Why does the 31P{1H} NMR spectrum of cis-[Mo(CO)2(dppe)2] show two signals?
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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.
I think it might be due to the carbonyl pi orbital electrons deshielding the phosphines with matching orbitals however I'm not sure if this is correct/explained correctly.
Apologies if I've made any errors in this post as it's my first post here.
![Structure of cis-[Mo(CO)2(dppe)2]](https://i.stack.imgur.com/taJNg.png)
inorganic-chemistry stereochemistry nmr-spectroscopy symmetry carbonyl-complexes
edited 4 hours ago
orthocresol♦
39.3k7114239
asked 4 hours ago
FunkFunk
282
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add a comment |
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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.
I think it might be due to the carbonyl pi orbital electrons deshielding the phosphines with matching orbitals however I'm not sure if this is correct/explained correctly.
Apologies if I've made any errors in this post as it's my first post here.
inorganic-chemistry stereochemistry nmr-spectroscopy symmetry carbonyl-complexes
edited 4 hours ago
orthocresol♦
39.3k7114239
asked 4 hours ago
FunkFunk
282
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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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
4 hours ago
$begingroup$
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♦
4 hours ago
$begingroup$
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
4 hours ago
1
$begingroup$
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♦
4 hours ago
add a comment |
$begingroup$
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.
I think it might be due to the carbonyl pi orbital electrons deshielding the phosphines with matching orbitals however I'm not sure if this is correct/explained correctly.
Apologies if I've made any errors in this post as it's my first post here.
inorganic-chemistry stereochemistry nmr-spectroscopy symmetry carbonyl-complexes
edited 4 hours ago
orthocresol♦
39.3k7114239
asked 4 hours ago
FunkFunk
282
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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.
I think it might be due to the carbonyl pi orbital electrons deshielding the phosphines with matching orbitals however I'm not sure if this is correct/explained correctly.
Apologies if I've made any errors in this post as it's my first post here.
inorganic-chemistry stereochemistry nmr-spectroscopy symmetry carbonyl-complexes
inorganic-chemistry stereochemistry nmr-spectroscopy symmetry carbonyl-complexes
edited 4 hours ago
orthocresol♦
39.3k7114239
asked 4 hours ago
FunkFunk
282
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
orthocresol♦
39.3k7114239
asked 4 hours ago
FunkFunk
282
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
orthocresol♦
39.3k7114239
edited 4 hours ago
orthocresol♦
39.3k7114239
edited 4 hours ago
orthocresol♦
39.3k7114239
39.3k7114239
asked 4 hours ago
FunkFunk
282
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
FunkFunk
282
asked 4 hours ago
FunkFunk
282
282
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Funk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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
4 hours ago
$begingroup$
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♦
4 hours ago
$begingroup$
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
4 hours ago
1
$begingroup$
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♦
4 hours ago
add a comment |
$begingroup$
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
4 hours ago
$begingroup$
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♦
4 hours ago
$begingroup$
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
4 hours ago
1
$begingroup$
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♦
4 hours ago
$begingroup$
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
4 hours ago
$begingroup$
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
4 hours ago
$begingroup$
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♦
4 hours ago
$begingroup$
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♦
4 hours ago
$begingroup$
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
4 hours ago
$begingroup$
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
4 hours ago
1
1
$begingroup$
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♦
4 hours ago
$begingroup$
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♦
4 hours ago
add a comment |
1 Answer
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$begingroup$
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]](https://i.stack.imgur.com/fCJOl.png)
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).
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
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1 Answer
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$begingroup$
In this complex there are two different 31P environments which are not related by symmetry:
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).
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
$endgroup$
add a comment |
$begingroup$
In this complex there are two different 31P environments which are not related by symmetry:
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).
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
$endgroup$
add a comment |
$begingroup$
In this complex there are two different 31P environments which are not related by symmetry:
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).
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
$endgroup$
In this complex there are two different 31P environments which are not related by symmetry:
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).
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
edited 4 hours ago
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
answered 4 hours ago
orthocresol♦orthocresol
39.3k7114239
39.3k7114239
add a comment |
add a comment |
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$begingroup$
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
4 hours ago
$begingroup$
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♦
4 hours ago
$begingroup$
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
4 hours ago
1
$begingroup$
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♦
4 hours ago