Free Radical Reactions
Monochlorination Products Of Propane, Pentane, And Other Alkanes
Last updated: December 20th, 2019 |
How Many Monochlorination Isomers Are Formed From Free-Radical Chlorination Of Alkanes?
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 the monochlorination of more complex alkanes?
Table Of Contents
- Free-Radical Halogenation Of Methane And Ethane Can Only Give One Mono-Chlorinated Product
- How Many Mono-Chlorinated Isomers Are Formed From Propane?
- A More Complex Example: Monochlorination of Pentane
- The Case of 2-Methylpentane
- If Chlorination Were Completely Random, What Yields Of 1-Chloropropane and 2-Chloropropane Would We Expect To See?
- Test Yourself
In the last post we discussed mono-chlorination of methane. It’s easy to talk about this reaction since all the hydrogens are equal. No matter which hydrogen you replace, you get the same product (CH3Cl).
Likewise, mono-chlorination of ethane also gives just one product: chloroethane.
(We can say that the hydrogens in methane and ethane are “homotopic“, but that’s a discussion for another day).
What about more complicated cases, like propane, butane, pentane, or even 2-methylpentane?
Let’s start with propane. 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).
These two products (1-chloropropane and 2-chloropropane) are constitutional isomers.
[Question to think about: If the chlorination of propane was completely random, what yields of 1-chloropropane and 2-chloropropane 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.
Here, we have 5 constitutional isomers possible (not counting stereoisomers). Again, it helps to break out your IUPAC nomenclature to double-check that there are no duplicates.
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.
5. If Chlorination Was Completely Random, What Yields Of 1-Chloropropane and 2-Chloropropane Would We Expect To See?
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 2 “methylene” (CH2) hydrogens. So you’d expect to see a ratio of 75% 1-chloropropane to 25% 2-chloropropane.
Is that what’s observed? No!
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).