Química Orgânica 2 · Condensação de Claisen e adição conjugada
Cyclohexanecarboxylic acid by malonic ester synthesis with 1,5-dibromopentane
Condições: NaOEt (2 equiv), EtOH; then H3O⁺, Δ
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O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 20 passos escritos
First enolate
Ethoxide removes one of the two CH2 hydrogens of diethyl malonate.
As setas curvas
- Ethoxide's lone pair takes an acidic hydrogen.
- The C–H pair stays on carbon.
First SN2 alkylation
The enolate carbon displaces one bromide of 1,5-dibromopentane, attaching a 5-bromopentyl chain.
As setas curvas
- The enolate carbon attacks the CH2 bearing bromine.
- Bromide leaves.
Second enolate
The second equivalent of ethoxide removes the remaining hydrogen between the esters, which is still acidic.
As setas curvas
- Ethoxide takes the last acidic hydrogen.
- The C–H pair stays on carbon.
Intramolecular SN2 closes the ring
The enolate carbon attacks the far CH2–Br of its own chain, closing a six-membered ring and expelling bromide.
As setas curvas
- The enolate carbon forms the ring C–C bond.
- The C–Br bond breaks and bromide leaves.
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.
Cyclohexanecarboxylic acid
The product is cyclohexanecarboxylic acid: the malonate carbon became a ring carbon of the cyclohexane closed by the double alkylation, and one carboxyl carbon left as CO2.
O que você vai praticar
Draw the malonic ester route to cyclohexanecarboxylic acid: two enolate alkylations that close the ring, hydrolysis of both esters, decarboxylation and tautomerization.
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