (1), which enol tautomer will be predominant?

Tautomerism in 1-phenylbutane-1,3-dione

Both carbonyl groups possess α-hydrogens, so can undergo keto-enol tautomerism. In particular, the carbonyl group on the right can enolise in two ways since there are two unique sets of α-hydrogens.

The answer is supposed to be enol 2. Why can't it be either 3 or 4?

  • 1
    $\begingroup$ You need to retain a conjugated carbonyl group which is more stable than two enols. [Acetone exists as a ketone with little enol.] The real question is why the given answer without considering the tautomeric form of the second structure. $\endgroup$
    – user55119
    Commented Sep 23, 2018 at 14:50
  • $\begingroup$ The question editor has modified the question, and maybe assumes that he has thought of every possible enol structure, but he didn't. $\endgroup$
    – mykhal
    Commented Sep 24, 2018 at 10:02
  • $\begingroup$ @mykhal Hmm, this is kind of two years old, but I merely copied the structures from the previous version of the question. It's in the revision history... Yes, there are missing enol forms as user55119 also suggests, but the original question did not have those either. $\endgroup$ Commented Dec 10, 2020 at 13:32

2 Answers 2


enter image description here

Two enol tautomers are possible for 1-phenylbutane-1,3-dione. Both are shown in the figure. Enol tautomer 1 is predominant due to more extended conjugation than Enol tautomer 2 which leads to more resonance structures give extra stability to 1.


enolic form

The compound is highly stabilised by higher Conjugation and is thus favored over the other enolic form. Another factor is intramolecular hydrogen bonding which stabilises the enolic form even more.

It can also be explained on the basis of acidity of H in keto from.

Keto form

Due to presence of 2 Electron withdrawing groups, the H atom is highly acidic and will favor the enolic form.

The reason (3) is not formed is because after the formation of (2), the H marked in the 2nd picture will not be acidic anymore, making the equillibrim highly towards (2) form. Also, formation of the allene leads to leads to loss of comjugation, since the pi orbitals are now perpendicular to each other, which makes the equillibrim even backwards (towards 2nd form).

  • $\begingroup$ I think this is not really answering the question. The diagrams in the question were not very clear, but I have edited it. There is a more fundamental reason why the allene 3 does not form, namely the loss of conjugation (as the pi orbitals are orthogonal). And you also don't explain why the terminal methyl group doesn't enolise, e.g. in 4. To me, it feels like you are just explaining why 1,3-dicarbonyls tend to enolise, without explicitly considering the alternative structures which OP asked about. $\endgroup$ Commented Sep 23, 2018 at 21:25
  • $\begingroup$ @orthocresol I have added explanation for (3). If the answer still doesn't answer the question, please inform me and i will delete it because i cannot answer the question any further than that. $\endgroup$
    – Groverkss
    Commented Sep 24, 2018 at 4:18

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