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  1. Problemas de práctica
  2. Orgánica 1
  3. Reacciones radicalarias y selectividad
  4. 1-Chloropropane and 2-chloropropane from propane

Química Orgánica 1 · Reacciones radicalarias y selectividad

1-Chloropropane and 2-chloropropane from propane

Dibuja el mecanismo · Aplicación

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.

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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
  1. 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.
  2. Propagation: abstract a primary hydrogen

    One chlorine radical removes a hydrogen from a terminal CH3, giving the primary propyl radical. Chlorine abstraction is exothermic with an early transition state, so the primary C–H reacts almost as readily as a secondary one.

    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.
  3. Propagation: abstract a secondary hydrogen

    The other chlorine radical removes a hydrogen from the central CH2 of a second propane, giving the more stable isopropyl radical. Per hydrogen this is only about four 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 secondary C–H bond pairs with it to complete the new bond to hydrogen.
    • The other secondary C–H electron stays behind as the unpaired electron of the new radical.
  4. Predict the ratio

    Product ratio = (number of hydrogens) × (relative reactivity per hydrogen). For Cl•, secondary : primary ≈ 3.9 : 1, so 2 × 3.9 : 6 × 1 ≈ 57 : 43.

  5. Propagation: the propyl 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 propyl 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.
  6. Propagation: the isopropyl 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 isopropyl 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.
  7. 1-Chloropropane, 2-chloropropane and HCl

    Both monochlorides form. Because chlorine is unselective, the ratio is close to the statistical 6:2 corrected by only a modest preference for the secondary hydrogens.

Qué vas a practicar

Draw both branches of propane chlorination, primary and secondary hydrogen abstraction, and explain why chlorine gives a mixture of 1-chloropropane and 2-chloropropane.

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