Química Orgânica 1 · Reações radicalares e seletividade
Brominate the benzylic carbon of toluene with NBS
Condições: Toluene with N-bromosuccinimide (NBS) and light (or a peroxide initiator) in CCl₄. Initiation has already supplied Br•. NBS reacts with the HBr formed in each cycle to release Br₂ at a low, steady concentration; one such Br₂ molecule is drawn. That ionic NBS/HBr exchange and succinimide are not drawn. Follow one propagation cycle.
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Abstract a benzylic hydrogen
Br• removes a hydrogen atom (H•), not a proton. The benzylic C–H bond (about 375 kJ/mol) is much weaker than an aromatic C–H bond (about 470 kJ/mol) because the benzyl radical it gives is resonance-stabilized.
Nota: The three methyl hydrogens are equivalent.
As setas curvas
- The bromine radical supplies one electron of the new H–Br bond.
- One electron of the benzylic C–H bond completes the H–Br bond.
- The other C–H electron stays on the benzylic carbon as its unpaired electron.
Delocalize the benzyl radical into the ring
This contributor puts the odd electron on an ortho carbon; further shifts place it at the other ortho carbon and at the para carbon. The single radical is spread over four carbons, which is why a benzylic radical is more stable than even a tertiary alkyl radical.
As setas curvas
- The unpaired electron on the benzylic carbon pairs into a new π bond to the ring.
- One electron of the adjacent ring π bond completes that exocyclic π bond.
- The other ring π electron is left on the ortho carbon as the unpaired electron.
Choose the carbon that bonds to bromine
Resonance contributors differ only in electron placement. The observed product comes from bonding at the carbon that lets the ring keep its aromatic sextet.
Transfer bromine to the benzylic carbon
Use the benzylic contributor of the same delocalized radical. The weak Br–Br bond (about 190 kJ/mol) makes this step fast and exothermic.
Nota: This is the same radical drawn with its benzylic contributor.
As setas curvas
- The benzylic unpaired electron supplies one electron of the new C–Br bond.
- One Br–Br bonding electron completes the C–Br bond.
- The other Br–Br electron regenerates the bromine radical chain carrier.
Benzyl bromide, HBr and a new Br radical
The benzylic carbon now carries bromine and two hydrogens: (bromomethyl)benzene. The ring keeps three π bonds. HBr formed in the first step goes on to react with NBS, supplying the next Br₂. The new bromine radical is neutral (three lone pairs, one unpaired electron) and starts the next cycle.
O que você vai praticar
Trace NBS-mediated benzylic bromination through a resonance-delocalized benzyl radical and explain why bromine ends on the side chain rather than the ring.
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