HRID55015
反应详情
EQUATION
反应方程式
REACTANTS
反应物
PRODUCTS
生成物
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.