Diels-Alder Stereochemistry Explained

Uploaded by lamechivanes on 12.08.2010


In this lesson, we're going to take a closer look
at how frontier molecular orbital theory
can help us explain the experimental observations
we observed in the last lesson as well as observations
about other cycloaddition reactions.
So we're going to begin with the Diels-Alder reaction
and first look at the transition state in orbital terms
and see how it explains the origin of
stereospecificity in the reaction.
So the first thing you should keep in mind
is that as the two reacting partners approach one another
in the Diels-Alder reaction,
they do so in a parallel fashion, so the two planes
in which the π systems reside
approach each other in a parallel manner.
Because the two bonds that form form at the same time,
the substituents off of each reacting partner
are forced to maintain their positions
throughout the reaction.
As a result,
their stereochemistry is maintained in the product.
The relative stereochemistry of substituents
is determined by their relative positions
in the transition state,
so notice that substituent B in the dienophile
and substituent E on the diene
are on opposite sides of the newly forming bond.
As a result, we see that substituents E and B
end up trans in the product.
Similarly, substituent C
is on the same side as substituent E in the diene,
and so we see that those two substituents end up cis
in the product.
This doesn't explain why the dienophile orients itself
the way it does; in fact, we'll get to that in a second,
but it does show how we can go from the position of the atoms
in the transition state to the position of atoms
in the stereochemistry in the product.
You'll want to have the ability to draw the stereochemistry
of any Diels-Alder product quickly and easily.
The easiest way to do this is to pick a transition state
and draw it from a top side view,
placing the substituents
on the same side of the newly forming bonds
with the same stereochemistry in the resulting product.
The schematic at the bottom of this slide
shows you how to do this.
In the reaction of this diene
with this α-β unsaturated aldehyde,
we can see that the transition state
will likely involve the diene
and the dienophile approaching each other in parallel planes,
one on top of the other, and that's drawn for you here.
Notice that we placed the aldehyde group
on the left-hand side
because the dienophile is likely to approach in an endo fashion
and this represents the endo approach of the dienophile.
Placing the reacting partners one on top of the other,
we see that the newly forming bonds are here and here.

And what you should notice is that the substituents
that are on the same side of these bonds
will end up cis in the product.
So in other words, these two hydrogens
will end up cis in the product;
and sure enough, we've draw them both with wedges.
And, the aldehyde and the methyl group
will end up cis in the product; and sure enough,
we've drawn both of these with dashes
in the final product.
We can look at the other end of the diene
and draw the same conclusion.
So, this H is on the same side as these two other hydrogens.
And so we would expect it to have the same stereochemistry
as those hydrogens in the product; and sure enough,
it, too, is drawn on the wedge
in our predicted Diels-Alder product.
For an example to work on your own,
see if you can draw the endo transition state
and the endo product
for the dimerization of cyclopentadiene
through the Diels-Alder reaction.
Now, let's discuss why the dienophile orients itself
the way it does in typical Diels-Alder reactions.
So there are two ways the dienophile
could theoretically orient itself, either exo or endo.
If the substituents on the dienophile
are far away from the double bonds of the diene,
then we call that the exo transition state;
whereas, if the substituents are near
the double bonds in the diene,
then we call that the endo transition state.
And a number of observations
have supported the idea that the endo transition state
is lower in energy than the exo transition state
because the endo product is often observed to be formed
faster than the exo product,
and you might think about how the transition state energy
relates to the rate of reaction for the endo and exo products.
The explanation that's typically offered for this
is that secondary FMO interactions
stabilize the endo transition state,
but not the exo transition state.
Notice that when the dienophile approaches
with its substituents endo,
any p orbitals on the substituents can interact
with the back p orbitals of the diene.
So for instance, the p orbitals in the carbonyl groups
of this anhydride here can interact with the p orbitals
on the diene that are not participating
in the bond forming event of the Diels-Alder reaction.
This secondary interaction,
while it doesn't form any chemical bonds necessarily,
does delocalize electrons, and as a result,
it stabilizes the endo transition state over the exo.
Of course, in the exo transition state,
those same p orbitals of the carbonyls
are much too far away to interact
with the orbitals on the diene.
Here you can see sort of a head-on view
where the orbitals participating in the bond forming events
are here in front, and the secondary orbital interactions
are labeled in the back with these red arrows. tion state,
those same p orbitals of the carbonyls