Química Orgánica 2 · Efectos de los sustituyentes y orientación
Nitration of m-xylene at C4
Condiciones: HNO3, H2SO4
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Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
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The ring attacks nitronium at C4
The nitronium ion, O=N+=O, is formed from nitric acid by sulfuric acid (protonation of the N–OH, loss of water); it is drawn here already formed. C4 is ortho to one methyl and para to the other, so both groups agree. The ring π pair forms the C–N bond at C4 and the cation appears right on a methyl-bearing carbon.
Las flechas curvas
- The ring π pair forms the new C–N bond to the nitronium nitrogen.
- One N=O π pair moves onto oxygen so nitrogen does not exceed four bonds; the new group is a neutral nitro group.
Both methyl groups share the charge
The positive charge can sit on either methyl-bearing carbon, so both methyls stabilize this σ complex by hyperconjugation. C2, between the two methyls, would give the same benefit but is too crowded.
Las flechas curvas
- The ring π pair shifts toward the cation, moving the positive charge to the other methyl-bearing carbon.
Why not C2?
C2 and C4 are electronically similar; sterics decide.
Hydrogen sulfate removes the ring proton
Hydrogen sulfate, the strongest base in mixed acid, removes the hydrogen on the sp3 ring carbon; the electrons flow back around the ring and aromaticity is restored.
Las flechas curvas
- A hydrogen sulfate oxygen lone pair takes the hydrogen on the sp3 ring carbon.
- The C–H bond pair becomes a ring π bond, starting to restore the aromatic sextet.
- A ring π pair shifts one bond along to make room for the new π bond.
2,4-dimethyl-1-nitrobenzene
Nitration gives 2,4-dimethyl-1-nitrobenzene (4-nitro-m-xylene) as the main product; C2 substitution is minor because of steric hindrance, and sulfuric acid is regenerated.
Qué vas a practicar
Draw nitration of m-xylene at C4 and show that its σ complex is stabilized by both methyl groups.
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