Química Orgânica 2 · Substituição eletrofílica aromática
2,4,6-Tribromoaniline from aniline in bromine water
Condições: Br2 (excess), H2O
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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 ring attacks bromine para to NH₂
The amino group donates its lone pair into the ring, making aniline so nucleophilic that Br₂ reacts without any Lewis acid. The para carbon attacks first.
As setas curvas
- The ring π pair at the carbon para to NH₂ forms the new C–Br bond; the neighboring carbon becomes cationic.
- The Br–Br pair moves onto the outer bromine, which leaves as bromide.
Move the charge to the amino-bearing carbon
Shifting a ring π pair places the positive charge on the carbon bonded to nitrogen. There the nitrogen lone pair can share the charge (an iminium contributor), which is why NH₂ is such a strong ortho/para activator.
As setas curvas
- The ring π pair shifts toward the cation, moving the charge onto the carbon that carries NH₂.
Return the charge beside the sp³ carbon
The arenium ion is one hybrid; redraw the contributor with the charge on the carbon next to the sp³ C–H. Its C–H electrons will re-form that π bond in the deprotonation.
As setas curvas
- The π pair shifts back toward the amino-bearing carbon, leaving positive charge beside the sp³ carbon.
Water removes the para proton
Water (the solvent) removes the proton from the sp³ carbon bearing bromine, and the C–H electrons restore the aromatic ring: 4-bromoaniline.
As setas curvas
- A water lone pair takes the ring proton.
- The C–H electrons re-form the ring π bond and remove the positive charge.
Why three bromines?
NH₂ activates the ring so strongly that every ortho/para position reacts in water; to stop at one bromine, the amine is first acetylated.
Brominate one ortho carbon
A second Br₂ is attacked by a carbon ortho to NH₂. The positive charge appears directly on the amino-bearing carbon, where the nitrogen lone pair can share it.
As setas curvas
- The ring π pair at the ortho carbon forms the new C–Br bond, placing the charge on the carbon bearing NH₂.
- The Br–Br pair leaves as bromide.
Water removes the ortho proton
Water removes the proton from the newly brominated ortho carbon; the C–H pair restores the aromatic ring.
As setas curvas
- A water lone pair takes the ring proton.
- The C–H electrons re-form the ring π bond.
Brominate the other ortho carbon
The last ortho position attacks a third Br₂. Its arenium ion is delocalized onto the amino-bearing carbon just like the first ortho attack.
As setas curvas
- The ring π pair at the second ortho carbon forms the new C–Br bond; the neighboring meta carbon becomes cationic.
- The Br–Br pair leaves as bromide.
Water removes the last ring proton
Water removes the final ring proton and the aromatic ring is restored, giving 2,4,6-tribromoaniline, which precipitates from water.
As setas curvas
- A water lone pair takes the ring proton.
- The C–H electrons re-form the ring π bond.
2,4,6-Tribromoaniline
All three positions ortho and para to the amino group now carry bromine, giving 2,4,6-tribromoaniline as a white precipitate. The three protons end up as hydronium and the other three bromines as bromide (3 HBr overall).
O que você vai praticar
Draw the three successive brominations of aniline in water (para, then both ortho positions) that give 2,4,6-tribromoaniline.
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