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IR Spectra of anhydrides

  1. Jun 27, 2010 #1

    Does anyone know why acetic anhydride has 2 IR bands at 1760 and 1820? If you look it up in a table it says they are due to the 2 carbonyl groups in an anhydride.

    But this is what I want to know:

    1) What specifically is causing the peaks? Is it asymmetric stretching of the carbonyl bond?
    2) In something like acetic anhydride which is symmetrical why aren't the 2 carbonyl groups 'equivalent' so that their stretching modes occur at the same frequency which would make them degenerate and appear as a single peak?

  2. jcsd
  3. Jun 28, 2010 #2
    You need to know the symmetry of the molecule. Then identify the allowed and forbidden vibrations for IR. If there is a change in the dipole moment in the molecule during vibration, then, those bands show up in IR. otherwise not.
    In general Asymmetric (both stretching and bending) vibrations are IR active and symmetric (vibrations and bending) are inactive.
  4. Jun 28, 2010 #3


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    Asymmetric and symmetric coupled C=O stretching. Usually around 1825 cm-1 and 1758 cm-1.

    Symmetric stretching is when the C=O bonds are both lengthening and shortening in unison. Asymmetrical stretching is when the C=O bonds are alternately lengthening and shortening... again in unison.

    Which do you think is of higher energy? Asymmetric or symmetric?
  5. Jun 28, 2010 #4
    Hi Chemisttree,

    I read somewhere that asymmetric vibrations occur at low cm-1 compared to symmetric ones (or may be other way, i really forgot).
    But is it all the time applicable ?
    just a cross question came to my mind!
  6. Jun 28, 2010 #5

    I don't know much about IR but in small molecules such as water and CO2 symmetric coupling isn't IR active because there is no net change in dipole moment. Presumably then you are saying symmetric coupling is IR active on complex molecules. I would imagine the asymmetric couplling would lead to a larger dipole but then i didn;t think symmetric coupling would be on the spectrum at all base on my primitive knowledge
  7. Jun 29, 2010 #6


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    There is no net dipole moment in carbon dioxide, as the molecule is linear, but I have a gut feeling anhydride is not that symmetrical.
  8. Jun 29, 2010 #7
    Hi Borek

    That's what I mean - you don't get a dipole from symmetrical stretching in linear molecules.

    Acetic anhydride is definitely symmetrical CH3 C=0 C-O -C=O -CH3

    It would seem then that you see symmetrical stretching in larger molecules as I've found another site that labels the peaks in an anhydride asn asymm and symm stretching of the carbonyl groups. And there are only 2 peaks whether your anhydride is symmetrical in structure or not so that has nothing to do with it which also surprises me as i thought it might confer som degeneracy.
  9. Jun 29, 2010 #8


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    -(C=O)COC(C=O)- is symmetrical, but not linear. Carbonyls can get longer at the same time (symmetrical) or alternating (asymmetrical). But I am not sure if I understand what you mean, so could be I am repeating obvious things.
    Last edited: Jun 29, 2010
  10. Jun 29, 2010 #9

    I know the anhydride is not linear but is symmetrical. It's confusing because there is the symmetry of the bond and the symmetry of a molecule.

    One of my points was I thought that if you had a symmetrical molecule which had 2 polar functional groups (in symmetrical positions) which have some vibrational mode (lets call it bending so we don't have to use the word symmetry again) then maybe this mode would only show up as one peak as the 2 peaks would be degenerate due to molecular symmetry
  11. Jun 29, 2010 #10


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    The C=O bond itself has a net dipole moment. Lengthening it increases that dipole... thus a net change in the dipole of that which is important, the dipole moment of the absorbing species C=O. Symmetrical stretching is where both carbonyls are lengthening at the same instant and compressing at the same instant. Asymmetric stretching is where one carbonyl is lengthing while at the same instant the other carbonyl is compressing. Remember we are describing a vibratory motion... lengthening and compressing along the internuclear axis between each C and O in the carbonyl group in this case.

    We are NOT talking about the net dipole moment of the entire molecule. We are talking only about the net dipole change of the IR 'chromophore'... each of the carbonyl groups.

    Carbon dioxide has a MASSIVE IR doublet absorbtion, BTW. I take pains to sparge the IR equipment with dry, CO2-free air ($$$) to get good looking spectra.

    Can you guess what types of stretching give rise to the doublet CO2 absorption?

    You should know that CO2 has a strong IR absorbtion if you know anything about the greenhouse effect and how CO2 affects global warming!
  12. Jun 29, 2010 #11
    For C02 is one peak the asymm stretch and the other peak the 2 degenerate bending modes?
  13. Jun 30, 2010 #12


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    Symmetric and asymmetric. Bending modes are at much lower energy.

    Are you sure that your understand what constitutes an asymmetric and a symmetric stretch?
  14. Jun 30, 2010 #13
    Maybe I don't because i thought the symmetric stretch in co2 wasn't ir active because there is in change in dipole moment. That's what my book says too but you say different?
  15. Jun 30, 2010 #14
    From book:

    The symmetrical stretch of CO2 is inactive in the IR because this vibration produces
    no change in the dipole moment of the molecule. In order to be IR active, a vibration
    must cause a change in the dipole moment of the molecule.(The reason for this
    involves the mechanism by which the photon transfers its energy to the molecule,
    which is beyond the scope of this discussion.) Of the following linear molecules,
    carbon monoxide and iodine chloride absorb IR radiation, while hydrogen, nitrogen,
    and chlorine do not. In general, the larger the dipole change, the stronger the intensity
    of the band in an IR spectrum.

    Only two IR bands (2350 and 666 cm–1) are seen for carbon dioxide, instead of four
    corresponding to the four fundamental vibrations. Carbon dioxide is an example of
    why one does not always see as many bands as implied by our simple calculation. In
    the case of CO2, two bands are degenerate, and one vibration does not cause a
    change in dipole moment.
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