Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    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
    Offline

    16
    ReputationRep:
    (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...
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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...:/
    Offline

    16
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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 but how does the TMEDA bind? Using the lone pairs on the Nitrogens give a 2+ complex :/

    Thanks

    Matt
    Offline

    16
    ReputationRep:
    (Original post by Chemhistorian)
    thats what /i thought 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?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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
    Offline

    16
    ReputationRep:
    (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.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.