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Inorg Chem Lab

by: Aurore Armstrong

Inorg Chem Lab CHEM 4100

Aurore Armstrong
GPA 3.55

Navamoney Arulsamy

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Navamoney Arulsamy
Class Notes
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This 3 page Class Notes was uploaded by Aurore Armstrong on Wednesday October 28, 2015. The Class Notes belongs to CHEM 4100 at University of Wyoming taught by Navamoney Arulsamy in Fall. Since its upload, it has received 40 views. For similar materials see /class/230370/chem-4100-university-of-wyoming in Chemistry at University of Wyoming.


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Date Created: 10/28/15
1H and 31P NMR of RuHOAcPPh33 RuHOAcPPh33 is a hexa coordinate complex One can write only two kinds of dissimilar structures meridional or facial Therefore the product must be either one of the isomers or a mixture of the two The 1H and 31F spectra measured for the hydrido compound are very useful in characterizing the product conclusively The isomers can be represented as follows P1 P2 and P3 represent the three PPh3 ligands O O represents the acetate ligand and H represents the hydrido ligand 1 H Spectru m of mer P O RuHOAcPPh33 First let us consider the 1H spectrum expected for the meridional isomer The spectrum U contains several closely spaced peaks in i the 7 to 85 ppm region due the presence of the phenyl protons a peak due to the acetate methyl protons 1 3 ppm region and another at 18 ppm due to the hydride The peak due to the hydride is highly informative In 3 P3 compounds that contain more than one Mendionallsomer Faciallsomer NMR amenable nuclei the proton signals are also split by the interaction of the other by the other nuclei In this case the phosphorus nuclei are capable of interacting with the hydride ligand The interaction splits the hydride peak in a certain manner which reveals the geometrical arrangement of the phosphine ligands with respect to the hydride ligand In general a phosphorus atom which is located trans to a hydride ligand splits the hydride signal by 100 Hz The splitting is known as coupling and is represented by the coupling constant J Phosphorus nuclei located cis to a hydride ligand split its peak with J values close to 25 Hz In the meridional isomer the three P atoms of triphenylphosphine ligands are arranged cis to the hydrido ligand Phosphine P2 is unique whereas P1 and P3 have identical ligand environment Therefore the magnetic interaction of P2 with H is not identical to the combined interaction of P1 and P3 However since all the three P s are cis to the H ligand the J values in each case are expected to be similar When the hydride ligand magnetically interacts with the two equivalent phosphines P1 and P3 the signal is split into a triplet Simultaneously the H ligand also interacts with the unique phosphine P2 and the interaction splits each of 1 31 the three peaks into a doublet The HtWO ClS Pcoupllng resultant pattern is three overlapping doublets and appears as a quadruplet as shown to the right with J 25 Hz 31 P Spectrum of mer rL RuHOAcPPh33 The 31F spectrum further sphttlng by the for the meridional isomer will contain N25HZ N25HZN25HZ one cis 3113 two peaks since there are two types of chemically different phosphines present One of them will be twice as intense as the other since there is one unique phosphine P2 and a pair of magnetically identical phosphines P1 and P3 The P signal of phosphorus P will be split into a triplet due to its interaction with the other two identical phosphorus nuclei P1 and P3 The P signal of P1 and P3 will be split into a doublet due to its interaction with P2 Just as explained for the proton spectrum above the phosphorus signals will also be split by their coupling with each other and those with other nuclei with nuclear spins in this case protons Though there are phenyl and acetate protons present they are located far away from the phosphine phosphorus nuclei Only the hydride ligand is suitably placed in the coordination sphere and could couple with the phosphorus nuclei However through instrumental manipulation once could circumvent this latter coupling and measure a proton decoupled phosphorus spectrum which in many cases will be simpler and easier to interpret If the product isolated is the meridional isomer the spectrum will be as follows and will contain a quartet and a triplet 31Pcis31P coupling 25 Hz V 1Hcis31P coupling consider the 1H spectrum of the facial isomer In this structure also two chemically different phosphines P1 is unique and P2 and P3 are chemically identical Since P1 is trans to the hydride H7770 the peak due to the hydride will be split into a doublet by the trans phosphorus and the J value will be 100 Hz The two cis P s P2 and P3 will further split the doublet into two triplets P 2 1 with J value close to 25 Hz Therefore the resultant pattern will be as follows Thus the 1H spectrum of the compound isolated will be 1 sufficient to assign its structure The P spectra will provide additional proof 1H Spectrum of facRuHOAcPPh33 Now let us 0 1H one trans31P coupling i w 1Htwo cis 31F coupling V V39I N100Hz gt stHZ NZSHZ NZSHz NZSHZ 31F Spectrum of facRuHOAcPPh33The proton decoupled 31F spectrum of the facial isomer will contain two peaks one of them due to P1 twice as intense as the other due to P2 and P3 The smaller peak will be split into a triplet due to the coupling of P1 with P2 and P3 The larger peak will be split into a doublet due to the coupling of P2 and P3 with P1 just as in the case of the meridional isomer However in the proton coupled 31P spectrum the triplet due to P1 will be further split into doublets with the J value of 100 Hz The doublet due to P2 and P will be split into two overlapping overlapping doublets of a doublet due to successive coupling of the rest of the two phosphorus nuclei The resultant pattern will be three triplets with J 25 Hz If proton coupling is allowed each of the triplets will further be split into doublets and the resultant spectrum will be as follows 31PicisslP coupling


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