Sodium Hydroxide → alpha-(2,2-Dichlorovinyl)isovaleric acid

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Reaction Steps

In 20 g of chloroform were dissolved 6.0 g of the ethyl α-bromo-α-(2,2,2-trichloroethyl)isovalerate and, after the addition of 3.5 g of 1,5-diazabicyclo[5,4,0]undecene-5(DBU), the solution was stirred at room temperature for 2 hours. The reaction mixture was then diluted with diethyl ether, washed with water and dilute aqueous hydrogen chloride and dried over anhydrous magnesium sulfate. The low-boiling fraction was removed by distillation, whereupon 4.6 g of oily product were obtained. Based on its NMR spectrum and GC-mass spectrum, this product was identified to be ethyl α-isopropyl-β-trichloromethylacrylate containing a certain amount of ethyl α-bromo-α-(2,2-dichlorovinyl) isovalerate. In 50 ml of ethanol were dissolved 4.6 g of the oily product obtained above and, following the addition of 0.4 g of 5% palladium-on-carbon, the solution was stirred in an atmosphere of hydrogen gas at 55° C. for 6 hours. Thereafter, the reaction mixture was filtered to remove the catalyst and the ethanol was distilled off. By the above procedure there were obtained 3.9 g of ethyl α-(2,2-dichlorovinyl)isovalerate as an oil. Then, 3.9 g of this ethyl α-(2,2-dichlorovinyl)isovalerate were dissolved in 20 ml of ethanol and a solution of 2.0 g of sodium hydroxide in 10 ml of water was added to the above ethanol solution. The mixture was stirred at room temperature for 10 hours, at the end of which time the ethanol was distilled off. The residue was neutralized with aqueous hydrogen chloride and extracted with diethyl ether. The ethereal solution yielded 2.9 g of α-(2,2-dichlorovinyl)isovaleric acid having the following NMR spectrum [yield: 84% based on ethyl α-bromo-α-(2,2,2-trichloroethyl)isovalerate].

Provenance & Evidence