# Formation of pi-allyl Pd cation complex

(from pg. 1334 of Clayden Organic Chemistry 1st ed.)

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?

See reference for general $\mathrm{S}_{N}2$ type oxidative addition. The only difference from your example is that the electrophile has a double bond that serves as an additional ligand for the palladium.
• That depends on the type of ligand. There are X ligands and L ligands. For X ligands, it's shared. For L ligands, they exclusively come from the ligand. Allyl is a XL ligand when it's in the $\eta_{3}$ mode where all three carbon atoms are coordinated. In this case, the X portion means a covalent-ish bond with the palladium. In reality, it's probably fairly delocalized. – Zhe Nov 1 '16 at 2:22