Let me start off with an apology to offended members, for inappropriate comments. Im sorry. I've done a lot of studying and read the link for
said reductions. However I have procured some pyridine and attempted its reduction with dissolving metal in ethanol. My first attempt. purity was
labeled 99% . I didnt dry it first. ethanol was fractionated and dried tho. Here is my questions, I see electro hydrogenation in sulfuric acid. Could
pyridine be slowly dissolved in excess con. sulfuric and then tin added to reduce? I dont have a parr hydrogenation app. Is anhydrous ammonia
required for the dissolving metal/ethanol? Whats the best way?
I will attempt the metal/ethanol again with dry pyridine. Instead of sodium I used lithium. It worked to some extent but was contaminated with
pyridine. size of rxn was .14M pyridine. thanks in advance for all the help.Number 9 - 27-6-2013 at 11:10
Don't work with Sn or Zn, even Mg doesn't work in a strong acid. You need a lead/PbO2 anode, the cathode is also Pb but you must have a diaphragm to
prevent oxidation since your organics get oxidize further to a tar-like substance. The electrolyt is 5% H2SO4. Current density is 1,5 kA/m2 which is
pretty high.
Source: Hamann, C.H., Vielstich, W., Hamnett, A. (1998). Electro- organic synthesis, 8.5, Electrochemistry, p. 323 - 324
Another way is to heat cadaverine monohydrochloride but the smell is terrible. The other compound which is then formed is ammonium chloride.
[Edited on 27-6-2013 by Number 9]Pickardjr - 28-6-2013 at 09:01
Yea I know about the cadaverine cyclization and other diamines, like to try Hexamethylenediamine. But honestly I think Im gonna build a small
hydrogenation app. I can wait a little longer. also I see that molybdenum disulfide is used as dry grease and is the catalyst used to hydrogenate
the ring . Im gonna try it, I know about removing the oxygen before H2 so just maybe I will live to do it twice.AndersHoveland - 28-6-2013 at 10:14
It would likely be rather hazardous, but what about using magnesium boride as the reducing agent? Slow addition of water would make toxic B2H6 that
would reduce some of the pyridine.
(Mg3B2 can be prepared by combusting Mg with boric acid in a sacrificial test tube)
One thing to mention though, the pyridine might passivate the Mg3B2 to reaction with water, if so, the reaction might be impractically hazardous if
you have to pass borane gas into the reaction.
[Edited on 28-6-2013 by AndersHoveland]smaerd - 28-6-2013 at 10:57
Reference for the synthesis of Mg3B2 using boric acid and magnesium metal?
Or that when it is added to water it forms diborane?woelen - 28-6-2013 at 11:22
You first have to heat the boric acid without magnesium in a metal crucible to make B2O3 (or something between HBO2 and B2O3, completely anhydrous
B2O3 is not easy to obtain). The B2O3 must be crunched and mixed with Mg-powder.
You can add so much Mg that just all oxygen ends up, bonded to the Mg. The boron then is turned into the element. With an aqueous acid you then can
leach out the MgO, remains of unreacted Mg and remains of unreacted B2O3 and (impure) boron is left.
You can also add a larger amount of Mg, such that both the oxygen and boron are bonded to the Mg and then you get a mix of MgO and Mg3B2 (and some
unreacted Mg and B2O3). This, however, hardly will make any B2H6 when added to water or an aqueous acid. B2H6 reacts with water to mainly boric acid
and hydrogen gas, only a small fraction ends up as B2H6. Making B2H6 requires anhydrous conditions and I'm not sure whether that can be done with
Mg3B2.DJF90 - 29-6-2013 at 00:12
Pyridine forms a stable adduct with borane, and is not reduced.AndersHoveland - 29-6-2013 at 00:26
Pyridine forms a stable adduct with borane, and is not reduced.
Are you sure? Because normally when these adducts are being formed, the BH3 is added in the form of its adduct with an ether. B2H6 is a bit more
reactive than BH3-ether adduct.
not sure, you may be right, but just suggesting the possibility that passing B2H6 directly into pyridine may result in something different.
And further more, even if the adduct does form, it might be possible to reduce it with more borane. What I mean is that it might first form
the adduct, then when all the pyridine is used up reduction will begin. So if the Py-BH3 were dissolved into some third inert solvent (not sure what
this would be) and B2H6 passed in
[Edited on 29-6-2013 by AndersHoveland]Number 9 - 30-6-2013 at 02:20
Alternatively, piperidine is obtained by reduction of pyridine with sodium metal in absolutely dry EtOH.
Ref. Organic Syntheses, Coll. Vol. 1, p.99 (1941), C.S. Marvel , W. A. Laziermadscientist - 1-7-2013 at 02:28
Pyridine forms a stable adduct with borane, and is not reduced.
Are you sure?
DJF90 is a skilled organic chemist, you are not, and you have failed to provide a reference to support your wild speculation (entirely unsurprising).
Na/EtOH is indeed the way to go here. The one other method that is convenient for the home chemist is to N-benzylate the pyridine, and hit it with
sodium dithionite. Nicodem has provided a nice writeup on the matter:
However, you now need to remove the benzyl PG, so this route can be more complicated.
With R ≠ Bn, the reduction stops at the 1,4-dihydropyridine.
[Edited on 1-7-2013 by madscientist]DJF90 - 3-7-2013 at 09:47
Cheers madscientist, I didnt realise I had a fan
Anders: Yes I'm sure. Try picking up a book sometime, it'll do you wonders. Sometimes I think you'd make a good ARTIST, what with all these crazy and
creative (read: made up) "ideas" of yours.
