Química Orgânica 2 · Ferramentas de oxidação e redução
LiAlH4 reduction of 2-piperidone to piperidine
Condições: LiAlH4 (2.0 equiv), THF, reflux 6 h; then H2O
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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 8 passos escritos
Hydride removes the N–H proton
The amide N–H is the most acidic hydrogen present. An Al–H pair bonds to that proton, releasing hydrogen gas and leaving an amide anion.
As setas curvas
- The Al–H pair forms an H–H bond to the N–H hydrogen.
- The N–H pair stays on nitrogen, which becomes an anion.
The carbonyl oxygen binds aluminium
A lone pair of the carbonyl oxygen bonds to the empty orbital of AlH3. The oxygen is now attached to aluminium, which will make it a leaving group later.
As setas curvas
- An oxygen lone pair forms the O–Al bond.
Hydride adds to the carbonyl carbon
A second AlH4− delivers hydride to the activated carbonyl carbon while the C=O π pair moves onto oxygen, which stays bonded to aluminium.
As setas curvas
- The Al–H pair carries the hydrogen to the carbonyl carbon.
- The C=O π pair moves onto oxygen.
Nitrogen expels the oxygen
The nitrogen lone pair forms a C=N π bond as the C–O bond breaks. The oxygen leaves still bonded to aluminium, and the ring is now a cyclic imine.
As setas curvas
- A nitrogen lone pair forms the C=N π bond.
- The C–O bond breaks; the oxygen leaves as an aluminate.
Which group leaves?
The tetrahedral intermediate could in principle lose nitrogen or oxygen. Oxygen bound to aluminium is the far better leaving group, and nitrogen's lone pair is the stronger donor, so the C–N bond is kept.
Hydride reduces the imine
A third AlH4− adds hydride to the imine carbon; the C=N π pair moves onto nitrogen, giving an amide anion.
As setas curvas
- The Al–H pair forms the second new C–H bond at the imine carbon.
- The C=N π pair moves onto nitrogen.
Water protonates nitrogen
On workup the nitrogen anion takes a proton from water, giving the neutral secondary amine.
As setas curvas
- A nitrogen lone pair takes a proton from water.
- The water O–H pair stays on oxygen as hydroxide.
Piperidine
The former carbonyl carbon is now a CH2 group bonded to nitrogen: the lactam has been reduced to the cyclic amine piperidine.
O que você vai praticar
Draw the LiAlH4 reduction of the lactam 2-piperidone: N–H deprotonation by hydride, activation of oxygen by aluminium, hydride addition, loss of the oxygen as an aluminate to form an imine, the second hydride addition and protonation, giving piperidine.
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