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Topic: NMR help  (Read 2923 times)

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Offline heidi_96

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NMR help
« on: June 25, 2014, 08:59:52 AM »
In a project, a complex formulated as [99RuCl2(dppe)2] was synthesised. The 31P {1H} NMR specturm of this shows one signal as a 1:1:1:1:1:1 sextet.
Account for this observation.

dppe is the bidentate ligand (C6H5)2PCH2CH2P(C6H5)2

I(31P) = 1/2
no coupling to Cl observed

Could anyone help me with this question, and explain how a 1:1:1:1:1:1 sextet is formed,
any help much appreciated,

Best Regards,
Heidi
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Offline Babcock_Hall

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Re: NMR help
« Reply #1 on: June 25, 2014, 09:28:23 AM »
According to forum rules, you must show an attempt first before anyone can help you.  Maybe you could clarify one thing as well:  do you mean that the P-31 was acquired coupled to H-1, or do you mean decoupled?
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Offline heidi_96

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Re: NMR help
« Reply #2 on: June 25, 2014, 09:37:53 AM »
Sorry this is my first post and i didn't realize,
As for an attempt i cant seen to figure out what one signal as a 1:1:1:1:1:1 sextant implies, whether its a triplet of doublets or that all atoms coupling to the Rh are of equal value... i've tried googling it with no success which is why i am trying on this!

Its a past exam question and doesnt state whether its coupled or decoupled... but i think its decoupled
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Offline Babcock_Hall

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Re: NMR help
« Reply #3 on: June 25, 2014, 10:01:10 AM »
What do you know about the nuclear spin quantum numbers?
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