- 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
IR spectra practice
Practice with identifying the compound that corresponds to an IR spectrum. Created by Jay.
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- Why don't amines establish hydrogen bonding, like the OH, and therefore have a broad signal as well?(18 votes)
- He mentions at1:40that if it was the amine, then there would be two distinct signals. This is due to the symmetric stretching and asymmetric stretching of the N-H bonds.(11 votes)
- What is the difference between an unconjugated and conjugated ketone?(4 votes)
- Conjugated means that there are p-orbitals that can interact with each other. This means that they can participate in resonance, usually making the molecule more stable and decreasing the individual bond strength. That's why we get the shift in the IR signal.(10 votes)
- For the second IR spectrum, cyclohexane is symmetric. doesn't this mean that there is no dipole and there should not be a c=c signal in IR spectrum?(4 votes)
- A vibrational mode involves the whole molecule, although it tends to be localized mostly on a functional group.
The C=C bond is symmetrical, but the rest of the molecule is attached to it, and the rest of the molecule is three-dimensional.
So there is usually a small dipole change during the vibration and a correspondingly weak but detectable IR signal.(7 votes)
- I understand how we used the presence of resonance in the conjugated ketone to conclude that the molecule we're looking at is the unconjugated ketone. But I would like to know if there would be any marked difference between the spectra of the conjugated and unconjugated ketones in the C-H region as well? Does that area of the spectrum give us useful info in this case too?(4 votes)
- There are some slight differences due to the fact that there are C-H bonds at different lengths from the carbonyl group and carbon hybridization that would differentiate an unconjugated and conjugated ketone from eachother, but the differences are subtle and may not appear all that great in the spectra. Thats why the peaks at the carbonyl and double bond is more useful because they have great peaks that point them out.
If you have done magnetic spectra before, you know that all H that are equivilent show up at the exact same point. The same is kinda true for IR except they tend to act like lone wolves and can get lost in the background noise so they are not all that dependable.(4 votes)
- From3:30~4:30, why does C=O bond have a higher signal intensity than C=C bond?
Also, is it true that the more of a particular bond (e.g. C-H bonds) a molecule has the greater the signal intensity of that bond? That is what I learned from Questions and Answers section under "Symmetric and asymmetric stretching" video.(2 votes)
- The C=O bond has a greater change of dipole moment during te stretch than the C=C bond does. You need a change in dipole moment for IR absorption to occur.
The more bonds of a given type, the greater the intensity of the absorption.(3 votes)
- For the last spectrum, would another clue be that there is a small, isolated peak above 3000 cm-? Would this peak be a result of the isolated sp3 C-H's to the RHS of the carbonyl?(1 vote)
- I would say it belongs to the sp2 hybridized C-H of the double bond, which is slightly higher in energy (or wavenumbers) than sp3 hybridized C-H bonds, like in the second example/spectrum.(2 votes)
- I wonder that ㅡ三ㅡ -> 2-butyne has no triple bond signal because it is symmetric?(1 vote)
- There must be a change in dipole moment during a vibration.
Since the stretching vibration does not change the dipole moment, it does not generate an infrared signal.(2 votes)
- In the last spectrum, I wonder why two peaks at ~3100 cm-1 and 2900 - 2800 cm-1 have the very small intensity. I expect that those peaks belong to C = C bond and C(sp3) - H but it's too small, compared to the other spectrum (such as the first and the second in the video). Why is this happen and does it relate to the structure of the ketone ?(1 vote)
- benzal aceton which one has more carbonyl vibration cis or trans form
or explain it by IR(1 vote)
- In the 3rd spectrum:
(#1) What are the peaks at 2900 cm-1 and 3050 cm-1? (I assume =C-H and -C-H, respectively.)
(#2) How would the peaks for =C-H and -C-H in the second resonance molecule differ?(1 vote)
- 1.you are correct, each H that is different and a different length from the C=O will show up as a peak
2. you would see 4 spikes like the 3 above, they may be smashed together in a broad peak from 2900-3100cm-1 so you may or may not be able to tell there are 4 peaks. Hydrogen can be pretty wild in IR spectra.(1 vote)
- [Voiceover] Let's look at some practice IR spectra, so here we have three molecules, a carboxylic acid, an alcohol, and an amine, and below there's an IR spectrum of one of these molecules. So let's figure out which molecule has this IR spectrum. So we could draw a line around 1,500 and ignore the stuff to the right and focus in on the diagnostic region. And here is your double bond region, and I don't see a signal at all in the double bond region. I certainly don't see a very strong carbonyl stretch, and so the carboxylic acid is out, so I don't so any kind of carbonyl stretch in here. We do see some signals over here to the left in the bond to hydrogen region. So I could draw a line about 3,000 and I know below that, we're talking about a carbon hydrogen bond stretch where you have an Sp3 hybridized carbon. That doesn't help us out here at all, but this other signal does, right? So we have another signal, centered on a higher wave number. And it's extremely broad, so whenever you see that you should think to yourself hydrogen bonding, and this is due to an O-H bond stretch. So immediately we know that we must be talking about an alcohol here. A carboxylic acid has a similar O-H bond stretch so it has a broad signal due to that, but there's no carbonyl so it couldn't possibly be this molecule. It also couldn't possibly be the amine, because even though we have nitrogen hydrogen bonds, a nitrogen hydrogen bond stretch is going to be in a similar region. We would expect two signals for this. We would expect a symmetric stretch signal and an asymmetric stretching signal, and it wouldn't be as broad as what we're talking about here for the alcohol, so it's definitely not the amine, so this spectrum is the alcohol. Let's look at three more molecules in a different spectrum. So let's look at the spectrum here. We start with 1,500, so we draw a line here. We look in the double bond region. Here's our double bond region. I do see a signal this time. And it doesn't look like it's a very strong signal, either. Let's see what the location of this signal is, so I drop down and the signal shows up between 1,600 and 1,700, so we'll say approximately 1,650, and that's not very strong. Both of those things, location, right, and the fact that it's not a very strong signal clue me in to the fact that this is probably a carbon carbon double bond stretch, that's what this is talking about here. So a carbonyl, we would expect that to be just past 1,700 and also much, much stronger. So we can rule out this molecule over here because I don't see any kind of a carbonyl stretch. Alright, so let's look in the triple bond region. So somewhere in here, I don't see any kind of a signal. So it couldn't possibly be this molecule. So we must be talking about cyclohexane here and if we look over in the bond to hydrogen region, and we draw a line, we can see that this signal just higher than 3,000, this must be talking about our carbon hydrogen bond stretch, where the carbon is Sp2 hybridized, so this is, of course, talking about our carbon hydrogen stretch where we're talking about an Sp3 hybridized carbon. And so cyclohexane is the only thing that makes sense with this IR spectrum. Let's do one more, so we have three molecules and an IR spectrum. Alright, so let's start analyzing. Draw our line around 1,500 right here, focus in to the left of that line, and this is our double bond region, so two signals, two clear signals in the double bond region. So let's look at this signal right here, so it's not as intense as the other one and it's pretty much between 1,600 and 1,700. So both those factors make me think carbon carbon double bond stretch. This is probably a carbon carbon double bond stretch here. The signal next to it, if this is 1,600, this is 1,700 so this signal is just past 1,700 and it's very strong, it's a very strong signal, so that makes me think carbonyl. So this makes me think carbonyl right here. So we can immediately rule out this one, right? So it couldn't possibly be that molecule and that brings us to this which is a conjugated ketone versus an un-conjugated ketone. So let's think about the un-conjugated ketone for a minute. So this carbonyl stretch, we talked about in an earlier video, we'd expect to find that somewhere around 1,715, so past 1,700. This ketone over here, this conjugated ketone, we have resonance, and we know what resonance does to the carbonyl, so it decreases the strength of the carbonyl, therefore it decreases the force constant k, that decreases the frequency of vibration and we would expect this carbonyl signal to have a lower wave number than 1,715, actually it moves it under 1,700, to somewhere around 1,680 is where we'd expect it to be. I don't know exactly where it is, but it's definitely less than 1,700. This is very clearly, let me go ahead and mark this here. This is very clearly the 1,700 line and our signal is past that, so this must be talking about the unconjugated ketone over here on the right, and so this spectrum corresponds to this molecule. So hopefully that gives you a little bit of insight into how to approach some simple IR spectra.