HRID2300624

反应详情

EQUATION

反应方程式

HRID 2300624 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 40 mg (0.092 mmol) of Intermediate 50 and 17 mg (0.092 mmol) of Intermediate 16 to yield 21 mg of Example 5: TLC (DCM/MeOH, 4/1): Rf=0.50; 1H NMR (DMSO-d6, 400 MHz) δ11.59 (d, 1H, J=9.8), 8.09 (d, 2H, J=9.0), 7.85 (d, 2H, J=8.8), 7.78 (m, 2H), 7.39 (m, 3H), 7.12 (d, 2H, J=8.0), 6.80 (d, 2H, J=8.0), 5.54 (s, 1H), 4.15 (t, 2H, J=6.6), 4.17 (m, 1H), 3.11 (m, 1H), 2.90 (t, 2H, J=6.6), 2.72 (m, 1H), 2.33 (s, 3H), 1.98 (m, 2H), 0.9 (t, 3H, J=7.5); low resolution MS (ES+)m/e 593.1 (MH+).