Answers

2016-03-16T13:02:07+05:30
Reaction can be taken place of benzamide to aniline.
with a Fourteen-Electron Iridium(I) Bis(Phosphinite).

To understand this you can see the reaction :
Anilines react with (POCOP)Ir(C6H5)(H), 12, (POCOP = 2,6-(OPtBu2)2C6H3) to yield equilibrium mixtures of 12, the Ir(I) σ-complexes (POCOP)Ir(NH2Ar), 13, and the Ir(III) oxidative addition adducts (POCOP)Ir(H)(NHAr), 14. Quantitative studies of these equilibria for a series of anilines were carried out. Anilines possessing electron-withdrawing groups favor the Ir(III) oxidative addition adduct over the Ir(I) sigma complex. Low temperature studies using p-chloroaniline show that the Ir(I) σ-complex is the kinetic product of reaction and is likely the precursor to the Ir(III) oxidative addition adduct. Reductive elimination of complexes 14 in the presence of ethylene led to the corresponding anilines and the ethylene complex (POCOP)Ir(C2H4). Kinetic analysis of these reactions for 14e,f,gbearing electron-withdrawing aryl groups (Ar- = p-CF3C6H4-, C6F5-, 3,5-bis(CF3)C6H3-) shows the rate is independent of ethylene concentration. The ΔG‡ values for these reductive eliminations fall in the range of 21–22 kcal/mol. X-Rayb analysis establishes 14f (Ar- = C6F5-) as a square pyramidal complex with the hydride occupying the apical site. Reaction of 12 with benzamides 21a,b yields quantitatively the Ir(III) oxidative addition adducts, (POCOP)Ir(H)(NHC(O)Ar), 22. X-Ray analysis of22b (Ar- = C6F5-) shows significant interaction of the carbonyl oxygen with Ir in the site trans to hydride. The barrier to reductive elimination of 22a, 29 kcal/mol, is substantially higher than for complexes 14e,f,g. #Copyright
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