IR spectra practice

- [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.