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HRID13700

反应详情

EQUATION

反应方程式

HRID 13700 的结构方程式

REACTANTS

反应物

2
结构：O=[N+]([O-])c1cccc2cc[nH]c12
HCID23396

Indole, 7-nitro-

C8H6N2O2

结构：CCC1(O)CCc2cc(F)ccc21
HCID59340428

未命名化合物

PRODUCTS

生成物

1
结构：CCC1(c2c[nH]c3c([N+](=O)[O-])cccc23)CCc2cc(F)ccc21
HCID11609506

3-(1-Ethyl-5-fluoro-2,3-dihydro-1H-inden-1-yl)-7-nitro-1H-indole

C19H17FN2O2

PROCEDURE

实验过程

Utilizing 1-ethyl-5-fluoro-indan-1-ol and 7-nitroindole, the title compound is prepared as in example 1. 0.37 g (42%). NMR (400 MHz, CDCl3): δ 0.85 (t, 3H), 2.18 (m, 2H), 2.35 (m, 1H), 2.53 (m, 1H), 3.00 (m, 2H), 6.83 (t, 1H), 6.92 (t, 1H), 7.01 (m, 2H), 7.08 (s, 1H), 7.44 (d, 1H), 8.09 (d, 1H), 9.76 (s, 1H, NH).