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For most Diels-Alder reactions, the product is endo because there are favorable interactions between the newly forming pi-bond and the electron withdrawing groups of the dienophile.

Why is the reaction between Furan and Maleic Anhydride an exception? From what I understand, the carbonyl groups of the anhydride are electron withdrawing.

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up vote 11 down vote accepted

Great question! It turns out that the rate of formation of the "expected" endo product is actually ~500 times faster than the rate of formation of the exo product. However, the Diels-Alder is a reversible reaction. In this case, the exo product is the thermodynamically favored product by about 1.9 kcal/mole over the endo product. So even though the predicted endo product is the kinetically favored product (and formed first), thermodynamics eventually takes over and the more stable exo-adduct is what you wind up with. Here's a link to the JOC paper that provides the detail. What a nice example of competing kinetic and thermodynamic pathways.

enter image description here

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Additionally, if this reaction is done "neat" (without any additional solvent), the exo product is less soluble (even at the higher temperature). As the exo product crystallizes, Le Châtelier drags the equilibrium toward exo. – Ben Norris Jun 7 '14 at 1:32

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