Química Orgânica 1 · Raciocínio ácido–base
Deprotonate phenol and spread the charge
Condições: Phenol with one equivalent of aqueous sodium hydroxide. First draw the proton transfer, then generate the phenoxide contributors one at a time: charge at an ortho carbon, then the para carbon, then the other ortho carbon. Na⁺ is a spectator.
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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 5 passos escritos
Remove the phenol proton
Phenol (pKa approximately 10) is a much stronger acid than water (pKa approximately 15.7), so hydroxide deprotonates it almost completely: K is roughly 10⁵ to 10⁶.
As setas curvas
- A hydroxide lone pair accepts the O–H proton of phenol.
- The O–H pair stays on oxygen, giving phenoxide.
Push the charge into the ring
Oxygen can donate a lone pair because the carbon bonded to it is part of the ring π system. That carbon must give up a π bond at the same time to keep its octet.
As setas curvas
- An oxygen lone pair forms a π bond to the ring carbon.
- The adjacent ring π pair becomes a lone pair on the ortho carbon, which now carries the negative charge.
Move the charge to the para carbon
The same two-arrow pattern moves the charge two atoms further around the ring.
As setas curvas
- The ortho carbon lone pair forms a π bond to the neighboring meta carbon.
- The adjacent π pair becomes a lone pair on the para carbon.
Move the charge to the other ortho carbon
This is the last new contributor. One more push from the negatively charged ortho carbon would restore its double bond to the oxygen-bearing ring carbon and put the charge back on oxygen.
As setas curvas
- The para carbon lone pair forms a π bond to the next meta carbon.
- The adjacent π pair becomes a lone pair on the other ortho carbon.
Phenoxide contributors
The four contributors of phenoxide, side by side (water and Na⁺ are left out of this summary): charge on oxygen, on either ortho carbon, or on the para carbon. The contributor with charge on oxygen is the most important, because oxygen is the most electronegative atom, but sharing the charge with three ring carbons is why phenol (pKa approximately 10) is far more acidic than cyclohexanol (pKa approximately 16 to 18). The meta carbons never carry the charge, which is also why phenoxide reacts with electrophiles at ortho and para positions.
O que você vai praticar
Connect the acidity of phenol to the four contributors of phenoxide by drawing each ring contributor in turn.
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