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HuagongsheHuagongshe Chemistry Print Report · For research reference only
HRID 1586281Generated: November 2, 2026

Iodoethane → N-Ethylphenothiazine

https://huagongshe.com/reaction/1586281

Reaction equation

Participants

RoleNameHCIDCharge
ReactantIodoethaneHCID 6340—
ReactantPhenothiazineHCID 7108—
ReactantPotassium CarbonateHCID 11430—
ReactantPotassium HydroxideHCID 14797—
CatalystTetrabutylammonium hydrogen sulfateHCID 94433—
SolventTolueneHCID 1140—
ProductN-EthylphenothiazineHCID 21917883.8%

Reaction Steps

Into a 1-liter reaction flask were introduced 25 g (0.126 mol) of phenothiazine, 50.0 g (0.757 mol) of 85% KOH, 4.3 g (0.013 mol) of tetra-n-butylammonium hydrogensulfate, 28.0 g (0.203 mol) of potassium carbonate, and 500 ml of toluene. Thereto was gradually added 39.0 g (0.250 mol) of ethyl iodide. The mixture was reacted at 60° to 70° C. for 5 hours and then filtered. The filtrate was washed twice with water, dried with magnesium sulfate, and then concentrated. As a result, 28.7 g of a crude reaction product was obtained. This crude product was recrystallized from 100 ml of isopropanol to obtain 23.99 g of 10-ethylphenothiazine (4a). Yield, 84.1%; m.p., 104°-105° C.

Provenance & Evidence

DOI10.6084/m9.figshare.5104873.v1
Patent numberUS05942615
Source datasetuspto-grants-1999_08
ORD IDord-77ae29dc13e64dd08ac732f57d06ac3f
Source noteuspto-grants-1999_08