Química Orgânica 1 · Ácidos, bases e ressonância: revisão para a prova
Carbonate deprotonates 4-nitrophenol
Condições: 4-Nitrophenol, Na₂CO₃ (1 equiv), H₂O, 25 °C
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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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Carbonate removes the phenol proton
A carbonate oxygen takes the O–H proton. 4-Nitrophenol (pKa 7.15) is a much stronger acid than bicarbonate (pKa 10.3), so log K = 10.3 − 7.15 ≈ 3.2 and the transfer is essentially complete.
As setas curvas
- A carbonate oxygen lone pair bonds to the phenol hydrogen.
- The O–H bonding pair stays on the phenol oxygen, which becomes negatively charged.
The charge moves to the para carbon
An oxygen lone pair forms a C=O π bond while two ring π pairs shift along the ring, putting the negative charge on the carbon that carries the nitro group.
As setas curvas
- An oxygen lone pair forms the C=O π bond.
- The ring π pair next to it shifts one bond along.
- The next ring π pair becomes a lone pair on the para carbon.
Why the para position matters
Only ortho and para positions receive the phenoxide charge, so only a para (or ortho) nitro group can take that charge onto its own oxygens.
The nitro group takes the charge
The lone pair on the para carbon forms a C=N π bond, and an N=O π pair moves onto oxygen. Both nitro oxygens now carry negative charge and nitrogen keeps its octet. This quinonoid contributor helps explain why 4-nitrophenoxide is yellow.
As setas curvas
- The para-carbon lone pair forms the C=N π bond.
- The N=O π pair moves onto oxygen so nitrogen does not exceed an octet.
Back to the aromatic phenoxide
The quinonoid contributor is important but not the major one: it has the same number of formal charges as the phenoxide, still with the negative charge on oxygen, yet it has given up the ring's aromatic sextet. Four pairs move back together to restore the aromatic ring with O⁻ on the phenoxide oxygen.
As setas curvas
- A nitro-oxygen lone pair re-forms the N=O π bond.
- The C=N π pair moves into the ring so nitrogen keeps its octet.
- A ring π pair shifts one bond toward the oxygen-bearing carbon.
- The C=O π pair returns to oxygen, which becomes the phenoxide O⁻.
4-Nitrophenoxide and bicarbonate
The aromatic contributor drawn here, with O⁻ on the phenol oxygen, is the major one; the negative charge is also shared, through the other contributors, by the ortho and para ring carbons and the nitro oxygens. Bicarbonate is the conjugate acid of carbonate.
O que você vai praticar
Draw carbonate removing the O–H proton of 4-nitrophenol, move the phenoxide charge through the ring onto an oxygen of the para nitro group, and return to the aromatic phenoxide contributor.
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