Chemical Properties and Facts of 15227-42-6

When you point to this article, it is believed that you are also very interested in this compound(15227-42-6)Formula: C10H10Cl2N2Pt and due to space limitations, I can only present the most important information.

Kukushkin, V. Yu.; Tkachuk, V. M. published an article about the compound: cis-Dichlorobis(pyridine)platinum(II)( cas:15227-42-6,SMILESS:[Cl-][Pt+2]([N]1=CC=CC=C1)([Cl-])[N]2=CC=CC=C2 ).Formula: C10H10Cl2N2Pt. Aromatic heterocyclic compounds can be classified according to the number of heteroatoms or the size of the ring. The authors also want to convey more information about this compound (cas:15227-42-6) through the article.

[Ph3PCH2Ph]2[PtXCl3] (X = Cl, NO2) are oxidized by Sb(BzCl)Cl5 (I) or Et3NCH2Ph[SbCl6] (II) in MeNO2, MeCN or DMF to give (Ph3PCH2Ph)2[PtXCl5]. SbCl5 oxidizes cis- and trans-Pt(py)2Cl2 to give cis- and trans-[Pt(py)2Cl4], resp. I oxidizes trans-[Pt(NH2CH2CO2H)2Cl2] to give trans-[Pt(NH2CH2CO2H)2Cl4] which on reaction with PCl5 in MeCN gives trans-[Pt(NH2CH2COCl)2Cl4]. [Pt(py)4]Cl2 reacted with I to give trans-[Pt(py)2Cl4]. cis- And trans-[PtL2Cl2] (L = PPh3, SMe2) are not oxidized by I or II.

When you point to this article, it is believed that you are also very interested in this compound(15227-42-6)Formula: C10H10Cl2N2Pt and due to space limitations, I can only present the most important information.

Reference:
Tetrahydroisoquinoline – Wikipedia,
1,2,3,4-Tetrahydroisoquinoline | C9H11N – PubChem