The scheme used to produce a somewhat more complex 5-a-reductase inhibitor relies on a chiral auxiliary to yield the final product as a single enantiomer. The first step in a sequence similar to that above starts with the reaction of bromotetralone () with R-a-phenethyl amine () to afford the enamine (). Reaction with methyl iodide adds the methyl group at what will be a steroid-like AB ring junction.
This product is then treated with acryloyl chloride. The initial step in this case probably involves the acylation of nitrogen on the enamine; conjugate addition then completes the formation of the lactam ring (). Treatment of that product with triethyl silane then reduces the ring unsaturation and cleaves the benzylic nitrogen bond on the auxiliary to yield (23-6) as the optically pure trans isomer. Displacement of bromine with the mercapto benzthiazole () completes the synthesis of izonsteride.