Question

2) Identify the products of the following reactions, and give the electron count and oxidation state of the metal in the reactant and product: PHE C R - co + CHE Oxidative Addition PHY HEÇ PPhg Php-Reco 1,1-insertion PP, PhyP Co PPhg CIOs CH2 B-elimination Reductive elimination CH3 PnP

Answer 1

Pla co midature addition a Rh + CHY CH3TH ARh - co PH3 PHA PH3 Reactant onidation State Rh is (+1) election count="1660, e 9+1+9+9+2=1600 117) a ca- Rh- co Rh ce co PH₃ PHA - Product (I) onidation State Rhis (+3) election count = 1860 here x-y group le (CH₂-H) is nonpolar hence onidature addition occur through concerted or three centred transition State In which firstly the incoming ligand bind as - Complex and then x-y bond break resulting in Shring back donetion of two electon from the metal into the o- antibanding molecular orbital of X-y ligandi Them Second steb really inualues the credation of metal because two election are formed bansferred to o- anti- bonding of X-Y

6 CH3 upph3 - PPhz 1) indertion pp 1 12))))) Phy 11 Re 1) inseation De co 1 ppha Phil-Rb - CO CH₂ PPh3 Ph₂p co Phy co Product onidation State is Reti) Reactant . Onidation state Reo) i In bl insertion reaction metal and the migrated ligand make the bonds to the same alm of the unsaturated ligand such as co. Election count for Reactent = 1868 Complex Electon count for product - 1860 Complet u H PPha ppha Belimination U ce - os - Cha Reductive CH3 elimination Per Се reactant a-Os Cha Phy a product Total eo count the onidation state Os(+3) Totale 16ee count onidation state 08 (43)