Química Orgânica 2 · Síntese e transformação de aminas
Aniline by Hofmann rearrangement of benzamide
Condições: Br2, NaOH, H2O; 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 12 passos escritos
Hydroxide removes an N–H proton
The amide N–H is weakly acidic (pKa ~17) because the anion is delocalised onto oxygen; hydroxide removes it.
As setas curvas
- Hydroxide takes the N–H proton.
- The N–H pair stays on nitrogen.
Nitrogen attacks bromine
The amide anion nitrogen attacks Br2, giving the N-bromoamide.
As setas curvas
- The nitrogen lone pair forms the N–Br bond.
- The Br–Br pair leaves as bromide.
Remove the second N–H proton
The N-bromoamide N–H is more acidic still, and a second hydroxide removes it.
As setas curvas
- Hydroxide takes the N–H proton.
- The N–H pair stays on nitrogen.
Phenyl migrates as bromide leaves
The phenyl group shifts from carbon to nitrogen with its bonding pair as bromide leaves; the nitrogen lone pair forms the C=N bond of phenyl isocyanate.
As setas curvas
- The C–C pair carries phenyl to nitrogen.
- The nitrogen lone pair forms the C=N π bond.
- The N–Br pair leaves as bromide.
Carbon count
One carbon is lost.
Hydroxide adds to the isocyanate
Hydroxide attacks the central isocyanate carbon and the C=N π pair moves onto nitrogen.
As setas curvas
- Hydroxide forms the C–O bond.
- The C=N π pair moves onto nitrogen.
Nitrogen takes a proton from water
The nitrogen anion takes a proton from water.
As setas curvas
- The nitrogen lone pair takes the proton.
- The O–H pair stays on oxygen.
Hydroxide deprotonates the carbamic acid
Hydroxide removes the carbamic acid O–H proton.
As setas curvas
- Hydroxide takes the O–H proton.
- The O–H pair stays on oxygen.
Lose CO2
The carboxylate oxygen forms the second C=O and the C–N bond breaks onto nitrogen, releasing CO2.
As setas curvas
- The oxygen lone pair forms the C=O π bond.
- The C–N pair stays on nitrogen.
Protonate nitrogen
The anilide nitrogen takes a proton from water, giving aniline.
As setas curvas
- The nitrogen lone pair takes the proton.
- The O–H pair stays on oxygen as hydroxide.
Hydroxide captures CO2
In the basic solution, hydroxide adds to CO2, giving bicarbonate.
As setas curvas
- Hydroxide forms the C–O bond.
- A C=O π pair moves onto oxygen.
Aniline
Aniline forms with bicarbonate and two bromide ions.
O que você vai praticar
Draw N-bromination of benzamide in base, migration of phenyl to nitrogen with loss of bromide to phenyl isocyanate, and hydrolysis with loss of CO2 to aniline.
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