Química Orgânica 2 · Substituição eletrofílica aromática
Electrophilic aromatic substitution — nitration
Condições: Benzene with nitric acid and sulfuric acid. Draw NO₂⁺ formation, ring substitution, and regeneration of H₂SO₄ from HSO₄⁻. Other acid equilibria and ion pairing are not drawn; sulfur is drawn with more than eight electrons, but nitrogen always has an octet.
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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.
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Protonate nitric acid
Transfer a proton from sulfuric acid to the OH group of nitric acid so water can leave next.
Nota: Protonate the OH oxygen of nitric acid, not either of its other oxygens.
As setas curvas
- The nitric acid OH oxygen takes a proton, turning OH into a water leaving group. The N–O bond stays intact for now.
- The O–H electrons stay on the sulfuric acid oxygen, forming HSO₄⁻ as nitric acid gains a positive charge.
Lose water and form nitronium
Draw both arrows together to make NO₂⁺. Nitrogen keeps its octet throughout this step.
Nota: The negatively charged oxygen forms the new π bond. The double-bonded oxygen keeps its π bond, and the protonated OH group leaves as water.
As setas curvas
- The N–OH₂ bond electrons go to oxygen, releasing water with the original nitric acid proton and the proton from sulfuric acid.
- A lone pair on the negatively charged oxygen forms the second N=O bond as water leaves. That oxygen becomes neutral and nitrogen keeps eight electrons.
The ring attacks nitronium
Form the C–N bond and open one N=O π bond together. Water and hydrogen sulfate do not react in this step.
Nota: Use the ring π bond touching the carbon with the drawn H to form the C–N bond at that carbon.
As setas curvas
- The ring π electrons form a C–N bond. The ring temporarily loses aromaticity, with positive charge on the adjacent carbon.
- An N=O π pair moves onto oxygen as C–N forms, keeping eight electrons around nitrogen.
Restore the ring and acid
Remove the proton from the carbon that gained NO₂. Its C–H electrons restore aromaticity in the same step.
Nota: Use the negatively charged oxygen of hydrogen sulfate to accept the proton from the carbon that gained NO₂. Sulfuric acid gets a ring proton back, not the proton it gave away.
As setas curvas
- The negatively charged oxygen of hydrogen sulfate takes the ring proton, regenerating sulfuric acid.
- The C–H electrons restore the ring π bond and remove the ring charge. The new C–N bond stays intact.
Nitrobenzene, water and sulfuric acid
Nitrogen and the two non-hydroxyl oxygens of nitric acid become the nitro group, with positive charge on nitrogen and negative charge on the singly bonded oxygen. The original hydroxyl oxygen leaves as water with a nitric acid proton and a sulfuric acid proton. Sulfuric acid is regenerated with the ring proton in place of the proton it donated; its other OH group is unchanged. All atoms remain present and the total charge stays zero.
O que você vai praticar
Make NO₂⁺ from nitric acid, add it to benzene, and regenerate sulfuric acid.
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