Química Orgânica 2 · Hidrólise e esterificação
Fischer esterification — complete acetic acid/methanol cycle
Condições: Acetic acid in excess methanol with a catalytic amount of concentrated H₂SO₄. Follow the forward esterification pathway; hydronium represents the strong-acid catalyst and carries protons between molecules. Counteranion omitted throughout, and equilibrium conversion is not asserted.
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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.
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Acidic reactant encounter
The explicit hydronium supplies the activation proton. Its conjugate water is retained as the proton shuttle throughout this pathway. The carbonyl-protonated acid (pKa about −6) is far more acidic than hydronium (pKa ≈ −1.7), so only a small fraction of acetic acid is activated at any moment; the catalyst is regenerated each cycle.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Activated carbonyl
Methanol adds through oxygen without losing its methyl group; the former carbonyl oxygen becomes a neutral hydroxyl.
As setas curvas
- The nucleophile uses a lone pair to form the new σ bond.
- The C=O π electrons move to oxygen so carbon keeps eight electrons.
First solvent-mediated proton exchange
Water removes a proton from the incoming oxygen. This separate transfer avoids an unsupported direct intramolecular proton jump.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Prepare the leaving group
Solvent-mediated proton delivery converts the selected leaving group into a neutral species on bond cleavage.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Tetrahedral intermediate in acid
The former carbonyl hydroxyl restores the pi bond while retaining its proton, so the acyl product is initially O-protonated. The leaving group is neutral after accepting the C–leaving-group pair.
As setas curvas
- An oxygen lone pair forms the C=O π bond again.
- The leaving group takes the electrons from the bond that breaks.
Resolve the activated product
Water removes the carbonyl-oxygen proton and restores hydronium. Every oxygen retains its identity through addition and elimination.
As setas curvas
- The base uses a lone pair to bond to the drawn hydrogen.
- The bond to hydrogen breaks, leaving its electrons on the atom that lost H.
Products
The incoming alcohol oxygen becomes the ester alkoxy oxygen; the original acid hydroxyl leaves as water. Hydronium is regenerated and every explicit proton is accounted for. The specified forward equilibrium branch does not imply quantitative conversion.
O que você vai praticar
Complete Fischer esterification from unactivated starting materials while tracking which oxygen leaves as water.
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