02-18-02 22:47
No 271478
I need some help from you,Bees!i`m going to do the NaCNBH3borch alkylation and now i would like to have some mor detailed information,the KrZ synthesis says "tryptamineHcl ans NaCNBH3in a mixture of MeOH and HCOOH,if you look at Heterocycles vol.49/1998 page 456 it says "A solution of 95% NaBH3CN in MeOH was added to a solution of trypt. in HCOOH,so do you first dissolve the tryp. in HCOOH and mix this with a sollution of NaCNBH3 in MeOH or do you first dissolve the HCOOH in MeOH and mix the tryp. and NaCNBH3 to this.
and than it goes "(KrZ)this sollution was cooled to 0°C,and in heterocycles they use roomtemp.,why? and what is the differance? and than over a steaddy stream of nitrogen.
The reaxtiontime in the KrZ Synth. is 60 H and in heterocycles only 3 H ,what makes the differance and why?
RHODIUM,LILLIENTHAL,ANYBODYcan you help me out,please
(Chief Bee)
02-18-02 23:04
No 271480
They are surely not using HCOOH (formic acid) but rather HCHO (formaldehyde), right? The different conditions are up to the chemist to decide. What is the yield difference between the two variations?
(Stranger)
02-21-02 14:18
No 271610
Yes!your right i mean acetic acid,not formic acid,sorry
(Stranger)
02-21-02 15:17
No 271630
I did some heavy stochiometric analysis of the KrZ synth and of the second one which you refer to from the rhodium archive:
First there is an error in the noting of how many moles of tryptamine HCl - it should be 51.02mmol, 62.4mmol at 10g is precisely what tryptamine freebase would be.
And yes, you are right, the NaCNBH3 should definitely be added to the mix last, because it will react with the water in the formaldehyde, the formaldehyde, and it will deaminate and/or quaternise the tryptamine.
Personally I don't think much of this reaction because of the very high quantity of water that is introduced into the mix via the formaldehyde solution.
As to the time factor, if you are looking at a proper scientific description of a reaction which says 3 hours, it's probably true. The cooling is likewise unneccessary (though probably decreases the reaction of the NaCNBH3 with the water and tryptamine while they are being mixed). Change the order and it becomes irrelevant. Likewise with the non-reactive atmosphere - possibly it might decrease the risk of hydrogen sparking, which would be released by the NaCNBH3 as it reacts with the water.
And finally, there is a bit of danger in this reaction that KrZ describes - though unlikely, there is (according to what is written in that second process) HCl ions floating around, and according to hazdat sheets about NaCNBH3, acids liberate the HCN - well, obviously it can't be weak acids they are referring to, rather it must be strong acids, and HCl ions might get a bit messy if they unhook. Though they wouldn't do that an awful lot, it is something to be aware of. Use freebase and you avoid that problem.
A side note, the first one has incorrect quantities of formaldehyde (30 instead of 45g) and incorrect quantities of NaCNBH3 (50 instead of 125g) I think the amounts of acetic acid are wrong too, but I don't know exactly what molar ratio of acetic acid is required to protect the tryptamine. In the second one it is almost 4x molar to the tryptamine, or 2x molar to the formaldehyde, so I would say that you actually need 60g in the first rxn.
This reaction would be much better if you could have the formaldehyde dissolved in anhydrous methanol rather than water. Perhaps the two could be mixed and the water salted out?
C12H16N2
(Stranger)
02-21-02 15:21
No 271631
I assume that formaldehyde and acetic acid are needed.
Just out of curiousity, could formic acid be substituted for acetic?
C12H16N2
(Stranger)
02-22-02 12:09
No 272142
Thank you,Urushibara for the comments could you more specify your sidenote on the quantities and ratios ans an other quation is in the heterocycles article they first dissolve the tryptamine Hcl in the glacial aceticacid and then mix this with a solution of NaCNBH3 in MeOH,and the N2 gas is probably used to push out any cyanidegas,I would like to do both reactionschemes but first think it be for i act!could you help me out with the amounds to start with? already thank for the help an i will all let you know how it all went!
(Stranger)
02-22-02 15:08
No 272222
Well, I it was suggested by someone on another site that probably 3 acetic acid molecules bond to tryptamine. Two molecules of formaldehyde are needed to provide the right number of final methyl groups. And I might be wrong on this one, but two NaCNBH3for each formaldehyde. Let me pose the question properly: does that oxygen on the formaldehyde transform into water, or does it find a fellow oxygen and turn into O2? I would think it turns into water, in which case you need 3 hydrogens (hydrides), one for the methyl group, and 2 for the oxygen.
Okay so here's the ratios
Say we start with 10g of tryptamine, I will give the weights of each material and the molar quantities.
Tryptamine: 1mol (~160g/mol) (10g - 0.0625mol)
Glacial Acetic Acid: 3mol (~60g/mol) (11.25g - 0.1875mol)
Formaldehyde: 2mol (~30g/mol) (3.75g - 0.125mol)
NaCNBH3: 3mol (~62.84g/mol) (11.78g - 0.1875mol)
Cyanide is not formed in the reaction. I'm not certain, but though NaCNBH3 is said to release cyanide in the presence of acids, I'm fairly sure they mean strong acids. The hazdat sheet I have seen didn't specify. Nitrogen atmosphere is usually used when hydrogen or some other flammable is released, or sometimes when oxygen is released, as it reduces the chance of ignition. Argon is used in welding for a similar purpose - as electric arcs can ionise the air so strongly that it can really make things go boom, almost anything at all. Perhaps the heterocycles writer believes that ozone comes out of the reaction? If that's the case, divide the quantity of NaCNBH3 by 3, as the O's take care of themself and don't need tying down. I am doubtful of this though, because an ozone is rare, though it might form if an O2 passed close to a reduction in process and the oxygen plugged into it. Then you would have ozone, which may make nitrogen atmosphere sensible. Otherwise, I can't see the sense. Nothing in the reaction is overly sketchy with oxygen that is likely slightly dissolved in the methanol and at 20% in the atmosphere around.
That HCl on the tryptamine could unhook from the tryptamine, and presumably must to allow the methylation reaction to occur, but free HCl is a no no with CNBH3. Methinks that's the reason for nitrogen perhaps. Better to start with freebase. The figures I gave above refer to tryptamine freebase.
C12H16N2