Química Orgânica 1 · Reações radicalares e seletividade
3-Bromo-3-methylpentane by tertiary selectivity
Condições: Br₂, hν
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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.
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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 hydrogen
The bromine radical removes the one tertiary hydrogen on the central carbon, giving a tertiary radical stabilised by hyperconjugation with three alkyl groups. Secondary and primary hydrogens react far more slowly with bromine.
As setas curvas
- The unpaired electron on the bromine radical supplies one electron of the new bond to hydrogen.
- One electron of the tertiary C–H bond pairs with it to complete the new bond to hydrogen.
- The other tertiary 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 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.
3-Bromo-3-methylpentane and HBr
Bromine replaced the tertiary hydrogen, giving 3-bromo-3-methylpentane. The central carbon carries two identical ethyl groups, so it is not a stereocentre.
O que você vai praticar
Draw the radical chain that brominates 3-methylpentane at its single tertiary carbon, giving 3-bromo-3-methylpentane.
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