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Free Radical Reactions

By James Ashenhurst

Selectivity In Free Radical Reactions

Last updated: December 20th, 2019 |

Free Radical Chlorination: Selectivity

This post is all about the selectivity 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 mean? (More on that topic in the next post).

Table of Contents

  1. Chlorination Of Propane Does Not Lead To  A Statistical Mixture Of Products. Why Not?
  2. Secondary C–H Bonds Are Weaker Than Primary C–H Bonds, And Result In More Stable Free Radicals
  3. How To Quantify “Selectivity” In Free-Radical Chlorination
  4. A Puzzle: What Is The Selectivity For 2-Bromopropane In This Reaction?

1. Chlorination Of Propane Does Not Lead To A “Statistical” Mixture Of 1-Chloropropane And 2-Chloropropane. Why Not?

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.

If we performed a free-radical chlorination on propane with 1 equiv Cl2 under normal conditions (25°C, initiated with light), what  be the expected ratio of 1-chloropropane and 2-chloropropane?

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 halogenation was statistically random, we would therefore expect a 75:25 ratio of 1-chloropropane to 2-chloropropane.


That’s not what happens!

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.

Why might this be?

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).


As we’d also mentioned earlier, this is also the ratio of C–H bond strengths: CH4 > R–CH3 > R–CH2–R > R3C–H . The more stable the free radical that is left behind, the weaker will be the C–H bond strength.

2. Secondary C–H Bonds Are Weaker Than Primary C–H Bonds. Breaking A Secondary C–H Bond Results In A More Stable Free Radical

Let’s turn to propane.

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.

All else being equal, which of these should be the easiest process?


Removing the secondary hydrogen, giving us a secondary free radical!

It’s for this reason that C–H 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.

3. How To Quantify “Selectivity” In Free Radical Chlorination

This deviation from a statistical mixture is referred to as selectivity. We say here that chlorination is more selective for the secondary carbon than for the primary carbon.

So how do we quantify selectivity?

If there were equal numbers of methylene (CH2) hydrogens and methyl (CH3) hydrogens, it would be  simple: 55:45 in this case, or 1.22 favoring methylene.

However, it’s not quite that simple. We have to adjust for the statistical bias in favor of the methyl groups.

Recall that there are three times as many methyl hydrogens as methylene hydrogens. 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.

So for this reaction, the free radical chlorination of propane  at 25°C, chlorine is 3.66 times more selective for secondary hydrogens than for primary hydrogens.


If we were examining a reaction where three 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.

4. A Puzzle: What Is The Selectivity For 2-Bromopropane In This Reaction?

Let’s leave off with yet another puzzle. Look at the exact same reaction, except using Br2 instead of Cl2 . Here, we observe a ratio of 97% 2-bromopropane to 1-bromopropane. 

What’s the selectivity for 2-bromopropane in this reaction?


And why might the selectivity for bromine be much higher than that for chlorine?

Answer next time!

Next Post: Selectivity In Free Radical Reactions – Bromine vs. Chlorine


Comment section

8 thoughts on “Selectivity In Free Radical Reactions

    1. You would get mixtures of replacement at C2, C3, and C4. There are twice as many hydrogens that would generate 2-chloropentane than would generate 3-chloropentane, so your dominant product would be 2-chloropentane, but you’d get a mixture of 1-chlropentane, 2-chloropentane, and 3-chloropentane.

  1. For the chlorination of propane in the presence of u.v light, what is the name of the longest possible
    non-chlorinated alkane product that can form (as a side product).

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