Phthalocyanine tin(IV) dichloride

CAS: 18253-54-8

Ref. 3D-FP36688

Product Information

• Name: Phthalocyanine tin(IV) dichloride
• Synonyms: (OC-6-12)-Dichloro[29H,31H-phthalocyaninato(2-)-κN²⁹,κN³⁰,κN³¹,κN³²]tin, 29H,31H-Phthalocyanine, tin complex, 31H-Phthalocyaninato(2-)-N29,N39,N31,N32]-Dichloro[29(Oc-6-12)-Ti, Dichloro(phthalocyaninato)tin, Dichloro[phthalocyaninato(2-)]-tin(IV), Dichloro[phthalocyaninato(2-)]tin, Dichlorotin phthalocyanine, Phthalocyanatodichlorotin iv, Phthalocyanine Tin(IV)Dichloride, SnCl2Pc, Tin(Iv) Phthalocyanine Dichloride, Tin, dichloro[29H,31H-phthalocyaninato(2-)-N²⁹,N³⁰,N³¹,N³²]-, (OC-6-12)-, Tin, dichloro[29H,31H-phthalocyaninato(2-)-κN²⁹,κN³⁰,κN³¹,κN³²]-, (OC-6-12)-, Tin, dichloro[phthalocyaninato(2-)]-, dichloro[29H,31H-phthalocyaninato(2-)-N29,N30,N31,N32]-,(OC-6-12)-Tin
• CAS: 18253-54-8

Description

Phthalocyanine tin(IV) dichloride is a compound that is formed by the reaction of boron nitride with hydroxyl groups. This compound is a light-sensitive disinfectant, and can be activated by exposure to ultraviolet light. Phthalocyanine tin(IV) dichloride has been shown to have strong bactericidal activity against Gram-positive bacteria, such as Bacillus subtilis, Streptococcus pyogenes, and Staphylococcus aureus. It also has an anti-inflammatory effect in mice models of colitis. The chemical structure of this compound consists of a polycarboxylic acid group and an alkylthio group that are reactive functional groups. These reactive functional groups react with hydroxy groups from polyhydroxy compounds to form crystalline structures containing divalent hydrocarbon chains.

• Brand: Biosynth

Chemical properties

• Molecular weight: 702.14 g/mol
• Formula: C₃₂H₁₆Cl₂N₈Sn
• Purity: Min. 95%
• InChI: InChI=1S/C32H16N8.2ClH.Sn/c1-2-10-18-17(9-1)25-33-26(18)38-28-21-13-5-6-14-22(21)30(35-28)40-32-24-16-8-7-15-23(24)31(36-32)39-29-20-12-4-3-11-19(20)27(34-29)37-25;;;/h1-16H;2*1H;/q-2;;;+4/p-2
• InChI key: InChIKey=BBRNKSXHHJRNHK-UHFFFAOYSA-L
• SMILES: Cl[Sn]1(Cl)n2c3c4ccccc4c2/N=C2\N=C(/N=c4/c5ccccc5/c(n41)=N/C1=N/C(=N\3)c3ccccc31)c1ccccc12