Química Orgânica 2 · Substituição eletrofílica aromática
Vilsmeier–Haack formylation of N,N-dimethylaniline
Condições: DMF, POCl3; then H2O
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O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 13 passos escritos
DMF attacks phosphorus
The carbonyl oxygen of DMF attacks phosphorus of POCl3 and a chloride leaves. DMF oxygen is unusually nucleophilic because the nitrogen lone pair pushes density onto it.
As setas curvas
- The DMF oxygen lone pair bonds to phosphorus.
- A P–Cl pair leaves as chloride.
Chloride adds to the activated carbon
Chloride attacks the iminium-like carbon of the O-phosphorylated DMF; the C=O⁺ π pair returns to oxygen.
As setas curvas
- A chloride lone pair bonds to the carbon.
- The C=O⁺ π pair returns to oxygen.
The Vilsmeier reagent forms
The nitrogen lone pair forms a C=N π bond and pushes out dichlorophosphate, giving the chloroiminium ion (Vilsmeier reagent), the electrophile.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with dichlorophosphate.
The para carbon attacks the chloroiminium ion
The ring carbon para to the dimethylamino group bonds to the iminium carbon; the C=N⁺ π pair returns to nitrogen. The chloroiminium ion is a weak electrophile, so only strongly activated rings react.
As setas curvas
- The para ring π pair bonds to the iminium carbon.
- The C=N⁺ π pair returns to nitrogen.
The amino group stabilises the arenium ion
A ring π bond shifts and the dimethylamino lone pair forms a C=N⁺ π bond, so every atom has an octet. This contributor explains the strong para activation.
As setas curvas
- A ring π bond shifts toward the cation.
- The nitrogen lone pair forms a π bond to the ring carbon.
Rearomatisation
Dichlorophosphate removes the ring proton; the π bonds shift back and the amino nitrogen regains its lone pair.
As setas curvas
- An oxygen lone pair of dichlorophosphate takes the ring proton.
- The C–H pair re-forms a ring π bond.
- The neighbouring π bond shifts along the ring.
- The C=N⁺ π pair returns to nitrogen.
Chloride leaves to form an iminium ion
The benzylic nitrogen lone pair pushes out chloride, giving an aryl iminium ion that water can attack in the workup.
As setas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–Cl pair leaves as chloride.
Water adds to the iminium carbon
In the aqueous workup water attacks the iminium carbon and the C=N⁺ π pair returns to nitrogen.
As setas curvas
- A water lone pair bonds to the iminium carbon.
- The C=N⁺ π pair returns to nitrogen.
Water removes the oxonium proton
A second water molecule removes a proton from the oxonium ion, giving the neutral N,O-hemiaminal and hydronium.
As setas curvas
- A water lone pair removes one oxonium proton.
- The O–H pair stays on oxygen.
Hydronium protonates the amine
Hydronium protonates the side-chain dimethylamino group, making it a leaving group.
As setas curvas
- The amine nitrogen lone pair takes the proton.
- The O–H pair stays on water.
Dimethylamine is expelled
The OH lone pair forms the C=O⁺ π bond and pushes out neutral dimethylamine.
As setas curvas
- An oxygen lone pair forms the C=O π bond.
- The C–N pair leaves with nitrogen as dimethylamine.
Dimethylamine removes the O–H proton
Dimethylamine, the strongest base present, removes the proton from the protonated aldehyde.
As setas curvas
- The amine lone pair takes the O–H proton.
- The O–H pair returns to oxygen.
4-(Dimethylamino)benzaldehyde
4-(Dimethylamino)benzaldehyde forms; dimethylammonium chloride and dichlorophosphoric acid are by-products.
O que você vai praticar
Draw formation of the chloroiminium (Vilsmeier) reagent from DMF and POCl3, its para attack by N,N-dimethylaniline, and the aqueous hydrolysis of the iminium ion to 4-(dimethylamino)benzaldehyde.
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