Infrared Spectroscopy Table - Online Tutor, Practice Problems & Exam Prep

Infrared (IR) spectroscopy involves understanding absorption frequencies and shapes related to various bonds. Key regions include the fingerprint region (below 1500 cm^-1), double bond region (up to 2000 cm^-1), and triple bond region (2210-3100 cm^-1). Notable absorptions include carbonyls around 1700 cm^-1 and specific functional groups like carboxylic acids showing dual absorptions. Conjugation lowers absorption values, while strained rings (banana bonds) increase them. Familiarity with these concepts aids in identifying molecular structures and their functional groups effectively.

Common IR Frequencies

Now we're going to dive right into the nitty gritty of IR spectroscopy. Basically, for every absorption, there are two things you need to know. First, you need to know the frequency or wavenumber at which it's absorbing, and secondly, the shape. What we're going to focus on now is just that first part: knowing exactly which wavenumber is going to correlate with different types of bonds. Before we begin, I need to make a disclaimer. In this section, I will need you to be a little bit flexible with me. I know it's hard to ask because many of you are perfectionists and want to ensure you have all the information before walking into the classroom. However, the problem is that analytical techniques are not taught in a perfectly standardized way. In fact, many professors' explanations of analytical techniques depend on their personal experience, interest in the subject, and teaching preference. I am going to default to the more complex explanation, which covers all the bases, and it's going to be your job to calibrate what I'm saying to what your exact professor wants. If they don't care about some of these values, that's okay, but I'm teaching them just in case. Also, if I happen to tell you that a specific absorption happens at 1750 and your professor says it happens at 1740, it's not the end of the world. We are going for the big picture here, not the exact specifics. I do not want you to get into an argument with your professor about how Johnny on Clutch Prep said it was 1750. We just want to understand the general overarching themes, and you can tailor this to your specific situation. This should work for most of the absorptions you ever need to know. So let's just look at this chart. As you can see, this is our familiar IR spectrum. We have the wavenumber on the x-axis and percent transmittance on the y-axis. But notice, I'm not focusing on shapes here; I don't have any stalactites. What I'm focusing on is memorizing the specific wavenumbers that occur with different types of bonds.

Now, what you can also recall is that this is still split into those regions we talked about. Notice that anything below 1500, I didn't even include because what is that called? The fingerprint region, which we're not going to discuss at Clutch Prep. Then, everything after 1500 follows the trend that we talked about: how we have our double bond region, which goes to about 2000. So you can see that right here. These are our double bonds. And then notice that from 2000 to about 2100, we have our triple bonds. And this is double bonds. Notice that pretty much everything after 2100 and beyond was our bonds to H. I'm just going to put a single bond to H. It doesn't necessarily have to be C; just anything single bonded to H. So we're keeping the same general picture but now we're going to memorize the specific absorptions. So first of all, you can already see a discrepancy if you guys were paying close attention. We're going to start at the very bottom at the lowest frequency; what you see is that I have alkene here. That's going to be the C double bond C in my double bond region resulting at 1600. In a previous lesson, I said 1650. It's a range. It's going to be somewhere between 1600 and 1650. Don't worry too much about it. As long as it's somewhere in that lower range, you know you have an alkene.

Now we get to this big section called the carbonyl region. There's a chance that your professor just wants you to know that carbonyls are around 1700, and that's it. That might actually be about half of you guys. The other half might need to know the exact types of carbonyls and all of their absorptions. And for that, I devised a little mnemonic, a little memory tool that you're not going to see anywhere else called "corn." Our favorite vegetable, this section is going to be "corn" because corn helps us remember the order in which these different carbonyls will present themselves. What are the different categories? Well, first we have acid chlorides, which result at the highest wavenumber of 1790. Then we have carboxylic acids and esters, resulting at 1750. Then we have aldehydes and ketones at 1710. Finally, we have amides at 1680. Where did I get "corn" from? Well, if you think about what is attached to the carbonyl in all of these; for acid chloride, I have a Cl. For carboxylic acid or ester, I have an O. For aldehydes and ketones, I have an R or an H. And then for amides, you have an N. Getting rid of that stupid little 'L,' I get the acronym CORN. That may help you guys memorize the order in which they are presented.

In general, conjugation, which is the ability of any of these carbonyls to resonate with something else, will lower the number by 20. For example, here I have a conjugated ketone. Notice that this conjugated ketone is resulting at 1690. What's the normal absorption of a ketone? 1710. Anytime you have conjugation present, it's going to lower your predicted value by about 20. You could apply that to the other ones as well, so a conjugated acid chloride would result at 1770. Awesome. Okay. So you guys are getting this. So another thing, this is getting even weirder: what the heck is a banana bond? Banana bonds happen in small strained rings. For example, cyclobutane has banana bonds. It just means that you have overlapping bonds, overlapping orbitals that have to shape kind of like bananas in order for them to overlap. Banana bonds, because of their spatial arrangement, are highly strained, and they're going to tend to result about 100 higher. You can imagine that a ketone that has highly strained bonds in it is going to result about 100 higher. So instead of 1710, it's around 1810. Now that's just a general rule. Your professor might have a slightly different number. But that's just something for you to remember: if it has these highly strained rings, that means like a 3-membered ring or 4-membered ring.

Major IR absorptions

Let's see if you got this first one right. For this molecule, how many different types of bonds were you able to spot? It turns out there's more than one type of bond here. There are actually 2 types of bonds. Let's go ahead and see. The first one that I'm looking at here is a CC single bond. That would be all of the bonds that are making up the alkane. The second kind of bond that I could spot is a CH bond because there are hydrogens everywhere, right? But we specifically have to include what the hybridization of that CH is because remember that the peak is going to change depending on what type of hybridization it is. So being an alkane, these would all be our sp³ CH bonds. Alright. So now I know what you're thinking. You're saying, Johnny, but I thought the CC bond didn't matter. And that's exactly right. You got it right. It turns out that this first bond, I'm just going to cross out because it doesn't matter. This is a single bond that's not to H. So where would it result? You got it. Fingerprints. For right now, I still want you to identify the fingerprint region bonds but don't worry about memorizing anything with them. I just want you to identify it and say that's irrelevant. Okay? So exactly, we did that right. This doesn't matter. I'm just going to cross it out like we did. The only one that really matters is this one. So, we've got one absorption and where would that absorption be? That would be around 29100. If you got that wave number, if you got that one peak, you're good. If you forgot to identify the single bond, that's okay. Don't worry about it. But for this next one, as we keep moving to the new examples, try to identify every single bond that there is and then throw out the ones that are in the fingerprint region so you can just focus on the functional group region bonds. All right. So move on to the second question.

Major IR absorptions

So how many different types of bonds did you guys find for this one? Well, it's more than 2. I'll tell you that. Let's go ahead and start off with the ones that don't matter. These are going to be single bonds to stuff that isn't H. So that would be, 1. I've got a CC single bond fingerprint region again. 2, I've actually got a C single bond O. That's also relevant because that's a single bond that's not a hydrogen. So I can immediately cross both of these out. Okay. Do we have anything else? Yes. Well, we have 3. We've got our CH bond again. Okay. And once again, this entire thing is an alkane, so it's going to be sp³ CH. Okay. And we do have a last fourth one which is OH. Does that one count? Yes. Because that is a single bond to hydrogen. So that means here you have to identify the wave numbers of 2 different bonds. So for our sp³ CH, we know that the value is again 2900. And what would be the value of the OH? You guys remember? It's actually a really big range. You can't just say one number. You can never say just one number for a broad peak because a broad peak by definition spans multiple wave numbers. The range you guys need to learn for alcohol is 3200 to 3600. Okay. So we're just always going to say 32 to 36 because it is such a broad signal. All right. So I think you guys are getting the hang of this. Let's move on to the next question.

Major IR absorptions

So here's another one. I hope that you guys were able to find 4 different types of bonds. Let's first of all talk about, I guess, the one that doesn't matter, which is C single bond C. We know that we see that around. But there are a few ones here that do matter. So for example, a new one that we had in this question is a C double bond C. Now remember that that gets its own signal. That's a double bond, so that would be in the double bond region. So that's something to hold onto. We'll do values in a second.

There was our sp3CH. That would be the entire alkane portion of this molecule. What's the 4th peak? Good job. So the 4th peak is actually going to be sp2CH. Why? Because there are 2 hydrogens on this molecule that are located on an alkene. Those hydrogens are going to have a higher wave number than the rest of the surrounding hydrogens. Why? Because they have a different hybridization. Now we have to talk about how many signals or how many different absorption frequencies do I need to state here? 3. We're not going to worry about the fingerprint. But what's the wave number that we have to memorize for an alkene? C C double bond, that would be 1600. We're just going to go with the value first of 16 as one stated value. It's a pretty sharp peak, so that's fine. For sp3CH, 2900. And lastly, for sp2CH. This should be easy if you remember that I just said it's going increase by 200 for each hybridization, so that means the sp2 would be 3,100. Got it? Awesome. So that was a little bit more tricky, but you guys got this. Let's move on to the last question.

Major IR absorptions

On this question, you might notice an added complication because I threw in what we called earlier a complex carbonyl. Remember I told you guys how complex carbonyls are the ones that are not just going to result in one area, they're going to result in 2. And we need to be able to identify that. First of all, let's go with the easy ones here. We can get rid of our C-C single bond. We know that there's going to be an sp³ CH bond. That would be just quickly identifying it. That would be like a hydrogen right there. That would be sp³ c h. But now we get to the part of the complex carbonyl. So 3, let's just talk about first of all the C double bond O. Where is that going to result? Now keep in mind, C double bond O is a carbonyl, so that means that it could result remember we have a big range for our carbonyl region. I said it was something like 1680 all the way up to 1840, so pretty big range there. But specifically because this is an aldehyde, because of corn, where would you expect to find it? Remember that an aldehyde would count as the R in corn because it could either be aldehyde or ketone R or H. That would result at 1710. Okay. So we've got the frequency for the aldehyde, but what else do we have to worry about? The CH bond because it's a complex carbonyl, so we're also going to have a bond to H to worry about. Specifically, the aldehyde H gets its own unique peak and that's going to be at 2700.

Hopefully, I'm not in the way there. You can see what I'm writing here. We've got the complex carbonyl with 2 different peaks. Just as a pop quiz, what's the other complex carbonyl that we're going to worry about in this course? Carboxylic acid because carboxylic acid also has a peak in the carbonyl region and has a very broad peak in the hydrogen region. So guys, that's it for this practice problem. I think you guys got the picture of what you need to do now. You're doing great. Let's move on to the next topic.

Answer each of the following questions based on the images below.

a) Which compounds show an intense peak ~ 1700 cm^-1? 

b) Which compound shows an intense, broad peak at ~ 3400 cm^-1? 

c) Which compound has a peak at ~1700 cm^-1, but no peaks at 2700 cm^-1? 

d) Which compound has no signal beyond the fingerprint region?

Hey guys, let's take a look at the following practice question. Here it says, "Answer each of the following questions based on the images below." Here, I'm giving us 5 images. We have an ether for example A. We have an ester for B. We have an alcohol for C, an aldehyde for D. And finally, we have what looks like an ether except I replaced all the hydrogens with fluorines. The first question says, which compounds show an intense peak at approximately 1700 cm-1? Remember, if we see a strong, intense peak around 1700, that means that we must have a carbonyl. Basically, we're looking for compounds that have carbonyls in them. If we take a look, the only 2 that fit that profile are B and D. They both have carbonyls within them. Next, which compound shows an intense broad peak at approximately 3,400 cm-1? Remember, broad is indicative of an alcohol. Alcohols are between around 3,200 cm-1 to about 3,650 cm-1. We're going to say here, this would be an alcohol, so this would be C. Next, we say which compound has a peak at around 1700 cm-1 but no peaks at 2,700 cm-1. So we know that the 1700 cm-1 portion is talking about the carbonyl. So that means the answer is going to be either B or D. Now, what's different about B and D? Since we're talking about the carbonyl signal, we have to look to see what's connected to the carbonyl. Here we have CH3 and an O. What we need to realize here is that in compound D, this is an aldehyde. Aldehydes are special because they have 2 types of signals from the carbonyl. The carbonyl signal itself would give us a signal around 1700 cm-1. But aldehydes are basically the only example where we have a carbonyl directly connected to a hydrogen. Here, this is a CH bond. This CH bond in the carbonyl gives us a signal around 2700 to around 2900 cm-1. What did the question say? The question says, "Which compound has a peak around 1700 but no peak around 2,700?" Basically, it's not an aldehyde because there is no signal around 2700. That means we're talking about the ester, not the aldehyde. The answer here would be B. Then finally, D says, which compound has no signal beyond the fingerprint region? That means all the signals will be below 1500 cm-1. So here, the answer would have to be E. Because here, we have no CH bonds, sp3, CH bonds. So there's no signal around 2850 to 2960 cm-1. There's no carbonyl. There's no alcohol. So, we're not going to get any of those signals. So, for the last question, the answer would have to be E. E would show us no signal above the fingerprint region. So all these signals will be below 1500 cm-1. We're going to say here, when it comes to a carbon halogen bond, we say usually the signal is around 700 cm-1, which is well within the fingerprint region.

Identify which carbonyl group will exhibit a signal at a lower wavenumber

• A. A
• B. B

Here’s what students ask on this topic:

What is the significance of the fingerprint region in IR spectroscopy?

The fingerprint region in IR spectroscopy, located below 1500 cm^-1, is significant because it contains a complex set of absorption bands unique to each molecule. This region is highly specific to individual compounds, much like a human fingerprint, making it invaluable for molecular identification. While the bands in this region are often difficult to assign to specific functional groups, they provide a unique spectral pattern that can be used to confirm the identity of a substance when compared to known spectra.

How do conjugation and strain affect IR absorption frequencies?

Conjugation and strain significantly impact IR absorption frequencies. Conjugation, which involves the resonance of π-electrons across adjacent double bonds or between a double bond and a lone pair, typically lowers the absorption frequency by about 20 cm^-1. This is because conjugation stabilizes the molecule, reducing the energy required for bond vibrations. On the other hand, strain, such as in small ring structures with banana bonds, increases the absorption frequency by approximately 100 cm^-1 due to the higher energy required to vibrate the strained bonds.

What are the characteristic IR absorption frequencies for carbonyl compounds?

Carbonyl compounds have characteristic IR absorption frequencies around 1700 cm^-1. Specific types of carbonyls absorb at slightly different frequencies: acid chlorides at 1790 cm^-1, carboxylic acids and esters at 1750 cm^-1, aldehydes and ketones at 1710 cm^-1, and amides at 1680 cm^-1. Conjugation with a double bond or aromatic ring can lower these frequencies by about 20 cm^-1, while strain in small rings can increase them by about 100 cm^-1.

How can you differentiate between an alcohol and an amine in an IR spectrum?

Alcohols and amines can be differentiated in an IR spectrum by their absorption shapes and frequencies. Alcohols typically show a broad, smooth absorption band around 3200-3600 cm^-1 due to O-H stretching. This broadness is due to hydrogen bonding. Amines, on the other hand, exhibit sharper, less intense peaks in the same region (3300-3500 cm^-1) due to N-H stretching. Primary amines show two peaks, while secondary amines show one peak. The distinct shapes and number of peaks help in distinguishing between these two functional groups.

What are the typical IR absorption ranges for alkanes, alkenes, and alkynes?

In IR spectroscopy, alkanes, alkenes, and alkynes have distinct absorption ranges. Alkanes (sp³ C-H) absorb around 2850-2960 cm^-1. Alkenes (sp² C-H) absorb slightly higher, around 3020-3100 cm^-1. Alkynes (sp C-H) have the highest absorption range, typically around 3300 cm^-1. Additionally, the C≡C triple bond in alkynes absorbs between 2100-2260 cm^-1. These ranges help in identifying the type of hydrocarbon present in a sample.