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Iridium, bis[bis(1,1-dimethylethyl)phenylphosphine]pentahydro- is a chemical with a specific purpose. Lookchem provides you with multiple data and supplier information of this chemical.

61663-67-0

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61663-67-0 Usage

Check Digit Verification of cas no

The CAS Registry Mumber 61663-67-0 includes 8 digits separated into 3 groups by hyphens. The first part of the number,starting from the left, has 5 digits, 6,1,6,6 and 3 respectively; the second part has 2 digits, 6 and 7 respectively.
Calculate Digit Verification of CAS Registry Number 61663-67:
(7*6)+(6*1)+(5*6)+(4*6)+(3*3)+(2*6)+(1*7)=130
130 % 10 = 0
So 61663-67-0 is a valid CAS Registry Number.

61663-67-0SDS

SAFETY DATA SHEETS

According to Globally Harmonized System of Classification and Labelling of Chemicals (GHS) - Sixth revised edition

Version: 1.0

Creation Date: Aug 13, 2017

Revision Date: Aug 13, 2017

1.Identification

1.1 GHS Product identifier

Product name pentahydridobis(phenyldi-t-butylphosphine)iridium(V)

1.2 Other means of identification

Product number -
Other names -

1.3 Recommended use of the chemical and restrictions on use

Identified uses For industry use only.
Uses advised against no data available

1.4 Supplier's details

1.5 Emergency phone number

Emergency phone number -
Service hours Monday to Friday, 9am-5pm (Standard time zone: UTC/GMT +8 hours).

More Details:61663-67-0 SDS

61663-67-0Relevant academic research and scientific papers

Ethylene hydrogenation by the hydride alkoxide species Ir(H)2(OCH2CF3)(PtBu2Ph)2. Ethylene-induced reductive elimination of alcohol from IrIII

Hauger, Bryan,Caulton, Kenneth G.

, p. 253 - 261 (2007/10/02)

Ir(H)2(ORf)P2 (P = PtBu2Ph, Rf = CH2CF3) reacts with ethylene at 25 deg C to give RfOH, ethane and Ir(PC)P(C2H4) (2) then Ir(PC)(C2H4)2 (1) and Ir(PC)H(ORf)P (3) (PC = η2-C6H4PtBu2).It is shown that 2 and 1 are in equilibrium by P and C2H4 addition/dissociation.Compound 3 is a product late in the reaction sequence, and results from H-ORf oxidative addition to 2.Since 3 reacts with ethylene to give 2, 2 and 3 are in thermal equilibrium.Compound 3 reacts readily with H2 to give IrH5P2 and RfOH.The reason why ORf and ethylene ligands seem to be mutually incompatible is discussed.

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