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?

Fig 1

  • $\begingroup$ A Diels-Alder reaction is a [4+2] cycloaddition, which is formally shown as a two electron process. I suspish that both ways of drawing the mechanism might be equivalent thouh. I was recently having a confusion about MS fragmentation products. Generally, I draw McLafferty rearrangements with three 2 electron arrows just like a Diels-Alder, but my book uses six 1 electron arrows. I don't think it really makes much of a difference. For Diels-Alder, though, I've never seen the six 1 electron arrow mechanism in a book, but that doesn't really mean it's wrong. $\endgroup$ – gannex Apr 23 '16 at 5:32
  • $\begingroup$ I'm don't think the mechanism on the right would necessarily require adherence to the Woodward-Hoffman rules. $\endgroup$ – SendersReagent Apr 23 '16 at 7:31
  • $\begingroup$ It's a little like with your question with benzene - both are only lame ways to try to represent what's really happening. $\endgroup$ – Mithoron Apr 23 '16 at 13:07

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.

This is a nice source for Diels-Alder mechanisms

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    $\begingroup$ This doesn't answer the question. Beyond that, it doesn't even point out anything useful. It mentions a lot of things that are related but definitely off-topic. The structure on the left uses three full arrows to represent six electrons. The structure on the right uses six half arrows to represent six electrons. Maybe the phrasing in the original isn't perfect, but look at the images and you clearly see that the question was asking if each electron acts alone vs each electron being part of a pair that moves togethers. $\endgroup$ – SendersReagent Apr 23 '16 at 8:13

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