The endo rule for the Diels–Alder reaction

It is probably easier to see this when both the diene and the dienophile are cyclic. All the double bonds are cis and the stereochemistry is clearer. In the most famous Diels–Alder reaction of all time, that between cyclopentadiene and maleic anhydride, there are two possible products that obey all the rules we have so far described. They are the only possible diastereo isomers of the product—although it has four stereogenic centres, any other diastereoisomers would be impossibly strained.

The two green hydrogen atoms must be cis in the product, but now there are two such com pounds, known as the exo and endo products. When the reaction is carried out, the product is, in fact, the endo compound. Only one diastereoisomer is formed, and it is the less stable one. How do we know this? Well, for cases in which the Diels–Alder reaction is reversible and therefore under thermodynamic control, the exo product is formed instead. The best-known example results from the replacement of cyclopentadiene with furan in reaction with the same dienophile.

Why is the exo product the more stable? Look again at these two structures. On the left-hand side of the molecules, there are two bridges across the ends of the new bonds (highlighted in black): a one-C-atom bridge and a two-C-atom bridge. There is less steric hindrance if the smaller (that is, the one-atom) bridge eclipses the anhydride ring. The endo product is less stable than the exo product and yet it is preferred in irreversible Diels–Alder reactions—it must be the kinetic product of the reaction. It forms faster because a bonding interaction between the electron-deficient carbonyl groups of the dienophile and the developing π bond at the back of the diene lowers the energy of the transition state, lead ing to the endo product.

The same result is found with acyclic dienes and dienophiles. Normally one diastereoisomer is preferred—the one with the carbonyl groups of the dienophile closest to the developing π bond at the back of the diene. Here is an example.

From our previous discussion (it’s a trans, trans diene) we expect the two methyl groups to be cis to each other and the only question remaining is the stereochemistry of the aldehyde group—up or down? The aldehyde will be endo—but which compound is that? The easiest way to fi nd the answer is to draw the reagents coming together in three dimensions. Here is one way to do this.

1. Draw the mechanism of the reaction and diagrams of the product to show what you are trying to decide. Put in the known stereochemistry if you wish. This we have just done (see above).

2. Draw both molecules in the plane of the paper with the diene on top and the carbonyl group of the dienophile tucked under the diene so it can be close to the developing π-bond. 3. Now draw in all the hydrogen atoms on the carbon atoms that are going to become stereogenic centres, that is, those shown in green here.

4. Draw a diagram of the product. Unfold the molecule to show the six-membered ring. All the substituents to the right in the previous diagram are on one side of the new molecule. That is, all the green hydrogen atoms are cis to each other.

 5. Draw a final diagram of the product with the stereochemistry of the other substituents shown too in the usual way. This is the endo product of the Diels–Alder reaction.

اخر الاخبار

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الأمين العام للعتبة العبّاسية المقدّسة يتفقد مجموعة مواكب أمّ البنين (عليها السلام) لخدمة الزائرين

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الآخبار التكنلوجية

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المزيد

مواضيع ذات صلة

The Hammett relationship

Systematic structural variation

Crossover experiments

Labelling experiments reveal the fate of individual atoms

Be sure of the structure of the product

Variation in the structure of the aldehyde

formation of an ester intermediate

formation of an intermediate dimeric adduct

Be sure of the structure of the product

Variation in the structure of the aldehyde

Proposed mechanism C: formation of an ester intermediate

Proposed mechanism B: formation of an intermediate dimeric adduct