Organic Chemistry 2 · Organometallic carbon–carbon bonds
Styrene by Stille coupling of iodobenzene and vinyltin
Conditions: Pd(PPh3)4 (cat.), toluene, heat
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The mechanism, step by step
Next plays each step's curved arrows and explains them. The written steps are below.
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Oxidative addition
Pd(0) inserts into the C–I bond of the iodobenzene in one concerted step, giving a palladium(II) complex with the aryl group and iodide on the metal. The halide-bearing ring carbon cannot do SN2, so the metal is what activates the C–I bond.
The curved arrows
- A palladium(0) lone pair forms the new Pd–C bond to the carbon bearing the halogen.
- The breaking C–I bond pair forms the Pd–I bond, so palladium goes from 0 to +2.
Transmetalation of the vinyl group
The vinyl group moves from tin to palladium while iodide moves from palladium to tin. Of the four groups on tin, the sp2 vinyl group transfers much faster than the sp3 butyl groups, so the three butyls act as non-transferable dummy ligands.
The curved arrows
- The Sn–C(vinyl) bond pair forms the new Pd–C bond.
- The Pd–I bond pair forms the I–Sn bond, giving tributyltin iodide.
Reductive elimination
The phenyl and vinyl groups couple on palladium, and the metal returns to Pd(0). Both carbons keep their sp2 geometry through the step.
The curved arrows
- The Pd–C(phenyl) bond pair forms the new C–C bond to the vinyl carbon.
- The Pd–C(vinyl) bond pair returns to palladium as a lone pair, regenerating Pd(0).
Styrene
The vinyl group is now attached to the ring carbon that carried iodine, giving styrene. Tributyltin iodide is the stoichiometric by-product, a toxic tin waste that is the main drawback of the Stille reaction.
What you'll practice
Draw one Stille cycle: oxidative addition of iodobenzene, transfer of the vinyl group from tributyl(vinyl)tin to palladium and reductive elimination to styrene.
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