Direct observation of reductive elimination of methyl iodide from a rhodium(III) pincer complex: the importance of sterics.

A rare case of directly observed alkyl halide reductive elimination from rhodium is reported. Treatment of the naphthyl-based PCP-type Rh(III) methyl complexes 2a,b [(C10H5(CH2PR2)2)Rh(CH3)(I)] (R = iPr 2a, R = tBu 2b) with CO resulted in facile reductive elimination of methyl iodide in the case of 2b, yielding the Rh(I) carbonyl complex [(C10H5(CH2PR2)2)Rh(CO)] 3b (R = tBu), while the less bulky 2a formed CO adducts and did not undergo reductive elimination, contrary to expectations based on electron density considerations. Moreover, 3b oxidatively added methyl iodide, while 3a did not. CD3I/CH3I exchange studies in the absence of CO indicate that reversible formation of (ligated) methyl iodide takes place in both systems. Subsequently, when CO is present, it displaces methyl iodide in the bulkier tBu system, whereas with the iPr system formation of the Rh(III) CO adducts is favored. Iodide dissociation followed by its attack on the rhodium-methyl group is unlikely.