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Topic: Anhydrides double peak in the infrared  (Read 11403 times)

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Offline Mimic

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Anhydrides double peak in the infrared
« on: April 06, 2018, 07:42:48 AM »
Anhydrides infrared spectra are recognizable because they have a double peak in the wave numbers between 2000 cm-1 and 1500 cm-1



Our teacher said the presence of these characteristic peaks is due to the fact that the molecules of anhydrides, taken as a whole, can vibrate in two different ways. So, if I have not misunderstood, it has a double peak because while a carbonyl bond is shortened, the other can be stretched?

Thanks in advance

Offline Arkcon

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Re: Anhydrides double peak in the infrared
« Reply #1 on: April 06, 2018, 01:59:48 PM »
That is a wide frequency separation.  I was not expecting something like that, so I now I have something new to look for.  So thank you for that.

I don't like your explanation 'tho, even if I don't have a better one.  That peak is for the stretch of the entire aromatic ring at once.  "Breathing", is what we call it some times. 

So, sometimes the phenyl ring is next to a carbonyl, and sometimes its next to an enol?  Is that a thing that happens?  Sure, but only an expert can tell us how that works.  Hopefully they will be here soon.

Side question for expert -- if that is the case, won't we get a different spectra for a solution, and a Nujol mull or dry powder on an ATR?  I only did Nujol mulls in college and only used the ATR at work.

Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Arkcon

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Re: Anhydrides double peak in the infrared
« Reply #2 on: April 06, 2018, 02:05:11 PM »
This page here: http://www.science.oregonstate.edu/~gablek/CH336/Chapter20/DerivIR.htm only says the two side groups interact with each other to produce two carbonyl stretches.  That makes more sense.  One symmetrical and one anti-symmetric, which makes sense to people who paid attention in school, not to me. :P
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Offline Mimic

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Re: Anhydrides double peak in the infrared
« Reply #3 on: April 06, 2018, 02:17:55 PM »
Fantastic! Thanks for the link!

Offline wildfyr

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Re: Anhydrides double peak in the infrared
« Reply #4 on: April 06, 2018, 02:30:19 PM »


Sure, but only an expert can tell us how that works.  Hopefully they will be here soon.

Side question for expert -- if that is the case, won't we get a different spectra for a solution, and a Nujol mull or dry powder on an ATR?  I only did Nujol mulls in college and only used the ATR at work.

Paging @Corribus... but I'll do my best. The reason you get two peaks is indeed because there is a symmetric and antisymmetric stretch of the cabronyls. Symmetric is when they are both C=O bonds are stretching out and contracting in sync, https://en.wikipedia.org/wiki/File:Symmetrical_stretching.gif

anti-symmetric is when they are exactly a half cycle off (as one reaches full stretch, the other is fully contracted).
https://en.wikipedia.org/wiki/File:Asymmetrical_stretching.gif


I think you misinterpreted what that Oregon State link said, they mean the carbonyls are interacting with one another, not the side groups.

For your side question, absolutely, yes. There are minor variations and shifts in peaks when you use different methods to get spectra (gas phase, nujol, ATR, KBr pellet etc). I saw this in grad school because we had KBr and ATR accessories.

Offline Arkcon

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Re: Anhydrides double peak in the infrared
« Reply #5 on: April 06, 2018, 02:39:16 PM »
Yeah, but the Oregon state link doesn't even mention the widely separated twin peaks of C=C aromatic stretch.  Why's that happening?  Aromatic sometimes in on state and sometimes another but what is the cause?
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Offline Corribus

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Re: Anhydrides double peak in the infrared
« Reply #6 on: April 10, 2018, 10:15:17 AM »
Although molecular vibration is a quantum phenomenon, you can take a few lessons from classical physics. Let's consider a molecule with two functional groups, one at each end, separated by a lot of uninteresting junk. Most largish molecules are "floppy", in the sense that if you twang one end of it, the vibration is mostly dampened out by insulating junk, such that the other end doesn't really feel it that much. Think of putting your hand on one side of a long table and having a friend tap on the other end - whether you feel the vibration depends on the strength of the source vibration (how hard is the tapping), the frequency/energy (we know bass is transmitted better than treble), and how well the material conducts the vibrational energy. A soft wood, like pine, may transmit the vibration much worse than a hard wood like mahogany. Molecules are not so different, and most of the aliphatic chains (much less, space between molecules) are poor transmitters of vibrations. In this sense, FTIR is in most cases a method to identify functional groups rather than entire molecules. It is why, also, most functional groups show up around the same energy, because they aren't usually strongly influenced by what else is around. (Not to say they aren't influenced at all - they are - but for the purpose of baseline organic analysis, the shifts aren't very significant or enlightening outside of identifying that molecule A is a ketone or amide or whatever.) If the amount of intermediate junk is reduced, or if the intermediate junk is made more rigid (e.g., via carbon-carbon conjugation) vibrations are transmitted more efficiently, and one vibrating functional group can influence the vibration of another. Although true in principle for all nearby vibrations (even those in only a through-space relationship), eventually we can consider two structurally-linked vibrations to be strongly coupled, in the sense that you can't even consider them functionally separate, because pouring energy into one invariably effects the vibrational state of the other. This is the case of the two carbonyl groups in an anhydride. The carbonyl stretch is very strong (it involves a large change in dipole moment) and two of them in so close proximity means their vibrations are bound to also interact strongly. In a way, you consider these to be a single functional group- in fact we may define a functional group from a spectroscopic standpoint as a unit that behaves as a single vibrating entity. The two carbonyls can either be vibrating in concert, or... whatever you would call the opposite of in concert. It isn't really so different from the vibrational states of carbon dioxide, which also have a symmetric and antisymmetric stretch. When both carbonyl groups are attached to the same carbon, you simply cannot stretch one carbonyl without impacting the other one.

Acid anhydrides are a fairly special case of coordination of two linked molecular vibrations because the carbonyl stretches are so spectroscopically significant. Although in principle any one vibration will transmit some energy to other nearby vibrations, standard (steady state) FTIR approaches aren't usually sensitive enough to deconvolute these physical relationships. Using techniques like 2D FTIR, however, you can selectively feed energy into one vibration and probe how that energy is transmitted into other nearby molecular regions. This has been immensely valuable in, say, polypeptide analysis, where the cross-talk physics between nearby vibrations (time delay and magnitude of energy transmission) provides a lot of detail on protein structure.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Arkcon

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Re: Anhydrides double peak in the infrared
« Reply #7 on: April 10, 2018, 01:26:14 PM »
Clearly, I need more than a brief refresher on how IR spectroscopy works.  However, doing the best I can:

Your average aromatic compound gives a sharp, strong peak at 1500 and maybe some shorter, sharp peaks at 1400 and 1600.  This is called C-C aromatic stretch.  https://orgchemboulder.com/Spectroscopy/irtutor/aromaticsir.shtml  The only thing I can find on Wikipedia is this page of cute animations: https://en.wikipedia.org/wiki/Infrared_spectroscopy#Theory  But they don't really have an aromatic C-C stretch, again, since its locked in a ring, we'd call that "breathing" -- if I understand correctly all C-C bonds stretch at that wave number.

Why are there two separated -- and separated widely -- C-C aromatic peak sets?  Why are there two vastly different environments for the aromatic group?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline wildfyr

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Re: Anhydrides double peak in the infrared
« Reply #8 on: April 10, 2018, 03:21:44 PM »
Arckon, not sure why you are focused on the C=C stretches, OP never mentioned them. Unless its just for your own learning :). I was taught that C=C stretches are roughly 1450, 1500, and 1600 cm^-1. And you can see 1, 2, or all of them in just about any combo, and a pattern for when you see what was never described to me. All the info I can find describes them as "skeletal planar vibrations." Even benzene has all 3 (1460, 1497, 1604).

Hopefully someone with a deeper knowledge of benzene IR spectra can comment on the significance of the # of C=C stretches and the reason for their significant separation.

Offline Corribus

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Re: Anhydrides double peak in the infrared
« Reply #9 on: April 10, 2018, 03:35:13 PM »
(Note - I didn't even see there was a spectrum and molecule shown in the OP until I loaded this on my phone.)

Arckon - the "carbon skeleton" (or breathing, as you call them) stretches of aromatics frequently come in multiplets around 1400-1600 cm-1, which are roughly diagnostic of the substitution pattern. In the most basic arene, benzene, as you can imagine most of the vibrational modes are degenerate, leading to a very simple FTIR spectrum. I believe the C-C stretching region is dominated by a single peak around 1480 or thereabouts. As soon as you break this degeneracy via substitution, some of the stretching modes go to higher frequency, and others to lower frequency. Often they appear as doublets, or doublets of doublets. Or sometimes for highly substituted species with low degrees of substitution the peaks are broad and complex. Moreover the C-C stretching modes can also be strongly coupled to other modes of appropriate symmetry. FTIR spectroscopy is complex for this reason, even for simple molecules, which is why it's better as a supporting diagnostic rather than a primary mode of identification.

You can see how the substitution pattern affects the C-C stretching region for some methyl-substituted benzenes here. (Look at about the third supplementary topic down - "Arene Absorption Features". The spectrum changes when you click on the molecule so you can compare.)

« Last Edit: April 10, 2018, 04:02:03 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline pgk

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Re: Anhydrides double peak in the infrared
« Reply #10 on: April 17, 2018, 11:28:27 AM »
1). Acyl peroxides RCO-OO-COR also have two different carbonyl peaks in their IR spectrum, obviously due to the same reasons.
2). Aromatic substitution in benzene rings can be distinguished by the characteristic peaks appearance between 950-650 cm-1, in comparison with IR spectra that can be found in any IR spectroscopy textbook.   
« Last Edit: April 17, 2018, 02:20:14 PM by pgk »

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