Hydrazines aminonitriles

Hurst <93AHC(58)215> has summarized the results of nitration of 1-phenyl-A -pyrazolines for instance, l,5-diphenyl-3-aryl-2-pyrazolines are nitrated by potassium nitrate in sulfuric acid to afford the 1-p-nitrophenyl derivative in quantitative yield. Thermal ring opening of A -pyrazolin-3-carboxylic acids has been used as a method of preparation of )8-aminonitriles <94T7543>, while base-promoted ring opening of 1,1 -disubstituted-3-amino-A -pyrazolinium salts affords either N,N-disubstituted hydrazines or a,)S-unsaturated amidrazones <86SC585>. 

A 3-aminopyrazole ring is formed on stirring an aminonitrile at ambient temperature with hydrazine hydrate. 

A fused 4-hydrazinopyrimidine ring is obtained when neighbouring imino and nitrile groups are annulated by means of an excess of a hydrazine [3372]. Two molecules of an aminonitrile sometimes react together under strongly basic... 

A -Cyano-A -(methoxycarbonylalkyl)amino groups attached to 1,3,5-triazines have been reacted with hydrazine to give imidazolyltriazines <89AKZ192>. The same author also reacted an aminonitrile with a chloroacetamide to give the same imidazolyltriazine structure with a phenyl substituent. Hydrazones (117) are available from hydrazinotriazines on reaction with benzaldehyde and have been converted into jS-lactams (118) by reaction with chloroacetyl chloride and base (Scheme 29) <91MI612-03>. 

Aminonitriles from unsym. hydrazines NNHg NGHgGHgGN... 

All of the reactions shown are fairly standard in type. The first is a Strecker synthesis the first formed product is the a-aminonitrile, which is then hydrolyzed to an amino acid. The second reaction is a standard oxime synthesis as the ketone is symmetric, there is no stereochemistry to worry about. In (c), an imine is formed the stereochemistry about the carbon-nitrogen double bond is determined by steric hindrance. Since the SM is racemic, two diastereoisomers will be formed, hence the wavy bond to the methyl group. In the final example, a hydrazine is prepared while it s not possible to be certain of the stereochemistry about the C=N double bond, it s logical to assume that the less sterically hindered isomer will predominate ... 

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