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Author: Subject: Synthesis of alkali nitrogen compunds.
vulture
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[*] posted on 31-8-2002 at 05:54
Synthesis of alkali nitrogen compunds.


Well, the title says it all, I want to find out how to synthesize alkali azides and amides without using elements.

Sodiumazide is made from sodiumamide (which I will come back to further in this post) but how?

Sodiumamide is a very interesting compound because of it being the precursor for sodiumazide, but also because it can form the strongly explosive oxides Na2N2O2 and Na2N2O3. These form when sodiumazide is exposed to air.

I wonder if there is a way to safely prepare and isolate them on a lab scale.

Normally, sodiumamide is prepared by reacting gaseous ammonia and sodium metal. It needs no further explanation that this reaction is almost impossible to perform by the amateur chemist.

Could it be produced from NH3 and Na2O?

Na2O + 2NH3 -> 2NaNH2 + H2O?




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[*] posted on 31-8-2002 at 10:09


Quote:
Na2O + 2NH3 -> 2NaNH2 + H2O


This reaction will not work, because NaNH2 is decomposed in contact with water.
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[*] posted on 31-8-2002 at 15:27
Liquid NH3


Sodium oxide Na2O reacts with liquid ammonia NH3 to form a mixture of sodium hydroxide and sodium amide.

Na2O + NH3 --> NaOH + NaNH2

First time I ever heard of this was in my chemistry dictionary.

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smile.gif posted on 9-9-2002 at 15:09


I think Na2N2O2 is hyponitrite of Na and it is feasible and isolable from common chemicals, but I don't remember where the info is!

About azides one of the simplies method uses the action of NaOH on butyl nitrite mixed with hydrazine!
The slow release of HNO2 due to hydrolyse of the nitrite ester of butanol react with hydrazine to make...
H2N-NH3NO2 --> H2N-NH-N=O +H2O --> H2N-N=N-OH --> HN3 + H2O!But this has already been explained in the tread about polyazachains!

The final exclusion of the weakly soluble NaN3 (precipitation) in the butanol/ester/water mix allows an easy way to get the desired compound!

They are other ways but they are even less simple!Starting from carbamide NH2-NH-CO-R!

PH Z:):D;):cool::o:P
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[*] posted on 14-10-2002 at 19:15


There are a few methods of prepairing hyponitrites, the main 2 are reduction of nitrites by sodium amalgam, and condensation of nitrites and hydroxylamine. You can make hydroxylamine rather usefully by acid hydrolysis of nitromethane. Nitrogen chemistry is something Ive spent a long time studying, as people following my theforum posts might have noticed.

You can make sodium amide from molten sodium metal and ammonia gas. Passing the gas over a raft in a pyrex tube is a common way, until hydrogen is nolonger present in the gas coming out. Somewhat troublesome to make certainly, but well within a modest home setup.
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[*] posted on 16-10-2002 at 11:01
Why bother


making NaN3. I have a trip set to the junk yard next week solely for acquiring NaN3 as one can obtain upto 100g per airbag. It also has some other stuff added like Fe2O3 and SiO2 for stabalizing purposes, but all of them can be easily seperated. NaN3 is in the form of small tablets inside a sealed Al case. While sawing it open ..... please use adequate amounts of some suitable liquid to prevent overheating and/or sparking, or just use the good old pliers very carefully.
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[*] posted on 21-10-2002 at 13:11


Azides can be prepared from the reaction of the metal amide with nitrous oxide gas.

For example:

NaNH2 + N2O ---> NaN3 + H2O

I'm not sure what temperature, pressure, etc. would be necessary...

David Hansen
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[*] posted on 27-10-2002 at 19:26
I am not too sure but ......


I came across a method of preparing CH3ONa by electrolysing NaCl in CH3OH. So if NH3 (dry) is passed through this soln ..... NaNH2 in CH3OH might be produced, which can just be let to dry to obtain the waxy NaNH2.



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[*] posted on 27-10-2002 at 20:10


Electrolyzing NaCl in CH3OH would be a mess. The liberated chlorine gas would react with the methanol, forming formaldehyde, and possibly formyl chloride and phosgene. You would get some NaCH3O, no doubt.

It's safer to just react NaOH with a large excess of CH3OH.

NaNH2 couldn't be prepared in a solution of methanol, because methanol is more acidic than ammonia.




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[*] posted on 19-2-2003 at 16:42


You can get amides from
2Na + 2NH3 -> 2NaNH2 + H2

Just get elemental sodium, and burn it in a stream of anhydrous NH3. i wouldn't recommend making ammonia from its constituent elements :).

If you want ultimately the azide either simply buy it or rape an old air bag. (sorry i couldn't help my self)
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cool.gif posted on 19-3-2003 at 07:50


It is much more interesting then to make metalic sodium by electrolysis of molten NaCl mixed with NH4Cl!
Molten NaCl will lead to Na vapour in one side and Cl2 on the other; assuming NH4Cl will liberate Cl2 on the same side as NaCl, the other way will get NH4(+) (NH3 + H(+)) that will thus free H2 and NH3 in a Na vapour phase!
The advantage of useing NH4Cl is that it will strongly lower the mp of the salt mix!
Both salts must be very dry!

Then
Na(g) + H2(g) --> NaH(s)
Na(g) + NH3(g) --> Na-NH2(s)
NaH(s) + NH3(g) --> Na-NH2(s) + H2(g)

So it is possible to make Na or Na-NH2 (trap the vapours of Na in a cold trap).




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[*] posted on 21-3-2003 at 07:15
I was wondering


if it was possible to prepare NaNH2 by reacting Na metal with molten urea in a Nitrogenous environment ? If it is .... both of these are easy to aquire.
Also ...... would refluxing a mix of NaNO2, NaOH and N2H4*Hydrate yeild NaN3. If not ... could anyone one tell me why ?:)




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[*] posted on 21-3-2003 at 13:02


doesnt NaNO2 + N2H4 --> NaN3 + 2H2O?



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[*] posted on 24-3-2003 at 10:16


Na will have more affinity for NH2-CO-NH2 than for NH3 because urea display more acidic H's!
NaNH-CO-NH2 + NaH are the resulting products.
But then you need refluxing with excess NH3 (l) to make the amidure!

NaNO2/NaOH/NH2-NH2 will not form NaN3 because: *NaOH is a much stronger base than NH2-NH2
*HNO2 can't form in strong base media but may be formed furtively when nitrite ester is hydrolised!

The use of nitrite ester allows slow amonolyse (here hydrazonolyse)!
NH2-NH2 + H2O + R-O-N=O --> NH2-NH-N=O + H2O + R-OH --> HN3 + 2H2O + R-OH
NaOH only serves to catalyses a bit the hydrolyse of the ester and catch/neutralise the HN3 as soon as it forms!




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