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For most Diels–Alder reactions, the major product is endo because there are favourable 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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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 thermodynamically favored over the endo product by about $\pu{1.9 kcal/mol}$.

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. More details are given in J. Org. Chem. 1978, 43 (3), 518. What a nice example of competing kinetic and thermodynamic pathways.

Exo vs endo Diels–Alder reaction between furan and maleic anhydride

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    $\begingroup$ 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. $\endgroup$ – Ben Norris Jun 7 '14 at 1:32

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