Química Orgânica 2 · Ligações carbono–carbono com organometálicos
Count both additions to an ester
Condições: Methyl acetate with two modeled equivalents of methyllithium; after both additions, introduce two waters for explicit proton accounting. Dry ether first, then the stated separate workup. R⁻ is only an ionic organolithium electron-accounting model, not a solution-speciation claim.
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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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Complete the dry carbon–carbon event
The carbon–metal reagent is represented by its bounded ionic electron-accounting form; no proton donor is present yet.
Nota: The two methyl donors are equivalent. Both ester oxygens remain through the workup.
As setas curvas
- The nucleophilic carbon pair forms a new C–C bond.
- The carbonyl π pair moves onto oxygen to preserve the carbon octet.
Resolve the first addition intermediate
An ester can lose alkoxide; this ketone intermediate is not the isolated endpoint in excess reagent.
As setas curvas
- The alkoxide pair restores a carbonyl.
- The acyl C–O pair leaves with methoxide.
Complete the dry carbon–carbon event
The carbon–metal reagent is represented by its bounded ionic electron-accounting form; no proton donor is present yet.
As setas curvas
- The nucleophilic carbon pair forms a new C–C bond.
- The carbonyl π pair moves onto oxygen to preserve the carbon octet.
Account for both workup proton transfers
Two waters enter after the dry reaction. This panel groups two independent proton transfers, not one concerted elementary event, and imposes no order between the two alkoxides. Both hydroxides are retained.
Nota: Show both independent workup proton transfers; their order is not tested.
As setas curvas
- One alkoxide accepts a proton from water.
- The first water retains its O–H bonding pair.
- The other alkoxide accepts a proton from the other water.
- The second water retains its O–H bonding pair.
Products
tert-Butanol contains both incoming methyl groups. Methoxide is retained and protonated to methanol rather than disappearing; two water molecules become two hydroxides in the bounded workup model.
O que você vai praticar
Continue ester addition through leaving-group loss, ketone capture, and both alkoxide protonations.
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