Química Orgânica 2 · Adição a aldeídos e cetonas
Wolff–Kishner reduction of cyclohexanone
Condições: H2NNH2 (hydrazine hydrate, 3 equiv), KOH (3 equiv), ethylene glycol, 200 °C
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O mecanismo, passo a passo
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Hydrazine adds to the carbonyl carbon
A lone pair on one nitrogen of hydrazine attacks the carbonyl carbon and the C=O π pair moves onto oxygen, giving a zwitterionic tetrahedral intermediate.
As setas curvas
- A nitrogen lone pair forms the new C–N bond.
- The C=O π pair moves onto oxygen.
Water protonates the alkoxide
The alkoxide takes a proton from water, which the hydrazine hydrate brings into the basic glycol solution (the glycol O–H could serve the same role).
As setas curvas
- An alkoxide lone pair takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Hydroxide deprotonates the ammonium nitrogen
Hydroxide removes a proton from the positively charged nitrogen, giving the neutral carbinolamine.
As setas curvas
- A hydroxide lone pair takes a proton from nitrogen.
- The N–H pair stays on nitrogen.
Loss of hydroxide gives the hydrazone
Hydroxide removes the remaining proton from the carbon-bound nitrogen while that N–H pair forms the C=N π bond and hydroxide leaves from carbon. The neutral hydrazone forms; at 200 °C the water is distilled out of the glycol, driving the condensation.
As setas curvas
- Hydroxide takes the N–H proton.
- The N–H pair becomes the C=N π bond.
- The C–O pair leaves with oxygen as hydroxide.
Hydroxide deprotonates the terminal NH2
At high temperature hydroxide removes a proton from the terminal NH2 of the hydrazone, giving a nitrogen anion conjugated with the C=N bond.
As setas curvas
- A hydroxide lone pair takes an N–H proton.
- The N–H pair stays on the terminal nitrogen.
The negative charge is shared with carbon
The anion is delocalised: pushing the nitrogen lone pair into an N=N bond moves the C=N π pair onto the former carbonyl carbon. This contributor shows why carbon, not nitrogen, is protonated next.
As setas curvas
- The nitrogen lone pair forms the N=N π bond.
- The C=N π pair moves onto carbon.
Water protonates carbon
The carbon end of the delocalised anion takes a proton from water, giving a neutral diazene with a new C–H bond.
As setas curvas
- The carbon lone pair takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Deprotonation expels nitrogen
Hydroxide removes the last N–H proton. That N–H pair forms the third bond of N≡N while the C–N bond breaks toward carbon, releasing nitrogen gas and leaving a carbanion on the former carbonyl carbon.
As setas curvas
- Hydroxide takes the N–H proton.
- The N–H pair forms the N≡N triple bond.
- The C–N pair stays on carbon as the carbanion.
Why can a carbanion form here?
Loss of N2, an extremely stable and gaseous molecule, pays for generating an unstabilised carbanion, and the gas escaping makes the step irreversible.
Water protonates the carbanion
The carbanion takes a proton from water, giving the CH2 group and regenerating hydroxide.
As setas curvas
- The carbanion lone pair takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Cyclohexane
The C=O group has been reduced all the way to CH2; both new C–H hydrogens came from water and the two nitrogens left as N2. Wolff–Kishner conditions suit substrates that tolerate strong base but not acid.
O que você vai praticar
Draw the formation of cyclohexanone hydrazone and its base-promoted reduction, with loss of nitrogen gas, to cyclohexane.
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