HRID55047

反应详情

EQUATION

反应方程式

HRID 55047 的结构方程式

PROCEDURE

实验过程

The procedure similar to that described in Example 1 was repeated, except that 223.5 mg (0.50 mmol) of Compound 89 obtained in Example 79 was used in place of Compound 24 and propyl iodide was used in place of methyl iodide. As a result, a free base of Compound 91 was obtained, which was then converted to the hydrochloride in the similar manner as in Example 41 to give 229.9 mg (yield: 87%) of the hydrochloride of Compound 91 as white crystals.