Química Orgánica 1 · Reacciones radicalarias y selectividad
(1-Bromoethyl)benzene by benzylic bromination
Condiciones: Br₂, hν
Dibuja este problema tú mismo
Arrastra desde los electrones que se mueven hasta donde van, o toca uno y luego el otro. Cada flecha se revisa; Pista te ayuda cuando te atascas. No necesitas cuenta.
El mecanismo, paso a paso
Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
Leer los 5 pasos 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.
Las flechas 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 a benzylic hydrogen
The bromine radical removes a hydrogen from the CH2 attached to the ring. The benzylic C–H is unusually weak (about 357 kJ/mol) because the radical formed is delocalized into the aromatic π system, onto the ortho and para carbons.
Las flechas curvas
- The unpaired electron on the bromine radical supplies one electron of the new bond to hydrogen.
- One electron of the benzylic C–H bond pairs with it to complete the new bond to hydrogen.
- The other benzylic C–H electron stays behind as the unpaired electron of the new radical.
Why the benzylic position?
Benzylic and allylic radicals are resonance-stabilized, so their C–H bonds are the weakest in the molecule and react first.
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.
Las flechas curvas
- The unpaired electron on benzylic 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-Bromoethyl)benzene and HBr
Bromine sits on the benzylic carbon. The planar benzylic radical is attacked from either face, so (1-bromoethyl)benzene forms as a racemate.
Qué vas a practicar
Draw the radical chain that brominates ethylbenzene at the benzylic CH2, giving (1-bromoethyl)benzene (Br2 and light on ethylbenzene).
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