In the Chain Propagation step of the reaction of propene ($\ce{CH3CH=CH2}$) and hydrogen bromide ($\ce{HBr}$), why is $\ce{CH3-\dot{\ce C}H-CH2-Br}$ free radical formed when a more stable, resonance stabilized allyl free radical ($\ce{\dot{\ce C}H2-CH=CH2}$) is possible?

enter image description here

Source : https://www.chemguide.co.uk/mechanisms/freerad/alkenehbr.html

  • $\begingroup$ Maybe because the electrons in the $\pi$ bond are higher in energy than the $\ce{C-H}$ $\sigma$ bond and radicals are so reactive that they just react kinetically instead of thermodynamically. $\endgroup$ Commented Feb 6, 2018 at 8:46
  • $\begingroup$ So, does it mean that allyl free radical is not formed at all or is it formed to a lesser extent? $\endgroup$
    – Rinish Sam
    Commented Feb 6, 2018 at 8:54
  • $\begingroup$ I actually don't know. I was just trying to guess. $\endgroup$ Commented Feb 6, 2018 at 8:59
  • $\begingroup$ I will not write an answer because I don't know if my guess is correct. $\endgroup$ Commented Feb 6, 2018 at 9:03
  • 2
    $\begingroup$ If you're confused, a good question to ask you would be: what is the mechanism of the initiation step? $\endgroup$
    – Zhe
    Commented May 7, 2018 at 13:08

1 Answer 1


The energy cost of bromine radical abstracting hydrogen from propene is 89 kcal/mole. The gain of forming $\ce{HBr}$ is 18-28 kcal/mol (for a $\ce{H-Br}$ bond from $\ce{H2}$ and $\ce{Br2}$; variation is due to solvent effects) + 23 kcal/mol (from the $\ce{Br2}$ which was fractured into 2 Br radicals by 46 kcal/mol bond energy); so the gain is 41-51 kcal/mol vs a cost of 89, or a net cost (net energy requirement) of 38-48 kcal/mol. This would be a slow reaction with an activation energy even higher than 41-51 kcal/mol. (All data from CRC Handbook)

The alternate reaction, of adding a Br radical (with 23 kcal/mol energy above ground state) to propene, loses the pi bond and its energy (84 kcal/mol) and gains a $\ce{C-Br}$ bond (70 kcal/mol). The energy cost of forming the bromoradical is 84 but the gain is 70 kcal/mol for $\ce{C-Br}$ plus the 23 kcal/mol from the Br radical, or 93 kcal/mol. Net gain is 9 kcal/mol. This reaction will go spontaneously while the first will not (rate determined by activation energies, of course).

It turns out that allylic stabilization is only part of the story, and is, in fact, rather small: the bond strength of $\ce{H - allyl}$ is 89 kcal/mol vs the bond strength of $\ce{H - n-propyl}$ which is 98 kcal/mol.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.