Química Orgânica 2 · Substituição eletrofílica aromática
4-Nitrotoluene from toluene with HNO₃ and H₂SO₄
Condições: HNO3, H2SO4, 0 °C
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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
Protonate nitric acid
Sulfuric acid, the stronger acid, protonates the OH oxygen of nitric acid so that it can leave as water.
As setas curvas
- The OH oxygen of nitric acid uses a lone pair to take a proton from sulfuric acid.
- The O–H electrons stay on the sulfuric acid oxygen, giving hydrogen sulfate.
Lose water to form nitronium
Water leaves as the negatively charged oxygen forms a second N=O bond, giving the linear nitronium ion, NO₂⁺ — the electrophile.
As setas curvas
- The N–O bond to the protonated oxygen breaks, releasing water.
- A lone pair on the negative oxygen forms the new N=O π bond so nitrogen keeps its octet.
The ring attacks nitronium
The carbon para to the methyl group attacks NO₂⁺. Toluene reacts about 25 times faster than benzene because the methyl group stabilizes the arenium ion.
As setas curvas
- The ring π pair at the carbon bearing the drawn hydrogen forms the new C–N bond; the neighboring carbon becomes cationic.
- One N=O π pair moves onto oxygen as the C–N bond forms, keeping eight electrons on nitrogen.
Delocalize the charge to the para carbon
Shifting a ring π pair puts the positive charge on the methyl-bearing carbon, a tertiary-like cation stabilized by hyperconjugation. Meta attack has no such contributor.
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.
Why para?
Para and ortho attack each give one contributor with the charge on the methyl-bearing carbon.
Hydrogen sulfate removes the ring proton
Hydrogen sulfate removes the proton from the sp³ carbon that gained the nitro group; the C–H pair restores aromaticity and sulfuric acid is regenerated.
As setas curvas
- The negative oxygen of hydrogen sulfate takes the ring proton, regenerating sulfuric acid.
- The C–H bond electrons form the ring π bond to the neighboring cationic carbon, restoring aromaticity.
4-Nitrotoluene, water and sulfuric acid
A nitro group has replaced the hydrogen para to the methyl group, giving 4-nitrotoluene. Water is released and sulfuric acid is regenerated, so it acts as a catalyst.
O que você vai praticar
Draw nitronium formation and nitration of toluene at the para position, including the contributor with charge on the methyl-bearing carbon.
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