Química Orgânica 2 · Substituição acílica e escolha do derivado
Reduce an ester through two hydride deliveries
Condições: Ethyl propanoate with lithium aluminum hydride in a dry ether medium, followed by a separate aqueous workup. Two AlH₄− donor equivalents each supply one drawn hydrogen; lithium counterions are omitted. Aluminum coordination, further donor consumption and complete quench chemistry are not shown. Retained AlH₃ and hydroxide show only these transfers, not isolated final aqueous aluminum species.
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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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First hydride delivery
One donor loses one hydrogen while the ester becomes a tetrahedral oxyanion. The original carbonyl oxygen and alkoxy oxygen are both retained; the second donor has not reacted.
Nota: The two donor equivalents are chemically equivalent. Each supplies one hydrogen; its three other hydrogens are included in the AlH₃ group.
As setas curvas
- The polarized Al–H bond supplies the pair for C–H bond formation; no free hydride ion is released.
- The carbonyl pi pair moves to oxygen as hydrogen adds, keeping the acyl carbon within its octet.
Collapse to the aldehyde
The oxygen lone pair restores C=O while ethoxide leaves with the acyl–oxygen sigma pair. The transferred hydrogen stays on carbon, so the acyl fragment is now propanal. This aldehyde is an intermediate that undergoes further reduction, not the isolated endpoint under these conditions.
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.
Second hydride delivery
The former carbonyl carbon now bears two transferred hydrogens. Propoxide and the previously expelled ethoxide are still anions; neither alcohol O–H bond has been supplied by the reducing agent.
As setas curvas
- The second donor supplies a different Al–H pair, making the second carbon-bound hydrogen.
- Opening the aldehyde pi bond gives an alkoxide without displacing the first transferred hydrogen.
Separate aqueous workup
The workup supplies an O–H proton, not a third carbon-bound hydrogen. Water protonates propoxide, and its O–H bond pair stays on the water oxygen. The resulting hydroxide is retained; full aluminum quench chemistry is not shown.
Nota: Two water molecules are introduced only now to protonate the two alkoxides.
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.
Protonate the departing alkoxide
The other workup water supplies ethanol’s O–H proton. The original ester O–ethyl bond survives throughout; protonating this fragment does not exchange the two original ester oxygens.
Nota: The second water protonates the ethoxide. Both original ester oxygens remain in their own organic fragments.
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
Propan-1-ol contains the original carbonyl oxygen and both transferred hydrogens on the former carbonyl carbon. Ethanol contains the original alkoxy oxygen and its intact two-carbon group. Both AlH₃ fragments and the two workup-derived hydroxides remain, preserving every introduced atom and net charge −2. These fragments show selected transfers, not complete aqueous aluminum speciation or a required minimum reagent stoichiometry.
O que você vai praticar
Follow two distinct Al–H transfers through an aldehyde, then map both ester oxygens into alcohols at a separate workup.
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