Química Orgánica 2 · Sustitución electrofílica aromática
4-Nitroanisole from anisole with HNO₃ and H₂SO₄
Condiciones: HNO3, H2SO4
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Arrastra desde los electrones que se mueven hasta donde van, o toca uno y luego el otro. Cada flecha se revisa; Pista te ayuda cuando te atascas. No necesitas cuenta.
El mecanismo, paso a paso
Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
Leer los 8 pasos escritos
Protonate nitric acid
Sulfuric acid, the stronger acid, protonates the OH oxygen of nitric acid so that it can leave as water.
Las flechas 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.
Las flechas 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 methoxy group attacks NO₂⁺. Anisole is strongly activated, so this step is much faster than for benzene.
Las flechas 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
The charge moves onto the methoxy-bearing carbon. Here an oxygen lone pair can form a C=O⁺ bond, giving a fourth contributor in which every atom has an octet — the reason OMe is a strong ortho/para activator.
Las flechas 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.
Las flechas curvas
- The π pair shifts back toward the para carbon, leaving the positive charge on the ortho carbon next to the sp³ C–H.
Oxygen stabilization
The oxonium contributor is the most important one for para attack because every atom has a full octet.
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.
Las flechas 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-Nitroanisole, water and sulfuric acid
The nitro group enters para to the methoxy group, giving 4-nitroanisole (with 2-nitroanisole as the other isomer). Sulfuric acid is regenerated.
Qué vas a practicar
Draw nitronium formation and para nitration of anisole, noting how the methoxy oxygen stabilizes the arenium ion.
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