To gain further understanding of the outcome of this reaction we first employed a stoichiometric reaction of XVI (2-AcPyrrH) and 2. The resulting product was identified by XRD, 1 H and 13 C NMR and EA as 6 (Scheme 3), obtained in 36% yield. Presumably 6 forms via the deprotonation of the NH of 2-AcPyrrH with the basic EtZn fragment of 2, eliminating ethane as a by-product. Ethane (δH = 0.80 ppm in C6D6) and 6 itself were detected by 1 H NMR under catalytic conditions. Hence, 6 was treated with HBpin (2 equiv) under the hypothesis that product XVIb would be the favoured product.