In the above reaction catalyzed by Pd, I'm not exactly sure what's going on at the second step. It says in the book that Pd is using its electrons to push out X, and the electrons are returned to Pd when Nu comes and pushes out Pd.
My understanding is that in the first intermediate, the double bond's 2 pi electrons are complexed to Pd. When X leaves, the 2 pi electrons (that are now spread out between 3 C's) are still complexed to Pd. By the drawing of the mechanism, it seems to me that Pd has also used 2 of its electrons to bond with the 3 C's. So in the second intermediate, are the Pd and the allyl cation sharing a total of 4 electrons? If so, shouldn't there be a double bond, not a single bond, from Pd to the allyl cation?