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Exo vs Endo Products In The Diels Alder: How To Tell Them Apart

Today we’ll cover how to tell endo– and exo- products apart in the Diels-Alder reaction. Shortcut below.  For the long winded walkthrough with lots of examples, read the post.

Exo and Endo Products in the Diels-Alder

In this series of articles on the Diels-Alder reaction, we’ve seen that:

  1. the Diels-Alder reaction always involves the breakage of 3 pi bonds and the formation of 3 new bonds (2 sigma, one pi), resulting in the formation of a new six-membered ring. [Intro]
  2. The stereochemistry of the dienophile is preserved. A cis arrangement of groups in a double bond will result in a cis arrangement in the new six-membered ring. And likewise,  trans →  trans
  3. For dienes, the two “outside” groups on the diene each end up on the same face of the new six-membered ring. Likewise, the two “inside” groups each also end up on the same face of the new ring.

(These last two parts are from the last post, where we went through the stereochemistry of the Diels-Alder reaction.)

Now let’s add the next layer of detail.  For certain diene/dienophile combinations,  we can end up with two products that fulfill all of those criteria. For instance:

These two products are stereoisomers. But since they are not non-superimposable mirror images (i.e. not enantiomers), they are diastereomers. [In fact, each of these molecules are achiral, due to the presence of a mirror plane – if you can’t see it, click here

These two molecules can be separated due to the fact that they have different physical properties.

An as it turns out, the major product is the one above left. The minor is the one above right.

Sometimes it can help to look at these things in perspective:

Here are the two products again (major and minor) built as models.

When Diels and Alder originally discovered this phenomenon, they assigned the name “endo” to the major product (where the dienophile is pointing “in”, towards the alkene) , and the term “exo” (outside, such as in “exoskeleton”) to refer to the minor product (where the dienophile is pointing “out”, away from the alkene).

In this particular example, identifying “exo” and “endo” seems straightforward based more on an appeal to intuition than anything else. But trust me when I say that there are plenty of examples where it isn’t so easy to assign.

In that vein, it’s helpful to have a standard “rule” that we can apply to each situation.

Here’s one way to do it:

Think of the endo as being the diastereomer where the “outside” group on the diene is on the same side of the new six-membered ring as the “electron-withdrawing group” on the dienophile.

The exo is the diastereomer where the outside group on the diene is on the opposite side of the new six-membered ring as the electron-withdrawing group on the dienophile. 

Let’s try applying those rules to the products of the first reaction we looked at. In the endo drawn below, the “outside” group on the diene (the hydrogens) points down, and the bond to the carbonyls on the dienophile point down as well. Same side of the new ring.

In the exo the outside group (hydrogen) is down and the bond to the electron-withdrawing group points up. Opposite sides of the new ring.

So how do you know if a given Diels-Alder is going to have endo and exo products?

Here’s a short answer. (A longer, more comprehensive answer follows at the bottom of the post). 

Let’s look at a few more examples.

Example 1: 2,4-hexadiene (diene) and acrolein (dienophile)

Note that the endo product has the methyl groups on the same face of the ring as the aldehyde (both dashes, as pictured, or both wedges, as in the enantiomer). The exo product has the methyl groups on the opposite face of the ring as the aldehyde.

Example 2: Cyclopentadiene and maleic anhydride

This is very similar to the original example, where the endo has the anhydride ring on the concave face, and the exo has the anhydride on the convex face.

Example 3: Cyclohexadiene and methacrylonitrile

Here, the nitrile is clearly the electron withdrawing group, and so it’s the functional group we judge endo and exo by, rather than the methyl group. I should note that when you look at a model of the product, the hydrogen is not really pointing “down” (it points out, more than anything else) but it’s certainly “down”, relative to the inside group (the CH2CH2 bridge).

Example 4.  Here’s a fun one: Cyclopentadiene and dimethyl fumarate

Our dienophile has two electron withdrawing groups placed on opposite sides of the pi bond. So which electron withdrawing group do we point to when determining exo or endo ?

We can’t distinguish them! Therefore there is no exo or endo in this case. The product will, however, exist as a mixture of enantiomers. [Note 2]


This post is just intended to give some examples of discerning exo and endo, and we can leave it at that.

It’s worth one last look, however, at how exo and endo might arise.

First,  let’s imagine how cyclopentadiene might come together with ethene in a Diels-Alder.

Cyclopentadiene can approach “head-first” (where C-7 floats over ethylene first) or “butt-first” (where C2-C3– floats over ethylene first). But because both sides of ethene are exactly the same, each of these approaches lead to the same product. Hence, there is no possibility of  exo and endo here.

Now, let’s look at how cyclopentadiene can approach a substituted dienophile like maleic anhydride.

Maleic anhydride does have a “head” and a “tail”, unlike ethylene. The two approaches of cyclopentadiene to maleic anhydride are therefore not equivalent, which results in two different transition states (diastereomeric transition states) which leads to two different products. [Note 3]

This is the origin of endo and exo!

But, you might ask, why is it that the endo product is favored over the exo by so much – especially when it seems like the endo product has much more steric hindrance than the exo?

That is a great question. Let’s look at it next time.

Note 1. In a pinch, CH3 or other groups that are not technically “electron withdrawing groups” can serve as placeholders for our exo-endo shortcut.

This frequently comes up in inverse electron-demand Diels-Alder reactions, where the “dienophile” is electron rich and the diene electron-poor.

Note 2. Dimethyl fumarate has a property called “C2-symmetry” , which is the property that turning it 180° in one direction leads to an indistinguishable molecule. (The letter “S” has that property).

Note 3. Things get even more complicated when either the diene or dienophile is non-symmetrical; this results in 4 possible products; two enantiomeric endo transition states, and two enantiomeric exo transition states.

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