Why do radical mechanisms favor the homolytic dissociation of a $\ce{C-H}$ bond over a $\ce{C-C}$ bond?

The $\ce{C-H}$ bond is stronger in magnitude (and it's also slightly polarized), so why would it break more easily as compared to the $\ce{C-C}$ bond?

My instructor explains this by claiming that the products of $\ce{C-H}$ bond dissociation are stabler than$\ce{C-C}$ bond dissociation.

I know that radicals are unstable and highly reactive, so why does it make any difference as to which radical is formed? Wouldn't the radical easiest to form be the one formed the most?


1 Answer 1


Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references for the organics and here for the halogens.



$$\ce{H3C-CH3 -> 2H3C.}\ \ \Delta H^\circ=377\ \text{kJ/mol}$$


$$\ce{H3CCH2-H -> H3CCH2. + H.}\ \ \Delta H^\circ=423\ \text{kJ/mol}$$

However, $\ce{C-C}$ and $\ce{C-H}$ homolysis steps are not common steps in radical mechanisms of alkanes. For example, the chlorination of ethane begins with the homolysis of $\ce{Cl2}$, which is much more favorable than homolysis of $\ce{C-C}$ or $\ce{C-H}$.

$$\ce{Cl2 -> 2Cl.} \ \ \Delta H^\circ = 243\ \text{kJ/mol}$$

The chlorine radicals then react with the alkane by abstraction, meaning that bonds are being formed as well as being broken.



$$\ce{Cl. + H3C-CH3 -> Cl-CH3 +H3C.}$$

$$\begin{array}{c|c|c|} & \text{Broken} & \text{Formed} \\ \hline \text{bond} & \ce{H3C-CH3} & \ce{Cl-CH3} \\ \hline \Delta H^\circ & 377\ \text{kJ/mol} & -350 \ \text{kJ/mol} \\ \hline \end{array}$$

$$\Delta_r H^\circ = +27 \ \text{kJ/mol}$$


$$\ce{Cl. + H3CCH2-H -> Cl-H +H3CCH2.}$$

$$\begin{array}{c|c|c|} & \text{Broken} & \text{Formed} \\ \hline \text{bond} & \ce{H3CCH2-H} & \ce{Cl-H} \\ \hline \Delta H^\circ & 423\ \text{kJ/mol} & -432 \ \text{kJ/mol} \\ \hline \end{array}$$ $$\Delta_r H^\circ = -9 \ \text{kJ/mol}$$

Thus, abstraction of $\ce{H}$ by $\ce{Cl.}$ is more favored because it is coupled to the formation of the $\ce{H-Cl}$ bond.

  • $\begingroup$ Okay, that explains the formation in the propagation and termination steps of the reaction. But why does it happen in the initiation of the radical reaction? $\endgroup$
    – tkhanna42
    May 3, 2015 at 10:54
  • 1
    $\begingroup$ It doesn't happen in initiation. Organic radical reactions almost exclusively are initiated by the homolysis (thermally or photochemically) of weaker $\ce{X-X}$ or $\ce{O-O}$ bonds, and not by the direct homolysis of $\ce{C-H}$ or $\ce{C-C}$ bonds. See Wikipedia page on radical initiators. $\endgroup$
    – Ben Norris
    May 3, 2015 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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