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Pd catalyzed SN2 (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?

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This is a classic oxidative addition. You're kind of outside the classic world of single and double bonds. You have 4 electrons spread of 3 carbon atoms that link stabilize a bonding interaction with the palladium. Two of those electrons came from the palladium, so we think of it as having been oxidized, hence the name of the process.

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.

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  • $\begingroup$ So when organometallic complexes are drawn, a bond from the metal to a ligand doesn't necessarily mean they only share 2 electrons right? They can share any amount of electrons? $\endgroup$ – carbenoid Nov 1 '16 at 1:54
  • $\begingroup$ 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. $\endgroup$ – Zhe Nov 1 '16 at 2:22

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