Consider a simple free radical substitution's propagation step for $\ce{CH4}$ to $\ce{CH3Cl}$:

$$ \begin{align} \ce{CH4 + Cl^. &-> CH3^. + HCl}\tag{R1}\\ \ce{CH3^. + Cl2 &-> CH3Cl + Cl^.}\tag{R2} \end{align} $$

My understanding is that since the free radical is regenerated at the end of the reaction, it must therefore be a catalyst of some sort. Is this understanding correct or flawed?

  • $\begingroup$ You may find useful these links for text formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic $\endgroup$
    – Poutnik
    Commented Jan 12, 2022 at 9:25
  • $\begingroup$ Yes, $\ce{Cl^•}$ acts as a catalyst here. $\endgroup$ Commented Jan 12, 2022 at 9:44
  • $\begingroup$ @Poutnik Thanks for letting me know! I was trying to find the command for the last 5 mins but couldn't find it and thus went for the last resort! $\endgroup$ Commented Jan 12, 2022 at 9:51
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    $\begingroup$ @Poutnik No need for \bullet, \cdot or other macro or Unicode chars. Just use ^., as it's stated in manuals for MathJax-mhchem, $\mathrm\LaTeX$-mhchem, or another package chemformula. $\endgroup$
    – andselisk
    Commented Jan 12, 2022 at 10:19
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    $\begingroup$ @Poutnik thank you for the help with formatting in chemicals. I find it very useful. $\endgroup$
    – user104941
    Commented Jan 12, 2022 at 13:32

2 Answers 2


Your interpretation is interesting. As you state, the free radical $\ce{Cl·}$ is regenerated at the end of the reaction. It behaves like Vanadium(V) which oxidizes $\ce{SO2}$ into $\ce{SO3}$. And once reduced into Vanadium(IV), it gets reoxidized by $\ce{O2}$ into Vanadium(V), according to the two equations $$\ce{V2O5 + 2 SO2 -> 2 VO2 + 2 SO3}$$ $$\ce{2 VO2 + O2 -> V2O5}$$ The main difference between the free radical substitution and the vanadium oxide catalysis is the fact that both vanadium oxides do exist as stable compounds. And the radical $\ce{Cl·}$ is not stable and cannot be isolated.

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    $\begingroup$ The question is, if we should consider transient, intermediate products in nonlinear complex reactions as catalysts. IUPAC Gold book seems to do so, but..... $\endgroup$
    – Poutnik
    Commented Jan 12, 2022 at 11:03

Well, consider this argument on the undoubtedly actual expanded system, which would commence with the action of strong UV on a Chlorine gas in presence of Methane, and seemingly would proceed based on random kinetics, as follows:

\begin{align} \ce{Cl2 + UV &-> Cl^. + Cl^.}\\ \ce{CH4 + Cl^. &-> CH3^. + HCl}\\ \ce{CH3^. + Cl2 &-> CH3Cl + Cl^.}\\ \ce{Cl^. + Cl^. &-> Cl2}\\ \ce{CH3^. + CH3^. &-> Products}\\ &\vdots \end{align}

where not all reactions are equally likely again based on kinetics.

In the above more complex system approaching reality, I would claim the Chlorine presence (certainly in the form of the $\ce{Cl^.}$) has catalytic aspects. Note, both the chlorine radical and the methane radical are cycled within the system, albeit, the methane radical self-reaction is a termination step, whereas the chlorine radical self-reaction restarts the system. However, Chlorine is consumed in the system so not by definition a catalyst.

  • $\begingroup$ Good point that chlorine is consumed. Only the radicals regenerate each other. I don't see what you mean by "kinetics", likely the wrong word. $\endgroup$
    – Karl
    Commented Jan 12, 2022 at 19:06
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    $\begingroup$ On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Commented Jan 12, 2022 at 20:11
  • $\begingroup$ @Karl won't the radicals be consumed if a chlorine radical hits a methyl radical? $\endgroup$
    – user104941
    Commented Jan 12, 2022 at 22:16
  • $\begingroup$ MaguireFan123: Yes, indeed .CH3 + .Cl = CH3Cl and the reverse may also be possible under UV light. Note: my use of dots to convene possible continuation of reactions. $\endgroup$
    – AJKOER
    Commented Jan 12, 2022 at 23:28

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