Addition of ethylene to mixtures of Cp*CrMe2(PMe3) (1) and excess methylaluminoxane (MAO), followed by aqueous workup of the resulting solutions, results in the formation of a distribution of odd-carbon alkanes. When B(C6F5)3 is added to 1, followed by THF, one obtains dark crystals. A single-crystal X-ray diffraction study revealed that the product of the reaction is [Cp*CrMe(PMe3)(THF)][MeB(C6F5)3] (2). Polyethylene is obtained when using 1 and 2 equiv of B(C6F5)3. The reaction of ethylene with a mixture of 1 + 2 B(C6F5)3 + 380 “AlMe3” produces a distribution of odd-carbon alkanes. If triethylaluminum is used, instead of trimethylaluminum, the product is a distribution of even-carbon alkanes. The product distribution shifts to lower molecular weight product, and lower activities are attained with increasing triethylaluminum concentration. Use of more dichoromethane instead of toluene or hexane increases the ethylene consumption. Triisobutylaluminum and trioctylaluminum result in considerably lower activities. A mechanism for the oligomerization of ethylene involving fast transmetalation reactions between chromium and aluminum is proposed.

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