In Diels-Alder reaction (or Diels-Alder cycloaddition), the atoms at the ends of the diene add to the dienophile double or triple bond (alkene or alkyne), thereby closing a ring (product is cyclohexene). The new bonds form simultaneously and stereospecifically. It does not include even an intermediate, it all happens in one step. This reaction is also called [4+2] cycloaddition because reaction takes place between four conjugated atoms containing four π electron react with a double bound containing two π electrons.
The Diels-Alder reaction takes place in one step. Both new carbon-carbon σ bonds and the new π bond form simultaneously, just as the three π bonds in the starting materials break. For such one-step reactions, we say they are concerted.
The mechanism of this reaction should not present a major problem in understanding. You just need to get closer diene to dienophile and add sp² carbon atoms of diene to sp² carbon atoms of dienophile. The arrows can be drawn proceeding in a clockwise fashion or vice versa. The end result is the same.
In this example, the simplest representatives of diene and alkene are used, 1,3-butadiene and ethene. But what happens if we these compounds are substituted? Then we must take care of stereochemistry!
Stereochemistry of the dienophile
When the dienophile doesn’t contain any substituents, the reaction is slow and the yield is low. But if dienophile contains an electron-withdrawing substituent such as a carbonyl group, the reaction will proceed more rapidly and with a much higher yield. These substituted dienophiles lead to the formation of chiral center(s). And because we must think about stereochemistry.
The stereochemistry at the original double bond of the dienophile is retained in the product. Specifically, a cis dienophile produces a cis disubstituted ring, and a trans dienophile produces a trans disubstituted ring.
Look at the following example. We have dienophile with two R groups which are cis to each other, so they are on the same side. Carbons of dienophile go from being sp² hybridized to being sp³ hybridized forming chiral centers. Since we have concerted movement of electrons, these two R groups end up on the same side.
If we look our dienophile and think about the groups on the left and the right side of the line, the groups on the right side will be always up in the product (drawn on the wedges). And groups on the left side of the line will be always down in the product (drawn on dashes).
But if think that diene approaches dienophile on the other side, our groups will be on the opposite side of the previous case. So, our R groups will be on the left side, and hydrogens will be on the right side. In the resulting compound, the R groups will be drawn on the dashes, and hydrogens on the wedges.
Now, if we have dienophile with two R groups which are trans to each other, in the product these R groups will be up and down, i.e. on a wedge and a dash. The R group which is drawn on the left side of the double bond, in the product will be on a dash, and the R group which is drawn on the right side of the double bond will be on a wedge.