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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Winga on October 02, 2006, 03:49:17 PM

Title: IR & NMR Analysis of Mo(dppe)2(N2)2
Post by: Winga on October 02, 2006, 03:49:17 PM

IR shows one characteristic absorption at 1979.8 cm^-1
It is N=N stretch.

³¹P NMR shows one sharp singlet peak at -69 ppm

The question is:
There are 2 isomers, which is the correct one?

When the two N2 ligands are trans to each other, P is appeared as one peak in NMR.
However, the two N2 ligands should have symmetric and asymmetric stretches as they can couple together in linear alignment.

When the two N2 ligands are cis to each other, they are orthogonal, so they should not couple to give sym. and asym. stretches in IR spectrum. That's why there is only one IR absorption.
However, in NMR, there should be 2 absorption peak as the chemical environment of each P of the dppe are different.

Title: Re: IR & NMR Analysis of Mo(dppe)2(N2)2
Post by: Dan on October 02, 2006, 07:21:56 PM

I think it's the trans isomer. That is the only way you can get that nmr spectrum, as you say.

Now, the sym NN stretch for the trans isomer will not be IR active (no change in dipole moment), so it must be the asym stretch that is the characteristic one.

The cis isomer would have two IR active NN stretches (sym and asym) since both change the dipole moment.

ie

N
^
|
N
MoN-->N

and

N
^
|
N
MoN<--N

Thats what I'd say, but it's been a while...

EDIT: sorry, the sym stretch for the cis isomer will be IR inactive too.