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Free Radical Reactions

By James Ashenhurst

Initiation, Propagation, Termination

Last updated: September 15th, 2020 |

Initiation, Propagation, and Termination In Free Radical Reactions

In the previous post on free radical substitution reactions we talked about why heat or light is required in free-radical reactions. In this post we’re going to go through the mechanism of a free-radical substitution reaction, which has three key types of steps: initiation, propagation, and termination.

Table of Contents

  1. The Mechanism For The Free-Radical Substitution Of An Alkane With Cl2
  2. A Step Where There Is A Net Increase In The Number of Free Radicals Is Called “Initiation”
  3. A Step Where There Is No Net Gain Or Loss Of Free Radicals Is Called “Propagation”
  4. Watch Out For This Common Mistake In Drawing Out Free-Radical Mechanisms
  5. There Are Two Propagation Steps In Free Radical Halogenation Of Alkanes
  6. A Step Where There Is A Net Decrease In Free Radicals Is Called, “Termination”
  7. The Full Mechanism Of Free-Radical Substitution Of An Alkane
  8. Summary: Free-Radical Substitution Reactions
  9. Notes
  10. (Advanced) References and Further Reading

1. The Mechanism For The Free-Radical Substitution Of An Alkane With Cl2

You may recall seeing this reaction in a previous post – it’s the free radical chlorination of methane with Cl2.

It’s a substitution on carbon because a C-H bond breaks and a new C-Cl bond forms. The byproduct is HCl.


Now that we know a little more about what free radicals are and their key properties, today we will answer,  “how does this reaction work?“.

We’re going to go through the key steps of this reaction and learn that they are composed of three key phases: initiation, termination, and propagation.

2. A Step Where There Is a Net Increase In The Number Of Free Radicals Is Called, “Initiation”

Free-radical reactions generally require heat or light to be applied. That’s because either of these energy sources can lead to homolytic cleavage of relatively weak bonds like Cl-Cl to give free radicals [i.e.  Cl•  ]

Every free radical reaction begins with a step where free radicals are created, and for that reason this initial step is called initiation. 

Here’s the equation for this initiation step. Two things to note:

  1. The reaction is an equilibrium – at any given time there is only a small concentration of the free radical present (but this will be enough, as we shall see)
  2. Note that there is a net increase of free radicals in this reaction. We’re going from zero (in the reactants) to two (in the products).


3. A Step Where There Is No Net Gain Or Loss Of Free Radicals Is Called “Propagation”

It’s only once the free radicals are present that our substrate (CHin our example) gets involved. Chlorine radicals are highly reactive, and can combine with a hydrogen from methane to give the methyl radical, •CH3

If you count the number of free radicals in this equation, you’ll note that there’s one in the reactants and one in the products. So there is no net increase in the number of free radicals. 

This type of step is referred to as “propagation”.


If you’re keeping score, by this point you should be able to see that there’s only one bond left to form before our reaction is complete. All we need to do is to form a C–Cl bond.

4. Watch Out For This Common Mistake In Drawing Out Free-Radical Mechanisms

It’s here where it’s easy to make a little mistake. Seeing that there’s two chlorine radicals formed in the initiation step, it would seem natural to bring together the methyl radical and the chlorine radical to form CH3–Cl . Right?????


Note the number of free radicals has decreased here, not stayed the same. It can’t be propagation! (It’s actually termination, which we’ll discuss in a minute).


5. There Are Two Propagation Steps In Free-Radical Substitution

In fact, we can do the proper “propagation” step this way: Take the methyl radical, and it reacts with the Cl2 still present. This gives us CH3Cl and the chlorine radical. Note that there has been no net change in the number of free radicals, so this is still a “propagation”.


Note again that we are forming a chlorine radical! What’s so crucial about this? It’s crucial because this chlorine radical can then perform Propagation Step #1 on a new molecule of our substrate (CH4), continuing the process. It’s a chain reaction – once generated, chlorine radical is catalytic. That’s why we only need a small amount of chlorine radical for this reaction to proceed.

6. A Step Where There Is A Net Decrease In Free Radicals Is Called, “Termination”

Can this chain reaction go on forever? No.

Let’s think about two limiting cases. If the concentration of Cl2 is low relative to CH4 (in other words, Cl2 is our limiting reagent) then the rate of Propagation Step #2 will slow down as its concentration decreases. Without any Cl2 to react with, our •CH3 radicals can just combine with another free radical (such as •Cl) to give CH3Cl, for example. There is essentially no barrier to this reaction. Note that here the number of free radicals decreases from 2 to zero. This is called termination. 


It’s also possible for two methyl groups to combine together to give CH3–CH3 ; this is also termination!

7. The Full Mechanism Of Free-Radical Substitution Of An Alkane

Let’s put all of these steps together so we can clearly see the initiation, propagation, and termination steps.



8. Summary: Free-Radical Substitution Reactions

These three types of steps are encountered in every free-radical reaction.

The bottom line here is that by counting the number of radicals created or destroyed in each step, you can determine if the step is initiation, propagation, or termination. 

  • Intiation -> net formation of radicals
  • Propagation -> no change in the number of free radicals
  • Termination -> net destruction of free radicals

We’ll leave with two teasers for future posts.

First… note that here we’re using CH4, where every C–H bond is identical. What might happen if we used an alkane where all the C–H bonds aren’t equal… like propane, or pentane, for example?

Secondly, this reaction fails spectacularly when Br2 is used instead of Cl2 for the reaction of CH4. However, we’ll see that Br2 can work in certain special cases.

More soon!

Next Post: Isomers From Free Radical Reactions


BONUS material.

We just talked about the situation where one equivalent of chlorine (Cl2) is used. What happens when we use multiple equivalents, or even a vast excess?

Think about it for a second. Imagine we had multiple equivalents of Cl2 in the presence of CH3Cl. What do you think might happen?

An atom of Cl• could react with CH3Cl to give •CH2Cl [and HCl], which could then react with Cl2 to give CH2Cl2 !


Likewise, if we still have an excess of Cl2, then we will observe conversion of CH2Cl2 to CHCl3.


Finally, given enough Cl2 we could then imagine the conversion of CHCl3 to give CCl4.


At this point there are no further C-H bonds to react with the chlorine radical, and thus our reaction would eventually terminate.

The bottom line here is that alkanes, given a large enough excess of Cl2, will eventually have all of their hydrogens replaced with chlorine.

This pathway is in fact how dichloromethane (CH2Cl2 – a common laboratory solvent) chloroform (CHCl3) and carbon tetrachloride (CCl4) are produced industrially. For many decades, CCl4 was produced on mega-ton scale for use as a refrigerant and dry cleaning solvent until studies implicated it and other CFC’s in depletion of the ozone layer.

(Advanced) References and Further Reading

  1. Chlorination of Methane
    T. McBee, H. B. Hass, C. M. Neher, and H. Strickland
    Industrial & Engineering Chemistry 1942, 34 (3), 296-300
    DOI: 10.1021/ie50387a009
    This paper shows that the chlorination of methane can be controlled to give any of the desired chloromethanes in high yield. This is of significance because CH3Cl, CH2Cl2, CHCl3, and CCl4 are all important feedstocks or solvents and this is how they are produced industrially.
    Robert N. Pease and George F. Walz
    Journal of the American Chemical Society 1931, 53 (10), 3728-3737
    DOI: 10.1021/ja01361a016
    This paper provides kinetic evidence that chlorination of methane is 2nd order (first order in both methane and Cl2).
    The Journal of Organic Chemistry 1941 06 (6), 818-829
    DOI: 10.1021/jo01206a005
    The nature of the free-radical chain reaction mechanism in the substitution of alkanes was not fully worked out until the 1940s. In this pioneering 1941 report, Kharasch proposes the chain mechanism we now read in textbooks:
    a) Br2 + hv –> 2 Br • (initiation)
    b) R–H + Br• –> R• + HBr (propagation step 1)
    c) R• + Br2 –> R-Br + Br•  (propagation step 2)
    Kharasch reports that free-radical substitution of cyclohexane with Br2 reacts very slowly in the dark, or in the absence of oxygen. In low concentration, oxygen can act as a free-radical initiator (forming Br• radicals from Br2)  but here Kharasch also observes that in high concentration oxygen can inhibit free-radical reactions. The selectivity of Br• to react with C-H bonds in the order tertiary > secondary > primary is also noted.


Comment section

23 thoughts on “Initiation, Propagation, Termination

  1. tanks for your information, it’s really useful.
    But i have a question.
    When methane react with chlorine, the products vary with the amount of chlorine in the mixture. So when equal amount of chlorine and methane are used, what happens? please help me out!

  2. I have two question.
    First, why bromine mostly fails to do these steps? Is it because bromine’s nature?

    Second, what if we reacted propene instead of methane? where will the position of chlorine be?

  3. In the formation of dichloromethane if an atom of Cl radical interacted with CH3Cl in the first propagation step how do you still have Cl2 in the second step? Wouldn’t you just have one Cl radical left?

    1. There’s an equilibrium between Cl2 and 2 Cl• . It lies heavily to the left, so there is actually a large amount of Cl2 present and very little Cl•

  4. Hi, how does the Cl radical ‘pull’ a hydrogen atom from CH4?
    Don’t talk about electroneg. difference, because then CH3 radical wouldn’t have been able to break the Cl-Cl bond.

    1. The unpaired electron of Cl interacts with the s orbital of the hydrogen and there is a transition state where there are partial bonds between Cl and H and carbon. The H-Cl bond is comparable in strength (102 kcal/mol) to the H-C bond in methane (105 kcal/mol) so the process is reasonably favorable. The resulting carbon-based radical then reacts with leftover Cl2. C-H bond dissociation energies in methane are quite high relative to other alkanes, so in other alkanes it will be even more favorable.

    1. Yes! Although the concentration of Cl• at any given time is very low, so it tends to be slow until the concentration of the species participating in propagation is depleted.

  5. When propanoic acid reacts with chlorine under UV light, why is 2-chloropropanoic acid formed and not 3-chloropropanoic acid? Why does the chlorine attach to the carbon next to COOH? Markovnikov’s rule can’t be applied as it is not electrophilic addition.

  6. Hey. I didn’t understand why it is necessary for the 2nd propogation step to occur. Why can’t the methyl radical and chlorine radical directly combine to give methyl chloride?

    1. The methyl radical and chlorine radical are present in very low concentrations (relative to CH4 and Cl2) and the rate should be low for most of the reaction since rate is proportional to concentration. If these radicals do meet each other, they will react of course, it’s just that it’s quite unlikely.
      The reaction of the alkyl radical with Cl2 is more favored simply because the concentration of cl2 is much, much higher than that of the Cl radical.

  7. propose a mechanism for the photo-chemical chlorination of 2-methylpropane. label the initiation, propagation and termination steps

  8. Hello! Thanks for all the info, the website is really great.
    I have one question if I may. Why in the first place does the Cl* radical “take” the hydrogen from, for example, the methane? Is the H-C bond in methane weaker than the Cl-C bond? Or should I ask if the Cl* radical least stable than the H3C* radical? Which one is the right question? And are there other basic thing to take into account?

    Thank you!

    1. The way to look at it is to compare bond enthalpies and calculate the difference in energy going from starting material to product.

      The bond strength of H-Cl is about 106-107 kcal/mol whereas the bond strength of H-CH3 is about 104 kcal/mol. Therefore the “gain” in energy is about 3kcal/mol from the chlorination of methane.

      (The C-Cl bond does not factor into this reaction, except as a possible termination step).

      That at least gives you an idea of whether or not the process will be energetically favorable overall.

      We don’t cover free-radical fluorination because the H-F bond is even stronger than C-H, and furthermore the reaction is so exothermic that combustion tends to take place.
      On the other hand free-radical iodination generally fails because H-I bonds are so weak, relative to C-H .

      So, in summary: bond strengths are a good guide.

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