Química Orgânica 1 · Radicais, álcoois e éteres
Why bromination is more selective than chlorination
Condições: Hydrogen abstraction, X• + R–H → H–X + R•. Bond dissociation energies (≈): H–Cl 431, H–Br 366 kJ/mol; alkane C–H bonds 400–440 kJ/mol.
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A pergunta
Why does a bromine atom discriminate between primary, secondary and tertiary hydrogens much more than a chlorine atom does?
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Resposta e explicação
Resposta correta Abstraction by Br• is endothermic, so its transition state is late and resembles the carbon radical; differences in radical stability are therefore felt almost fully.
By the Hammond postulate an endothermic step has a product-like transition state, so the activation energies differ nearly as much as the radical energies do.
Esta não Abstraction by Cl• is endothermic, so the transition state is late and all hydrogens look alike.
H–Cl is about as strong as a C–H bond, so the chlorine step is roughly thermoneutral or exothermic, with an early transition state. A late transition state would increase selectivity, not reduce it.
Esta não The bromine atom is larger, so it can reach only the tertiary hydrogen.
A larger atom would favour the more exposed primary hydrogens. Selectivity follows bond strength, not size.
Esta não The Br–Br bond is stronger than the Cl–Cl bond, so fewer bromine atoms form and each one chooses carefully.
Br–Br (about 193 kJ/mol) is weaker than Cl–Cl (about 242 kJ/mol), and the number of radicals does not decide which hydrogen each one removes.
Solução explicada
Compare the step energies: Br• + R–H is uphill because H–Br is weaker than C–H, whereas Cl• + R–H is close to thermoneutral. An uphill step has a late transition state that looks like R•, so a more stable radical means a lower barrier almost one for one. An early transition state barely feels the radical's stability, so chlorine is unselective.
O que você vai praticar
Explain the different selectivities of chlorination and bromination with the Hammond postulate.
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