“…Chromatography of the residue with 1.3-1.5% methanol in chloroform afforded 0.57 g (81%) of an oil: TLC (V) Rf = 0.15, (VI) Rf = 0.57; TO NMR (CDC13) 8 8.57-8.43 (m, 1H), 7.63-7.51 (m, 1H), 7.38-6.98 (m, 12 H), 5.20-5.00 (4 d, total 2 H), 3.80-3.52 (m, 2 H), 3.38-3.18 (m, 1H), 3.08-2.57 (envelope, 5 H), 2.86,2.82 (2 s, total 3 H), 2.30, 2.18 (2 dd, total 1 H); MS m/e (M + H)+ 417.…”