Last time we covered a comparatively simple reaction: free-radical chlorination of methane to (CH4) to give chloromethane (CH3Cl) and saw that the reaction proceeds through three stages – initiation (where free radicals are created), propagation (the main “product-forming” step of the chain reaction, where a chloroalkane is created without net formation of new free radicals) and termination (where radicals combine, resulting in a net reduction of the number of free radicals).
There’s one simple extension of this reaction I’d like to cover in this post. We just covered the simple molecule CH4. What happens when we move beyond CH4 to more complex alkanes?
All the hydrogens in CH4 are equal. If we replace a hydrogen with a single chlorine, we will get CH3Cl no matter what. Likewise for ethane, where chloroethane is the only possibility. What about longer alkanes?
The first step here, if it isn’t immediately obvious, is to be aware of the “hidden” or “implicit” hydrogens on the line diagrams shown above – it’s important to be able to expand out a line diagram to a condensed formula.
The next step is to go about drawing the various possibilities as we replace a single hydrogen on propane with chlorine. For propane, hopefully it should be clear that there are only two possiblities: chlorination at C-1 or C-2 (chlorination at C-3 would give the same product as that at C-1).
[Question to think about: If the chlorination of propane was completely random, what yields of 1-propane and 2-propane would you expect to see? Answer below]
Similarly, what do we get for the slightly more complicated example of pentane? There’s no mathematical formula for figuring this out, but as with many things in organic chemistry, it can pay dividends to be systematic. Start at one end and work towards the other. One thing that can help is applying IUPAC nomenclature to each possibility – it can assist in realizing if you’ve drawn a duplicate.
We have 3 possibilities for constitutional isomers: 1-chloropentane, 2-chloropentane, and 3-chloropentane. [note that I said “constitutional isomers” – can you see possibilities for stereoisomers in any of these molecules? [*answer below]
Finally let’s look at another slightly more complex example: 2-methylpentane.
It is impossible to capture the variety of potential questions with these three examples, but the general thrust is the same. It also helps to have a systematic approach – starting the exchange of hydrogen for chlorine at one side of the molecule, and gradually working to the other side.
Question answer: If the chlorination of propane was completely random, what yields of 1-propane and 2-propane would you expect to see?
Well, there’s 6 methyl hydrogens, and two “methylene” (CH2) hydrogens. So you’d expect to see a ratio of 75% 1-chloropropane to 25% 2-chloropropane.
Instead, experiments show that free-radical chlorination of propane in the gas phase at 25°C give 45% 1-chloropropane to 55% 2-chloropropane!
Why that might be? We’ll talk about that in the next post.
Next Post: Selectivity In Free Radical Reactions
Here’s some more practice problems. How many constitutional isomers would you expect to see for the mono-chlorination of each of these molecules?
*hopefully you can see that 2-chloropentane has a chiral center, so can exist as either (R)-2-chloropentane or (S)-2-chloropentane. Under free radical conditions we will obtain a racemic mixture of these two compounds (i.e. 50% mixture of (R) and (S).