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Identification of Stereochemical Isomers of [Mo(CO) 4 (L) 2 ] by Infra-Red Spectroscopy

Identification of Stereochemical Isomers of [Mo(CO) 4 (L) 2 ] by Infra-Red Spectroscopy. AKA Group Theory Emma Kate Payne, Courtney Arnott, and Julia Schmitz. Background Information. As an electron drops from a higher energy state, it creates both a magnetic and electric vector

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Identification of Stereochemical Isomers of [Mo(CO) 4 (L) 2 ] by Infra-Red Spectroscopy

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  1. Identification of Stereochemical Isomers of [Mo(CO)4(L)2] by Infra-Red Spectroscopy AKA Group Theory Emma Kate Payne, Courtney Arnott, and Julia Schmitz

  2. Background Information • As an electron drops from a higher energy state, it creates both a magnetic and electric vector • Only those molecules that create changing magnetic vectors are absorbed by IR • IR absorption will occur if the vibration between two atoms has a magnetic oscillating frequency that is the same frequency as the photon of light emitted • The more symmetric a molecule the less vibration and less absorption by the IR

  3. How to Determine Expected IR Absorption • Use symmetry operations and a Group Theory Flow Chart to find point group of molecule • Then use a characteristic table to determine the number of bands expected to show up on IR • Try to distinguish between cis-Mo(CO)4(PPh3)2 and trans-Mo(CO)4(PPh3)2

  4. Group Theory Flow Chart

  5. E C2 sv(xz) sv(yz) A1 1 1 1 1 z x2, y2, z2 A2 1 1 -1 -1 Rz xy B1 1 -1 1 -1 x, Ry xz B2 1 -1 -1 1 y, Rx yz Characteristic Tables C2v FIND THAT cis will have four bands

  6. E 2C4 C2 2C2' 2C2'' i 2S4 sh 2sv 2sd A1g 1 1 1 1 1 1 1 1 1 1 x2 + y2, z2 A2g 1 1 1 -1 -1 1 1 1 -1 -1 Rz B1g 1 -1 1 1 -1 1 -1 1 1 -1 x2 - y2 B2g 1 -1 1 -1 1 1 -1 1 -1 1 xy Eg 2 0 -2 0 0 2 0 -2 0 0 (Rx, Ry) (xz, yz) A1u 1 1 1 1 1 -1 -1 -1 -1 -1 A2u 1 1 1 -1 -1 -1 -1 -1 1 1 z B1u 1 -1 1 1 -1 -1 1 -1 -1 1 B2u 1 -1 1 -1 1 -1 1 -1 1 -1 Eu 2 0 -2 0 0 -2 0 2 0 0 (x, y) Characteristic Tables (cont.) D4h Find that trans has one band total

  7. IR Identification • In IR the carbon monoxide peak of the cis molecule will have four peaks • The carbon monoxide peak of the trans molecule will have one peak because it is more symmetrical

  8. Reaction Scheme

  9. Synthesis of Mo(CO)4(pip)2 • All reactions carried out under N2 • 1 g MoCO6 suspended in dry toluene • 10 ml piperidene added • Reflux 2 hrs  yellow-orange solution, yellow precipitate • Filter, dry on vacuum line • Take weight, IR, melting point

  10. Synthesis of Mo(CO)4(PPh3)2 • 0.5 g MoCO6 suspended in dry CH2Cl2 • 0.75 g PPh3 added • Reflux 15 min  orange solution, filtered • Filtrate reduced on rotovap, 15 mL MeOH added, cooled in freezer  yellow crystals • Filter, dry on vacuum line • Take weight, IR, melting point

  11. Thermal Isomerization • 0.5 g Mo(Co)4(PPh3)2 dissolved in dry toluene • Reflux 15 min  dark solution • Rotovap to yield an off-white product • Rinse with CH2Cl2 • Filter, dry on vacuum line • Take weight, IR, melting point

  12. Experimental Error • During our second synthesis, our solvent boiled off • No condenser tube • Thus we thermally isomerized our reaction straight to the trans configuration

  13. What We Obtained

  14. IR of Mo(CO)4Pip2

  15. IR Mo(CO)4(PPh3)2 (Cis)

  16. IR of Mo(CO)4(PPh3)2 (Trans)

  17. Conclusions • Made both the Cis and Trans Compounds • Group Theory correctly predicted the number of bands in both IRs • Improvements: • Condenser column on all reactions • Better N2 conditions • More careful monitoring of heat/evaporation

  18. We love our seniors! Only 2 days til Junior Banquet… :-D

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