Química Orgânica 2 · Ferramentas de oxidação e redução
Wolff–Kishner reduction of 1-tetralone to tetralin
Condições: H2NNH2·H2O (3 equiv), KOH (3 equiv), ethylene glycol, 130 °C then 200 °C (distil off water)
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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 11 passos escritos
Hydrazine adds to the ketone
A nitrogen lone pair of hydrazine attacks the ketone carbon and the C=O π pair moves onto oxygen, giving a zwitterionic tetrahedral adduct.
As setas curvas
- A hydrazine nitrogen lone pair forms the new C–N bond.
- The C=O π pair moves onto oxygen.
Hydroxide removes an N–H proton
Hydroxide takes a proton from the positively charged nitrogen, leaving a neutral nitrogen and an alkoxide.
As setas curvas
- A hydroxide lone pair takes the N–H proton.
- The N–H pair stays on nitrogen.
Water protonates the alkoxide
The alkoxide takes a proton from water, giving the neutral carbinolamine and regenerating hydroxide.
As setas curvas
- An alkoxide lone pair takes a proton from water.
- The water O–H pair stays on oxygen as hydroxide.
Loss of hydroxide gives the hydrazone
Hydroxide removes the remaining proton on the carbon-bound nitrogen; that N–H pair becomes the C=N π bond as hydroxide is expelled from carbon. The product is the hydrazone.
As setas curvas
- A hydroxide lone pair takes the N–H proton.
- The N–H pair becomes the C=N π bond.
- The C–O bond breaks and hydroxide leaves.
Hydroxide deprotonates the NH2 group
At high temperature hydroxide removes a proton from the terminal NH2 of the hydrazone, giving an anion delocalised over nitrogen and carbon.
As setas curvas
- A hydroxide lone pair takes an N–H proton.
- The N–H pair stays on the terminal nitrogen.
Water protonates the carbon
The terminal nitrogen lone pair forms an N=N bond while the C=N π pair takes a proton from water onto carbon. The former carbonyl carbon now carries its first new hydrogen, and the nitrogen fragment is a diazene.
As setas curvas
- The nitrogen lone pair forms the N=N π bond.
- The C=N π pair forms the new C–H bond to the water hydrogen.
- The water O–H pair stays on oxygen as hydroxide.
Hydroxide removes the last N–H proton
Hydroxide deprotonates the diazene N–H, giving a diazenide anion poised to lose nitrogen.
As setas curvas
- A hydroxide lone pair takes the N–H proton.
- The N–H pair stays on nitrogen.
Nitrogen gas leaves
The nitrogen lone pair forms the N≡N triple bond as the C–N bond breaks with its pair staying on carbon. N2 escapes and a benzylic carbanion is left behind.
As setas curvas
- The nitrogen lone pair forms the third N–N bond.
- The C–N pair stays on carbon as the N2 molecule departs.
What drives the reduction?
Forming N2, a very stable molecule that leaves as a gas, pays for creating the carbanion and makes the sequence irreversible.
Water protonates the carbanion
The benzylic carbanion takes a proton from water, completing the CH2 group.
As setas curvas
- The carbanion lone pair takes a proton from water.
- The water O–H pair stays on oxygen as hydroxide.
Tetralin
The ketone carbon has become a CH2 group: 1-tetralone has been deoxygenated to 1,2,3,4-tetrahydronaphthalene (tetralin), with N2 and water as by-products.
O que você vai praticar
Draw the Wolff–Kishner reduction of 1-tetralone: hydrazone formation under basic conditions, then the deprotonation, carbon protonation and N2 loss that convert the C=O carbon into a CH2 group of tetralin.
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