HRID55015

反应详情

EQUATION

反应方程式

HRID 55015 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 (0.63 mmol) of Compound 24 was used and 4-nitrobenzyl bromide was used in place of methyl iodide. As a result, 145.0 mg (yield: 38%) of Compound 11 was obtained as pale yellow crystals.