Química Orgânica 2 · Substituição eletrofílica aromática
1,3,5-Trimethyl-2-nitrobenzene from mesitylene
Condições: HNO3, H2SO4
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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 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
Mesitylene has three equivalent ring C–H positions, each flanked by two methyl groups and para to the third, so it reacts very rapidly with NO₂⁺.
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 moves the charge to the para methyl-bearing carbon. In fact all three arenium contributors put the charge on a methyl-bearing carbon, so the methyl groups reinforce each other.
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.
How many products?
Symmetry makes all three ring C–H positions identical.
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.
1,3,5-Trimethyl-2-nitrobenzene
The nitro group occupies one of the three equivalent positions, giving a single product, 1,3,5-trimethyl-2-nitrobenzene; sulfuric acid is regenerated.
O que você vai praticar
Draw nitration of mesitylene and show that every arenium contributor puts the charge on a methyl-bearing carbon.
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