IR Spect:Drawing Spectra - Online Tutor, Practice Problems & Exam Prep

Understanding infrared (IR) spectroscopy involves recognizing the shapes and positions of absorption peaks. Key functional groups include alcohols, which exhibit broad peaks from 3210 to 3610 cm^-1, and amines, which show weaker, sharper peaks around 3310 and 3410 cm^-1. Carbonyl groups, such as ketones and esters, produce strong, sharp absorptions at 1710 and 1750 cm^-1, respectively. Aldehydes present a double peak around 2700 cm^-1. Recognizing these patterns is crucial for identifying organic compounds through IR spectroscopy.

Drawing Hydrocarbons

Now that we understand the frequencies of absorptions a lot better, it's time to move on to the second part of what it means to really understand an absorption, which is the shape. Remember that we need to know, is it going to be broad? Is it going to be sharp? In this course, you may be asked to actually hand draw an IR spectrum from scratch. At the very least, if your professor doesn't ask you to do that, they're going to want you to be able to recognize what the peaks look like on an IR spectrum and be able to understand the different shapes and identify what they are. What we're going to do in this next section is I'm just going to walk you through drawing every single important peak that you need to know and that's going to help you to familiarize yourself with the shapes that are important. Let's start off with the basic molecules which is just hydrocarbons. We're going to start off with alkanes, which is just our number one go-to simplest molecule to draw. I'm going to ask us to forgive my drawing here. These are not meant to be perfect representations but they should give you a pretty good idea of what we're looking for. We said that in a normal alkane, how many different types of bonds are there? Well, there are two. There's a CC bond and there's an sp3CH bond. How many of these do we actually draw? Just the second one. We're only going to draw the sp3CH. Okay? So just so you guys know, sp3CH, we know one thing about them. They result around 29100, But what is their shape? Their shape is called choppy. Now notice that this doesn't follow the normal criteria of first word, second word. I'm not saying sharp. I'm not saying broad. I'm just saying choppy. The reason is that typically, there are so many of these signals of these absorptions in a molecule because think about it; there are hydrogens everywhere, right? They're all going to overlap with each other and make kind of like a clustered mess around 29100. So instead of giving it a specific shape, we just call it choppy. Now you're wondering what is this going to look like? Let's go ahead and start drawing. You might want to watch me do this one first and then just pause the video and copy it. What I'm going to do is I'm really just going to neglect. As you can see, this graph starts my 1500 starts here, so I'm never going to draw anything significant below there. I'm just going to draw like a line. I'm literally just making this part up and I'm not going to do anything until I get to right below 3,000. This is my 3,000 mark, so right before there, I'm going to start to draw something. And what I'm going to draw is a choppy peak. Now when I get to 3,000, I abruptly stop and I continue. Okay? Now notice what I just did there. It's got like a few little edges. It's kind of slanted. That's going to totally change based on the exact IR spectrum. But in general, that's what we're looking for. We're looking for something that's a little bit more than halfway down in terms of transmittance. And it's something that has a jagged edge. You're going to see me draw these a lot during the course.

That was easy. Let's move on to our second situation, which is alkenes. Now notice that alkenes had more that we had to worry about because we still have that CC, which we're not going to draw. But now we had two different peaks that we have to draw. We had the C double bond C, which is here and here. And we also have the sp2CH. So that's in addition to the rest of the hydrocarbon that still has sp3CH. So what I'm trying to say here is that we've got these two extra peaks that we have to worry about that are complicating this more than just a regular alkane. So how do we draw this? These are all going to have their own peaks that you need to know. I'm sorry. I just gave it away. I'm on automatic mode right now. We just talked about how a C double bond C would result in the double bond region. That's going to be 1600. And what is that going to look like? Well, that is going to be weak and sharp. The peak at 1600 is expected to be weak and sharp. Then we would expect both of these to be choppy. And we know their ranges. We talked about how sp3 is 29100 and we also discussed how sp2 is 31100. So basically, we're going to have a lot of choppiness everywhere from between 29 and 3001100. This time, what's going to be very distinct about this graph is that it's going to go past 3,000 and it's going to continue to be choppy after the 3,000 point. Once again, probably better for you to just let me draw this and then you can copy it down. What we're going to do here is once again, blah, blah, blah. No one cares about 1500. But wait, this time when I cross 1500, I immediately have to draw a weak sharp peak. That's going to represent my CC double bond. This is my CC double bond. Okay? So that's my weak sharp peak. Now once again, nothing really happens for a while. I'm bored. But I get to 29100 and I have to start paying attention because now what I'm going to have is I'm going to use different colors. I'm going to use red for the alkane, blue for the alkene. I'm going to have my choppy peak once again. And usually, if this is just an alkane, I would stop right here and that would be it. Don't draw this. Now because there's an alkene component, there are these sp2 hydrogens, that means that some of this choppiness is going to continue past the 3,000 mark. So the chop, the little spikes that happen after 3,000 are for the sp2 component and the spikes before are for the sp3 component. Does that make sense? So suddenly, this graph just got a little bit more complicated than the one that we drew before. Cool so far? Awesome. Now let's move on to terminal alkynes.

Now remember that I stated earlier that it's very important that you only use terminal alkynes here because you must have a hydrogen on it to get that sp peak. Let's look at this molecule. This would be a typical alkyne that you might have to draw and let's identify once again like we've always done, all the different bonds here. Well, we've got once again CC, who cares? I've also got my sp3CH. By the way, heads up, you're always going to be drawing that sp3CH. Get used to it. He's around to stay because remember that alkanes are the backbone of all organic molecules. What are the chances that there's no alkane component? Very little. You're almost always going to draw an sp3CH absorption. Okay? So we're crossing out this one. This one we're definitely going to draw, but what else do we have? Well, now we have a C triple bond C that we're adding and we've also got an sp1CH bond which is the hydrogen that's right here. This just got interesting. We know that the peak at the sp3 is going to be 29100. We know it's going to be choppy. But what do these other guys look like? Well, C C triple bond, if you had to guess where it's going to be, triple bond region, that's going to be 22100. And once again, similar to our alkene, it's going to be weak and sharp. It might be medium. It's somewhere between weak and medium. It really depends on the drawing. But for the purposes of memorizing, I'm totally fine with you drawing it just like you would draw an alkene because it really depends on the exact spectrum. So 22100 weak sharp and then we've got the sp1CH, which because of what I told you guys about hybridization and the wave numbers, this one's actually going to be strong and sharp. Okay? So you might be wondering, Johnny, why is this hydrocarbon, why is this H showing up as sharp when all the other H's that I've been drawing have been choppy? The answer actually lies in the drawing. Notice that the drawing, I only have one. There's only one H. So why would it be choppy? There's only one absorption that's going to happen. It's from that exact hydrogen. So that can't be choppy. It's going to be just one lone sharp peak at 33100. Let's go ahead and give this a whirl. I'm going to start off. I completely ignore 1500. By the way, just so you know, I've been ignoring it but it might as well look like this. You know, there might be stuff going on but the whole point is that I don't care. Okay? So it could have spikes. It could not have spikes. It doesn't really matter. Point is that nothing's going to really happen until I get to 22100. When I get to 22100, I'll give my weak sharp peak. Peak. So that is going to represent my C triple bond C in the triple bond region. From there, I'm going to go ahead and draw my alkane. My C CH, that is sp3 hybridized. As you can see, I pretty much always draw these peaks the same. They're kind of choppy. They're kind of irregular looking. Now in this case, because there's kind of a gap between 2930 300. They're not going to be back to back. So I actually would expect this to kind of descend all the way back down. But then I'm going to have my strong, sharp peak coming from my alkyne. This is my sp1CH. It's choppy. But now this is my strong and sharp sp1CH hydrogen peak. Okay? So notice that there's actually a separation here. Okay? Now what might get even more confusing is if you had a molecule that not only had a triple bond in it, but let's say it also had a double bond in it. So now you had Hs there, then what would happen? Well, then that just means that anything that's sp2 would wind up going in here. This would be your sp2 range. So in that case, there might be a little bit less separation. But we've definitely gone over pretty much all the possibilities that can happen just with these straight hydrocarbon chains. I hope that's making sense so far. We still have a lot of functional groups to go but this is your basis. This is kind of your fundamental groups and now we're going to start learning new functional groups that you're going to add to this. Let's move on to the next video.

Drawing Alcohols and Amines

Now we're going to move on to drawing and recognizing functional groups with hydrogen in them. These are going to be alcohols and amines. The first thing we want to focus on is probably the granddaddy of all absorptions and that's alcohol. Alcohols, as we said, are one of the broadest absorptions there is, and they absorb from 32100 all the way to 36100 in their absorption. So this thing is massive. Remember that I told you that an alcohol almost looks like a parabola. That 'OH' is going to look huge. The official name for this type of absorption is that it's strong and broad. That should tell you strong—it's moving all the way down to the floor and broad—it's very, very wide. Now on top of that, because of the fact that I'm always including an alkene component, what other peak should we always be drawing? We should always be drawing our sp³ CH bonds as well because all of these molecules are going to have those. But notice that I have C-C single bonds which I'm excluding. I also have C-O single bond which I'm excluding as well. I don't have to worry about anything else. I'm just going to draw those 2 absorptions. Let's go ahead. I've got my fingerprint. Who cares? Now I move all the way to 29100 and once again, I'm going to draw my sp³. At this point, you guys should be pretty good at drawing these sp³. Go crazy. You know, I expect different shapes coming up at this point. Okay? You got your choppy sp³ peaks. But what's going to happen? Well, when we get to 3200, it gets real. This alcohol is going to be huge. It's going to do something like this. Boom. Okay? Now in this, spectrum, I actually went past 36100 so I probably drew it a little bit too big but it's not a big deal. The whole point is that I just want you to know that it's a huge massive peak that takes up pretty much the entire 3,000 area and that can actually make it kind of challenging because alcohols are so large in terms of their absorption that they tend to block out things that are behind them. Remember that I told you, for example, what's an absorption that you can think of that happens between 3,203,600? Remember our triple bonds, are 33100 alkyne. So imagine that you have an alcohol, but you also have a terminal alkyne that has a sharp peak around 33100. You might barely see it. In fact, what it might look like is you might have an alcohol that looks like this. It's coming down and then it has this sharp little thing and then it keeps going. Because you could almost think of an IR spectrum as a silhouette. It's almost like a shadow of all the peaks. So you can't see multiple layers or multiple dimensions here. If there truly was a sp hybridized CH along with an alcohol, you might only see a little peak like that popping out, just popping right out of the alcohol peak. My whole point here is that alcohols can make reading the 3,000 region a little bit challenging because they take up so much space and there's really nothing we can do about it. One of the limitations of IR spectroscopy.

Now let's move on to amines. What you'll notice is that I've got primary amines and secondary amines and they're going to result differently. Let's start off with primary amines and then move on to secondaries. Then I'll explain why tertiaries are not on this page. The general rule for amines is that you're going to draw as many peaks, same number of absorptions as hydrogens that you have in your amine. So a primary amine has how many hydrogens? 2. That means that we would expect it to have 2 absorptions. In general, these absorptions are going to be weaker and sharper than alcohol. Okay. So if you guys recall, the range of amines was somewhere between like 33030-3500. That means that amines could completely get covered by an alcohol if it was present. Now thankfully, your class is not going to go into that much complexity where you have to figure out what's behind the alcohol. But just letting you know that there's a lot of overlap between these ranges and really the only way that you could tell the difference is by the shape because it's weaker and it's sharper. Now you might be wondering why does it have these 2 different absorptions. I'm just going to tell you quickly this is beyond the scope of the course. You don't necessarily need to know this. But basically, there's 2 types of there's 2 types of vibrations that are going to be happening with your amine. When we talked about how there's stretching, wagging, rocking and all of those are under the umbrella of vibrations, well, when you have those 2 hydrogens present on the amine, you're going to have a type of vibration called symmetrical stretching and you're going to have a type of vibration called asymmetrical stretching. I'll tell you where I'm going with this in a second. Symmetrical looks like this, that both of them are stretching exactly the same. Asymmetrical is where one is stretching in a different direction than the other. Because you have these 2 different vibrations that are kind of vibrating at different frequencies, one is going to be higher than the other. The symmetrical stretch is going to be right around 33100. The asymmetrical stretch is going to be a little higher. It's going to be around 34100. That's why when you have 2 hydrogens present on the amine, we're going to expect double peaks. That's enough for now. Let's go ahead and draw it and then I'll explain why this is important. We're going to go ahead and do nothing, nothing, nothing, nothing. We get to 29100 and we draw our choppy sp³. So much fun. And then we get to around 33100 and this is where we draw our amine peaks. So we're going to get a weak, weaker, sharper peak around 33100 and then a

Drawing Simple Carbonyls

So now we're going to throw some carbonyl bonds into the mix, and we're going to start off with the easiest scenario which is just learning how to draw simple carbonyls. Now recall that my definition of a simple carbonyl was just a carbonyl that's only going to result in one absorption. It's a carbonyl that you only have to worry about one time. It doesn't come back to haunt you a second time. We're not going to cover all of them here but we've got 2 good examples. We're going to cover ketones and esters. I'm going to explain why they're simple carbonyls. We already know that we should be getting used to the peaks there being our sp₃ CH bonds, right? But we know that we have a C double bond O. Okay. So now remember that C double bond O's are always going to result in the carbonyl region and there are specific ranges that relate to different functional groups within that region. Ketone would count as my R of CORN, of the acronym CORN. So that means that hopefully you guys remember these ranges that this is going to result at 1710. Now carbonyls are probably the second most distinctive or easy to recognize absorption in IR spec because they're extremely strong and sharp. That's going to apply to all carbonyls, not just ketones, not just the R. That's going to write to every single letter of corn. They're always going to be strong and sharp, meaning that I want it to almost hit the bottom of the spectrum and I want it to be very, very sharp. Let's go ahead and draw this guy. I'm going to do nothing up until 1500. Now right when I get to 1700, I'm just going to go for it. I'm going to draw the sharpest thing I can imagine. It goes almost all the way to the bottom and I try to make it as sharp as possible. Then I go ahead and I do nothing until I get to 29100 and I complete my chops. Now this one's just it's getting uglier and uglier but it doesn't really matter because that's not the point. The point is that we have this new peak at 1710. Okay. So there's no other bonds here that I have to worry about, so that's really it. That's going to be the full spectrum for a ketone. Now notice that there isn't a second type of bond I have to worry about. There's no bonds to hydrogen, extra bonds or anything like that. So we're done. Now what's going to be the biggest difference between drawing a ketone and drawing an ester? Well, let's look at the different types of bonds we have in an ester. Once again, we have our sp₃ CH, so we're always going to draw that. Now we also have a carbonyl. Now this carbonyl happens to be an ester which is the O in CORN. So that means would you expect it to be higher than 1710 or lower? Higher. These results, the O's result at 1750 or around 1750. So I'm going to expect 1750. Now has anything changed in the shape? Is it going to be a different looking carbonyl that I'm just like, Oh, this is definitely an ester because of how it looks? Guys, no. Every single carbonyl looks pretty much identical. It's always going to be strong and sharp. That part really hasn't changed. Actually, this looks exactly the same as the one we just drew except it's up by 40 wave numbers. Now if you look at the units that I have on this x-axis, 40 is almost an imperceivable difference with how big I have these units, meaning that I could probably copy and paste the exact spectrum that I drew at the top and it would apply to this one because it's going to look identical. But we do have this extra bond. Notice that in an ester, you have a C O bond that's created. Where does C single bond O result? C single bond O. Guys, this is a fingerprint. Meaning that an ester is a simple carbonyl. It's not going to create a second bond you have to worry about. You don't even draw this. Meaning that the whole point of this exercise is that I want you guys to know that esters and ketones pretty much look exactly the same. We're going to do this. Then instead of waiting till 1710, I'm going to wait till 1750. But as you can see, I mean that is hard to draw. That is hard to figure out. Okay. And then finally I have my sp₃ blah, blah, blah. I drew them both ugly. Perfect. Notice that if I look at this one back to back with this one, the ketone and the ester are almost impossible to differentiate. The biggest difference is that if a problem that your professor gave you said 1750, if it actually stated the number like right here, then you would know, oh, this is an ester. Or if it's stated 1710, then you would think, oh, this is an aldehyde or ketone. Get what I'm saying? The numbers are really important here when it comes to these little tiny differences in wave number. A more specific number is probably necessary for you to be able to tell the difference. Now you guys understand what a simple carbonyl is. Let's move on to more complicated.

Drawing Complex Carbonyls

So now that we know how to draw simple carbonyls, we have to take the next logical step and move on to complex carbonyls. Complex carbonyls, remember, are those carbonyls that have 2 absorptions. Complex carbonyls are going to be those molecules that you wind up kicking yourself about later after the exam because you drew the first absorption and you forgot to draw the second. Or maybe you didn't know what the second absorption was about. That's what we're going to go over now because these are a little bit more complicated. Aldehydes are a great example of this. We have a CO double bond which, in this case, is an aldehyde. So where would I expect that to result? I know you got this. That's the R in corn, aldehydes, and ketones. That's going to be around 1710 again. And we know the shape doesn't change; it's strong and sharp. But notice that now we also have this other bond that you have to worry about, just a second, this other bond that you have to worry about which is CH, but it's specifically an aldehyde CH, specifically aldehyde. This one is tricky. This one has its own set of peaks that we need to know. And actually, it turns out that aldehydes, the CH, happens to have a double peak. What we're going to see is double peak or double absorption at 2700-2800. Okay. Usually, I mean when I told you guys earlier, I just said 27 because that's where it's going to be centered around, but typically it's going to kind of blend into the 2900 peak that you get from alkanes, right? Sometimes the real only thing that you're going to see is this 2700 because the 28 is kind of close to 29. But the biggest point here is that if you see something at 2700, that's very distinctly an aldehyde absorption. There's nothing else that results in that range with that type of peak. Let's go ahead and draw this. We get to 1710 and we draw our very aggressive absorption. That's our carbonyl, so that's going to be our C=O. And then we get over to our aldehyde. Our aldehyde is going to look like a double peak around 2700. So I'm going to draw that like this. I'm going to draw that like a peak here and a peak here. But notice that the second one is going to kind of blend into the choppiness that happens around 3000. Okay? So notice that, I mean in this case, I did a pretty good job of conveying my point which is that sometimes all you're going to see is the 2700 that comes from the aldehyde CH. And it's important that you recognize that sometimes you're going to see a double peak there but sometimes you might just see that one peak because it kind of blends in. Okay? But technically, both of those peaks came due to the aldehyde. And then finally, we have all of our sp³ CH over here. Cool. Makes sense so far? So complex carbonyl, there's a little bit more that we have to worry about. Now let's move on to carboxylic acid. Carboxylic acid is also a complex carbonyl. We've got our sp³ CH. We've got our C=O that in this case would result where? I heard you say it. 1750. It's the O in corn. Good. 1750. But then we've also got this other peak we have to worry about, the OH. Now what did I tell you guys about OH bonds previously? I told you that these are very broad peaks that will be around 3100 to 3600. But be very careful not to call this an alcohol because it's not. This is not an alcohol. This is a carboxylic acid. The OH is going to change its properties when it's next to a carbonyl. This OH is actually going to have a very broadpeak. In fact, I'm even going to put the word here very because this is the most broad peak that there is. And it's going to start around 2500 and sometimes it ends around 3000 which can happen. Sometimes it can even go all the way up to about 3300. This thing is a mess. Kind of imagine that it's like you took the alcohol peak and you just shifted it down and stretched it out, and that's what you're going to get. Now think about it. Do you know any other peaks that result between 2500-3300? Almost everything. A ton of stuff results there. So you can imagine that this absorption is going to block the view of a lot of these other peaks. So we're going to have to think about that. Remember, this is supposed to be drawn like a silhouette, so I'm going to be drawing my carboxylic acid and there's going to be some other stuff sticking out of it. So let's go ahead and start off. I go to 1750 and I draw my carbonyl spike, then when I get to 2500, I start drawing this really incredibly broad peak that kind of looks like this. It's going to just start off. It's not quite as deep as strong as an alcohol, but it definitely is broad. We got this thing going on, but it's not going to end here. Remember that I have these 2900 peaks to draw. Right? I'm going to draw that kind of sticking out. It's going to be kind of like this, and then it's going to come back, and then I'm going to finish off with the end like that. So what you wind up getting is kind of like an overlap of peaks that is very common with carboxylic acid. Sometimes it clouds the view of your choppier alkanes and alkenes and stuff like that, which makes it kind of problematic. That would be another example of a complex carbonyl since I had to worry about my C=O. But I also had to worry about specifically the OH that was involved in the carbonyl. Then once again, I always had the sp³ CH as before. Now you guys got it. You got simple and you got complex carbonyls. Let's move on to the next topic.

Drawing Concealed Functional Groups

This is just a quick little video to address a subset of functional groups that you might not usually think of as related. These functional groups are all going to be what we call concealed functional groups because they don't show up in IR. Let's go ahead and talk about these and why exactly that happens and what they look like. These groups are as follows. We've got alkyl halides, ethers, and tertiary amines. These groups, even though they really seem like they have nothing to do with each other, they actually have something really big in common which is that you're not going to see these in the functional group region. They have no absorptions in the functional group region. Why is that? Well, because remember that in order to have an absorption in the functional group region, you need basically three different things. One of three things. You either need a double bond, or you need a triple bond, or you need some kind of bond to hydrogen.

So now we look at these three and what we wonder is, well, what kind of bonds do they have? What we're going to notice is that they all have CH sp3 bonds. So all of them have that. You really pretty much every single molecule in the world is going to have that. But then besides that, what else do they have? Well, one has CX, another has CO, and the last has CN. What's kind of the common theme here? None of these extra bonds are going to result because these are all fingerprint bonds. These are all bonds that would show up below 1500. That means that when I draw the IR for these molecules, what are they going to look like on IR? These molecules are all going to look like alkanes because the only bond that they have to H happens to be the alkane, not anything else. Nothing else has a bond to H. The only reason I'm going to see these things at all, there's going to be any absorption, has to do with the H's over here. It doesn't have to do with the functional group at all. That means that these three functional groups are going to be imperceivable from alkanes. You're going to need to use other clues to figure out if you have them.

Let's just go ahead and draw this really quick. That means that I would expect that the spectra for these guys is literally going to look just like an alkane and that's it. Nothing else. It might as well it could have been well, I don't want to do that. It could have been cyclobutane or something. It could have just been an alkane. But it happens to be an alkyl halide or it happens to be an ether, you're never going to tell an IR. That's another limitation of IR that IR doesn't do a great job of picking out these different functional groups that don't show up in the functional group region.

Now I mean a quick note that some professors might say that you can use the fingerprint region to tell if you have these molecules or not. The truth is it's complicated. The truth is it's not as straightforward as they might say. Most courses in organic chemistry are not going to cover that. That's why like I said, I'm ignoring the fingerprint region. But for right now, we're just going to assume that you can't see these functional groups on an IR spectrum. I hope that made sense, guys. Let me know if you have any questions. Let's move on to the next topic.

Here’s what students ask on this topic:

What are the key differences between the IR spectra of alcohols and amines?

Alcohols and amines have distinct IR spectra due to differences in their functional groups. Alcohols exhibit a broad, strong absorption peak ranging from 3210 to 3610 cm^-1, often resembling a parabola. This broad peak is due to the O-H stretching vibration. In contrast, primary amines show two weaker, sharper peaks around 3310 and 3410 cm^-1 due to symmetrical and asymmetrical N-H stretching. Secondary amines, having only one N-H bond, display a single peak around 3310 cm^-1. Tertiary amines, lacking N-H bonds, do not show any significant absorption in this region. Recognizing these patterns helps in distinguishing between alcohols and amines in IR spectroscopy.

How do you draw the IR spectrum of a simple ketone?

To draw the IR spectrum of a simple ketone, follow these steps: First, ignore the region below 1500 cm^-1 (the fingerprint region). Next, draw a strong, sharp absorption peak at 1710 cm^-1, representing the C=O stretching vibration. This peak should almost reach the bottom of the spectrum. Finally, include the sp³ C-H stretching vibrations around 2910 cm^-1, which appear as choppy peaks. The resulting spectrum will have a prominent sharp peak at 1710 cm^-1 and choppy peaks around 2910 cm^-1, characteristic of a ketone.

What is the significance of the double peak in the IR spectrum of aldehydes?

The double peak in the IR spectrum of aldehydes is significant because it is a distinctive feature that helps identify the presence of an aldehyde group. This double peak appears around 2700 and 2800 cm^-1 and is due to the C-H stretching vibrations of the aldehyde group. The peak at 2700 cm^-1 is particularly important as it is unique to aldehydes and does not overlap significantly with other functional groups. Recognizing this double peak, along with the strong, sharp C=O peak at 1710 cm^-1, is crucial for identifying aldehydes in IR spectroscopy.

Why do tertiary amines not show up in the functional group region of an IR spectrum?

Tertiary amines do not show up in the functional group region of an IR spectrum because they lack N-H bonds. The functional group region (above 1500 cm^-1) typically requires the presence of double bonds, triple bonds, or bonds to hydrogen to produce significant absorptions. Since tertiary amines only have C-N bonds and no N-H bonds, their absorptions fall in the fingerprint region (below 1500 cm^-1), making them indistinguishable from alkanes in the functional group region. Therefore, tertiary amines are considered 'concealed' functional groups in IR spectroscopy.

How do you differentiate between the IR spectra of ketones and esters?

Ketones and esters can be differentiated in IR spectra primarily by the position of their C=O stretching vibrations. Ketones exhibit a strong, sharp absorption peak at 1710 cm^-1, while esters show a similar strong, sharp peak at a slightly higher frequency, around 1750 cm^-1. Although the shapes of these peaks are nearly identical, the exact position helps distinguish between the two. Additionally, esters have a C-O bond, but its absorption falls in the fingerprint region, making it less useful for differentiation. Therefore, the key difference lies in the precise wave number of the C=O peak.