SYNTHESIS: A crude solution of methoxyethylamine hydrochloride was prepared from 17.7 g methoxyethylamine and 20 mL concentrated HCl with all volatiles removed under vacuum. This was dissolved in 75 mL MeOH and there was added 4.45 g of 3,4-methylenedioxyphenylacetone (see under MDMA for its preparation) followed by 1.3 g sodium cyanoborohydride. Concentrated HCl in MeOH was added as required to maintain the pH at about 6 as determined with external, dampened universal pH paper. About 4.5 mL were added over the course of 5 days, at which time the pH had stabilized. The reaction mixture was added to 400 mL H2O and made strongly acidic with an excess of HCl. After washing with 2×100 mL CH2Cl2 the aqueous phase was made basic with 25% NaOH, and extracted with 4×75 mL CH2Cl2. Removal of the solvent under vacuum yielded 6.0 g of an amber oil that was distilled at 110-120 °C at 0.2 mm/Hg. There was obtained 4.7 g of a crystal-clear white oil that was dissolved in 30 mL IPA and neutralized with 45 drops of concentrated HCl producing a heavy mass of spontaneous crystals that had to be further diluted with IPA just to be stirred with a glass rod. These were diluted with 200 mL of anhydrous Et2O, removed by filtration, and washed with additional Et2O. After air drying there was obtained 4.9 g of 3,4-methylenedioxy-N-(2-methoxyethyl)amphetamine hydrochloride (MDMEOET) with a mp of 182.5-183 °C. Anal. (C13H20ClNO3) N.
DOSAGE: greater than 180 mg.
EXTENSIONS AND COMMENTARY: This is another example of the replacement of a neutral atom out near the end of a chain, with a more basic and a more polar one. MDMEOET would be called an isostere of MDBU in that it has the same shape, with a methylene unit (the CH2) replaced by an oxygen atom. No activity turned up with either compound, so nothing can be learned from this particular example of change of polarity.