Initiation, Propagation, Termination

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 (See post: Free Radical Reactions – Why Is Heat or Light Required). 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. Quiz Yourself!
11. (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 (CH₄ in 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?????

Nooooo!

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. It is not unusual for 10⁴ or more of these cycles to proceed before termination occurs. (See this reference below)

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

Notes

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.

Quiz Yourself!

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

Become a  MOC member to see the clickable quiz with answers on the back.

(Advanced) References and Further Reading

1. Walling, C. in Free Radicals In Solution, Wiley and Sons, New York 1957 p. 352
2. 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.
3. KINETICS OF THE THERMAL CHLORINATION OF METHANE
   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 2^nd order (first order in both methane and Cl₂).
4. THE BROMINATION OF CYCLOHEXANE, METHYLCYCLOHEXANE, AND ISOBUTANE
   M. S. KHARASCH, WILLIAM HERED, and FRANK R. MAYO
   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.