HRID1954013

反应详情

EQUATION

反应方程式

HRID 1954013 的结构方程式

PROCEDURE

实验过程

Compound 20 (3.21 g, 4.23 mmol) was prepared from 18 (3.66 g, 4.49 mmol) in 94% yield by the method used for the preparation of 19. For 20: mp 180-182° C.; 1H NMR (DMSO-d6 /D2O) d 3.69, 3.81 (AB, Jgem =19 Hz, 2H, CH2S), 4.61 (s, 2H, OCH2CO), 4.93, 5.20 (AB, Jgem =14 Hz, 2H, CH2O), 5.20 (d, J=5.0 Hz, 1H, HC(6)), 5.77 (d, J=5.0 Hz, 1H, HC(7)), 6.96-7.53 (m, 14H, 4×ArH); IR (nujol): 3270-3400 (OH, NH, CO2H)), 1788 (β-lactam), 1725 (ester), 1680 (amide), 1620-1635 (C=O) cm-1. Anal. (C37H27N2O12SCl) C, H, N, S, Cl.