Реакция #58555

Уравнение реакции

product

5-Benzyl-4,5,6,7,7a,8-hexahydro-3bH-3-thia-5-aza-cyclopenta[α]indene

K2CO3

Ace-Cl

→

subtitle compound

Выход 66.0%

4,5,6,7,7a,8-Hexahydro-3bH-3-thia-5-aza-cyclopenta[α]indene

Выход 66.0%

Растворители

dichloroethane

Условия реакции

Обработка

1. 1
   ФильтрацияNext, the reaction mixture was filtered
2. 2
   Концентрированиеconcentrated
3. 3
   workup.DISSOLUTIONthe residue was dissolved in anhydrous MeOH (25 ml)
4. 4
   КонцентрированиеThe MeOH was concentrated
5. 5
   Другоеtriturated with ether producing

Методика

The product from step f) (409 mg, 1.52 mmol) in dichloroethane (8 ml) was treated with K2CO3 (839 mg, 6.08 mmol) and Ace-Cl (658 μl, 6.08 mmol) at 80° C. for 16 hours. Next, the reaction mixture was filtered, concentrated and the residue was dissolved in anhydrous MeOH (25 ml) and stirred for 1 hour at 22° C. The MeOH was concentrated and triturated with ether producing to give 215 mg (66%) of the subtitle compound as its HCl salt. MS calculated for C10H13NS+H 180, observed 180.

Источник