Química Orgânica 1 · Reações radicalares e seletividade
1-Bromo-1-methylcyclohexane by tertiary ring bromination
Condições: Br₂, hν
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: split Br₂ with light
Light breaks the weak Br–Br bond homolytically: each bromine atom keeps one of the two bonding electrons, giving two neutral bromine radicals and no ions. A second Br₂ molecule waits for the propagation steps.
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 the tertiary ring hydrogen
Of the fourteen hydrogens, only one is tertiary: the one on the ring carbon that also holds the methyl group. Bromine removes it to give a tertiary radical, the most stable radical available in this molecule.
As setas curvas
- The unpaired electron on the bromine radical supplies one electron of the new bond to hydrogen.
- One electron of the tertiary ring C–H bond pairs with it to complete the new bond to hydrogen.
- The other tertiary ring C–H electron stays behind as the unpaired electron of the new radical.
Why this hydrogen?
Radical stability follows tertiary > secondary > primary > methyl, and the weakest C–H bond gives the most stable radical. Bromine is far more selective than chlorine because its abstraction step is endothermic, with a late, product-like transition state.
Propagation: take Br from a second Br₂
The carbon radical attacks Br₂, forming the C–Br bond and releasing a new bromine radical. That radical starts the next cycle, so one initiation event drives many cycles.
As setas curvas
- The unpaired electron on tertiary ring C 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.
1-Bromo-1-methylcyclohexane and HBr
Bromine replaced the tertiary hydrogen, giving 1-bromo-1-methylcyclohexane as the major product. The new C–Br carbon is not a stereocentre because two ring paths around it are identical.
O que você vai praticar
Draw the radical chain that brominates methylcyclohexane at the ring carbon bearing the methyl group, giving 1-bromo-1-methylcyclohexane.
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