1 g 2C-H*HCl (4,6 mmol)
1,4 g Oxone (2,3 mmol)
10 ml MeOH
5 ml water
2C-H*HCl was dissolved in the mixture of water nad MeOH in a 50 ml rb flask equipped with a magnetic stirbar and the oxone added in three portions during 5 minutes. The solution gradually became more yellow during 10 minutes. No change in temperature was noted. After 30 minutes 25 mmol HOAc was added and the solution extracted with 2x15 ml DCM which removed most of the color. The acidic aqueous phase was then made strongly alkaline with NaOH, which caused the solution to become dark amber in color, then extracted with 2x15 ml DCM. Most of the color was now in the DCM extracts.
According to PIHKAL none of the hydrochlorides of the 4-halogenated 2,5-DMPEA`s are particualry soluble in water, while 2C-H*HCl is quite soluble. So I decided to perform a simple test to see if any 2C-C was made. To the DCM phase 20 ml conc HCl was added in one portion. Within 10 seconds the acidic aqueous phase was cloudy with white crystals. The phases was separated and 40 ml IPA was added to the aqueous phase and the crystals removed by filtration and dried to constant weight.
Yield 370 mg of, most likely, 2,5-dimethoxy-4-chlorophenethylamine hydrochloride (2C-C*HCl)
The yield is terrible low. But the theory seems to be correct. Next time I´ll try this without the MeOH and use a two-phase system consisting of water/DCM instead. Dissolve 2C-H*HCl in 10 ml water, add 10 ml DCM, then oxone in portions. When the reaction is over add NaOH until pH 12, separate and 20% HCl until pH 2-3 is reached (perhaps wash the DCM phase with aq NaHSO3 before acidification). Then it should just be a matter of remove the crystals, wash then a little and dry them to constant weight.
Catalytic hydrogenation freak