Post by Anders Hoveland on Mar 31, 2011 12:20:02 GMT -8
This topic is about preparation of 4-amino,5-nitro-1,2,3-triazole,
which, while being itself a very stable and powerful explosive, may be of greater value as a precursor to several other advanced explosives. 4-amino,5-nitro-1,2,3-triazole (decomp 297C) is also more thermally stable than 3-amino,5-nitro-1,2,4-triazole.
Bubbling nitrogen dioxide into a mixture of glacial acetic acid and acetic anhydride should form some 2-nitroacetic acid. To add a second nitro group, a modification of the procedure below may be employed.
This procedure is for oxidizing nitroethane into 1,1-dinitroethane, but could no doubt be used on nitroacetic acid instead:
To a stirred solution of 2.66 g (66.5 mmoles) of sodium hydroxide in 15 ml of water at 20° was added 5.0 g (66.5 mmoles) of nitroethane. When all the nitroethane dissolved, the solution was cooled to 5°-7° in an ice-water bath.
The sodium salt of nitroethane was prepared as above. To the stirred solution was added a solution of 20 g (288 mmoles) of sodium nitrite in 50 ml of water, followed by a solution of 4.4 g (13.4 mmole) of potassium ferricyanide in 25 ml of water. Finally, 16.0 g of solid sodium persulfate (67 mmoles) was added all at once. The reaction temperature, moderated by an ice-water cooling bath, increased to 50°. The orange mixture was stirred at 25° for 1 hour and then cooled to 10°. Urea, 20 g (0.33 mole), was added, followed by 10 ml of glacial acetic acid. The mixture was extracted with three 25 ml portions of ether and the combined extracts were washed with brine and dried. The crude product was distilled to give 4.2 g (52% yield) of 1,1-dinitroethane, b.p. 87°-89° (16 mm), identified by its nmr spectrum. Patent 4910322
With the dinitroacetic acid thus obtained, the ethyl ester can be formed. This is somewhat complicated by the fact that any concentrated strong acid could risk hydrolyzing the nitro groups away. It may be better to react with ethyl bromide instead, using propylene carbonate (or some other such solvent) to ensure the HBr formed stays very dilute.
Here is the main procedutre:
4-amino, 5-nitro, 1,2,3-triazole was synthesized by condensing acetaldehyde with ethyl 2,2-dinitro-acetate in the presence of sodium azide, to form 3-methyl, 5-nitro, 1,2,3-triazole.
The cyclization reaction with the azide ion is very comlex. To get some idea of the intermediate steps, see the attachment at the bottom of the page. For step #3, a nitrous acid is pulled out under the alkaline conditions, leaving a double bond between carbons. (cyclize is the American spelling)
CH3CH=O (NO2)2CHC(=O)OCH2CH3. I think that (NO2)2CHC(=O)O(-) converts to (-)NO2=C(NO2)H and CO2.
The (-)NO2=C(NO2)H would then condense with CH3CH=O, in a Michael-type addition, and the aldehyde would disproportionate, being reduced while oxidizing another free CH3CH=O under the alakine conditions. Then an azide ion would displace one of the nitro groups. So now the intermediate is CH3CH*C(=NO2(-))(--N=N=N). Where the * signifies a radical. This then spontaneously cyclizes, the electron moves toward a nitrogen in the ring, and then a hydrogen ion stick on. This is all just my own conjecture. So the methyl group on the resulting ring almost certainly comes from the methyl on the acetaldehyde. The carboxyl groups comes off as CO2. The original ester gets hydrolyzed by the NaN3, and in so doing acts as a dehydrating agent.
which, while being itself a very stable and powerful explosive, may be of greater value as a precursor to several other advanced explosives. 4-amino,5-nitro-1,2,3-triazole (decomp 297C) is also more thermally stable than 3-amino,5-nitro-1,2,4-triazole.
Bubbling nitrogen dioxide into a mixture of glacial acetic acid and acetic anhydride should form some 2-nitroacetic acid. To add a second nitro group, a modification of the procedure below may be employed.
This procedure is for oxidizing nitroethane into 1,1-dinitroethane, but could no doubt be used on nitroacetic acid instead:
To a stirred solution of 2.66 g (66.5 mmoles) of sodium hydroxide in 15 ml of water at 20° was added 5.0 g (66.5 mmoles) of nitroethane. When all the nitroethane dissolved, the solution was cooled to 5°-7° in an ice-water bath.
The sodium salt of nitroethane was prepared as above. To the stirred solution was added a solution of 20 g (288 mmoles) of sodium nitrite in 50 ml of water, followed by a solution of 4.4 g (13.4 mmole) of potassium ferricyanide in 25 ml of water. Finally, 16.0 g of solid sodium persulfate (67 mmoles) was added all at once. The reaction temperature, moderated by an ice-water cooling bath, increased to 50°. The orange mixture was stirred at 25° for 1 hour and then cooled to 10°. Urea, 20 g (0.33 mole), was added, followed by 10 ml of glacial acetic acid. The mixture was extracted with three 25 ml portions of ether and the combined extracts were washed with brine and dried. The crude product was distilled to give 4.2 g (52% yield) of 1,1-dinitroethane, b.p. 87°-89° (16 mm), identified by its nmr spectrum. Patent 4910322
With the dinitroacetic acid thus obtained, the ethyl ester can be formed. This is somewhat complicated by the fact that any concentrated strong acid could risk hydrolyzing the nitro groups away. It may be better to react with ethyl bromide instead, using propylene carbonate (or some other such solvent) to ensure the HBr formed stays very dilute.
Here is the main procedutre:
4-amino, 5-nitro, 1,2,3-triazole was synthesized by condensing acetaldehyde with ethyl 2,2-dinitro-acetate in the presence of sodium azide, to form 3-methyl, 5-nitro, 1,2,3-triazole.
The cyclization reaction with the azide ion is very comlex. To get some idea of the intermediate steps, see the attachment at the bottom of the page. For step #3, a nitrous acid is pulled out under the alkaline conditions, leaving a double bond between carbons. (cyclize is the American spelling)
CH3CH=O (NO2)2CHC(=O)OCH2CH3. I think that (NO2)2CHC(=O)O(-) converts to (-)NO2=C(NO2)H and CO2.
The (-)NO2=C(NO2)H would then condense with CH3CH=O, in a Michael-type addition, and the aldehyde would disproportionate, being reduced while oxidizing another free CH3CH=O under the alakine conditions. Then an azide ion would displace one of the nitro groups. So now the intermediate is CH3CH*C(=NO2(-))(--N=N=N). Where the * signifies a radical. This then spontaneously cyclizes, the electron moves toward a nitrogen in the ring, and then a hydrogen ion stick on. This is all just my own conjecture. So the methyl group on the resulting ring almost certainly comes from the methyl on the acetaldehyde. The carboxyl groups comes off as CO2. The original ester gets hydrolyzed by the NaN3, and in so doing acts as a dehydrating agent.