Química Orgânica 2 · Rotas de síntese: reagentes, grupos protetores e planejamento em várias etapas
Benzoic acid to methyl benzoate through benzoyl chloride
Condições: SOCl2, Δ; then CH3OH
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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.
Ler todos os 10 passos escritos
The carbonyl oxygen attacks sulfur
The carbonyl oxygen of benzoic acid attacks the sulfur of thionyl chloride and a chloride leaves. This converts the carboxyl oxygen into part of a chlorosulfite, an excellent leaving group.
As setas curvas
- A carbonyl oxygen lone pair bonds to sulfur.
- The S–Cl pair leaves as chloride.
Chloride adds to the activated carbonyl carbon
Chloride attacks the now strongly electrophilic carbon and the C=O+ π pair returns to oxygen, giving a neutral tetrahedral intermediate.
As setas curvas
- A chloride lone pair forms the C–Cl bond.
- The C=O+ π pair moves onto oxygen.
Collapse expels the chlorosulfite
The OH oxygen pushes a lone pair back toward carbon and the C–O bond to the chlorosulfite group breaks, releasing the chlorosulfite anion and the protonated acyl chloride.
As setas curvas
- The OH lone pair forms a C=O+ π bond.
- The C–O bond pair leaves with the chlorosulfite.
Chlorosulfite falls apart
The chlorosulfite anion loses chloride to give sulfur dioxide gas; losing SO2 makes the whole activation irreversible.
As setas curvas
- The oxygen lone pair forms an S=O bond.
- The S–Cl pair leaves as chloride.
Why go through the acid chloride?
Reactivity ladder of acid derivatives.
Methanol adds to the protonated acyl chloride
Methanol attacks the carbonyl carbon of the protonated benzoyl chloride and the C=O+ π pair returns to oxygen, giving a neutral O–H on the tetrahedral carbon.
As setas curvas
- The methanol oxygen lone pair bonds to the carbonyl carbon.
- The C=O+ π pair moves onto the protonated oxygen.
Methanol removes the oxonium proton
A second methanol molecule takes the proton from the added methanol oxygen, giving a neutral tetrahedral intermediate.
As setas curvas
- The methanol oxygen lone pair takes the proton.
- The O–H electrons stay on the methoxy oxygen.
Chloride leaves
The OH oxygen pushes a lone pair back toward carbon and chloride, the best leaving group, departs.
As setas curvas
- The OH lone pair forms a C=O+ π bond.
- The C–Cl pair leaves as chloride.
Chloride removes the last proton
Chloride takes the proton from the protonated ester, giving methyl benzoate and HCl gas, which escapes.
As setas curvas
- The chloride lone pair takes the proton.
- The O–H electrons stay on the carbonyl oxygen.
Methyl benzoate
Methyl benzoate forms with SO2 and HCl as by-products; the acid was carried up the reactivity ladder to the activated acyl chloride and then down to the ester.
O que você vai praticar
Convert benzoic acid to methyl benzoate by drawing acyl chloride formation with SOCl2 and its methanolysis.
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