Química Orgánica 2 · Sustitución electrofílica aromática
4-bromotoluene from toluene with Br₂ and FeBr₃
Condiciones: Br2, FeBr3
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 6 pasos escritos
The ring attacks bromine
FeBr₃ (not drawn) polarizes Br₂. The ring carbon para to the methyl group attacks the outer bromine; aromaticity is lost in this slow step.
Las flechas curvas
- The ring π pair at the carbon bearing the drawn hydrogen forms the new C–Br bond, leaving a positive charge on the neighboring carbon.
- The Br–Br bond pair moves onto the far bromine, which leaves as bromide.
Delocalize the charge to the para carbon
Shifting a ring π pair puts the positive charge on the carbon bearing the methyl group. This tertiary-like contributor is stabilized by hyperconjugation with the methyl C–H bonds, which is why toluene brominates para (and ortho), not meta.
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 para?
Para (and ortho) attack puts positive charge on the methyl-bearing carbon in one contributor; meta attack never does.
Lose the proton and restore aromaticity
Bromide (released as FeBr₄⁻, regenerating FeBr₃) removes the proton from the sp³ carbon that gained bromine, and the C–H electrons restore the aromatic ring.
Las flechas curvas
- A lone pair on the bromide ion takes the ring proton, forming HBr.
- The C–H bond electrons form the ring π bond to the neighboring cationic carbon, restoring aromaticity.
4-bromotoluene and HBr
Bromine has replaced the hydrogen para to the methyl group, giving 4-bromotoluene; 2-bromotoluene forms as the other major isomer, and almost no meta product forms.
Qué vas a practicar
Draw bromination of toluene at the para position, showing the arenium contributor with the charge on the methyl-bearing carbon.
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