HRID2300627

反应详情

EQUATION

反应方程式

HRID 2300627 的结构方程式

PROCEDURE

实验过程

The title compound was prepared (as described above for the preparation of Example 2) from 500 mg (1.56 mmol) of Intermediate 46 and 315 mg (1.56 mmol) of Intermediate 32 to yield 412 mg of Example 9: TLC (DCM/MeOH, 4/1): Rf=0.53; 1H NMR (DMSO-d6, 300 MHz) δ11.53 (d, 1H, J=9.3), 7.99 (m, 2H), 7.95 (m, 2H), 7.39 (t, 2H, J=8.7), 7.26 (t, 2H, J=8.7), 7.17 (d, 2H, J=8.1), 6.86 (d, 2H, J=8.1), 5.60 (s, 1H), 4.20 (t, 2H, J=6.6), 4.12 (m, 1H), 3.21 (m, 2H), 2.94 (t, 2H, J=6.6), 2.78 (dd, 1H, J=13.8, 8.4), 2.38 (s, 3H), 2.05 (m, 2H), 0.96 (t, 3H, J=7.5); low resolution MS (ES+)m/e 561.4 (MH+).