The previous post on radical halogenation of alkanes discussed what a radical is, radical stability, and the radical halogenation of methane. If you missed it, you can read it by clicking here. In this post, we will discuss selectivity in both radical chlorination and bromination reactions.
To start, let's move away from just methane and add some other alkanes in the mix. For example, have a look at 2-methylbutane. When looking at this structure, what types of carbons are in the molecule with respect to primary, secondary, and tertiary?
In 2-methylbutane there are primary, secondary, and tertiary carbons. Remember, that in terms of radical stability, tertiary is the most stable. Since tertiary radicals are the most stable, they should form the fastest. You can also view this in terms of bond dissociation energy (BDE). Tertiary C-H bonds have a BDE of about 403 KJ/mol, while secondary C-H bonds have a BDE of 413 KJ/mol, and primary C-H bonds have a BDE of about 423 KJ/mol. From these numbers we can see that it costs less energy to break a tertiary C-H bond, hence, there would be a lower activation energy for the process. Therefore, we can expect that we would have a major product that is the result of the formation of the tertiary radical.
Remember, the mechanism is the same as we discussed in the earlier post on radical reactions. The reaction will proceed through an initiation step, followed by propagation steps, and ending with a termination step. Half arrows will still be used throughout this mechanism. Note for yourself how the mechanism remains the same below.
When comparing chlorination and bromination, chlorination tends to be the least selective. The chloride radical is less stable than the bromide radical so it is therefore more reactive and leads to more product mixtures. The bromide radical is more stable and can "wait around" looking for the easiest bond to break. So in chlorination, the tertiary product might be around 60% of the total mixture with the secondary and primary products making up the rest. With bromination, the mixture would be above 90% tertiary product.
How does bromination work? It works exactly the same way with initiation, propagation, and termination. However, we do typically use a different starting material to gain the radical. N-bromosuccinimide (NBS) is commonly used with an initiator such as azobisisobutyronitrile (AIBN).
NBS produces small amounts of Br2 in solution (through an equilibrium process). The low amount of Br2 has the advantage of keeping the radical concentration low, helping to prevent unwanted side reactions. But, what about the AIBN? What is the purpose of AIBN? AIBN is one of many types of radical initiators. AIBN can be used with either heat or light to start the initiation process. Here is how AIBN works:
One of the great things about using NBS/AIBN is that we can selectively add to allylic or benzylic positions. However, that is a topic for another day since this post is about radical halogenation of alkanes. We can take a look at a side-by-side example to see the selectivity in bromination. Note, these examples are from the following textbook (Wade, L. G; Simmer, J. W. Organic Chemistry, Pearson, New York, New, 2017; pp 176-180).
Thanks for reading. Check back soon for more posts!