Química Orgánica 2 · Sustitución electrofílica aromática
4-Nitroacetanilide from acetanilide
Condiciones: HNO3, H2SO4, 0 °C
Dibuja este problema tú mismo
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 acetamido group attacks NO₂⁺. The amide nitrogen still donates into the ring, though less than a free NH₂ because its lone pair is shared with the carbonyl.
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 nitrogen-bearing carbon, where the amide nitrogen lone pair can help stabilize it. That extra stabilization is available only for ortho/para attack.
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.
Why acetylate?
Acetylation tames the amine: the ring stays activated but is not over-reactive, nitrogen is not oxidized, and it is not protonated to a meta-directing anilinium ion.
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-Nitroacetanilide, water and sulfuric acid
The nitro group enters para to the acetamido group, giving 4-nitroacetanilide. Hydrolysis of the amide would then give 4-nitroaniline.
Qué vas a practicar
Draw nitronium formation and para nitration of acetanilide, showing why the acetamido group is a moderate ortho/para director.
Acabas de dibujar cada flecha de este problema.
Cada uno de los 5349 problemas de Orgánica 1 y 2 funciona así: comentarios en cada flecha, pistas cuando te atascas y un modo examen sin pistas. Gratis con una cuenta, sin tarjeta.
¿Atascado con tu propia tarea?
Tómale una foto: el primer escaneo es gratis, sin registrarte
¿Escaneas cada semana? Scan Standard: por unos problemas