Química Orgânica 2 · Acetais, iminas e equilíbrios relacionados
N-Methylbenzylamine by reductive amination
Condições: NaBH3CN, MeOH, pH 5 (AcOH buffer)
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 8 passos escritos
The nitrogen adds to the carbonyl carbon
The methylamine nitrogen lone pair attacks the electrophilic carbonyl carbon of benzaldehyde and the C=O π electrons move onto oxygen. At pH 5 an acetic acid buffer (assumed) supplies the protons; NaBH3CN does not reduce the aldehyde at this pH.
As setas curvas
- The nitrogen lone pair forms the new C–N bond to the partially positive carbonyl carbon.
- The C=O π pair moves onto oxygen, which becomes an alkoxide.
Check
Answer before you continue.
Protonate the alkoxide
The tetrahedral intermediate carries O⁻ next to N⁺. Acetic acid, the acid catalyst, protonates the negative oxygen.
As setas curvas
- The alkoxide lone pair takes the acidic proton.
- The O–H bond pair stays on acetate.
Remove the N–H proton
The acetate removes a proton from the positively charged nitrogen, giving the neutral carbinolamine: one carbon carrying both OH and N.
As setas curvas
- A lone pair on acetate takes the N–H proton.
- The N–H bond pair returns to nitrogen as its lone pair.
Protonate the OH group
OH⁻ is a poor leaving group in an acidic medium, so the acid protonates the hydroxyl oxygen to make water a leaving group.
As setas curvas
- A hydroxyl lone pair takes the proton from the acid.
- The O–H bond pair stays on acetate.
Water leaves as the C=N⁺ forms
The nitrogen lone pair pushes into the C–N bond while water departs, so no free carbocation forms; the result is an iminium ion with a full octet on every atom.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O bond pair leaves with the oxygen as neutral water.
Hydride reduces the iminium ion
The B–H bond of cyanoborohydride delivers hydride to the iminium carbon; the C=N π pair returns to nitrogen as a lone pair.
As setas curvas
- The B–H bond pair forms the new C–H bond.
- The C=N π pair moves onto nitrogen.
N-Methylbenzylamine
The C=O carbon is now a CH2 bonded to NHCH3: a secondary amine, made without the over-alkylation that alkyl halides cause.
O que você vai praticar
Draw iminium formation from benzaldehyde and methylamine in a pH 5 buffer, then hydride transfer from NaBH3CN to give N-methylbenzylamine.
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