Química Orgânica 2 · Ferramentas de oxidação e redução
Wolff–Kishner reduction of acetophenone to ethylbenzene
Condições: NH2NH2, KOH, diethylene glycol, 200 °C
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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 next to the 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 hydroxide.
As setas curvas
- The alkoxide lone pair takes a proton from water.
- The 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.
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, giving 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 a 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 N2 departs.
What drives the reduction?
Forming N2, a very stable molecule that leaves as a gas, pays for creating the carbanion.
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.
Ethylbenzene
The ketone carbon has become a CH2 group: acetophenone has been deoxygenated to ethylbenzene, with N2 and water as by-products.
O que você vai praticar
Draw hydrazone formation from acetophenone under basic conditions, then the N–H deprotonations, carbon protonations by water and loss of N2 that turn the C=O carbon into the CH2 group of ethylbenzene.
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