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 completely 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.
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.
This is referred to as selectivity. We say here that chlorination is more selective for the secondary carbon.
Now, if the number of methylene (CH2) hydrogens were equal to the number of methyl (CH3) hydrogens, it would be easy to compare the selectivity: 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.
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!