Química Orgânica 1 · Caminhos de substituição
Invert a secondary alcohol with PBr₃
Condições: (S)-Butan-2-ol with PBr₃ in dry ether at 0 °C. Follow one alcohol-conversion cycle to (R)-2-bromobutane and HOPBr₂; HOPBr₂ can convert two more alcohol molecules, which is not followed. In the first drawing oxygen and the ethyl group lie in the page, the methyl wedge points toward you and the hydrogen on the stereocenter points away.
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Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
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Oxygen attacks phosphorus
Phosphorus in PBr₃ is electron-poor, bonded to three electronegative bromines. The alcohol oxygen replaces one of them. The alcohol C–O bond is untouched here, so the configuration at the stereocenter is unchanged.
As setas curvas
- A lone pair on the alcohol oxygen bonds to phosphorus.
- A P–Br pair leaves with bromine as bromide.
Bromide attacks from the back
The oxygen, now bonded to phosphorus, is part of a good leaving group. Bromide enters exactly opposite it, so the ethyl, methyl and hydrogen groups flip to the other side of the stereocenter like an umbrella in the wind. No carbocation forms, so there is no rearrangement and no loss of configuration.
As setas curvas
- A lone pair on bromide bonds to the carbon bearing oxygen from the opposite side.
- The C–O pair stays on oxygen, which leaves as neutral HOPBr₂.
Inverted secondary bromide
Br now bonds to the stereocenter opposite the departing oxygen, reversing the tetrahedral arrangement of ethyl, methyl and H by backside substitution. The product is (R)-2-bromobutane. Br has the same priority rank that OH had, so here the R/S letter changes along with the real inversion. The alcohol oxygen left as HOPBr₂, still bonded to its original hydrogen.
O que você vai praticar
Use oxygen attack on PBr₃ to make a leaving group without carbocation formation, then predict inversion at the alcohol carbon from the bromide backside attack.
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