Química Orgânica 2 · Substituição eletrofílica aromática
tert-Butylbenzene from benzene and tert-butyl chloride
Condições: AlCl3
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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 8 passos escritos
Chlorine bonds to AlCl₃
The Lewis acid AlCl₃ accepts a lone pair from the chlorine of tert-butyl chloride, turning chloride into a much better leaving group.
As setas curvas
- A chlorine lone pair forms a bond to aluminum, the Lewis acid.
Ionize to the tert-butyl cation
The C–Cl bond breaks, giving the tertiary tert-butyl cation and AlCl₄⁻. A tertiary cation forms readily and cannot rearrange to anything more stable.
As setas curvas
- The C–Cl bond pair moves onto chlorine, leaving the tertiary carbocation and AlCl₄⁻.
The ring attacks the carbocation
A ring π pair bonds to the cationic carbon; the ring loses aromaticity 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 tert-butyl 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.
Alkylation limits
The alkylated product is more reactive than benzene, so excess benzene is used to limit polyalkylation.
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.
tert-Butylbenzene, HCl and AlCl₃
The tert-butyl group has replaced a ring hydrogen, giving tert-butylbenzene. HCl is released and AlCl₃ is regenerated, so it is a catalyst.
O que você vai praticar
Draw AlCl₃ activation of tert-butyl chloride, formation of the tert-butyl cation, ring attack, and deprotonation by AlCl₄⁻ to give tert-butylbenzene.
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