Dienes and MO Theory
Regiochemistry In The Diels-Alder Reaction
Last updated: September 12th, 2019 |
Regiochemistry (“Regioselectivity”) In The Diels-Alder Reaction
The Diels-Alder is an onion, and we just keep peeling back the layers. Today’s post is about regiochemistry in the Diels-Alder. Quick summary:
When non-symmetrical dienes react with non-symmetrical dienophiles, two regioisomers are possible. Dienes with substituents on the terminus (“1-substituted dienes”) tend to give “1,2” products (“ortho”). Dienes with substituents on the 2-position (“2-substituted dienes”) tend to give the “1,4” product (“para”). In general, “1,3” products (“meta”) are only minor byproducts.
Table of Contents
- Regiochemistry In The Diels-Alder Reaction
- Flashback: Markovnikov’s Rule and “Regioselectivity”
- The Diels-Alder Reaction Is Regioselective
- Summary: Regioselectivity In The Diels-Alder Reaction
- Notes (more on the origin of regioselectivity)
So far in the Diels-Alder, we’ve seen examples of:
- symmetrical dienes with symmetrical dienophiles
- unsymmetrical dienes with symmetrical dienophiles
- symmetrical dienes with unsymmetrical dienophiles
These three situations (laid out in the image below) each have the potential to form stereoisomers (i.e. diastereomers and/or enantiomers). But all the products have exactly the same connectivity.
This brings us to a particularly challenging case. What happens when we have an unsymmetrical diene reacting with an unsymmetrical dienophile?
Why is this situation different from the first three?
Because there are two different ways for the diene and dienophile to “line up! “. In the example above, the diene and dienophile can come together two ways:
- “head to head” such that the methyl group on the diene and the C=O bond point in the same direction (giving the top product);
- “head to tail” such that they point in the opposite direction (bottom product).
How are these products related to each other, overall (not counting stereochemistry)?
Since they have the same molecular formula but different connectivity, they’re constitutional isomers.
Now comes the big question. In this kind of a Diels-Alder, are these products formed in roughly equal ratios… or is there a preference for one type of connectivity over another?
The short answer is, “yes, the Diels-Alder has a preference for one type of connectivity.”
In other words, the reaction has regioselectivity.
Wait. What’s regioselectivity? Quick review:
Where have we seen this type of situation before?
With alkenes! We saw that adding an acid like HCl to an alkene like 1-butene, we could obtain two possible products: 1-chlorobutane or 2-chlorobutane. [See: Markovnikov’s Rule]
These products have the same molecular formula, but different connectivity, which makes them constitutional isomers.
However, we saw that the reaction doesn’t give an equal ratio of products. Instead, there’s about a 4:1 preference for 2-chlorobutane over 1-chlorobutane, which we saw was due to a greater preference for the more stable carbocation intermediate.
This preference goes by the name “Markovnikov’s Rule”:
The tendency of alkenes to obey “Markovnikov’s rule” in these reactions is an example of regioselectivity. The reaction is selective in forming one constitutional isomer over another.
(We say “selective” and not “specific” because more than one product is formed. Use of the prefix “regio” comes from the observation that the chloride tends to attack one region of the double bond, and the proton, another. )
It’s worth noting that 2-chlorobutane is formed as a racemic mixture of enantiomers. So while the reaction is regioselective, it is not enantioselective.
Like the reaction of acids with alkenes, the Diels-Alder reaction is regioselective. Two main cases will illustrate the point.
Case 1: 1-substituted dienes
The first important case concerns a diene with a substituent on the “1” position of the diene, such as 1-methoxybutadiene. (I know that if a methyl group were present instead, it would technically be the “4” position according to IUPAC – bear with me on this).
Consider the Diels-Alder of 1-methoxy butadiene with methyl methacrylate. There are two ways that the diene and dienophile can connect.
- Line it up one way (“head to head”, below), and you get a new six-membered ring where two substituents are on two adjacent carbons (a “1,2” relationship)
- Line it up another way (“head to tail”), and the two substituents have a “1,3” relationship.
These two products have different connectivity and are therefore constitutional isomers (“regioisomers”).
By analogy to aromatic nomenclature, the “1,2” and “1,3” patterns are nicknamed ortho- and meta- respectively. [I say “nicknamed” because these are terms of convenience, nothing else. Don’t tell IUPAC!]
So which of the two Diels-Alder products is favored? The “ortho” or the “meta” ?
Here’s what’s experiments tell us:
The ortho product is major and the meta product is minor.
This holds for a large number of 1-substituted dienes; I’ll just show two. Hans Reich at UW-Madison has a longer list of examples – see here.
- With 1-methoxybutadiene, the ortho is the only product!!
- When a methyl group is in that position, the ortho outnumbers para by about 8:1
Case 2: 2-substituted dienes
The second important case is when there’s a substituent on the 2-position of the diene, such as 2-methylbutadiene.
Again, there are two ways it can line up, except this time it’s to provide para (1,4) and meta (1,3) products.
So which of these products is dominant?
The para product is major and the meta product is minor.
- 2-methoxybutadiene favors the para product by about 8:1
- 2-methylbutadiene favors the para product by about 2:1
What if there’s substituents on both the 1- and 2- positions? In these cases, it turns out that the substituent on the 1-position is more powerful at directing the products. [See footnote]
Avoid the meta- product. That’s really it.
So why does the Diels-Alder wind up this way?
For the answer, read on…..
Think of the diene as a nucleophile and the dienophile as the electrophile.
The dominant product will be the one where the most nucleophilic carbon on the diene lines up with the most electrophilic carbon on the dienophile.
So how do we determine what these carbons are?
- Look for the second-best resonance form of the diene and the dienophile! (sometimes known as Grossman’s rule)
- Line up the negative charge on the diene with the positive charge on the dienophile.
This will get you to the right result!
Note how this results in the ortho– product and not the meta- product, just like what’s observed in experiment.
What about 2-substituted dienes?
Same deal. Draw out the “second-best” resonance form. Now line up the negative charge on the nucleophile with the positive charge on the “2nd best” resonance form of the electrophile:
Note how this correctly predicts the para product will be favored over the meta product.
Now… is this really the best way to do it?
For our purposes, yes.
For more advanced purposes… we rely on molecular orbital calculations. In advanced courses, we talk a lot about the size of “coefficients” on the HOMO or LUMO of the diene/dienophile, and we’re not going there. If you are interested in this topic, I highly recommend Ian Fleming’s Frontier Orbitals and Organic Chemical Reactions. Classic book.
What About 1,2-Substituted Dienes?
What’s observed is that the 1-position on the diene has a greater influence on the product than the 2-position.