Química Orgánica 1 · Reacciones radicalarias y selectividad
Allyl bromide (3-bromoprop-1-ene) from propene with NBS
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 the methyl group next to the C=C, giving the allyl radical. That radical is resonance-stabilized (the odd electron is shared by both terminal carbons), which makes the allylic C–H bond the weakest one in propene.
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 allyl radical takes a bromine atom from Br₂. Its two ends are equivalent, so bonding at either terminal carbon gives the same allyl bromide, and a new bromine radical continues the chain.
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.
Allyl bromide, HBr and Br•
The double bond survived and bromine replaced an allylic hydrogen: 3-bromoprop-1-ene (allyl bromide). The HBr formed reacts with NBS to release the next Br₂.
Qué vas a practicar
Draw the NBS radical chain that brominates propene at its methyl group, giving allyl bromide (3-bromoprop-1-ene).
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