Diels Alder Reaction of dienes and dienophiles

by James

Description: The Diels Alder reaction converts a diene and an alkene (usually electron-poor, called a “dienophile”) into a six-membered ring containing an alkene (cyclohexene).

Notes: X here is usually an electron withdrawing group such as a ketone, ester, or CN.


Notes: Example 1 shows a simple Diels-Alder.

Example 2 shows the reaction with a cyclic diene, to form a bicyclic compound.

Example 3 shows how stereochemistry on the dienophile is conserved (the groups cis to each other on the double bond will end up cis on the ring). That is to say that the reaction is stereospecific. 

Examples 4 and 5 show that the groups on the “outside” of the diene end up on the same side of the ring to each other.

Example 6 shows what happens when we have substitution on both the diene and the dienophile. A mixture of diastereomers is formed, called the “endo” and the “exo” products. Generally speaking the “endo” product is favored.

Example 7 shows an alkyne being used as a dienophile.

Examples 8 and 9 show what happens when we have electron donating groups attached to the diene. We always line up the more “electron-rich” carbon of the diene with the more “electron-poor” carbon of the dienophile.

Mechanism: The mechanism of this reaction is concerted meaning that all the bond forming and bond-breaking happens at the same time. The reaction proceeds in a 1-step cycloaddition (Step 1, arrows A, B and C).

Notes: The arrows could just as correctly been drawn proceeding in a clockwise fashion. The end result is the same.



{ 5 comments… read them below or add one }

Sayan Roy December 24, 2011 at 2:53 pm

very very helpful…thanks


Joseph September 12, 2012 at 4:01 pm

Extremely articulate. Thank you!


Adam Moyer February 27, 2014 at 9:24 pm

Note: Pertaining to example 6, the endo product is kinetically favored, while the exo product is thermodynamically favored.
It is a little tricky to explain, but basically the orbitals in the endo product line up well, so when there is not a lot of energy in the system, the reaction will go toward the endo product which has a lower transition energy. However, when there is more energy in the system the product will go toward the exo product because it is a lower energy state overall. You can see that the exo product would be more stable because there are less steric interactions.


James February 28, 2014 at 11:52 am

Thank you for clarifying that point!


Andres Vanegas April 29, 2014 at 11:48 pm

In regards to exo v. endo.- I was under the impression that while the exo is stericlly favored, the endo is favored as the major product due to secondary stabilization provided by overlap of pi bonds during the transition state.

Second, are the arrows actually only able to be represented as clockwise since the diene is symmetrical(final example), therefore having equal partial negative charges on each end from which electron density will be dispersed to the partial positive region on the dienophile? If the diene were not symmetrical, arrows would need to be from the partial negative regions to the partial positive regions, correct?


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