Química Orgânica 2 · Condensação de Claisen e adição conjugada
3-Phenylpropanoic acid by malonic ester synthesis
Condições: NaOEt, EtOH; then H3O⁺, Δ
Desenhe este problema você mesmo
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 19 passos escritos
Form the malonate enolate
Ethoxide removes one of the doubly activated hydrogens of diethyl malonate (pKa ≈ 13).
As setas curvas
- Ethoxide's lone pair takes the acidic hydrogen.
- The C–H pair stays on carbon, delocalized onto both esters.
SN2 alkylation with benzyl bromide
The enolate carbon attacks the benzylic CH2 of benzyl bromide from the back side, displacing bromide.
As setas curvas
- The enolate carbon forms the new C–C bond.
- The C–Br bond breaks and bromide leaves.
What will leave as CO2?
Planning the malonic ester synthesis.
Protonate the carbonyl of the first ester
Hydronium protonates the carbonyl oxygen of the first ester, making its carbon much more electrophilic toward weakly nucleophilic water.
As setas curvas
- A carbonyl oxygen lone pair takes a proton from hydronium.
- The O–H pair returns to oxygen, releasing water.
Water adds to the carbonyl carbon
Water attacks the activated carbonyl carbon, giving a tetrahedral intermediate with a positively charged oxygen.
As setas curvas
- A water lone pair forms the new C–O bond.
- The C=O π pair moves onto the protonated oxygen, neutralizing it.
Deprotonate the added water
Water removes a proton from the oxonium oxygen, giving a neutral tetrahedral intermediate.
As setas curvas
- Water's lone pair takes the oxonium proton.
- The O–H pair stays on oxygen.
Protonate the ethoxy oxygen
Hydronium protonates the ethoxy oxygen, turning it into a good leaving group (neutral ethanol).
As setas curvas
- The ethoxy oxygen lone pair takes a proton from hydronium.
- The O–H pair returns to oxygen as water.
Expel ethanol
An OH lone pair re-forms the C=O bond and pushes out ethanol.
As setas curvas
- An oxygen lone pair forms the C=O π bond.
- The C–O bond breaks and ethanol leaves.
Deprotonate to the carboxylic acid
Water removes the proton from the protonated carbonyl, giving the neutral carboxylic acid and regenerating hydronium.
As setas curvas
- Water's lone pair takes the proton.
- The O–H pair stays on oxygen, giving C=O.
Protonate the carbonyl of the second ester
Hydronium protonates the carbonyl oxygen of the second ester, making its carbon much more electrophilic toward weakly nucleophilic water.
As setas curvas
- A carbonyl oxygen lone pair takes a proton from hydronium.
- The O–H pair returns to oxygen, releasing water.
Water adds to the carbonyl carbon
Water attacks the activated carbonyl carbon, giving a tetrahedral intermediate with a positively charged oxygen.
As setas curvas
- A water lone pair forms the new C–O bond.
- The C=O π pair moves onto the protonated oxygen, neutralizing it.
Deprotonate the added water
Water removes a proton from the oxonium oxygen, giving a neutral tetrahedral intermediate.
As setas curvas
- Water's lone pair takes the oxonium proton.
- The O–H pair stays on oxygen.
Protonate the ethoxy oxygen
Hydronium protonates the ethoxy oxygen, turning it into a good leaving group (neutral ethanol).
As setas curvas
- The ethoxy oxygen lone pair takes a proton from hydronium.
- The O–H pair returns to oxygen as water.
Expel ethanol
An OH lone pair re-forms the C=O bond and pushes out ethanol.
As setas curvas
- An oxygen lone pair forms the C=O π bond.
- The C–O bond breaks and ethanol leaves.
Deprotonate to the carboxylic acid
Water removes the proton from the protonated carbonyl, giving the neutral carboxylic acid and regenerating hydronium.
As setas curvas
- Water's lone pair takes the proton.
- The O–H pair stays on oxygen, giving C=O.
Decarboxylate through a cyclic transition state
On heating, the β-carbonyl oxygen takes the carboxyl proton while CO2 leaves, all in one six-membered cyclic step. This gives carbon dioxide and an enol.
As setas curvas
- The C=O π pair of the β-carbonyl bonds to the carboxyl hydrogen.
- The O–H pair becomes the second C=O of carbon dioxide.
- The C–C bond pair becomes the C=C π bond of the enol.
Protonate the enol on carbon
Hydronium protonates the enol at its α-carbon, giving a protonated carbonyl.
As setas curvas
- The enol oxygen lone pair forms a C=O π bond as the C=C π pair leaves.
- The enol C=C π pair bonds to a proton from hydronium.
- The O–H pair returns to oxygen as water.
Deprotonate oxygen
Water removes the O–H proton, completing the tautomerization to the carboxylic acid.
As setas curvas
- Water's lone pair takes the O–H proton.
- The O–H pair stays on oxygen, forming the C=O bond.
3-Phenylpropanoic acid
The product is 3-phenylpropanoic acid: the malonate supplied the CH2–COOH unit and benzyl bromide the benzyl group; one carboxyl carbon left as CO2.
O que você vai praticar
Draw the malonic ester synthesis of 3-phenylpropanoic acid: benzylation of the malonate enolate, hydrolysis of both esters, decarboxylation and tautomerization.
Você acabou de desenhar cada seta deste problema.
Cada um dos 5349 problemas de Orgânica 1 e 2 funciona assim: feedback em cada seta, dicas quando você travar e um modo prova sem dicas. Grátis com uma conta, sem cartão.
Travou na sua própria lição?
Tire uma foto: o primeiro scan é grátis, sem cadastro
Você escaneia toda semana? Scan Standard: por cerca de problemas