Química Orgânica 2 · Síntese e transformação de aminas
N,N-Dimethylbenzylamine by Eschweiler–Clarke methylation
Condições: HCHO (excess), HCOOH, 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.
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Amine adds to formaldehyde
The benzylamine nitrogen attacks the very electrophilic formaldehyde carbon and the C=O π pair moves onto oxygen.
As setas curvas
- The nitrogen lone pair forms the N–C bond.
- The C=O π pair moves onto oxygen.
Formic acid protonates the alkoxide
Formic acid, the strongest acid present, protonates the alkoxide oxygen.
As setas curvas
- The alkoxide takes the formic acid proton.
- The O–H pair stays on oxygen as formate.
Formate removes an N–H proton
Formate takes a proton from the ammonium nitrogen, giving the neutral hemiaminal.
As setas curvas
- Formate takes the N–H proton.
- The N–H pair returns to nitrogen.
Formic acid protonates the OH
Formic acid protonates the hemiaminal OH so it can leave as water.
As setas curvas
- The OH oxygen takes the formic acid proton.
- The O–H pair stays on oxygen as formate.
Water leaves; the iminium forms
The nitrogen lone pair forms the C=N π bond as water leaves, giving an iminium ion.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with water.
Formate delivers hydride
Formate gives the hydride on its carbon to the iminium carbon; its O⁻ lone pair forms the second C=O, so CO2 is released. This makes the first N-methyl group.
As setas curvas
- The formate O⁻ lone pair forms a C=O bond.
- The C–H pair moves as hydride to the iminium carbon.
- The C=N π pair moves onto nitrogen.
Why it stops at the tertiary amine
Iminium formation needs an N–H.
Second round: add formaldehyde again
The secondary amine N-methylbenzylamine is still nucleophilic and adds to a second formaldehyde.
As setas curvas
- The nitrogen lone pair attacks formaldehyde.
- The C=O π pair moves onto oxygen.
Protonate the alkoxide
Formic acid protonates the new alkoxide.
As setas curvas
- The alkoxide takes the formic acid proton.
- The O–H pair stays on oxygen.
Remove the last N–H proton
Formate takes the remaining N–H proton, giving the neutral hemiaminal.
As setas curvas
- Formate takes the N–H proton.
- The N–H pair returns to nitrogen.
Protonate the OH again
Formic acid protonates the hemiaminal OH.
As setas curvas
- The OH oxygen takes the proton.
- The O–H pair stays on oxygen as formate.
Lose water to the second iminium
The nitrogen lone pair pushes out water, giving an iminium ion with no N–H.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with water.
Second hydride transfer
A second formate delivers hydride to the iminium carbon and leaves as CO2, installing the second methyl group.
As setas curvas
- The formate O⁻ lone pair forms a C=O bond.
- The C–H pair moves as hydride to the iminium carbon.
- The C=N π pair moves onto nitrogen.
N,N-Dimethylbenzylamine
N,N-Dimethylbenzylamine forms with two CO2 and two water molecules. Loss of CO2 makes each hydride transfer irreversible.
O que você vai praticar
Draw both rounds of formaldehyde addition, iminium formation and hydride transfer from formate (with CO2 loss) that convert benzylamine into N,N-dimethylbenzylamine.
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