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I'm trying to do the math on the hydrolysis of N-MethylAcetamide (abbriviating it to NMA) in order to get the methylamine.
I would be using basic hydrolysis, so there will be sodium acetate and methylamine in solution. The sodium acetate likely will not interfere with a mercury amalgam reduction of various ketones, so I'm not concerned about removal.
I want to know how much of each reagent to react so as to get a 40% solution of methylamine.
Say one has 100 grams of NMA; how much water and sodium hydroxide will be needed to produce a 40% methylamine solution?
I would be using basic hydrolysis, so there will be sodium acetate and methylamine in solution. The sodium acetate likely will not interfere with a mercury amalgam reduction of various ketones, so I'm not concerned about removal.
I want to know how much of each reagent to react so as to get a 40% solution of methylamine.
Say one has 100 grams of NMA; how much water and sodium hydroxide will be needed to produce a 40% methylamine solution?