Assuming 1Kg post o2 wacker “mostly” mdp2p, ready for distillation, is 15% mdp3p (roughly 150g or .84 mol.), how would a quality conscious bee go about removing it? Throw in 163g or .84 mol. K2CrO4 and stir?
Looking at Post 438347 (https://www.thevespiary.org/talk/index.php?topic=10543.msg43834700#msg43834700)
(Rhodium: "3-(3,4-Methylenedioxyphenyl)-propyl-methylamine", Newbee Forum) hints at oxidation with potassium chromate or manganese dioxide. UTFSE yields no proposed or executed reaction details.
#1) Should bees use a molar equivalent of the oxidizer? Slight molar excess?
#2) Which oxidizer is the most forgiving and safest for bees? K2CrO4 cas#7789-00-6 mol wgt 194.17 or MnO2 cas#1313-13-9 mol wgt 86.94?
#3) How should bees carry out the oxidation? Add and stir? Predisolve oxidizer? Heat? How long?
Higher purity benefits everyone!
Alcoholic wackers have been done in the past with various yeilds and from what swim remembers, the ones in MeOH weren't repeatable or very high yeilding for that matter. If swim were to try this again he wouldn't waste his time doing the same runs many have done in the past. He would do 1 or both of these experiments:
1.try using polyethylene glycol(PEG) 400 as solvent as per https://www.thevespiary.org/rhodium/Rhodium/pdf/wacker.peg.pdf (https://www.thevespiary.org/rhodium/Rhodium/pdf/wacker.peg.pdf)
or
2.try cupric acetate as co-catalyst in AcNMe2 as solvent https://www.thevespiary.org/rhodium/Rhodium/pdf/wacker.o2.cupric.acetate.pdf (https://www.thevespiary.org/rhodium/Rhodium/pdf/wacker.o2.cupric.acetate.pdf)