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HuagongsheHuagongshe Chemistry Print Report · For research reference only
HRID 1586288Generated: November 3, 2026

Iodoethane → 10-ethylphenoxazine

https://huagongshe.com/reaction/1586288

Reaction equation

Participants

RoleNameHCIDCharge
ReactantIodoethaneHCID 6340—
ReactantPotassium CarbonateHCID 11430—
ReactantPotassium HydroxideHCID 14797—
ReactantPhenoxazineHCID 67278—
CatalystTetrabutylammonium hydrogen sulfateHCID 94433—
SolventTolueneHCID 1140—
Product10-ethylphenoxazineHCID 328622778.5%

Reaction Steps

Into a reaction flask were introduced 3.0 g (16.4 mmol) of phenoxazine, 6.5 g (98.5 mmol) of 85% KOH, 3.7 g (26.8 mmol) of potassium carbonate, 577 mg (1.7 mmol) of tetra-n-butylammonium hydrogensulfate, and 100 ml of toluene. Thereto was gradually added 5.1 g (32.7 mmol) of ethyl iodide with a dropping funnel (slight heat generation occurred). Subsequently, the mixture was reacted at 60° to 80° C. for 7 hours. After the disappearance of the phenoxazine was ascertained by TLC and gas chromatography, the resultant toluene solution was cooled, washed with water three times, dried (with MgSO4), and then concentrated. The residue was distilled under vacuum to obtain 2.72 g of 10-ethylphenoxazine.

Provenance & Evidence

DOI10.6084/m9.figshare.5104873.v1
Patent numberUS05942615
Source datasetuspto-grants-1999_08
ORD IDord-abe2815eef714fcaade5117f81ac6d2d
Source noteuspto-grants-1999_08