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Topic: 1HNMR Problem  (Read 3141 times)

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Offline ajones51

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1HNMR Problem
« on: October 25, 2008, 10:07:06 PM »
The cation [OsH(CO)2(PPh3)3]+ exhibits the following spectroscopic features:
Infrared   υ(CO) at 1985(vs)
1H nmr      A multiplet centred at -9.30 ppm (below, with line separations in Hz). 







Use this information to determine which diastereomer of this complex is present. Explain how your assignment fits the data. Please see attached photo for NMR Spectrum. 

Any ideas on how to solve it? I think the spectrum represents a doublet of triplets. 187Os has a spin number of 1/2 while 189Os (which has a greater natural abundance) has a spin number of 3/2. P has a spin number of 1/2. Since there is only one band for the IR, this suggests to me that two CO ligands are chemically equivalent, but I am not sure about this. Any insight or help to solve this problem would be great.



Offline banana

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Re: 1HNMR Problem
« Reply #1 on: October 27, 2008, 04:12:25 PM »
The picture is wrong, because the intensity ratio of the components of a triplet are 1:2:1, while the picture shows something different. Apart from that, you can explain the splitting with the coupling H-P. For example, there is one trans-P (doublet) and 2 cis-P (triplet). The distance bteween H and P is two bonds only, therefore they are certainly coupled.

The cation [OsH(CO)2(PPh3)3]+ exhibits the following spectroscopic features:
Infrared   υ(CO) at 1985(vs)
1H nmr      A multiplet centred at -9.30 ppm (below, with line separations in Hz). 







Use this information to determine which diastereomer of this complex is present. Explain how your assignment fits the data. Please see attached photo for NMR Spectrum. 

Any ideas on how to solve it? I think the spectrum represents a doublet of triplets. 187Os has a spin number of 1/2 while 189Os (which has a greater natural abundance) has a spin number of 3/2. P has a spin number of 1/2. Since there is only one band for the IR, this suggests to me that two CO ligands are chemically equivalent, but I am not sure about this. Any insight or help to solve this problem would be great.




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