Regarding the complex $\ce{[RuH2(CO)(PPh3)3]}$, there are three possible octahedral stereochemical configurations,

a) Draw these isomers indicating which, if any, are chiral and identify the spin systems in each case

b) Sketch the expected appearance of the $\ce{^{31}P\{^1H\}}$ ($\ce{^1H}$ decoupled) spectra

c) Sketch the expected appearance of the $\ce{^1H}$ spectra

My attempt:

I think I have completed (a) but would appreciate on any advice for (b) and (c)

Three possible octahedral stereochemical configurations of [RuH2(CO)(PPh3)3]

  • For the isomer on the far left I think the decoupled phosphorus NMR would show a doublet and a triplet.

  • The middle one I think it might too be a doublet and a triplet.

  • The isomer on the right I'm leaning towards doublet of doublets but I'm unsure.

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    $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I improved the formatting of your post with MathJax; for more information on how to do so yourself, check out the help center, this meta-post or this one. Also note that your question qualifies as a homework question as per our homework policy. Therefore, please show your work, e.g. the results you have for a) and how you would approach b) and c). $\endgroup$ – Jan Nov 22 '15 at 0:30
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    $\begingroup$ (cont’d.) Otherwise your question may get closed. $\endgroup$ – Jan Nov 22 '15 at 0:30

Let’s call the isomers mer-trans, mer-cis and fac from left to right for easier addressing.

What you didn’t indicate yet is which symmetry elements the different ones have and if they are chiral; this is helpful in determining the look of the NMR spectra because it helps us identify enantiotopic and homotopic atoms which give identical spins. The mer-trans isomer is clearly the one with the highest degree of symmetry: I judge its point group to be $C_\mathrm{2v}$. The other two should be $C_\mathrm{s}$, if I did the looking correctly.

Once we have identified the symmetry, we need to find out which atoms are chemically and/or magnetically equivalent. For the two $C_\mathrm{s}$ compounds it is pretty easy: If the plane of symmetry transforms one atom onto the other, they are chemically equivalent but enantiotopic (so magnetically non-equivalent). If the plane of symmetry contains the atoms in question, they must be chemically non-equivalent, thus have different spins. For the mer-trans isomer it may look slightly more tricky, but note that spinning around the $C_2$ axis will always place one atom of the same kind in question on top of the other thus again making them chemically equivalent and homotopic. (Also note that the three phosphorus atoms can never all be chemically equivalent to each other since one always occupies a special position by symmetry.)

With our sets of chemically equivalent or non-equivalent atoms in hand — to check: I arrived at sets of 1, 1 and 2 hydrogens and always two phosphorus; in one phosphorus set we have a pair chemically equivalent atoms, the other two contain a pair of enantiotopic ones — we can predict the look of the spectra and what will couple to what. Note that the $\ce{^{31}P\{^1H\}}$ spectrum will not contain any couplings to hydrogen atoms.

  • For mer-trans, your reasoning is sound and I also expect the result to be a doublet and a triplet; the coupling constants must be identical. If I had to guess, I would say that those closer to $\ce{CO}$ should have a higher chemical shift due to $\ce{CO}$’s electron withdrawing nature; but it could be due to a push-pull mechanism that its influence is greater in the para-position.

  • mer-cis should also be a doublet and a triplet for more or less the same reasons. The chemical shifts must be non-identical but could be closer to each other due to every phosphorus being a neighbour of $\ce{CO}$ (which is still the single most differentiating species in the compound).

  • fac cannot be a doublet of doublets since that would be a single signal for all three; the three are, however, chemically non-equivalent. The single special phosphorus will again display a triplet because it is coupling to two distinct phosphorus atoms. The other two should show up as a doublet. I’ll let you do the reasoning.

When proceeding to the hydrogen spectra, remember that $\ce{^{31}P}$ is also a magnetically active, spin ½, $100~\%$ abundance nucleus. Thus, you must consider the coupling of the protons with each other (if applicable) plus the coupling of the protons with the phosphorus atoms (the magnetically non-equivalent ones). In no particular order you should arrive at:

  • One spectrum containing two distinct proton shifts, each with two doublet couplings and a triplet coupling. I cannot say a priori which one should be larger or smaller.

  • One spectrum containing a single proton shift which is split up into a doublet and a triplet. Again, I cannot say which coupling is stronger.

  • One spectrum containing a single proton shift, split into a doublet of doublets of doublets (ddd). At least here I don’t have to say which coupling is stronger because they are all distinct.

Retroactively assinging proton spin systems, I get $\ce{AX, A2}$ and $\ce{AA'}$; for phosphorus I arrive at $\ce{AX2, AA'X}$ and $\ce{AX2}$, again in no particular order for all six spin systems (i.e. you would still need to assign them to the correct structures). If you wish to combine the spin systems into one entire one per structure, use $\ce{M}$ rather than $\ce{A}$ in the hydrogen ones (or $\ce{K}$ and $\ce{R}$ instead of the hydrogen $\ce{AX}$). Note however that I am terrible at spin systems and that you should take those with a grain of salt.

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  • $\begingroup$ I hope that someone good at spin systems can confirm that part of my answer to be correct. $\endgroup$ – Jan Nov 23 '15 at 17:29

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