HRID55024

反应详情

EQUATION

反应方程式

HRID 55024 的结构方程式

PROCEDURE

实验过程

The procedure similar to that described in Example 1 was repeated, except that 300 mg (0.67 mmol) of Compound 22 was used in place of Compound 24 and allyl bromide was used in place of methyl iodide. As a result, 187.8 mg (yield: 58%) of Compound 53 was obtained as white crystals.