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After extraction added sodium bicarbonate to acid/water mix and it solidified. Unreacted glycidic acid? Never thought such a simple reaction can be such a mess. I made a mistake in a post above: it was 20g of H2SO4

Another experiment, 80ml water, 10g sulphuric acid, 20g powder, 2ml P2P. It makes me mad.

Can anyone check their 5449 pH (water solution) please? Mine is strongly alkaline. Is this normal?

Another experiment was done on 10g of powder: Dissolved in a hot water, heated & stirred fro 3 hours at 90 degrees with HCl. Got 2ml of very viscous oil with smell of phenylacetone. This viscousity is from unreacted glycidic acid or polymerization?

So the question is: when I dissolve 5449 in a water , then add H2SO4 I have immediate layer separation. The oil stays on top. Is this p2p? Or it is a glycidic acid? It doesn't smell p2p. Do I need to heat it & mix ffor decarboxylation?

I tried to dissolve powder in cold water, then added some amount of H2SO4 conc. and brown oil immediately separated on the top. Is this normal behaviour for 5449?

Yes, with condenser. No quenching, extracted with DCM, washed with water, ecaporated DCM. Everyone says it must be bubbles during rxn but i didn't see any

you can try. it is exotermic so beaker need to be cooled or h2o2 will decompose. but Hg+Acetic+H2O2 -> Hg.Acetate Hg.Acetate + HCl -> HgHl2 Is proven to work, it is quick, and no smell

Have you tried phosphoric route? Do I need to add dry powder straight to hot acid with stirring or powder has to be dissolved in water prior to addition?

Separated from sodium :). Actually I am struggling now with 5449. Tried bunch of methods and so far got 1ml from 10g. Will try phosphoric acid at 110. When I add acid to 5449 it turns to oil very quickly but it is not p2p. Or may be my 5449 is not real 5449

With H2SO4 layers separate instantly but that doesn't mean you got P2P. It is just glycidic acid separated. I just stried it and got two layers with no smell.

No this is unrealistic. What you can do if you need a chloride: 1. Take metallic mercury in a beaker (from a termometer) 2. Add 5ml of glacial acetic acid and 5ml of hydrogen peroxide (30..50%) 3. Stir until all mercury turns into white powder 4. Slowly add hydrochloric until clear solution is...

I read that phosphoric acid or mineral+organic acid mix give better yeld. However I decided to try pure aqueous HCl hoping that I can get at least SOME yeld. Wanted to check if 5449 I bought is actually what it is. So I heated up 20ml of water to 60 degrees, slowly disolved 10g of bmk glycidate...

1. Borohydride or Al/Hg depending on the amount of P2P. Al/Hg doesn't require dry conditions but it is a messy workup (Al sludge) and loading more than 20..25g of P2P are hard to handle (too exotermic). Can do many Al/Hg flasks in parallel. Borohydride is cleaner procedure but water can hurt...

No, you need anhydrous calcium chloride. It is a white powder. It works just by consuming water from aqueous HCl so gas is released.

you need to boil water out of it as much as possible. you don't need it to be clean.

You need 1 mol of NaOH per 1 mol of Methylamine.HCl, NaCl is not a problem, No you can't distill. What you can do if you really need to get rid of NaCl is to drip solution of methylamine.hcl over wet NaOH. It will generate gas which you can bubble thru water. Then you need to find out %% of...

Oh I forgot. Very simple method and it is a high yelding: drip your HCL acid over anhydrous calcium chloride. It will produce alot of HCL gas, but it also needs to be dried thru calcium chloride packed tube.

You can generate HCl gas from aqueous HCl by heating it. High temperature will push gas from the solution. This method is low yelding. Also there will be water vapours going along with gas. You need a flask with muriatic acid, heating, then dry your gas thru calcium chloride tube, then bubble it...

Oh sorry didn't read your post carefully. Al/Hg system produces a gas, reaction is strongly exotermic, HCl is not needed. In my opinion it is worse than formaldehyde route. Your methylamine gas however will be dissolved in the reaction mixture.