Problem

The reaction of with tri- o –tolylphosphine in ethanol at 25 °C gives a blue-green compl...

The reaction of with tri- o –tolylphosphine in ethanol at 25 °C gives a blue-green complex I (C42H42P2Cl2Rh) that has at 351 cm-1 and μeff = 2.3 BM. At a higher temperature, a diamagnetic yellow complex II that has an Rh:Cl ratio of 1:1 is formed that has an intense band near 920 cm-1. Addition of NaSCN to II replaces Cl with SCN to give a product III having the following 1H NMR spectrum:

Treatment of II with NaCN gives a phosphine ligand IV with the empirical formula C21H19P and a molecular weight of 604. IV has an absorption band at 965 cm-1 and the following 1H NMR spectrum:

Determine the structural formulas of compounds I through IV , and account for as much of the data as possible. (See M. A. Bennett, P. A. Longstaff, J. Am. Chem. Soc. , 1969 , 91 , 6266.)

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Solutions For Problems in Chapter 14