Química Orgânica 2 · Ligações carbono–carbono com organometálicos
Open a cyclic ester with excess carbon donor
Condições: Gamma-butyrolactone with two modeled equivalents of methyllithium, followed by two explicit water proton transfers. Dry ether first, then the stated separate workup. R⁻ is only an ionic organolithium electron-accounting model, not a solution-speciation claim.
Desenhe este problema você mesmo
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.
Ler todos os 5 passos escritos
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 cyclic acyl intermediate
The leaving alkoxide remains tethered to the carbon chain; ring opening does not create a separate carbon fragment.
As setas curvas
- The oxygen lone pair restores the ketone carbonyl.
- The acyl–oxygen bond opens the lactone ring.
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
The lactone ring becomes an open-chain diol. The former carbonyl oxygen is the tertiary OH and the original ring oxygen is the terminal OH; neither end of the tether is lost.
O que você vai praticar
Retain the tethered leaving oxygen through lactone double addition and produce a diol rather than an alcohol plus a detached alkoxy group.
Você acabou de desenhar cada seta deste problema.
Cada um dos 5349 problemas de Orgânica 1 e 2 funciona assim: feedback em cada seta, dicas quando você travar e um modo prova sem dicas. Grátis com uma conta, sem cartão.
Travou na sua própria lição?
Tire uma foto: o primeiro scan é grátis, sem cadastro
Você escaneia toda semana? Scan Standard: por cerca de problemas