HRID2300623

反应详情

EQUATION

反应方程式

HRID 2300623 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 75 mg (0.18 mmol) of Intermediate 51 and 41 mg (0.18 mmol) of Intermediate 24 to yield 48 mg (43% yield) of Example 3: TLC (DCM/MeOH, 4/1): Rf=0.42; 1H NMR (DMSO-d6, 400 MHz) δ11.5 (d, 1H, J=8.9), 7.95 (d, 2H, J=7.2), 7.77 (d, 2H, J=8.7), 7.72 (d, 2H, J=8.2), 7.1 (d, 2H, J=8.4), 6.97 (d, 2H, J=8.9), 6.79 (d, 2H, J=8.5), 5.58 (s, 1H), 4.65 (m, 1H), 4.11 (t, 2H, J=6.5), 4.1 (m, 1H), 3.16 (dd, 1H, J=13.7, 3.6), 2.84 (t, 2H, J=6.5), 2.74 (dd, 1H, 13.7, 9.0), 2.28 (s, 3H), 1.67 (s, 3H), 1.25 (d, 6H, J=6.0); low resolution MS (ES+)m/e 637.1 (MH+);