4-Methyl-5-(2-hydroxyethyl)thiazole → 5-(Acryloyloxyethyl)-4-methylthiazole

HRID 726812Notes

Reaction equation

HRID 726812 reaction equation
[OH:1][CH2:2][CH2:3][C:4]1[S:8][CH:7]=[N:6][C:5]=1[CH3:9].C(N(CC)CC)C.[C:17](Cl)(=[O:20])[CH:18]=[CH2:19]>C1C2NC3C(=CC=CC=3)SC=2C=CC=1.C(Cl)(Cl)Cl>[C:17]([O:1][CH2:2][CH2:3][C:4]1[S:8][CH:7]=[N:6][C:5]=1[CH3:9])(=[O:20])[CH:18]=[CH2:19]

Participants

6

Reactant 3

HCID 11364-Methyl-5-(2-hydroxyethyl)thiazoleHCID 8471TriethylamineHCID 13140Acryloyl chloride

Product 1

HCID 212992875-(Acryloyloxyethyl)-4-methylthiazoleYield 82%

Reagents, catalysts & solvents 2

HCID 7108PhenothiazineCatalystHCID 6212ChloroformSolvent

Procedure

A one liter, three necked flask was equipped with overhead stirrer, nitrogen atmosphere and an addition funnel. The flask was charged with 100 g of 5-(2-hydroxyethyl)-4-methylthiazole (0.70 mol.), 71 g of triethylamine (0.70 mol.), 500 g of chloroform and 3 g of phenothiazine. Next, 63 g of acryloyl chloride (0.70 mol.) was slowly added to the reaction over 1.5 hour via the addition funnel. The reaction was stirred for 2 hours at room temperature. The reaction mixture was extracted with NaCl(aq.), NH4OH(10%), NaCl(aq.) and dried over MgSO4. The chloroform was evaporated with a rotoevaporator to yield 113 g (82%) of a dark brown liquid. The product was confirmed by IR and 13C NMR.

Workup

  1. customA one liter, three necked flask was equipped with overhead stirrer
  2. extractionThe reaction mixture was extracted with NaCl(aq.), NH4OH(10%), NaCl(aq.)
  3. dry with materialdried over MgSO4
  4. customThe chloroform was evaporated with a rotoevaporator

Sources & evidence

ORD record
ord-b3f17101f56b48979494396b1a5ca876
Source dataset
uspto-grants-1998_01
Patent number
US05710281

Reaction notes

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