How to Decide Between Elimination or Substitution

What to do if deciding between the “2” mechanisms

A simple tip…

Start thinking about substitution and eliminations reactions as what I call the “1’s” and the “2’s”. This will help you to find patterns that will make it easier for you to ace these questions on your exam.

In this post, we will be specifically looking at SN2 versus E2 (the “2’s”) in terms of the groups attached to the carbon that is being attacked.

Here is a reaction that demonstrates both:

Let’s first start with the mechanism of SN2:


In order for a reaction to proceed through an SN2 mechanism, the nucleophile must be able to attack 180° to the leaving group. When this occurs, the reaction goes through a transition state where all five atoms are associated with the carbon being attacked. Finally, the nucleophile adds and the leaving group leaves simultaneously. The SN2 mechanism DOES NOT involve a carbocation. It proceeds in a single step.

Adding groups (other than hydrogen) to the carbon being attacked


As we begin to increase the number of non-hydrogen groups, the reaction begins to move towards an E2 mechanism. Why is this the case? Think about the non-hydrogen groups as being much bigger than hydrogen:

Remember that in an SN2 mechanism, the nucleophile must be able to attack 180° to the leaving group. As the groups get larger, the room for a nucleophile to come in an attack gets smaller. Here is another view point we can look at:

See how it is getting harder and harder to see the carbon atom as more and more groups are added. This means it is harder and harder for the nucleophile to see and attack the carbon.


Stereochemistry with Substitution Reaction

Ok, so we now feel more comfortable with SN2 versus E2, but what about stereochemistry with SN2? Is that something we need to worry about? Well…it depends on the starting material that is used. In order for a compound, in this case, to be chiral, all four groups must be different. If we look at the first example, this is not the case:

We do not need to worry about chirality or stereochemistry. Let’s look at a structure with four different groups and see if we can figure this out:

If we look at the conversion from an alkyl halide to an alcohol shown above, we can see that the groups on the starting alkyl halide are not located in the same place in space as the groups on the alcohol. The groups are obviously the same, but the compound has gone through an inversion. This is like turning your shirt inside out; it is the same shirt, but the inside and outside are now switched.

Priorities of the Groups

If we look at the priorities of the groups, the alkyl halide and alcohol are the highest priority. We can just say for argument sake that R1, R2, and R3 are the priority assignment for the R groups:

If we push the –X and –OH groups into the paper, we can draw that in the following manner:

Now, it is easy to see that the relative position of the groups has changed.


A look at the E2 Elimination Reaction and Mechanism

Ok, so let’s finish up by looking a bit more at the E2 reaction and mechanism:

In this case, I have placed a methyl group, R1, and R2 on the central carbon. We can think of all these as large groups that block the central carbon from attack. The hydroxide ion is not only a good nucleophile but a great base. Since it does not have the space to attack, it will look for a proton to pull off. Where does it look to pull off a proton? It looks for protons that are attached to the carbon adjacent to the carbon with the alkyl halide. Say what??  Let me illustrate:

Here are a few other examples:

Now that we have that squared away, let’s move on to the mechanism:

The hydroxide comes in and pulls off one of the adjacent hygrogens we talked about earlier. Now, the electrons of the bond can now swing down and kick out the halide, forming a double bond. This reaction proceeds in one step; it does not form a carbocation. The above reaction and the below reaction are the same. I just wanted to write it in two different ways; some prefer to see all the atoms and some prefer line drawings.


When I talk about the more groups we add the harder it gets to ‘see’ the carbon, we are going from a methyl group to a primary carbon to a secondary carbon to a tertiary carbon.

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