- Introduction to infrared spectroscopy
- Bonds as springs
- Signal characteristics - wavenumber
- IR spectra for hydrocarbons
- Signal characteristics - intensity
- Signal characteristics - shape
- Symmetric and asymmetric stretching
- IR signals for carbonyl compounds
- IR spectra practice
How hydrogen bonding affects the shape of an IR signal (causing it to become broader and less sharp). Created by Jay.
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- Why is the wavenumber for an sp3 oxygen-hydrogen bond much larger than an sp3 carbon-hydrogen bond when an oxygen nucleus has a higher mass than a carbon nucleus and intuitively would result in a smaller wavenumber?(8 votes)
- Well, that's because Oxygen is more electronegative than carbon, hence an O-H bond is stronger than a C-H bond. For example, a =C-H is stronger than a -C-H bond because the carbon in the former group is sp2 hybridized which is more electronegative than the sp3 hybridized C atom in the latter group.(5 votes)
- At approximately3:17he states the peak at 1100cm^-1 corresponds to a C-O single bond. However, in prior videos he described the area below 1500 as the "fingerprint" area, and the area above 1500 as the diagnostics area. Can you explain the significance of this?(3 votes)
- Some peaks are so strong and so characteristic that you can identify them even they are in the fingerprint region.(3 votes)
- In the first two examples, it looks like the intensity of the O-H bond is less than the intensity of the C-H bond. But doesn't O-H have a larger dipole moment than C-H? Why is it that the C-H bond shows a higher intensity?(2 votes)
- What if you had the O-H bond, but also a sp2 C-H bond, which is in the >3000 wavelength?(1 vote)
- The C-H bond is not that important since most molecules will have it. Therefore, the O-H bond is far more significant in determining the molecular structure of a molecule.(4 votes)
- Can't NH2 also form hydrogen bond? Why it's shape is not as broad as OH(2 votes)
- as mentioned in the previous videos,signal strength depends on dipole moment. NH2 is obviously less polar than OH bond.(2 votes)
- Are steric hindrance & the molecule's ability to form H-bonds the only factors that change the peak shape at ~3600-3500cm-1 for -OH?(1 vote)
- In these examples yes, the steric hindrance stops the H-bonds from occurring. The H-bonding would cause there to be a lot of variance in the strength of the O-H bond of the and because the wavelength that is absorbed is partially dependent on the strength of the bond, this would cause there to a broad amount of wavelengths absorbed, thus the broad peak around 3500.
Because the steric hindrance stops the H-bonding, all the O-H bonds are very close in strength so they just absorb a very narrow wavelength, thus the sharp peak around 3500.(3 votes)
- I saw a similar question and answer but I still don't get why the OH isn't always creating a broad peak? I guess I'm not sure I understand the difference between (a) the existing two OH bonds and (b) the bond that doesn't form between two OH groups... Can someone shed some light?(1 vote)
- If the solution is very dilute, there is little chance of two OH-containing molecules coming together and forming H-bonds.
Then you get a sharp O-H stretching peak.(2 votes)
- How to find the difference between the signal of an alcohol having an SP2 hybridized C-H bond with the signal of a carboxylic acid at wave number 1700 (which signifies a double bond )? Wouldn't both the compounds show a similar spectroscopy?(1 vote)
- Are you asking about distinguishing a molecule that is both an alcohol and an alkene from a carboxylic acid?
If so, the OH signal will broader and at a lower wavenumber in the carboxylic acid and there will also be a strong carbonyl signal somewhere between 1710 and 1780.
In the "ene-ol" there may be a signal from the C=C bond but it will be below 1680.
This table should be helpful:
- At around 10 min, he starts talking about molecules either stretching in phase (lower in E) or out of phase (higher in E), my question is how come there isn't a "broad" band here instead of two separate, but close, bands? Because why would there only be in phase or out of phase stretching? Wouldn't there be all degrees of intermediates between the two? Thank you in advance for the clarification.(1 vote)
- Yes, the vibrations are all happening at once.
But they are all just oscillations of vector forces, and we can break all these vectors down into "fundamental" vibrations that match the symmetry of the molecule.
These are the "symmetric" and "asymmetric" vibrations that you see in the spectrum.(1 vote)
- Why can C-C or C=C bonds appear on the IR spectroscopy graph? Isn't there zero difference in electronegativity between bonds of the same atom and hence zero dipole moment?(1 vote)
- Why does there need to be a dipole to show up on IR? I think you’ve confused it with microwave spectroscopy.(1 vote)
- The shape of the signal can also be important, especially when you're talking about alcohols. So here, I have a generic alcohol. And we're concerned about the bond between the oxygen and the hydrogen. We know oxygen is more electronegative, so this oxygen here gets a partial negative, and this hydrogen here gets a partial positive. Same thing for this alcohol over here. That sets up the opportunity for hydrogen bonding. So this partially positive hydrogen can be attracted to this partially negative oxygen here. And this attractive force-- if this hydrogen is being attracted, that's going to weaken this oxygen-hydrogen bond. So hydrogen bonding weakens the oxygen-hydrogen bond, and we know if we're weakening the strength of a bond, that's like decreasing the force constant, or decreasing the spring constant. We saw in an earlier video, if you decrease k, you're going to decrease the frequency, or decrease the wavenumber, and so the signal is going to change on your IR spectrum. At any moment in time, different alcohol molecules are going to have different amounts of hydrogen bonding. Some molecules might have a little bit of hydrogen bonding, so k decreases a little bit, and the wavenumber decreases a little bit. But other molecules might have a lot of hydrogen bonding, and so we can decrease k even more, therefore we are going to decrease the wavenumber even more. You get a range of wavenumbers, and since you get a range of wavenumbers for the OH bond, when hydrogen bonding is present, you get a very broad signal on your IR spectrum. So, if we go over here in this region, so we're talking about the IR spectrum for 1-hexanol, this is the region for bonds to hydrogen. So we draw a line at 3000, and we know that just below 3000, we're talking about a carbon-hydrogen bond stretch, where the carbon is sp3-hybridized. But, this over here, this very broad signal right here, this is due to the OH. So let me go ahead and highlight that. This bond right here, this oxygen-hydrogen bond, gives us a very broad signal on our IR spectrum because of hydrogen bonding. So we get this very broad signal because of the different wavenumbers. And usually you're going to see this somewhere around 3500 to 2900. So if I find this is 31, 32, 33, 34, 35... so usually in this range, maybe even a little bit higher than that, you're going to find this very broad signal. In this case, the oxygen-hydrogen bond. And so you know immediately to think about the possibility of an alcohol functional group in your molecule. Also, we can draw a line at 1500 here, and this signal actually, so somewhere around 1100 wavenumbers, this is actually the carbon-oxygen single bond. Let me go ahead and highlight that. So we have a carbon-oxygen single bond, and this is the single bond region in here. And that's where-- that's this stretch right here. Not always going to be super useful to you, but it's just thinking about what we talked about in the earlier video, I think we calculated the approximate wavenumber for a carbon-oxygen single bond. So that's what the typical spectrum for an alcohol is going to look like. Look for that broad signal there. Alright, let's compare this alcohol to another one here. So, this molecule is butylated hydroxytoluene, or BHT, and I drew two BHT molecules in there for a reason. Let's think about why. So, you might think at first, "OK, I have another opportunity for hydrogen bonding." So, here's an opportunity for hydrogen bonding, so we're going to get a broad signal for this OH bond. So I'm going to highlight it here. I might expect, since I have hydrogen bonding, that weakens this OH bond. So I might get another broad signal for my OH. But, in this case, we have so much steric hindrance from these tert-butyl groups, so there's tons of steric hindrance here. And then we have these big ones over here too, and that's going to prevent the hydrogen bonding from taking place. And so, because of steric hindrance, these molecules can't get close enough to each other for hydrogen bonding to occur. So, we don't get any hydrogen bonding. If we go over here to the IR spectrum for BHT, as usual, it helps to draw a line around 3000 or so, and then, we don't see this broad signal. So this broad signal up here is missing down here. But what we do have is a sharp signal. So let me go ahead and highlight that here. So we have a sharp signal right about-- let's drop down and see where we are for wavenumbers, so this would be 31, 32, 33, 34, 35, 36. And so somewhere around 3600, we see this sharp signal, and this is that OH bond. So this is that OH bond, and we don't see a broad signal because we don't have hydrogen bonding to worry about. And so we don't see that broad shape. We see a sharp signal for the OH. So this allows you to think about where you would find this signal here. So if you have an oxygen-hydrogen bond with no hydrogen bonding, you expect to see it around 3600. If you have an oxygen-hydrogen bond stretch, and there is hydrogen bonding, look for this broad signal here over this large area. Alright, let's do one more molecule where we're thinking about hydrogen bonding. And, this time we're talking about a carboxylic acid. So here's our carboxylic acid over here. And let's analyze the IR spectrum for this carboxylic acid. So we see this OH here, so we think to ourselves, "Ah, hydrogen bonding can occur." So, where would that signal be? Before, I said it would be somewhere around 3500 to 2900, somewhere in that range, so if we look, we see an ever broader signal here, even broader than the range we talked about before. And that's because carboxylic acids have more hydrogen bonding. So, if I go down here, let me show you-- here's some hydrogen bonding for a carboxylic acid. So, we have opportunity for a hydrogen bond here, and we have an opportunity for hydrogen bonding here as well. So, a large amount of hydrogen bonding makes the signal even broader, when you're talking about the OH on a carboxylic acid. So this very broad signal is talking about this oxygen-hydrogen bond stretch. And once again, if we draw a line at 3000, so if we draw a line right here, we can see this little signal right in here, and that's actually the carbon-hydrogen stretch. With the carbon, we're talking about an sp3-hybridized carbon here. So, the broad signal is often centred around 3000, so that partially obscures that carbon-hydrogen bond stretch that we've talked about before. So, there's another hint that you're talking about a carboxylic acid. But of course, the biggest hint is when you also see the very strong signal for the carbonyl, occurring somewhere around 1700. So, let's go ahead and identify that. So, we draw a line right here, and then in the double bond region, we see this really intense signal right here, approximately around 1700. So, usually, I think it's a tiny bit higher than that. But, this is due to our carbon-oxygen double bond stretch. So, we're talking about our carbonyl here. So if you see this really broad signal, which tells you OH, and then this really strong signal, which tells you carbonyl, put an OH and a carbonyl together, and you get a carboxylic acid. And so it's pretty easy to identify the carboxylic acid using an IR spectrum here. And just a quick note about hydrogen bonding, we talked about hydrogen bonding weakening a carbon-oxygen bond above, a similar thing happens here. So let me go ahead and highlight that. So, if we have hydrogen bonding right in here, that's going to weaken our carbonyl, so that's going to decrease the double bond character a little bit. So that's actually going to change where we find the signal. So it actually changes the signal, if it's weakening the bond a little bit, you're going to decrease the wavenumber, and so this carbonyl is a tiny bit lower, in terms of wavenumber, where you find the signal, than what you would expect. I'll briefly mention that in a later video, when we talk about carbonyl chemistry and more about IR spectra. So, for the shape of the signal, remember broad, think hydrogen bonding if it's a broad signal.