Química Orgânica 2 · Enóis, enolatos e reações em alfa
Deprotonate between two carbonyls
Condições: Pentane-2,4-dione with one equivalent of sodium ethoxide in ethanol. Remove a hydrogen from the central carbon, then draw an enolate contributor with charge on one carbonyl oxygen and, in one move, the contributor with charge on the other. Na⁺ is a spectator; enol tautomers are not considered.
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Arraste dos elétrons que se movem até onde eles vão, ou toque em um e depois no outro. Cada seta é verificada; Dica ajuda quando você travar. Não precisa de conta.
O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 4 passos escritos
Remove the central proton
Pentane-2,4-dione (pKa approximately 9) is far more acidic than ethanol (pKa approximately 16), so ethoxide deprotonates it essentially completely (K roughly 10⁷). The terminal CH₃ hydrogens are much less acidic, because only one carbonyl can stabilize the anion they would leave.
As setas curvas
- An ethoxide lone pair accepts a proton from the carbon between the two carbonyls.
- The central C–H bond electrons stay on carbon as a lone pair.
Move the charge onto a carbonyl oxygen
This is the usual enolate pair: the carbon lone pair and the carbonyl π pair trade places.
As setas curvas
- The central carbon lone pair forms a π bond to a carbonyl carbon.
- That carbonyl π pair becomes a lone pair on oxygen.
Move the charge to the other oxygen
Three arrows carry the charge across the whole O–C–C–C–O system in one redrawing. Every carbon keeps four bonds throughout.
As setas curvas
- A lone pair on the negatively charged oxygen rebuilds its carbonyl π bond.
- The C=C π pair forms a π bond between the central carbon and the other carbonyl carbon.
- The remaining carbonyl π pair becomes a lone pair on its oxygen.
One charge on the central carbon and both oxygens
The three contributors side by side (ethanol and Na⁺ are left out of this summary): charge on the central carbon or on either oxygen. The two oxygen contributors are the major ones, and they are equivalent, so the real ion is symmetric with its charge mostly on the two oxygens. This extra delocalization is why pentane-2,4-dione (pKa approximately 9) is about ten powers of ten more acidic than acetone (pKa approximately 19 to 20).
O que você vai praticar
Explain the high acidity of a 1,3-dicarbonyl compound by generating all three contributors of its conjugate base.
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