Química Orgânica 2 · Ligações carbono–carbono com organometálicos
Propiophenone from benzonitrile and ethylmagnesium bromide
Condições: EtMgBr, Et2O; then H3O+, heat
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 10 passos escritos
The ethyl carbon adds to the nitrile carbon
The nucleophilic carbon of the C–Mg bond attacks the electrophilic nitrile carbon. One π pair of C≡N moves onto nitrogen, giving an imine anion paired with MgBr+.
As setas curvas
- The C–Mg bond pair forms the new C–C bond to the nitrile carbon.
- One π pair of the triple bond moves onto nitrogen.
Why only one addition?
An ester reacts twice with excess Grignard reagent; the nitrile does not.
Hydronium protonates the imine anion
Aqueous acid is added. The basic imine nitrogen takes a proton from hydronium, giving the neutral imine.
As setas curvas
- A nitrogen lone pair takes a proton from hydronium.
- The O–H pair stays on water.
Protonation activates the imine
A second hydronium protonates the imine nitrogen, giving an iminium ion whose carbon is strongly electrophilic.
As setas curvas
- The imine nitrogen lone pair takes a proton from hydronium.
- The O–H pair stays on water.
Water adds to the iminium carbon
A water molecule attacks the iminium carbon and the C=N π pair moves onto nitrogen, giving a tetrahedral intermediate with a protonated oxygen.
As setas curvas
- A water lone pair forms the new C–O bond.
- The C=N π pair moves onto nitrogen, neutralising it.
Water removes a proton from the oxonium ion
A solvent water molecule deprotonates the new oxonium oxygen, giving a neutral carbinolamine.
As setas curvas
- A water lone pair takes a proton from the oxonium oxygen.
- The O–H pair stays on the carbinolamine oxygen.
Hydronium protonates the nitrogen
Protonating the amino group turns it into a neutral ammonia leaving group.
As setas curvas
- The nitrogen lone pair takes a proton from hydronium.
- The O–H pair stays on water.
The oxygen lone pair expels ammonia
A lone pair on the hydroxyl oxygen forms the C=O π bond as the C–N bond breaks, releasing ammonia and leaving a protonated ketone.
As setas curvas
- An oxygen lone pair forms the C=O π bond.
- The C–N pair leaves with nitrogen as ammonia.
Water deprotonates the protonated ketone
Water removes the proton from the carbonyl oxygen, giving neutral propiophenone and regenerating hydronium.
As setas curvas
- A water lone pair takes the proton from the carbonyl oxygen.
- The O–H pair returns to oxygen as a lone pair.
Propiophenone
The nitrile carbon has become the ketone carbon and now carries the new ethyl group: propiophenone (1-phenylpropan-1-one). Ammonia is released and is protonated to ammonium in the acidic solution.
O que você vai praticar
Draw the single addition of ethylmagnesium bromide to benzonitrile and the acidic hydrolysis of the imine to propiophenone.
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