The E2 Reaction Mechanism

 

The E2 reaction mechanism is a 1-step mechanism that results in the formation of a carbon-carbon double bond.

 

Keep it Simple

Just like substitution reactions, the “2” in E2 reaction mechanism does not refer to the number of steps but rather to rate law of the reaction, which depends on both the concentration of the substrate and the base.

 

In the E2 reaction mechanism, the leaving group on the substrate leaves and a neighboring hydrogen is deprotonated. The electrons from the carbon-hydrogen bond fill in to form an alkene between the carbon that held the hydrogen and the carbon that was bonded to the leaving group.

Below is an example of an E2 reaction mechanism. The rate law is also shown below.

 

This figure illustrates the E2 reaction mechanism and the E2 rate law.

In an E2 reaction mechanism, the leaving group (the bromine here) leaves the molecule. At the exact same time, a base deprotonates a neighboring hydrogen. The neighboring hydrogen is a hydrogen that is connected to a carbon neighboring the carbon that is bonded to the leaving group (see figure below).

 

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Note that the base takes away the hydrogen and not any of the electrons of the carbon-hydrogen bond. The electrons in the carbon-hydrogen bond fill in to form the double bond between the carbon that had the hydrogen and the carbon that had the leaving group attached.

 

Keep it Simple
The base will NEVER deprotonate a hydrogen on the same carbon as the leaving group in an elimination reaction. This is a common mistake students make.

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Example
What is the product of the following E2 reaction? Draw the mechanism as well.

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Answer

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The first thing we must do is identify the leaving group. There are 2 halogens on the molecule, but remember from the Mechanism Preparation chapter, fluorine cannot be a leaving group because it cannot properly stabilize a negative charge; therefore, the leaving group is the chlorine.

After finding the leaving group, we then look at the neighboring carbons to see if there are hydrogens present. There are 3 hydrogens on the right neighboring carbon and none on the left neighboring carbon.

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All 3 hydrogens on the neighboring carbon on the right give the same product, so we can show the mechanism using any of the 3. We can now draw our E2 reaction mechanism and product:

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Anti-Periplanar

We just learned that the hydrogen that is deprotonated must be on a neighboring carbon; however, there still is 1 more stipulation about the deprotonated hydrogen: it must also be anti-periplanar to the leaving group. To understand what that means, let’s break down the term anti-periplanar.

Anti: Think back to Newman projections. Remember the anti conformation? This was the most stable position for 2 substituents as they were as far apart as possible. In E2 reactions, the hydrogen that will be deprotonated and the leaving group must be anti to one another.

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Periplanar: The neighboring hydrogen and the leaving group must also be in the same plane as one another. Using the same molecule in the Newman projection above redrawn into bond line drawing, we can see that both the leaving group and the hydrogen are in the plane of the page.

 

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The leaving group and the hydrogen can also be periplanar even if they are not in the plane of the page. Here’s an example below where the leaving group and the hydrogen are anti-periplanar, while being on dashes and wedges.

 

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To briefly summarize, the hydrogen that is deprotonated during an E2 reaction must be both on a neighboring carbon and anti-periplanar to the leaving group.

 

Example
Identify any hydrogen atom(s) that could be deprotonated if the molecule below underwent an E2 reaction.

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Answer

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In an E2 reaction mechanism, the deprotonated hydrogen must be (1) on a neighboring carbon and (2) anti-periplanar to the leaving group. Therefore, we must identify all the hydrogen atoms in this molecule that are neighboring and anti-periplanar.

The first step is to draw all the hydrogen atoms neighboring the leaving group as these are the hydrogen atoms we must further assess. The leaving group in this molecule is the chlorine.

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After explicitly drawing out each hydrogen, there is one hydrogen that is clearly anti-periplanar, shown in red below.

 

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But remember, there is free rotation around single bonds. So the molecule can rotate to make more neighboring hydrogens anti-periplanar. Try to do this with your model kit to see how each neighboring hydrogen can be rotated into the anti-periplanar position.

Because of free rotation around single bonds, all the neighboring hydrogens in this molecule can be rotated into an anti-periplanar position with the leaving group.

 

 

Determining the neighboring, anti-periplanar hydrogen(s) in a chair configuration can be very difficult. In fact, there might not even be a neighboring, anti-periplanar hydrogen. If there is no neighboring, anti-periplanar hydrogen, then the E2 reaction cannot occur. See below for an example of finding the anti-periplanar hydrogen atoms of a cyclohexane chair.

 

Example
Identify the neighboring, anti-periplanar hydrogens(s) in the molecule below.

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Answer

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First, let’s draw all neighboring hydrogens, remembering where the substituents are located in a chair conformation.

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Now we must see which, if any, of these hydrogens are anti-periplanar. As the bromine is pointed directly up, we need our hydrogen to be pointing straight down for it to be anti-periplanar. Using this approach, we can immediately see that there are 2 hydrogens pointing straight down, so these are the neighboring, anti-periplanar hydrogens.

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Because this molecule has anti-periplanar hydrogens, it would be able to undergo an E2 reaction in the presence of the right base. If this is difficult to see, I would highly recommend using a model kit because chair conformations can be difficult to fully understand spatially.

 

When dealing with E2 reactions using cyclohexane, you must draw the chair conformation to see if there are any neighboring anti-periplanar hydrogens. Even if you are explicitly given the 2-D cyclohexane, you should convert it into the 3-D chair prior to solving the E2 reaction.

 

Example
What is the E2 reaction product of the following? Draw the complete mechanism.

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Answer

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First, we must identify any neighboring, anti-periplanar hydrogens. To do this, we start by drawing out the implied hydrogens on the neighboring carbons:

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Now we assess each of these hydrogens to see if any are anti-periplanar. This reveals that none of these hydrogens are anti-periplanar (use a model kit if you are having trouble seeing this). Since an E2 reaction cannot occur without a neighboring, anti-periplanar hydrogen, does that mean no reaction will occur here? Normally yes, but since we are using a cyclohexane chair, we can perform a chair flip:

 

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After the chair flip, our chlorine is pointing straight up, so our anti-periplanar hydrogens must therefore be pointing directly down. After assessing the neighboring hydrogens, we can see we have 2 neighboring anti-periplanar hydrogens, both of which would give the exact same product:

 

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Keep it Simple
It can be difficult to tell whether groups are on the same side or on opposite sides of the double bond when doing an elimination reaction. For example, you might not be sure which of the products below the following elimination reaction will form.

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The two products are different molecules, so it’s important we know how to differentiate these two.

To solve difficult questions like these, I’d recommend rotating the leaving group and the neighboring hydrogen so they are anti-periplanar and in the plane of the page.

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From here, we perform the E2 reaction mechanism. In the final product, the groups that were on wedges will be on the same side of the double bond, while the groups that were on dashes will be on the other side of the double bond. Therefore, the final product looks like this:

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When does an E2 reaction occur?

E2 reactions occur when a strong base is present. Just like an SN2 reaction, E2 reactions prefer for the leaving group to be located on a primary carbon rather than a tertiary carbon; however, the preference is more flexible– E2 reactions can occur on a tertiary carbon if the base is strong enough. In the next chapter, we’ll talk more about assessing the reaction conditions to decide if a substitution reaction or an elimination reaction will occur, so don’t worry about this too much for now.

 

Summary Points for E2 Reactions

  • The rate of the E2 reaction depends on the concentration of both the substrate and the base.
  • Occurs with strong bases. There is a preference for the leaving group to be on a primary carbon, but, if the base is strong enough, an E2 reaction can still take place on a tertiary carbon. There will be much more on this next chapter.
  • There must be a neighboring anti-periplanar hydrogen to the leaving group in order for an E2 reaction to occur.