HRID37029

反应详情

EQUATION

反应方程式

REACTANTS

反应物

结构：O=C(CCCCc1c2ccccc2nc2ccccc12)Oc1c(F)c(F)cc(F)c1F

HCID19934285

未命名化合物

结构：O=C(CCCCc1c2ccccc2nc2ccccc12)N1C[C@H](O)C[C@H]1CO

HCID22853587

未命名化合物

PRODUCTS

生成物

HCID9215

Acridine

C13H9N

HCID22853587

未命名化合物

收率 92%

PROCEDURE

实验过程

A procedure similar to that described in Example XXIV was used for the preparation of 21. 1.28 g (3.00 mmol) of acridine TFP ester 20 gave 1.04 g (92% yield) of 21 as a pale yellow solid foam:

来源与证据

来源类型: Open Reaction Database
ORD 记录: ord-11b09e838dbb4d5b8db53e3e9d14a692
来源数据集: uspto-grants-1995_05
DOI: 10.6084/m9.figshare.5104873.v1
专利: US05419966
来源说明: uspto-grants-1995_05

反应 SMILES

[CH:1]1[C:14]2[C:5](=[N:6][C:7]3[C:12]([C:13]=2[CH2:15][CH2:16][CH2:17][CH2:18][C:19]([N:21]2[C@H:25]([CH2:26][OH:27])[CH2:24][C@@H:23]([OH:28])[CH2:22]2)=[O:20])=[CH:11][CH:10]=[CH:9][CH:8]=3)[CH:4]=[CH:3][CH:2]=1.C1C2C(=NC3C(C=2CCCCC(OC2C(F)=C(F)C=C(F)C=2F)=O)=CC=CC=3)C=CC=1>>[CH:1]1[C:14]2[C:5](=[N:6][C:7]3[C:12]([CH:13]=2)=[CH:11][CH:10]=[CH:9][CH:8]=3)[CH:4]=[CH:3][CH:2]=1.[CH:1]1[C:14]2[C:5](=[N:6][C:7]3[C:12]([C:13]=2[CH2:15][CH2:16][CH2:17][CH2:18][C:19]([N:21]2[C@H:25]([CH2:26][OH:27])[CH2:24][C@@H:23]([OH:28])[CH2:22]2)=[O:20])=[CH:11][CH:10]=[CH:9][CH:8]=3)[CH:4]=[CH:3][CH:2]=1