Química Orgânica 1 · Reações radicalares e seletividade
Allyl bromide (3-bromoprop-1-ene) from propene with NBS
Condições: NBS, hν, CCl₄
Desenhe este problema você mesmo
Arraste dos elétrons que se movem até onde eles vão, ou toque em um e depois no outro. Cada seta é verificada; Dica ajuda quando você travar. Não precisa de conta.
O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 5 passos 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.
As setas 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.
As setas 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.
As setas 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.
O que você vai praticar
Draw the NBS radical chain that brominates propene at its methyl group, giving allyl bromide (3-bromoprop-1-ene).
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