Química Orgânica 1 · Substituição e eliminação: revisão para a prova
E1 dehydration of 2,2-dimethylcyclopentanol with a methyl shift
Condições: 85% H3PO4 (cat.), 150 °C; alkene distilled out
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Arraste dos elétrons que se movem até onde eles vão, ou toque em um e depois no outro. Cada seta é verificada; Dica ajuda quando você travar. Não precisa de conta.
O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 6 passos escritos
Phosphoric acid protonates the hydroxyl group
An oxygen lone pair takes the acidic proton of H₃PO₄, turning OH into an alkyloxonium ion that can leave as water.
As setas curvas
- An alcohol oxygen lone pair bonds to the acidic hydrogen.
- The O–H bonding pair stays on oxygen as dihydrogen phosphate.
Water leaves
The C–O bond breaks and water departs, leaving a secondary carbocation in the ring next to the carbon that carries two methyl groups.
As setas curvas
- The C–O bonding pair leaves with oxygen as water.
What does the secondary cation do next?
The neighbouring ring carbon has two methyl groups and no hydrogen. Moving one methyl onto the cationic carbon leaves the positive charge on a carbon with three carbon substituents.
1,2-Methyl shift
One methyl group moves with its C–C bonding pair from the quaternary carbon to the cationic carbon. The carbon it left keeps one methyl and two ring bonds, so it becomes a tertiary cation.
As setas curvas
- The C–CH₃ bonding pair carries the methyl to the cationic carbon.
Loss of a proton gives the tetrasubstituted alkene
Dihydrogen phosphate removes the hydrogen on the ring carbon that received the methyl group. That C–H pair becomes the π bond to the tertiary cationic carbon, giving the most substituted alkene available.
As setas curvas
- A dihydrogen phosphate oxygen lone pair bonds to the β-hydrogen.
- The C–H bonding pair becomes the new C=C π bond.
1,2-Dimethylcyclopentene
Each alkene carbon now carries a methyl group: 1,2-dimethylcyclopentene, the tetrasubstituted alkene. Phosphoric acid is regenerated, so it is a catalyst.
O que você vai praticar
Draw the E1 dehydration of 2,2-dimethylcyclopentan-1-ol: protonation, loss of water, the 1,2-methyl shift to a tertiary cation and loss of a proton giving 1,2-dimethylcyclopentene.
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