反应详情

Compound 155 (0.52 g, 51%) was prepared from 154 (1.5 g, 2.6 mmole), BCl3 (5M solution in CH2Cl2, 14 mL), CH2Cl2 (30 mL) and MeOH (20 mL) by the method described for 166 which on recrystallization from EtOAC gave 155: m.p. 155°; 1H NMR (DMSO-d6): d 5.76 (s, 2H, C-1'), 4.63 (t, 2H, exchanges with D2 O, 2×OH), 3.54-3.48 (m, 1H, C-3'), 3.43-3.25 (m, 4H, C-4' and 5'H). Anal. Calcd. for C11H9N2Cl5O3 : C, 33.46; H, 2.28; N, 7.09. Found: C, 33.60; H, 2.15; N, 6.91.

1. customdescribed for 166 which on recrystallization from EtOAC
2. customgave 155