Química Orgânica 2 · Substituição eletrofílica aromática
Friedel–Crafts acetylation of anisole
Condições: Anisole, acetyl chloride (1.1 equiv), AlCl3 (1.1 equiv), CH2Cl2, 0 °C → 25 °C; then ice/HCl workup
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Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
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Acetyl chloride binds aluminium chloride
A chlorine lone pair of acetyl chloride coordinates to the empty orbital on aluminium, weakening the C–Cl bond.
As setas curvas
- The chlorine lone pair forms a dative bond to aluminium.
The acylium ion forms
The C–Cl bond breaks and the chlorine leaves as part of tetrachloroaluminate. The acylium ion is stabilised by the oxygen lone pair (C≡O⁺ contributor), so unlike alkyl cations it never rearranges.
As setas curvas
- The C–Cl bonding pair moves onto chlorine, which stays on aluminium as AlCl4⁻.
The para carbon of anisole attacks
A ring π bond at the carbon para to the methoxy group forms the new C–C bond to the acylium carbon, giving an arenium ion.
As setas curvas
- The π electrons at the para carbon bond to the acylium carbon.
Methoxy donation stabilises the arenium ion
Because attack was para, the positive charge can reach the ring carbon bearing oxygen, and an oxygen lone pair forms a C=O⁺ π bond. In this contributor every atom has an octet, which is why the methoxy group directs ortho/para.
As setas curvas
- The ring π bond shifts toward the cationic carbon, moving the positive charge to the carbon bearing oxygen.
- An oxygen lone pair forms a π bond to the ring, so oxygen carries the positive charge.
Why para?
Only ortho or para attack lets the methoxy oxygen share the positive charge; para is preferred over ortho for steric reasons.
Tetrachloroaluminate removes the ring proton
A chlorine of AlCl4⁻ takes the proton on the sp3 ring carbon and leaves aluminium as HCl, regenerating AlCl3. The C–H electrons, the ring π bond and the C=O⁺ π bond shift together to restore the aromatic ring and neutral methoxy group.
As setas curvas
- A chlorine lone pair takes the ring proton.
- The Al–Cl pair returns to chlorine, so HCl leaves and neutral AlCl3 is regenerated.
- The C–H pair forms a ring π bond.
- The neighbouring π bond shifts along the ring.
- The C=O⁺ π pair returns to oxygen as a lone pair.
4-Methoxyacetophenone
The aryl ketone forms together with HCl and regenerated AlCl3. The AlCl3 is not free to turn over, because the ketone oxygen binds it as a Lewis-acid complex; a full equivalent is therefore used, and the free ketone is released by the aqueous workup.
O que você vai praticar
Draw the formation of the acylium ion from acetyl chloride and AlCl3, its attack by the para carbon of anisole with the methoxy-stabilised arenium ion, and the deprotonation that gives 4-methoxyacetophenone.
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