6
$\begingroup$

I've tried searching for literature references that explain this, finding only a single reference which refers to the two signals as originating from phosphine environments cis/trans to carbonyls, however I'm having trouble rationalising why this occurs.

Structure of cis-[Mo(CO)2(dppe)2]

$\endgroup$
  • 1
    $\begingroup$ I just realised why, I'm a total idiot 😂 - I blame this on my lack of sleep. To ask a question that requires an actual answer @orthocresol I see you added a far better representation of the complex, do you have any tips on how to draw better looking chelating ligands? $\endgroup$ – Funk Mar 4 '19 at 23:04
  • 2
    $\begingroup$ Not really! I just drew a regular octahedral complex in ChemDraw (from Templates > Stereocentres), made it a bit bigger (or else the PPh2 clashes into the CO), filled in the PPh2's and CO's, then drew a few bonds between phosphorus. Yours was perfectly fine, just a bit large, and I was drawing it for my answer anyway so it was trivial to add it into the question itself. $\endgroup$ – orthocresol Mar 4 '19 at 23:08
  • $\begingroup$ Are you using ChemDraw's default document settings? If I try to follow a similar process the ligand bonds end up being too long so I end up with disproportionately large ligands. $\endgroup$ – Funk Mar 4 '19 at 23:14
  • 2
    $\begingroup$ Nope, for all the diagrams on SE, I use a version of the Trauner group ChemDraw template where everything is scaled by 0.75x. If you use the ACS Document 1996 template in ChemDraw it gives stuff which look pretty similar, so you could try that out. $\endgroup$ – orthocresol Mar 4 '19 at 23:16
9
$\begingroup$

In this complex there are two different 31P environments which are not related by symmetry:

Inequivalent phosphorus nuclei in cis-[Mo(CO)2(dppe)2]

The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted.

As extra proof, consider that the green P is cis to both carbonyl ligands whereas the purple P is cis to one and trans to the other.

They therefore have different chemical shifts and in the spectrum you would expect to see two different peaks. Presumably they would show coupling to each other, so I would expect two triplets, if we ignore satellites from all other nuclei (e.g. 13C).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.