Química Orgánica 1 · Reacciones radicalarias y selectividad
3-Bromocyclohexene by NBS allylic bromination
Condiciones: NBS, hν, CCl₄
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 5 pasos escritos
Initiation: make a bromine radical
Light splits a trace of Br₂ (kept at low concentration by NBS, which converts HBr back into Br₂) into two bromine radicals. NBS and succinimide are not drawn; a second Br₂ molecule is drawn for the propagation.
Las flechas curvas
- One electron of the Br–Br bond stays on the first atom.
- The other electron of the Br–Br bond stays on the second atom, so the bond splits evenly into two radicals.
Propagation: remove an allylic hydrogen
The bromine radical removes a hydrogen from a CH2 next to the double bond. The allylic radical formed is delocalized over three carbons, so the allylic C–H (about 360 kJ/mol) is the weakest in the molecule; vinylic C–H bonds are much stronger.
Las flechas curvas
- The unpaired electron on bromine supplies one electron of the new bond to hydrogen.
- One electron of the allylic C–H bond pairs with it to complete the new bond to hydrogen.
- The other allylic C–H electron stays behind as the unpaired electron of the new radical.
Why NBS?
NBS reacts with the HBr produced in each cycle to release Br₂ slowly, so Br₂ stays at a very low concentration. Addition of Br₂ across the C=C needs Br₂ and is suppressed, while the allylic chain uses only one Br₂ per cycle and keeps running.
Propagation: trap the allylic radical with Br₂
The allylic radical takes a bromine atom from Br₂. Its odd electron is shared by both ends of the allylic unit, but in cyclohexene the two ends are equivalent, so bromine at either end gives the same 3-bromocyclohexene.
Las flechas curvas
- The unpaired electron on the allylic carbon supplies one electron of the new bond to Br.
- One electron of the Br–Br bond completes that new bond.
- The other Br–Br electron stays on the far Br, regenerating the chain-carrying radical.
3-Bromocyclohexene, HBr and Br•
Bromine now sits next to the intact double bond. The new stereocentre formed from a planar radical, so 3-bromocyclohexene is racemic. HBr reacts with NBS to make the next Br₂.
Qué vas a practicar
Draw initiation and both propagation steps that convert cyclohexene into 3-bromocyclohexene with NBS and light, keeping the C=C intact.
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