Química Orgânica 2 · Síntese e transformação de aminas
Staudinger reduction to butan-1-amine
Condições: PPh3, THF; 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.
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Phosphine attacks the terminal nitrogen
The phosphorus lone pair of triphenylphosphine is a soft nucleophile and adds to the terminal azide nitrogen, while one N≡N π pair moves onto the central nitrogen. This gives a phosphazide.
As setas curvas
- The phosphorus lone pair forms the P–N bond.
- One N≡N π pair moves onto the central nitrogen.
Close the four-membered ring
The anionic carbon-bound nitrogen attacks the positive phosphorus, forming a four-membered P–N–N–N ring with a five-coordinate phosphorus.
As setas curvas
- The nitrogen lone pair bonds to phosphorus.
Lose N2
The ring breaks apart: N2 leaves and the carbon-bound nitrogen becomes doubly bonded to phosphorus, giving the iminophosphorane (aza-ylide).
As setas curvas
- The P–N bond pair becomes the third N–N bond of N2.
- The N–N bond pair becomes the P=N π bond.
Water protonates nitrogen
The iminophosphorane nitrogen is strongly basic (its P=N bond is largely P⁺–N⁻), so it takes a proton from water.
As setas curvas
- The nitrogen lone pair takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Hydroxide adds to phosphorus
Hydroxide adds to the electrophilic phosphonium phosphorus; the P=N π pair moves back onto nitrogen, giving a neutral five-coordinate hydroxyphosphorane.
As setas curvas
- A hydroxide lone pair bonds to phosphorus.
- The P=N π pair moves onto nitrogen.
A second water protonates nitrogen
Water puts a second proton on the nitrogen, turning it into a neutral-amine leaving group.
As setas curvas
- The nitrogen lone pair takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Hydroxide removes the P–OH proton
Hydroxide deprotonates the phosphorus-bound OH, giving an oxyanion that can push out the amine.
As setas curvas
- Hydroxide takes the O–H proton.
- The O–H pair stays on oxygen.
Form P=O and release the amine
The oxyanion forms the very strong P=O bond while the P–N bond breaks onto nitrogen, releasing neutral butan-1-amine. The strength of the P=O bond drives the whole reaction.
As setas curvas
- The oxygen lone pair forms the P=O π bond.
- The P–N bond pair stays on nitrogen, giving the neutral amine.
Driving force
Phosphorus chemistry is often driven by one bond.
Butan-1-amine and triphenylphosphine oxide
Butan-1-amine forms under neutral, mild conditions, so groups that LiAlH4 would reduce survive. Triphenylphosphine oxide and N2 are the by-products.
O que você vai praticar
Draw phosphine attack on the azide, N2 loss through a four-membered ring to the iminophosphorane, and its hydrolysis to butan-1-amine and triphenylphosphine oxide.
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