HRID55008

反应详情

EQUATION

反应方程式

HRID 55008 的结构方程式

PROCEDURE

实验过程

The procedure similar to that described in Example 1 was repeated, except that 334.6 mg (0.7 mmol) of Compound 24 was used and butyl iodide was used in place of methyl iodide. As a result, 228.9 mg (yield: 61%) of Compound 4 was obtained as pale yellow crystals.