The Vespiary
The Hive => Methods Discourse => Topic started by: psytech on May 27, 2004, 03:28:00 AM
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On Rhodiums sites mdma from nabh4 synth uses a ratio of 2 moles methylamine to 1 mole mdp2p. Then for the synth for meth uses a ratio of 3.2 moles methylamine to 1 mole p2p. My question is why the big diffrence.
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Correct your post's title, will you (for TFSE's sake): methylamine.
On Rhodiums sites mdma from nabh4 synth uses a ratio of 2 moles methylamine to 1 mole mdp2p. Then for the synth for meth uses a ratio of 3.2 moles methylamine to 1 mole p2p. My question is why the big diffrence.
It 's just excess. With Le Chateliers principle in mind, excess methylamine (obviously the cheap reagent) is used to drive the equilibrium (imine formation) to the right. Molecular sieves (absorption of the water ) are used for the same reason. It 's all a matter of equilibrium. You wouldn't like to reduce the propanone, would you?
Aztec fanclub member
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I know its in excess, but why isn't the same ratio applied to both ketones. And (btw) I tried the TFSE first. Here at the Hive its all about improving methods, and understanding the principles behind those methods. So dont get so uptight.Thanks for the input but you really did'nt help increase knowledge base. The insight I was inquiring about was, in decreasing the amount of methylamine in NaBH4 Reductive Alkylation of P2P.If you havent't performed the reduction, you really have no experience, theory is all well and fine, but I live in the real world, and all reagents are precious.
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I believe the difference is due to how much methylamine the particular author had on hand (compared to the ketone) rather than an actual requirement for each of the ketone. The methods on my page are mostly syntheses "found to work" rather than being "fully optimized" in regards to reactant/solvent ratios, reaction times etc.
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Thanks Rhodium, so I'll have to try it myself, I'll report back.