Реакция #736458

ord-9979805180de437ba2fcdc635d24a6df

Уравнение реакции

Oc1ccc(O)cc1
hydroquinone
O=C(Cl)c1ccc([N+](=O)[O-])cc1C(=O)Cl
4-nitrophthalic acid dichloride
→
O=C1c2ccc([N+](=O)[O-])cc2C(=O)c2c(O)ccc(O)c21
6-nitro-1,4-dihydroxy-9,10-anthraquinone
Выход 75.0%

Реактанты

Oc1ccc(O)cc1
hydroquinone
O=C(Cl)c1ccc([N+](=O)[O-])cc1C(=O)Cl
4-nitrophthalic acid dichloride

Реагенты

Нет

Растворители

CCO
EtOH

Условия реакции

Подробные условия
See reaction.notes.procedure_details.

Методика

The method was similar to that in Example 1 but using 11.0 g (0.1 mols) of hydroquinone and 24.7 g (0.1 mols) of 4-nitrophthalic acid dichloride. The product was 21.4 g (75% yield) of 6-nitro-1,4-dihydroxy-9,10-anthraquinone (5), m.p.=226°-228° C. (EtOH). The compound had the following characteristics:

Источник

DOI: 10.6084/m9.figshare.5104873.v1Патент: US04739062uspto-grants-1988_04