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What is the predominant enol tautomer of 1-phenylbutane-1,3-dione?
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Milan Woodson
What is the predominant enol tautomer of 1-phenylbutane-1,3-dione?
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.
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.
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.
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).
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.
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).
. 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.More
@orthocresol I have added explanation for (3). If the answer still doesnt answer the question, please inform me and i will delete it because i cannot answer the question any further than that.More
does not form, namely the loss of conjugation (as the pi orbitals are orthogonal). And you also dont explain why the terminal methyl group doesnt enolise, e.g. in More
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 More
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.
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.
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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.
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).
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.
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).
More
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