{"id":5623,"date":"2012-08-28T14:28:46","date_gmt":"2012-08-28T18:28:46","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=5623"},"modified":"2025-08-27T14:55:44","modified_gmt":"2025-08-27T19:55:44","slug":"walkthrough-of-elimination-reactions-1","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2012\/08\/28\/walkthrough-of-elimination-reactions-1\/","title":{"rendered":"Elimination Reactions (1): Introduction And The Key Pattern"},"content":{"rendered":"

Introduction to Elimination Reactions<\/strong><\/p>\n

\"Elimination<\/a><\/p>\n

As you’ve probably noticed by now, organic chemistry is a lot different from physics. 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.<\/p>\n

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.<\/strong><\/p>\n

Furthermore, to use an analogy, 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<\/strong>.<\/p>\n

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….).<\/p>\n

I say this because this post is the first in a series on elimination reactions<\/strong>, the third of the “four most important” reactions in first semester organic chemistry.\u00a0 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.<\/p>\n

1. Elimination Reactions Often Accompany Substitution Reactions<\/h2>\n

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\u00a0not<\/strong> substitution products.<\/p>\n

\"alkenes<\/p>\n

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.<\/p>\n

\"the<\/p>\n

2. Elimination Reactions: The Key Bond Forming\/Breaking Pattern<\/h2>\n

Note the key pattern here. In each case, we’re forming a new C\u2013C\u00a0\u03c0 bond, and breaking two single bonds to carbon.\u00a0<\/strong>So this is a completely different pattern than acid-base reactions or substitution reactions, because it involves two adjacent<\/strong> carbon atoms. This class of reactions are called “elimination” reactions.<\/p>\n

However, it’s not *completely* without comparison to reactions we’ve seen before:<\/p>\n