{"id":7573,"date":"2013-09-23T08:00:10","date_gmt":"2013-09-23T12:00:10","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=7573"},"modified":"2026-05-07T11:29:07","modified_gmt":"2026-05-07T16:29:07","slug":"selectivity-in-free-radical-reactions","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2013\/09\/23\/selectivity-in-free-radical-reactions\/","title":{"rendered":"Selectivity In Free Radical Reactions"},"content":{"rendered":"

Free Radical Chlorination: Selectivity<\/strong><\/p>\n

This post is all about the\u00a0selectivity\u00a0<\/em>of free-radical halogenation: what does “selectivity” mean, anyway? And how do we calculate it? It’s often said that chlorination is less “selective” than bromination. What does that<\/em> mean? (More on that topic in the next post<\/a>).<\/p>\n

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

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

    \n
  1. Chlorination Of Propane Does Not Lead To\u00a0 A Statistical Mixture Of Products. Why Not?<\/a><\/li>\n
  2. Secondary C\u2013H Bonds Are Weaker Than Primary C\u2013H Bonds, And Result In More Stable Free Radicals<\/a><\/li>\n
  3. How To Quantify “Selectivity” In Free-Radical Chlorination<\/a><\/li>\n
  4. A Puzzle: What Is The Selectivity For 2-Bromopropane In This Reaction?<\/a><\/li>\n
  5. Quiz Yourself!<\/a><\/li>\n
  6. (Advanced) References and Further Reading<\/a><\/li>\n<\/ol>\n
    \n

    <\/a>1. Chlorination Of Propane Does Not Lead To A “Statistical” Mixture Of 1-Chloropropane And 2-Chloropropane. Why Not?<\/h2>\n

    In the last post we showed some examples of how different isomers might be formed in free-radical halogenation of alkanes. And I left off with a question.<\/p>\n

    If we performed a free-radical chlorination on propane with 1 equiv Cl2<\/sub>\u00a0under normal conditions (25\u00b0C, initiated with light), what \u00a0be the expected ratio of 1-chloropropane and 2-chloropropane?<\/p>\n

    Notice that there are six primary (“methyl”) hydrogens (removal of any of which would give rise to 1-chloropropane) and two secondary (“methylene”) hydrogens. That’s a 3:1 ratio. If<\/strong><\/span> halogenation was statistically random<\/strong>, we would therefore expect a 75:25 ratio of 1-chloropropane to 2-chloropropane.<\/p>\n

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

    That’s not what happens!<\/p>\n

    Instead, experiment tells us that we obtain a 55:45 ratio of 2-chloropropane to 1-chloropropane. So clearly there is more to this product distribution than pure randomness.<\/p>\n

    Why might this be?<\/p>\n

    Recall our earlier discussion of free radical stabilities. We observed that free radicals increase in stability as the number of carbon substituents increases, from methyl (least stable) < primary < secondary < tertiary (most stable).<\/p>\n

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

    As we’d also mentioned earlier, this is also the ratio of C\u2013H bond strengths: CH4<\/sub> > R\u2013CH3<\/sub> > R\u2013CH2<\/sub>\u2013R > R3<\/sub>C\u2013H . The more stable the free radical that is left behind, the weaker will be the C\u2013H bond strength.<\/strong><\/p>\n

    <\/a>2. Secondary C\u2013H Bonds Are Weaker Than Primary C\u2013H Bonds. Breaking A Secondary C\u2013H Bond Results In A More Stable Free Radical<\/h2>\n

    Let’s turn to propane.<\/p>\n

    In the free radical chlorination reaction, chlorine radical may abstract a hydrogen from either from one of the methyl groups of propane, or from the methylene.<\/p>\n

    All else being equal, which of these should be the easiest process?<\/p>\n

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

    Removing the secondary hydrogen, giving us a secondary free radical!<\/p>\n

    It’s for this reason that C\u2013H bonds are not broken randomly (and hence we don’t get a 75:25 ratio). The stability of the free radical we create is a very important factor.<\/p>\n

    <\/a>3. How To Quantify “Selectivity” In Free Radical Chlorination<\/h2>\n

    This deviation from a statistical mixture is referred to as selectivity<\/strong>. We say here that chlorination is more\u00a0selective<\/strong> for the secondary carbon than for the primary carbon.<\/p>\n

    So how do we quantify selectivity?<\/p>\n

    If there were equal numbers of methylene (CH2<\/sub>) hydrogens and methyl (CH3<\/sub>) hydrogens, it would be\u00a0 simple: 55:45 in this case, or 1.22 favoring methylene.<\/p>\n

    However, it’s\u00a0not quite that simple.\u00a0<\/strong>We have to adjust for the statistical bias in favor of the methyl groups.<\/p>\n

    Recall that there are three times as many methyl hydrogens as methylene hydrogens<\/strong>. In order to make this a true ratio, we need to divide the number we obtained for methyl by 3. This gives us a ratio of 55:15 , or 3.66.<\/p>\n

    So for this reaction, the free radical chlorination of propane \u00a0at 25\u00b0C, chlorine is 3.66 times more selective for secondary hydrogens than for primary hydrogens.<\/p>\n

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

    If we were examining a reaction where\u00a0three<\/strong> different types of hydrogens were present (or even more) we would likewise adjust each yield by a factor equal to the number of hydrogens of each type, and then compare directly.<\/p>\n

    <\/a>4. A Puzzle: What Is The Selectivity For 2-Bromopropane In This Reaction?<\/h2>\n

    Let’s leave off with yet another puzzle. Look at the exact same reaction, except using Br2<\/sub> instead of Cl2<\/sub>\u00a0. Here, we observe a ratio of 97% 2-bromopropane to 1-bromopropane.\u00a0<\/strong><\/p>\n

    What’s the selectivity for 2-bromopropane in this reaction?<\/p>\n

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

    And why might the selectivity for bromine be much higher than that for chlorine?<\/p>\n

    Answer next time!<\/p>\n

    Next Post: Selectivity In Free Radical Reactions – Bromine vs. Chlorine<\/a><\/strong><\/p>\n

    \n

    <\/a>Notes<\/h2>\n

    Related Articles<\/strong><\/p>