Química Orgânica 2 · Substituição eletrofílica aromática
Isopropylbenzene (cumene) from benzene and 1-chloropropane
Condições: AlCl3
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 8 passos escritos
Chlorine bonds to AlCl₃
AlCl₃ accepts a lone pair from the chlorine of 1-chloropropane, making the C–Cl bond weak and polarized.
As setas curvas
- A chlorine lone pair forms a bond to aluminum, the Lewis acid.
Hydride shift as chloride leaves
A free primary cation is too unstable to form. Instead, a hydride from the middle carbon shifts to the terminal carbon as the C–Cl bond breaks, giving the secondary isopropyl cation directly.
As setas curvas
- The C–H bond pair on the middle carbon moves with its hydrogen to the terminal carbon losing chlorine.
- The C–Cl bond pair moves onto chlorine, giving AlCl₄⁻; the middle carbon is left cationic.
Why rearrange?
Cation stability: secondary > primary.
The ring attacks the isopropyl cation
A ring π pair bonds to the secondary carbocation; aromaticity is lost in this slow step.
As setas curvas
- The ring π pair at the carbon bearing the drawn hydrogen forms the new C–C bond to the cationic middle carbon.
Delocalize the charge to the para carbon
The arenium ion carries its positive charge on the two ortho carbons and the para carbon.
As setas curvas
- The neighboring ring π pair shifts toward the cationic carbon, moving the positive charge to the carbon para to the new bond.
Return the charge beside the sp³ carbon
The arenium ion is one hybrid; redraw the contributor with the positive charge back on the carbon next to the sp³ C–H. That contributor shows which π bond the C–H electrons will re-form in the deprotonation.
As setas curvas
- The π pair shifts back toward the para carbon, leaving the positive charge on the ortho carbon next to the sp³ C–H.
AlCl₄⁻ removes the ring proton
A chloride of AlCl₄⁻ removes the proton from the sp³ carbon bearing the alkyl group, giving HCl, regenerating AlCl₃ and restoring aromaticity.
As setas curvas
- The Al–Cl bond pair forms a new Cl–H bond, so AlCl₄⁻ releases chloride as it takes the proton.
- The C–H bond electrons form the ring π bond to the neighboring cationic carbon, restoring aromaticity.
Isopropylbenzene (cumene), HCl and AlCl₃
The ring is bonded to the middle carbon of the former propyl chain, giving isopropylbenzene (cumene) as the major product. HCl is released and AlCl₃ is regenerated.
O que você vai praticar
Draw AlCl₃ activation of 1-chloropropane with a 1,2-hydride shift to the isopropyl cation, ring attack and deprotonation to give isopropylbenzene (cumene).
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