I have a question about the mechanism of the Diels-Alder reaction. Is this a two-electron or one-electron process as described by Fig (a) or (b) respectively or does not matter? How do we know that?
First of all, full arrow represents a movement of 2 electrons. Half arrows represent movement of 1 electron.
Diels-Alder is a 6π electron pericyclic reaction. The diene provides 4π electrons and the dienophile 2π electrons to the formation of the bonds. The Diels-Alder reaction belongs to [4+2] cycloadditions according to the Woodward-Hoffmann nomenclature.
A purely formal way of determining the participating electrons consists of simply counting the electron flow arrows that are used to illustrate the mechanism and multiplying by two. Each arrow moves two electrons. You have 3 arrows representing a reaction, so 3*2=6 electrons
The number of electrons participating in the cycloaddition determines wether the reaction follows either a supra-supra (s+s) or a supra-antara (s+a) mechanism.