MushroomMan
Rising Star
I looked at a few MHRB STB teks and one of them says to use 1g of NaOH per 1g of MHRB and 15mL of water per gram. This makes no sense to me at all. I don't know how much DMT and related alkaloids are present in MHRB but I'll take a guess and say 5%, that would be 50mg of alkaloids per gram. DMT reacts with NaOH in a 1:1 ratio. DMT has a molar mass of about 180g/mol and NaOH about 40g/mol so in 1g of NaOH there are 1g/40g/mol = 25mmol and the number of moles in 50mg of DMT is 0.05g/180g/mol = 0.28mmol.
So unless I did my calculations wrong, the NaOH is present in a 97:1 ratio to the DMT. Secondly, why so much water? NaOH has a solubility of 1.11g/mL at room temperature. Another tek I looked at uses the same ratio of NaOH to MHRB but about half as much water. I'm new to all this but I'm a chem major so I've done extractions on different compounds before, this is the first time I've seen such excesses of solvent and reagent being used, is there something I'm not taking into consideration here?
So unless I did my calculations wrong, the NaOH is present in a 97:1 ratio to the DMT. Secondly, why so much water? NaOH has a solubility of 1.11g/mL at room temperature. Another tek I looked at uses the same ratio of NaOH to MHRB but about half as much water. I'm new to all this but I'm a chem major so I've done extractions on different compounds before, this is the first time I've seen such excesses of solvent and reagent being used, is there something I'm not taking into consideration here?