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Orgabometallics reaction help please

Hi guys,

The question i'm stuck on is about heating Cr(CO)6 and reacting it with Me2NCH2CH2NMe2

Cpd A is the product of this reaction and has MF C10H16CrN2O4

Furthermore, cpd A gives IR bands at 2007 cm-1, 1885cm-1, 1873 cm-1 and 1852 cm-1

I've gotten this far:

Heating Cr(CO)6 removes two CO groups to give Cr(CO)4 which has 2 free binding sites.

The IR bands say that there are both terminal CO groups and u2 bridging CO ligands (the u is meant to be mew).

So I've got all the pieces together, however I'm finding it hard to see how the Me2NCH2CH2NMe2 binds to the complex to give both the given IR data and the molecular formula - the only way Ive managed to get the molecular formula is to bind the free sites using the lone pair on the Nitrogens - however this gives an ionic cpd with only terminal CO ligands :/

Any help would be greatly appreciated!!

Thanks
Reply 1
Original post by Chemhistorian
Hi guys,

The question i'm stuck on is about heating Cr(CO)6 and reacting it with Me2NCH2CH2NMe2

Cpd A is the product of this reaction and has MF C10H16CrN2O4

Furthermore, cpd A gives IR bands at 2007 cm-1, 1885cm-1, 1873 cm-1 and 1852 cm-1

I've gotten this far:

Heating Cr(CO)6 removes two CO groups to give Cr(CO)4 which has 2 free binding sites.

The IR bands say that there are both terminal CO groups and u2 bridging CO ligands (the u is meant to be mew).

So I've got all the pieces together, however I'm finding it hard to see how the Me2NCH2CH2NMe2 binds to the complex to give both the given IR data and the molecular formula - the only way Ive managed to get the molecular formula is to bind the free sites using the lone pair on the Nitrogens - however this gives an ionic cpd with only terminal CO ligands :/

Any help would be greatly appreciated!!

Thanks


Is C10H16CrN2O4 the molecular formula or the empirical formula? Because bridging ligands can't happen with only 1 metal centre...
Reply 2
Original post by illusionz
Is C10H16CrN2O4 the molecular formula or the empirical formula? Because bridging ligands can't happen with only 1 metal centre...


Molecular - also in my notes I have down that u2 ligands give bands 1850-1750 so I could argue the IR data says u1 only...:/
Reply 3
Original post by Chemhistorian
Molecular - also in my notes I have down that u2 ligands give bands 1850-1750 so I could argue the IR data says u1 only...:/


If it is the molecular formula then it's pretty simple. Two adjacent sites occupied by the TMEDA and 4x CO. Octahedral overall. The IR data confirms this.
Reply 4
Original post by illusionz
If it is the molecular formula then it's pretty simple. Two adjacent sites occupied by the TMEDA and 4x CO. Octahedral overall. The IR data confirms this.


thats what /i thought :smile: but how does the TMEDA bind? Using the lone pairs on the Nitrogens give a 2+ complex :/

Thanks

Matt
Reply 5
Original post by Chemhistorian
thats what /i thought :smile: but how does the TMEDA bind? Using the lone pairs on the Nitrogens give a 2+ complex :/

Thanks

Matt


Why would it be a 2+ complex?
Reply 6
Original post by illusionz
Why would it be a 2+ complex?


Because the Nitrogens would have 4 bonds each (2 methyls, 1 to the (CH2)2 and one to the metal)

Or do the CO groups take electrons when theyer removed so that the complex charge balances to 0?

Thanks
Reply 7
Original post by Chemhistorian
Because the Nitrogens would have 4 bonds each (2 methyls, 1 to the (CH2)2 and one to the metal)

Or do the CO groups take electrons when theyer removed so that the complex charge balances to 0?

Thanks


You are starting with Cr (0) and 6x neutral CO ligands. React with neutral TMEDA and displace 2 of the CO. Everything is neutral. There are no charges beforehand so there can be no charge afterwards.

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