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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: pantsboy on December 09, 2009, 10:46:26 PM

Title: Why can only HBr undego free-radical addition to an alkene?
Post by: pantsboy on December 09, 2009, 10:46:26 PM
Why can't Iodide or Chloride?
Title: Re: Why can only HBr undego free-radical addition to an alkene?
Post by: Dharmalas on December 10, 2009, 12:04:51 PM
The reason why H-Cl or H-I cant undergo radical addition is because of the kinetics of the reaction and Hammond's postulate. Hammond's postulate simply states that in a exothermic reaction the transistion states of the reactants resemble the starting materials, while in a endothermic reaction the transistion states are closer to the products. H-Br under radical addition is a exothermic so this supports the idea that multiple products are made between each radical. H-I and H-Cl have endothermic reactions so it is rather difficult for the transistion state to look like all the possible products.
Title: Re: Why can only HBr undego free-radical addition to an alkene?
Post by: Ranadeep on December 10, 2009, 12:56:04 PM
H-Cl free radical initiation is more endothermic
H-I after initiation I2 is immediately formed
Title: Re: Why can only HBr undego free-radical addition to an alkene?
Post by: g-bones on December 10, 2009, 03:16:58 PM
The issue arises from the radical abstraction from HCl, which is endothermic by about 5kcal/mol (energy of breaking the H-Cl bond).  so the chain transfer wouldn't be as favorable, the alkyl radical would more preferentially combine with another alkyl radical than abstract a radical H from HCl.  this recombination would lead to oligomers.  Refer to Ansyl & Dougherty page 571 for more.

Hope this helps
 
Title: Re: Why can only HBr undego free-radical addition to an alkene?
Post by: bromidewind on December 16, 2009, 09:18:51 PM
This is right out of Wade's Organic Chemistry Sixth Edition. Periods indicate a free radical.
Quote
The reversal of orientation in the presence of peroxides is called the peroxide effect. It occurs only with the addition of HBr to alkenes. The reaction of an alkyl radical with HCl is strongly endothermic, so the free radical chain reaction is not effective for the addition of HCl.

   Cl-CH2-CH2. + HCl -> Cl-CH2-CH2-H + Cl. ΔH°=+42 kJ (+10 kcal)

Similarly, the reaction of an iodine atom with an alkene is strongly endothermic, and the free-radical addition of HI is not observed. Only HBr has just the right reactivity for each step of the free-radical chain reaction to take place.

   I. + H2C=CH2 -> I-CH2-CH2. ΔH°=+54 kJ (+13 kcal)

This is why bromine is my favorite halide. It's just right :)