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Author: Subject: MDD, methanediammonium dinitrate?
artem
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[*] posted on 15-3-2006 at 09:39
MDD, methanediammonium dinitrate?


There are some mentions about this compound:
MDD:melting point~130C, >130C - HNO3 evaporation, at~175C releasing of NH3(g), HNO3(g) /РЕР 1990-03-77/. The subj is very usefull as precursor of some nitramines.
Evaluation shows that it may be good HE itself (Q~TNT, brisance>TNT), but I failed to know something about synthesis of the subj and its properties.
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[*] posted on 15-3-2006 at 09:48


Maybe nickname for formamidine dinitrate?

H2N-CH=NH.2HNO3 (aka O3N-NH3-CH=NH2-NO3)




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[*] posted on 15-3-2006 at 12:49


Quote:
Originally posted by PHILOU Zrealone
Maybe nickname for formamidine dinitrate?

No, this is CH2(NH2)2*2HNO3
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Axt
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[*] posted on 15-3-2006 at 17:50


Does it reference into anything? Maybe you can attach the article?

Best I could find was (presumably) the prep for diaminomethane dihydrochloride which should be found in:
<i>P. Knudsen, Chem. Ber., 47, 2698 (1914)</i>.
Though I dont have it, nor have the means to get it right now. "Chem. Ber." I think being "Chemishe Berichte" .

I dont think you can distill the free base, so maybe AgNO3 on the hydrochloride will give you what you want, maybe not due to formation of complex salt. You may even be able to buy the dihydrochloride commercially.

[Edited on 16-3-2006 by Axt]
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smile.gif posted on 16-3-2006 at 10:27


I don't think the first diammonium to be correct...
maybe must be something like:
Methylenediamine dinitrate
Methylenediaminium dinitrate

or as Axt said Diaminomethane dinitrate

There is a mention on Google of methylendiaminediformamide...
H-CO-NH-CH2-NH-CO-H

By use of a diluted acid like HCl or H2SO4 one migt get the hydrolyse to the diamine salt..strong oxydisers like conc HNO3 or HClO4 are no go!

H-CO-NH-CH2-NH-CO-H = CH2(NH-CH=O)2

CH2(NH-CH=O)2 + 2 HCl --> CH2(-NH2Cl-CH=O)2 -->
CH2(-NH2Cl-CH=O)2 + 2H2O --> CH2(NH3Cl)2 + 2HCO2H

With a exchange resin or by methathesis from Pb(NO3)2, NaNO3 one might get the dinitrate or the diperchlorate...
:cool::cool::cool:

[Edited on 16-3-2006 by PHILOU Zrealone]




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[*] posted on 16-3-2006 at 12:59


Quote:
Originally posted by Axt
Does it reference into anything? Maybe you can attach the article?
Best I could find was (presumably) the prep for diaminomethane dihydrochloride which should be found in:
<i>P. Knudsen, Chem. Ber., 47, 2698 (1914)</i>.
Though I dont have it, nor have the means to get it right now. "Chem. Ber." I think being "Chemishe Berichte" .
[Edited on 16-3-2006 by Axt]

The PEP article is attached. Alack I don't have the possibility to see your reference too :(
Sometimes methanediamin is mentioned in patent literature without detailes. There is an article (I can not to attach it) about RDX-synthesis where different sources of -CH2- were compared with (CH2)6N4 (for example, (CH2O)3, CH2(ONO2)2, different cyclic and linear precursors), and CH2(NH3NO3)2 was giving highest reaction rate and very good yield, more than (CH2)6N4 (the last < 64% on -CH2-groups).
It's interesting, is it possiible to get such salts by ammonolysis of (CH2)6N4, for example?

Attachment: PEP90_03_77.djvu (64kB)
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[*] posted on 12-4-2006 at 02:36


Quote:
Originally posted by PHILOU Zrealone
There is a mention on Google of methylendiaminediformamide...
H-CO-NH-CH2-NH-CO-H

By use of a diluted acid like HCl or H2SO4 one migt get the hydrolyse to the diamine salt..


Wasnt looking for it but I did just come across that in "Sedgwick's Organic Chemistry of Nitrogen" at least confirming that it does work.

<i>Methylenediamine, H2N-CH2-NH2, is very unstable; its salts are obtained by the hydrolysis of methylenediformamide with acid:
HCHO + 2H-CO-NH2 --> CH2(HN-CHO)2 --> H2N-CH2-NH2.2HCl
Its derivatives with the nitrogen atoms fully alkylated, R2N-CH2-NR2 are more stable.</i>

Thats all it had.

[Edited on 13-6-2007 by Axt]
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[*] posted on 13-6-2007 at 00:25


Found an alternate preparation for the sulphate, so with barium nitrate should leave a solution of the methylenediamine dinitrate. Extracted from JACS; 70; 3659-3664 (1948)

<b>Preparation of Henry Solution</b>
Into 25 g. (0.31 mole) of 37% formalin at 0° was passed 6.5 g. (0.38 mole) of gaseous ammonia. To the cold solution was added 8 g. of potassium carbonate with stirring at 10° until two layers separated. The top layer was decanted after centrifuging, weight 14.5 g. To this was added 5 g. of potassium carbonate. The mixture was stirred thoroughly, cooled in the refrigerator , again centrifuged and the top layer decanted. After repetition with 5 g. more potassium carbonate, the top layer (wt. 4.8 g.) was much more viscous than a hexamine solution of the same formaldehyde content, nor did it freeze to a crystalline mass at -30°, as did a hexamine solution. Analyses carried out by the method of Carmack, Kuehl and Leavitt' checked within 3%.

<b>Preparation of Methylenediamine Sulfate</b>
To 8 cc. of 60% sulfuric acid cooled to 5" was added 5.0 cc. (0.025 mole) of Henry solution (analyzed as 43% cyclotrimethylenetriamine) over a five-minute period. The temperature was maintained at 10° to 15°. A crystalline product, which formed at once, was filtered after one hour at 0°, washed with 20 cc. each of ether, ethanol and ether. The product dried at 50°, weighed 0.47 g. (13%) and melted at 183-193" (dec.. soft at 158°).


The following is the prep from the formamide derivative.


<b>Preparation of Methylenediformamide</b> from US patent #2713594.
A mixture of 70 grams of hexamine (0.5 mole) and 270 grams(6 moles) of freshly distilled formamide were heated to 140°C and maintained by regulation for four hours. On cooling the crystals were separated. Yield 87.8g (29%). The mother liquor was retained, and to it was added 23.5 grams (0.185 moles) of hexamine and 90 grams (2 moles) of formamide. The mixture was maintained at 140°C for five hours. The mixture was cooled and the crystals separated. Yield 79.3g (total 41%).. and repeat ect. ect.

<b>Preparation of Methylenediamine Dihydrochloride</b> extracted from J. Chem. Soc.; 266 - 271 (1964).
Methylenediformamide (41.5 g.) was dissolved in conc. hydrochloride acid (270 ml.) and set aside at room temperature until crystals of methylenediamine dihydrochloride had separated (30.5 g., 63%), m. p. 260° (decomp.) (Found: C, 10.2; H, 6.85; N, 23.35% ; Equiv. 59.0; CH,Cl,N, requires C, 10.1; H, 6.7; N, 23.55%; Equiv. 59.5).

[Edited on 13-6-2007 by Axt]
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[*] posted on 13-6-2007 at 11:24


Very interesting, Axt, thanks.
1)Probably CH2(NH2)2 forms easy in the excess of NH3. (Problem is to separate water and avoid hexamine formation).
So, hexamine+NH3 -> CH2(NH2)2?
2)Probably, in both cases HNO3 (instead H2SO4 or HCl) can be used to form dinitrate directly?
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[*] posted on 14-6-2007 at 19:06


Yeh I dont know, since "Henry solution" is a mixture of the intermediates of hexamine, and I'll attach the article which mentions its irreversibility.

Though it does seem like salts of methylenediamine have quite limited solubility, so I'd try with HNO3 and see what happens. Have a melting point for it (130°C) so should be able to tell what you end up with.

Quote:
There is an article (I can not to attach it)


Why not? if its too big can you email it to scimadfiles@gmail.com

[Edited on 15-6-2007 by Axt]

Attachment: The Reaction between Formaldehyde and Ammonia.pdf (763kB)
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[*] posted on 6-7-2007 at 11:44


irreversibility of hexamine+NH3:
and what about hexamine+NH4X?
(not the CH3NH3(+) formation, but, for example,
long fusion of dry (CH2)6N4 with AN or AP) at >120C)
What is known about the reaction of CH2O with excess of AN without heating?

by the way, from US Patent 5415785
''...Generally speaking, more diaminomethane will be produced when the ratio of NH group equivalents to aldehyde group equivalents is 2 to 1, while increasing amounts of amine capped polyoxymethylene ethers will be produced when the ratio of NH group equivalents to aldehyde group equivalents is less than 2 to 1. In fact, diaminomethanes can be prepared substantially free of amine capped methylene ether by slowly adding the aldehydes to a solution containing the amines such that there are 2 NH group equivalents per aldehyde equivalent. The slow addition of aldehydes and the 2 to 1 ratio of NH group equivalents to aldehyde equivalents almost completely eliminates polyoxymethylene ether formation.
Polycyclic amines are preferentially formed by reacting ammonia with formaldehyde in ratios from about 2 to about 1 and 1 to 1 of NH group equivalents per aldehyde equivalent. For tetramethylenetetramine, the preferred ratio is 2 to 1. Higher reaction temperatures in conjunction with aldehydes other than formaldehyde are conducive to the formation of enamines..."

Quote:
Originally posted by Axt
Why not? if its too big can you email it to scimadfiles@gmail.com

no, the trouble is the origin of the article
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