HRID2300630

反应详情

EQUATION

反应方程式

HRID 2300630 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 (0.177 mmol) of Intermediate 51 and 39 mg (0.194 mmol) of Intermediate 32 to yield 70 mg of Example 15: TLC (DCM/MeOH (4:1): Rf=0.55; 1H NMR (DMSO-d6, 300 MHz) δ11.53 (d, 1H, J=9.6), 7.89 (dd, 2H, J=8.8, 5.7), 7.85 (d, 2H, J=8.8), 7.26 (t, 2H, J=8.8), 7.18 (d, 2H, J=8.5), 7.05 (d, 2H, J=8.8), 6.86 (d, 2H, J=8.5), 5.60 (s, 1H), 4.73 (hept, 1H, J=6.0), 4.19 (t, 2H, J=6.7), 4.13 (br s, 1H), 3.19 (m, 1H), 2.92 (t, 2H, J=6.6), 2.80 (dd, 1H, J=14.0, 9.2), 2.36 (s, 3H), 2.07 (m, 2H), 1.34 (d, 6H, J=6.0), 0.99 (t, 3H, J=7.5); low resolution MS (ES+)m/e 601.1 (MH+).