Walkthrough of Elimination Reactions (1)

by James

in Alkenes, Organic Chemistry 1, Organic Reactions

As you’ve probably noticed by now, organic chemistry is a lot different from physics. Why do I say that? Because when we’re looking to predict what reaction might occur in a given situation, we don’t have a handy series of equations we can simply refer to. No, it’s messier than that. We take the experimental results, work backwards, and then try to rationalize what is happening in terms of the key concepts at work.

Furthermore, organic chemistry isn’t “digital”. It’s extremely uncommon that one set of conditions will give you 100% yield of one product, and a different set of conditions will give you 100% yield of another. Rather, organic chemistry is “analog”. Think of each reaction as having a series of knobs we can turn, which can “tune” a reaction toward a different result. (Some examples of “knobs” – temperature, solvent, type of substrate, type of leaving group, type of nucleophile….).

I say this because this post is the first in a series on elimination reactions, the third of the “four most important” reactions in first semester organic chemistry. Elimination reactions often occur under similar conditions to substitution reactions, which means that we will have to learn how to think about how these reactions compete with each other.

Let’s start with a concrete example. In these two reactions, substitution is the major product. However, the yield of substitution products is not 100% . As it turns out there are actually some minor byproducts in this reaction that are not substitution products.
Remember that the key pattern of bond forming/bond breaking in a substitution reaction is a simple swap of a C-(leaving group) bond for a C-(nucleophile) bond? Clearly, in the boxed products, this isn’t occurring here. Let’s break down these patterns in more detail.

Note the key pattern here. In each case, we’re forming a new C–C π bond, and breaking two single bonds to carbon. So this is a completely different pattern than acid-base reactions or substitution reactions, because it involves two adjacent carbon atoms. This class of reactions are called “elimination” reactions.

However, it’s not *completely* without comparison to reactions we’ve seen before:

  1. Note how we’re breaking a C-H bond and forming a new base between H and something else (in the case above, oxygen). We’ve seen this before. This is an “acid-base” reaction. So one component of the elimination reactions we will see is the involvement of a strong base.
  2. In addition, we’re breaking a bond between carbon and a leaving group. We’re familiar with leaving groups by now; remember how good leaving groups are weak bases? The same principle applies to elimination reactions as well.
What’s slightly new here is that we’re forming a C-C π bond in addition to the above. So when you add up the number of species in solution, we’re going from two species in solution (base and substrate) to three species (product, conjugate acid, and leaving group)

That’s the simple pattern for this reaction. As we’ll see, there are going to be a lot of little wrinkles with the mechanics of how this reaction works. But as I’ve said before, knowing “what” [what bonds formed, what bonds broke] is the most important question you can ask yourself when learning a new reaction. In the coming few posts, we can then start to make more observations and start to ask “why”?

Next Post: Walkthrough of Elimination Reactions (2) – Zaitsev’s Rule

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{ 1 comment… read it below or add one }

sg

Hi James,
first and foremost, wonderful post !! I just have one question that, if there would have been an alkyl halide, in which the carbon bearing the leaving group (halogen atom), which is the alpha carbon, is also having a H-atom attached to it & one H-atom is attached to the β-carbon as well, then in my view the H-atom attached to the alpha carbon is more acidic than the one attached to the β-carbon. So then why isn’t the most acidic proton abstracted by a Base ??

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