Química Orgânica 1 · Eliminação e caminhos competitivos
Dehydrate without a carbocation
Condições: Cyclopentanol with phosphorus oxychloride (POCl₃) in pyridine, 0 °C. Follow the dehydration to cyclopentene. Phosphorus is drawn with the conventional expanded-valence P=O bond, and substitution at phosphorus is shown as one concerted step. Two pyridine molecules are shown: either can remove the O–H proton, and the remaining neutral pyridine is the E2 base. In the flat ring the C–O bond is on a wedge and the explicit neighboring hydrogen on a hash, so they are trans and can be anti-periplanar.
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Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 4 passos escritos
Attach oxygen to phosphorus
Phosphorus in POCl₃ is electron-poor, bonded to oxygen and three chlorines. The alcohol oxygen replaces one chlorine, the first step in turning OH into a good leaving group. The C1–O bond is untouched; only bonds at phosphorus change.
As setas curvas
- A lone pair on the alcohol oxygen bonds to phosphorus.
- One P–Cl pair leaves with chlorine as chloride.
Pyridine removes the O–H proton
The positively charged oxygen is strongly acidic, and pyridine is basic enough to remove its proton. The product is a neutral alkyl dichlorophosphate, R–O–P(=O)Cl₂.
As setas curvas
- The nitrogen lone pair of either pyridine bonds to the O–H hydrogen.
- The O–H pair stays on oxygen, which becomes neutral.
Eliminate in one step
–OP(O)Cl₂ is a good leaving group because its negative charge is spread over the phosphoryl group. Pyridine removes the anti β hydrogen as the C–O bond breaks. Because no carbocation forms, there is nothing to rearrange, and the mild basic conditions suit acid-sensitive molecules.
As setas curvas
- The nitrogen lone pair of the remaining neutral pyridine bonds to the hydrogen drawn on a hashed bond next to the carbon bearing oxygen.
- The C2–H pair becomes the C1=C2 π bond.
- The C1–O pair leaves with oxygen as the dichlorophosphate anion.
Cyclopentene
C1 and C2 now share the π bond in cyclopentene. The alcohol oxygen left carbon and stays on phosphorus in dichlorophosphate, ⁻O–P(=O)Cl₂. Each pyridine gained one proton to become pyridinium; their two positive charges balance chloride and dichlorophosphate.
O que você vai praticar
Convert an alcohol OH into a dichlorophosphate leaving group with POCl₃, then complete a pyridine-promoted E2 so that no carbocation ever forms.
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