Wednesday, April 9, 2014

NMR problems about transition metal hydrides

I have just done some NMR practice and I thought I should write about them.

So, the problem asks to assign four different Pt and Pd complexes to each spectrum given. (only high field is given)

Here are the spectra and the complexes ( I will explain the reasons below the figure):

1st complex and its spectrum: There is a bidentate ligand so the complex has to have cis geometry. This makes two phosphines non-equivalent and the complex should give two sets of doublets (doublet of doublets). 

2nd complex and its spectrum: So, there are two doublets of triplets. Triplets are due to the cis phosphines and the doublet is the result of the trans phosphine. The weaker resonances on each side of the spectrum are called satellites and maybe I should write a paragraph about them in a future post. (Only Pt-195 isotope has a spin and its abundance is 33%.) 

3rd complex and its spectrum: There are two equivalent phosphines (trans). So, this complex should just give a triplet. No satellites, because the metal is Pd.

4th complex and its spectrum:  There are two equivalent phosphines and we expect to see a triplet. We can see the satellites again due to Pt metal center.

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