Química Orgânica 2 · Enolatos: alquilação, condensações e adições conjugadas
4-Methylpentanoic acid by malonic ester synthesis with 1-bromo-2-methylpropane
Condições: NaOEt, 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 19 passos escritos
Form the malonate enolate
Ethoxide removes a hydrogen from the CH2 between the two ester carbonyls (pKa ≈ 13), giving a doubly stabilized enolate.
As setas curvas
- Ethoxide takes the acidic C–H proton.
- The C–H pair stays on the central carbon.
SN2 alkylation
The enolate carbon attacks the CH2 carbon of 1-bromo-2-methylpropane from the back side and bromide leaves. Branching one carbon away slows this SN2, but a primary bromide still reacts.
As setas curvas
- The carbanion lone pair forms the new C–C bond.
- The C–Br bond breaks and bromide leaves.
Protonate the first ester carbonyl
In the acidic workup, hydronium protonates one ester carbonyl oxygen, activating it toward water.
As setas curvas
- The carbonyl oxygen takes a proton.
- The O–H pair returns to water.
Water adds
Water adds to the activated carbonyl carbon and the C=O π pair moves to the protonated oxygen.
As setas curvas
- A water lone pair bonds to the carbonyl carbon.
- The π pair moves onto oxygen.
Deprotonate the added water
Water removes a proton from the oxonium oxygen, giving the neutral tetrahedral intermediate.
As setas curvas
- Water takes the proton.
- The O–H pair stays on oxygen.
Protonate the ethoxy oxygen
Hydronium protonates the ethoxy oxygen so that ethanol, not ethoxide, can leave.
As setas curvas
- The ethoxy oxygen takes the proton.
- The O–H pair returns to water.
Ethanol leaves
An OH lone pair re-forms the C=O bond and ethanol departs.
As setas curvas
- The oxygen lone pair re-forms the carbonyl.
- The C–O bond breaks and ethanol leaves.
Deprotonate to the carboxylic acid
Water removes the proton from the protonated carbonyl, giving the first carboxylic acid.
As setas curvas
- Water takes the proton.
- The O–H pair stays on oxygen.
Protonate the second ester
The same sequence hydrolyses the second ester: hydronium protonates its carbonyl oxygen.
As setas curvas
- The carbonyl oxygen takes the proton.
- The O–H pair returns to water.
Water adds again
Water adds to the second activated carbonyl carbon.
As setas curvas
- A water lone pair bonds to the carbonyl carbon.
- The π pair moves onto oxygen.
Deprotonate
Water removes the proton from the new oxonium oxygen.
As setas curvas
- Water takes the proton.
- The O–H pair stays on oxygen.
Protonate the second ethoxy group
Hydronium protonates the remaining ethoxy oxygen.
As setas curvas
- The ethoxy oxygen takes the proton.
- The O–H pair returns to water.
Second ethanol leaves
An OH lone pair re-forms C=O and the second ethanol leaves.
As setas curvas
- The oxygen lone pair re-forms the carbonyl.
- Ethanol leaves.
Form the diacid
Water removes the proton from the protonated carbonyl, giving the substituted malonic acid.
As setas curvas
- Water takes the proton.
- The O–H pair stays on oxygen.
Decarboxylation
On heating, the β-dicarboxylic acid loses CO2 through a six-membered cyclic transition state: one carbonyl oxygen takes the other acid’s O–H proton while the C–C bond breaks, giving CO2 and an enol.
As setas curvas
- The carbonyl oxygen lone pair takes the carboxylic acid proton.
- The O–H pair forms the second C=O of carbon dioxide.
- The C–C bond pair becomes the C=C π bond of the enol.
- The C=O π pair moves onto oxygen, which keeps its new O–H.
Why heating removes only one CO2
Why heating removes only one CO2.
Enol to acid: protonate carbon
The enol tautomerizes in acid: hydronium protonates the α-carbon while an OH lone pair forms C=O+.
As setas curvas
- An oxygen lone pair pushes toward carbon.
- The C=C π pair takes a proton from hydronium at the α-carbon.
- The O–H pair returns to water.
Deprotonate oxygen
Water removes the proton from the protonated carbonyl, giving the neutral carboxylic acid.
As setas curvas
- Water takes the proton.
- The O–H pair stays on oxygen.
4-Methylpentanoic acid
4-Methylpentanoic acid is the product: the CH2CO2H unit comes from diethyl malonate and the 2-methylpropyl group from the alkyl bromide. Ethanol, CO2 and bromide are by-products.
O que você vai praticar
Draw the malonic ester synthesis of 4-methylpentanoic acid: malonate enolate SN2 on 1-bromo-2-methylpropane, acid hydrolysis of both esters, decarboxylation and tautomerization.
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