Química Orgânica 2 · Enóis, enolatos e reações em alfa
2-Methylcyclohexanone by Stork enamine methylation of cyclohexanone
Condições: pyrrolidine, TsOH (cat.), −H2O; then CH3I; then H3O⁺
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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 14 passos escritos
TsOH protonates the ketone
The catalytic acid protonates the carbonyl oxygen, making the carbonyl carbon a much better electrophile for the neutral amine.
As setas curvas
- The carbonyl oxygen lone pair takes the acid proton.
- The O–H electrons stay on the sulfonate.
Pyrrolidine adds to the carbonyl carbon
The nitrogen lone pair of the secondary amine attacks the activated carbonyl carbon; the C=O π electrons move onto oxygen, giving a neutral OH.
As setas curvas
- The nitrogen lone pair forms the new C–N bond.
- The C=O π electrons move onto oxygen.
The ammonium loses its proton
A second molecule of pyrrolidine (used in excess) removes the proton from the positive nitrogen, giving the neutral carbinolamine.
As setas curvas
- The nitrogen lone pair of a second pyrrolidine (excess amine) takes the N–H proton.
- The N–H electrons return to nitrogen.
Protonate the OH
TsOH protonates the hydroxyl so that it can leave as water.
As setas curvas
- The hydroxyl oxygen lone pair takes the acid proton.
- The O–H electrons stay on the sulfonate.
Water leaves, pushed by nitrogen
The nitrogen lone pair forms a C=N π bond as water departs, giving an iminium ion.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O electrons leave with water.
Lose an α-proton to give the enamine
The iminium nitrogen has no N–H to lose, so another pyrrolidine molecule removes an α-hydrogen instead; the electrons form the C=C bond of the enamine and nitrogen becomes neutral.
As setas curvas
- A pyrrolidine nitrogen takes an α-proton.
- The C–H electrons form the C=C π bond.
- The C=N π electrons move onto nitrogen.
Why an enamine?
The enamine is a neutral, enolate-like nucleophile at the α-carbon.
The enamine carbon attacks methyl iodide
The nitrogen lone pair pushes into the C=C, and the α-carbon attacks the methyl carbon in an SN2 reaction, giving a C-methylated iminium ion.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C=C π electrons form the new C–C bond.
- Iodide leaves with the C–I electrons.
Water adds to the iminium ion
In the aqueous acid workup, water attacks the iminium carbon and the C=N π electrons move onto nitrogen.
As setas curvas
- A water lone pair forms the new C–O bond.
- The C=N π electrons move onto nitrogen.
Water removes a proton from the oxonium
The added oxygen carries a positive charge; a water molecule removes one of its protons.
As setas curvas
- A water lone pair takes the O–H proton.
- The O–H electrons stay on oxygen.
Protonate the nitrogen
Hydronium protonates the nitrogen so that pyrrolidine can leave as a neutral amine.
As setas curvas
- The nitrogen lone pair takes the proton from hydronium.
- The O–H electrons stay on water.
Oxygen pushes out pyrrolidine
The OH lone pair forms a C=O π bond and expels the neutral amine, giving the protonated ketone.
As setas curvas
- An oxygen lone pair forms the C=O π bond.
- The C–N electrons leave with the amine nitrogen.
Deprotonate the ketone oxygen
Water removes the proton from the protonated carbonyl, releasing 2-methylcyclohexanone and regenerating hydronium.
As setas curvas
- A water lone pair takes the proton.
- The O–H electrons stay on the carbonyl oxygen.
2-Methylcyclohexanone
One methyl group has been added at the α-carbon; the new stereocentre is racemic. Pyrrolidine is released (as its ammonium salt in the acidic workup) and the ketone oxygen now comes from the workup water.
O que você vai praticar
Draw enamine formation from cyclohexanone and pyrrolidine, C-methylation of the enamine with methyl iodide, and hydrolysis of the iminium ion to 2-methylcyclohexanone.
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