{"id":7528,"date":"2013-09-06T15:33:39","date_gmt":"2013-09-06T19:33:39","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=7528"},"modified":"2025-07-08T09:28:42","modified_gmt":"2025-07-08T14:28:42","slug":"initiation-propagation-termination","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2013\/09\/06\/initiation-propagation-termination\/","title":{"rendered":"Initiation, Propagation, Termination"},"content":{"rendered":"

Initiation, Propagation, and Termination In Free Radical Reactions<\/strong><\/p>\n

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<\/a><\/em><\/span>). 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.<\/p>\n

\"summary<\/a><\/p>\n

Table of Contents<\/strong><\/p>\n

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

    <\/a>1. The Mechanism For The Free-Radical Substitution Of An Alkane With Cl2<\/sub><\/h2>\n

    You may recall seeing this reaction in a previous post – it’s the free radical chlorination of methane with Cl2<\/sub>.<\/p>\n

    It’s a substitution<\/strong> on carbon because a C-H<\/strong> bond breaks and a new C-Cl<\/strong> bond forms on the same carbon. The byproduct is HCl.<\/p>\n

    \"in<\/a><\/p>\n

    Now that we know a little more about what free radicals are and their key properties, today we will answer,\u00a0 “how does this reaction work?<\/strong>“.<\/p>\n

    We’re going to go through the key steps of this reaction and learn that they are composed of three key phases: initiation<\/strong>, termination<\/strong>, and propagation<\/strong>.<\/p>\n

    <\/a>2. A Step Where There Is a Net Increase In The Number Of Free Radicals Is Called, “Initiation”<\/h2>\n

    Free-radical reactions generally require heat or light to be applied.<\/a> 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. \u00a0Cl\u2022 \u00a0]<\/p>\n

    Every free radical reaction begins with a step where free radicals are created, and for that reason this initial step is called\u00a0initiation.\u00a0<\/strong><\/p>\n

    Here’s the equation for this initiation step. Two things to note:<\/p>\n

      \n
    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)<\/li>\n
    2. Note that there is a net increase of free radicals\u00a0<\/strong>in this\u00a0reaction. We’re going from\u00a0zero<\/strong> (in the reactants) to\u00a0two<\/strong> (in the products).<\/li>\n<\/ol>\n

      \"example<\/a><\/p>\n

      <\/a>3. A Step Where There Is No Net Gain Or Loss Of Free Radicals Is Called “Propagation”<\/h2>\n

      It’s only once the free radicals are present that our substrate (CH4\u00a0<\/sub>in our example) gets involved. Chlorine radicals are highly reactive, and can combine with a hydrogen from methane to give the methyl radical, \u2022CH3<\/sub><\/p>\n

      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\u00a0no net increase in the number of free radicals.\u00a0<\/strong><\/p>\n

      This type of step is referred to as “propagation”.<\/p>\n

      \"first<\/a><\/p>\n

      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\u2013Cl bond.<\/p>\n

      <\/a>4. Watch Out For This Common Mistake In Drawing Out Free-Radical Mechanisms<\/h2>\n

      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<\/em> natural to bring together the methyl radical and the chlorine radical to form CH3<\/sub>\u2013Cl . Right?????<\/p>\n

      Nooooo!<\/p>\n

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

      \"common<\/a><\/p>\n

      <\/a>5. There Are Two Propagation Steps In Free-Radical Substitution<\/h2>\n

      In fact, we can do the proper “propagation” step this way: Take the methyl radical, and it reacts with the Cl2<\/sub> 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”.<\/p>\n

      \"proper<\/a><\/p>\n

      Note again that we are forming a chlorine radical! What’s so crucial about this? It’s crucial because\u00a0this chlorine radical can then perform Propagation Step #1 on a new molecule of our substrate (CH4<\/sub>),\u00a0<\/strong>continuing the process. It’s a chain reaction<\/strong> – once generated, chlorine radical is\u00a0catalytic.\u00a0<\/strong>That’s why we only need a small amount of chlorine radical for this reaction to proceed. It is not unusual for 104<\/sup> or more of these cycles to proceed before termination occurs. (See this reference<\/a> below<\/em><\/span>)<\/p>\n

      <\/a>6. A Step Where There Is A Net Decrease In Free Radicals Is Called, “Termination”<\/h2>\n

      Can this chain reaction go on forever? No.<\/p>\n

      Let’s think about two limiting cases. If the concentration of Cl2<\/sub> is low relative to CH4<\/sub> (in other words, Cl2<\/sub> is our limiting reagent) then the rate of Propagation Step #2 will slow down as its concentration decreases. Without any Cl2<\/sub> to react with, our \u2022CH3<\/sub> radicals can just combine with another free radical (such as \u2022Cl) to give CH3<\/sub>Cl, 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\u00a0termination.\u00a0<\/strong><\/p>\n

      \"termination<\/a><\/p>\n

      It’s also possible for two methyl groups to combine together to give CH3<\/sub>\u2013CH3<\/sub> ; this is also termination!<\/p>\n

      <\/a>7. The Full Mechanism Of Free-Radical Substitution Of An Alkane<\/h2>\n

      Let’s put all of these steps together so we can clearly see the initiation, propagation, and termination steps.<\/p>\n

      \"-combined<\/a><\/p>\n

       <\/p>\n

      <\/a>8. Summary: Free-Radical Substitution Reactions<\/h2>\n

      These three types of steps are encountered in every free-radical reaction.<\/p>\n

      The bottom line here is that\u00a0by counting the number of radicals created or destroyed in each step, you can determine if the step is initiation, propagation, or termination.\u00a0<\/strong><\/p>\n