49575-10-2 Usage
Description
1-(2-AMINOETHYL)-2-METHYL-5-NITROIMIDAZOLE DIHYDROCHLORIDE MONOHYDRATE is an off-white crystalline solid that serves as an interesting intermediate in the synthesis of various antibacterial and antitumor agents. Its unique chemical structure allows it to be a key component in the development of pharmaceuticals targeting bacterial and cancerous cells.
Uses
Used in Pharmaceutical Industry:
1-(2-AMINOETHYL)-2-METHYL-5-NITROIMIDAZOLE DIHYDROCHLORIDE MONOHYDRATE is used as an intermediate compound for the synthesis of [application reason] antibacterial and antitumor agents. Its role in the development of these pharmaceuticals is crucial, as it contributes to the creation of effective treatments against bacterial infections and various types of cancer.
Used in Antibacterial Applications:
In the field of antibacterial research and development, 1-(2-AMINOETHYL)-2-METHYL-5-NITROIMIDAZOLE DIHYDROCHLORIDE MONOHYDRATE is used as a key intermediate for creating new and potent antibacterial agents. These agents aim to combat drug-resistant bacteria and provide novel treatment options for various bacterial infections.
Used in Antitumor Applications:
1-(2-AMINOETHYL)-2-METHYL-5-NITROIMIDAZOLE DIHYDROCHLORIDE MONOHYDRATE is also utilized as an intermediate in the synthesis of antitumor agents. Its incorporation into the development of these pharmaceuticals helps in the creation of drugs that can effectively target and treat various types of cancer, potentially improving patient outcomes and survival rates.
Check Digit Verification of cas no
The CAS Registry Mumber 49575-10-2 includes 8 digits separated into 3 groups by hyphens. The first part of the number,starting from the left, has 5 digits, 4,9,5,7 and 5 respectively; the second part has 2 digits, 1 and 0 respectively.
Calculate Digit Verification of CAS Registry Number 49575-10:
(7*4)+(6*9)+(5*5)+(4*7)+(3*5)+(2*1)+(1*0)=152
152 % 10 = 2
So 49575-10-2 is a valid CAS Registry Number.
InChI:InChI=1/C6H10N4O2/c1-5-8-4-6(10(11)12)9(5)3-2-7/h4H,2-3,7H2,1H3/p+1