Química Orgánica 1 · Reacciones radicalarias y selectividad
1-Chloro-2-methylpropane and 2-chloro-2-methylpropane from isobutane
Condiciones: Cl₂, hν
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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 7 pasos escritos
Initiation: split Cl₂ with light
Light breaks the Cl–Cl bond homolytically into two chlorine radicals. Two alkane molecules and two more Cl₂ molecules are drawn so that both product-forming branches can be followed.
Las flechas curvas
- One electron of the Cl–Cl bond stays on the first atom.
- The other electron of the Cl–Cl bond stays on the second atom, so the bond splits evenly into two radicals.
Propagation: abstract a primary hydrogen
A chlorine radical removes one of the nine methyl hydrogens, giving the primary isobutyl radical. There are nine such hydrogens, so this branch wins on numbers.
Las flechas curvas
- The unpaired electron on chlorine supplies one electron of the new bond to hydrogen.
- One electron of the primary C–H bond pairs with it to complete the new bond to hydrogen.
- The other primary C–H electron stays behind as the unpaired electron of the new radical.
Propagation: abstract the tertiary hydrogen
The other chlorine radical removes the single tertiary hydrogen of a second isobutane, giving the tert-butyl radical. Per hydrogen this is about five times faster than primary abstraction.
Las flechas curvas
- The unpaired electron on chlorine 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.
Numbers versus reactivity
Relative amount = hydrogens × reactivity: primary 9 × 1 = 9, tertiary 1 × 5 = 5, so about 64 % primary and 36 % tertiary chloride.
Propagation: the isobutyl radical takes chlorine
The first carbon radical attacks Cl₂, forming its C–Cl bond and releasing a chlorine radical that continues the chain.
Las flechas curvas
- The unpaired electron on the isobutyl carbon supplies one electron of the new bond to Cl.
- One electron of the Cl–Cl bond completes that new bond.
- The other Cl–Cl electron stays on the far Cl, regenerating the chain-carrying radical.
Propagation: the tert-butyl radical takes chlorine
The second carbon radical reacts the same way with another Cl₂. Both chlorine-transfer steps are fast and exothermic; the product ratio was already fixed when the hydrogens were removed.
Las flechas curvas
- The unpaired electron on the tert-butyl carbon supplies one electron of the new bond to Cl.
- One electron of the Cl–Cl bond completes that new bond.
- The other Cl–Cl electron stays on the far Cl, regenerating the chain-carrying radical.
Both monochlorides and HCl
1-Chloro-2-methylpropane (major) and 2-chloro-2-methylpropane (minor) form together. With Br2 instead, the tertiary bromide would be essentially the only product.
Qué vas a practicar
Draw primary and tertiary hydrogen abstraction in the chlorination of isobutane (2-methylpropane) and account for the mixture of 1-chloro-2-methylpropane and 2-chloro-2-methylpropane.
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