Química Orgânica 2 · Condensação de Claisen e adição conjugada
Crossed Claisen condensation — ester alpha-formylation
Condições: Ethyl acetate and ethyl formate with sodium ethoxide in ethanol, then a separate workup with dilute aqueous HCl (drawn as hydronium). Follow the displayed alpha-formylation branch. Counterions are omitted throughout. Display the neutral keto tautomer without predicting its equilibrium abundance.
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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.
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Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 6 passos escritos
Remove an alpha hydrogen
Follow the specified donor site. The carbon-centered enolate is a resonance contributor, not a claim of a localized free carbanion. Initial enolate formation need not be quantitative; later product trapping drives this sequence. An ester α C–H (pKa ≈ 25) is far less acidic than ethanol (pKa ≈ 16), so ethoxide keeps only a tiny enolate fraction (K ≈ 10⁻⁹) at any moment. Ethoxide matches the ethyl esters, so any acyl substitution by the base simply regenerates the same ester.
Nota: The two drawn alpha hydrogens are equivalent.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Connect the donor and acceptor
The donor carbonyl stays intact. C–C bond formation occurs at the electrophilic acyl carbon, not at an alkyl group on an ester oxygen.
As setas curvas
- The nucleophile uses a lone pair to form the new σ bond.
- The C=O π electrons move to oxygen so carbon keeps eight electrons.
Resolve the acyl-addition intermediate
Carbonyl restoration expels ethoxide. The beta-dicarbonyl product still has a hydrogen between the two carbonyls, allowing deprotonation to drive the reaction forward.
As setas curvas
- An oxygen lone pair forms the C=O π bond again.
- The leaving group takes the electrons from the bond that breaks.
Remove an alpha hydrogen
The hydrogen between the two carbonyl groups is more acidic than the original donor alpha hydrogen. Its removal traps the condensation product as a stabilized anion and protonates the leaving alkoxide. With pKa about 11 or lower for this β-dicarbonyl C–H against 16 for ethanol, the deprotonation is favorable by roughly 10⁵ and pulls the reversible earlier steps toward product; that is why a full equivalent of alkoxide is used.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Separate acidic workup
Only now is hydronium introduced for a separate workup. Protonate the carbon-centered contributor to display the neutral keto tautomer; subsequent tautomer ratios are not part of the assignment. Hydronium (pKa ≈ −1.7) is a far stronger acid than the β-dicarbonyl product (pKa about 11), so this protonation goes to completion.
Nota: Hydronium is now introduced for the workup.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Products
One new C–C bond and a restored acceptor carbonyl create the beta-dicarbonyl framework. The two original donor hydrogens moved to two ethanol molecules; the final alpha hydrogen comes from workup. The neutral keto tautomer is displayed without claiming it is the only solution tautomer. All counterions are consistently outside the reacting model.
O que você vai praticar
Use an ester with a carbonyl-bound hydrogen as acceptor and preserve that hydrogen through acyl transfer and product trapping.
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