IR Spect: Frequencies

Concept: Concept: Common IR Frequencies

Video Transcript

Now we're going to dive right into the nitty-gritty of IR spectroscopy. Basically, for every absorption, there's two things you need to know. You need to know first of all what's the frequency or wave number it's absorbing at and two, what the shape is. What we're going to focus on right now is just that first part of knowing exactly which wave number is going to correlate with different types of bonds.
Before we begin, you already see something pretty complicated in front of you. I need to make a disclaimer for you guys, which is this, on this section I'm going to need to ask you guys something which is I'm going to need for you to be a little bit flexible with me. I know that that's hard to ask because a lot of you guys out there are super perfectionists and you want to make sure that you have all the information even before you walk into the classroom. I think that's who you are.
Anyway, the problem with that is that analytical techniques are not taught in a perfectly standardized way. In fact, many professors explanations of analytical techniques will depend based on their personal experience, based on their personal interest in the subject, and based on their teaching preference.
What I'm going to do is I'm going to default to the more complex explanation, the one that covers really all of the bases and it's going to be your job to calibrate what I'm saying to what your exact professor wants. If he doesn't care about some of these values, that's okay, but I'm going to teach 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're going for big picture here. We're not going for the exact specific. I do not want you to get into an argument with your professor about how Johnny on Clutch Prep said that it was 1750. We just want to know the general overarching themes and you can tailor this to your specific situation. That being said, this should really work for most of the absorptions you ever need to know.
Let's go ahead and just look at this chart. As you can see, this is our familiar IR spectrum. We have wave number on the x-axis. We have percent transmittance on the y-axis. But notice that I'm not focusing on shapes here. I don't have any stalactites. What I'm focusing on is the actual numbers and memorizing the specific wave numbers that happen with different types of bonds.
What you also can recall is that this is still split into those regions that we talked about. Notice that anything below 1,500, 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 1,500 follows the trend that we were talking about, how we have our double bond region, which goes to about 2,000, so you can see that right here, these are our double bonds. Then notice that from 2,000 to about 2,500, we have our triple bonds and this was double bonds. Notice that pretty much everything after 2,500 and beyond was our bonds to H. I'm just going to put a single bond H. It doesn't necessarily have to be C, just anything single bonded to H.
We're keeping the same general picture, but now we're going to memorize the specific absorptions. 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 in 1,600. On 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 1,700 and that's it. But actually, might be about half of you guys might be that lucky. The other half might need to know the exact types of carbonyls and all of their absorptions. 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 to remember the order in which these different carbonyls will present themselves.
Let's just talk about this. What are the different categories? First we have acid chlorides. We have acid chlorides and those are resulting at the highest wave member of 1790. Then we have carboxylic acid. Let me just get a different color there. And esters, which are resulting at 1750. I'll explain CORN in a second. Then we have aldehydes and ketones and 1710. Then finally, we have amides at 1680.
Now, where the heck did I get CORN. If you think about the things that are attached to the carbonyl on 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 and maybe that helps you guys to memorize the order in which they're at. It doesn't tell you the exact numbers, but maybe you can remember acid chloride is higher than amide because of CORN. You're going to sound like a total vegan this semester.
What else do we have going on here? Notice that I have these two bullet points, which have to do with general rules. In general, conjugation, which is basically the ability of any of these carbonyls to resonate with something else, so I'm just going to put here, in case you don't know the definition of conjugation, that's okay. That just means can it resonate with something else.
The presence of conjugation is going to lower the number by 20. That means that, for example, here I have a conjugated ketone. It's right by my head, so I'm going to move out of the way a little bit. Notice that the reason I'm calling this a conjugated ketone is because I have a ketone with a double bond that could resonate with it. If you drew resonance structures, you'd be able to draw charges and all that stuff.
Now notice that this conjugate ketone is resulting at – I covered it – 1690. What's the normal absorption of a ketone? 1710. Anytime you have conjugation present, it's going to lower your predicted value by 20, about 20. Is that cool? You could apply that to the other ones as well, so a conjugated acid chloride would result at 1770. Awesome. You guys are getting this.
Other thing, this is getting even weirder. What the heck is a banana bond? A banana bond – again, this is so that I can just be comprehensive. I'm just trying to make sure that I cover all of my bases.. 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 so that they can overlap.
Now we've got vegetables, we've got fruits. What the heck is going on? It feels like I'm walking through a produce aisle. But anyway, banana bonds because of their spatial arrangement they're highly strained, they're going to tend to result about 100 higher. You could imagine that a ketone that has highly strained bonds in it is going to result about 100 higher, so instead of 1710, it's at 1810.
That's just a general rule. Your professor might have a slightly different number. That's just something for you to remember that if it has these highly strained rings, that would mean like a three-membered ring or a four-membered ring, these have banana bonds and they're going to get a higher absorption frequency. Cool so far? That was the carbonyl region. Maybe that was too much information depending on your class, but now you know.
Then we're going to close off the double bond region with one more molecule that I did mention earlier. Cumulenes. A cumulene is just two double bonds that are back-to-back. It could be any cumulene. It doesn't just have to be n double bond n, double bond n as I have in this example. It could be CON, it could be – there's a lot of different molecules that fall into this cumulene area. They're going to result around 1,900 to 2,000. Cool? They still belong in the double bond region because they have only double bonds present.
Now let's move on to the triple bond region where we have alkynes and nitriles sitting between 2,200 and 2,300. Just remember that nitriles result a little bit higher than alkynes, but typically the way that we can differentiate an alkyne from a nitrile is there even a nitrogen in this molecule. If there are not any nitrogens in the molecule, you know you can't have a nitrile. It's usually not a very difficult task to figure out if you have one of the two.
Now we have bonds to H and boy, there's a lot of them. Let's start off with carboxylic acid. Carboxylic acid is a bond of OH, but it's a bond of OH that happens on a carbonyl. Carboxylic acid would look like this. A carboxylic acid is an example of something I'm going to be referring to as a complex carbonyl. Why is that? Because a carboxylic acid doesn't just have one absorption. It has two. Notice that this double bond is an absorption that I already mentioned. It's right above my head actually. The double bond between the C and the O, because it's attached to an O, should result at 1750.
We just talked about that, so then why am I bringing it up here? Because notice that carboxylic acids also have an OH bond. This OH is also going to result somewhere and it turns out it's going to be a very broad peak between 2,500 and 3,000. It's literally going to take up a huge swath of the spectrum. A complex carbonyl is when you get more than one absorption from a carbonyl. We're going to be looking at these and we're going to have to make sure that we can kind of locate them and recognize them because this is going to happen a lot with IR spectroscopy.
Our other example of a complex carbonyl actually occurs in the same range. Aldehyde. Have we talked about aldehyde already? Yes, we did. We talked about how an aldehyde should result where according to my carbonyl region? Should result at 1710. But it also has a bond to H. That bond to H is actually going to result right here at 2700. That's another thing to be aware of that both of these are examples of complex carbonyls and we need to look out for both of their peaks, one in the carbonyl region and one in the hydrogen region, single bond to H region. Sounds good so far? Awesome.
Now let's move on to the other bonds to H. We've got our hydrocarbons. We've got our – basically, let me write this here – we've got our alkanes. We've got our alkenes and we've got our alkynes. Notice that the terminology I'm using for this is sp3, sp2, and sp. Why am I referring to it like that? That's just the shortest way that I could tell you that sp3 means that it's going to have four groups, so it should be an alkene. It should have a single bond with – a carbon with four bonds around it. That would be an alkane.
Alkenes are sp2 hybridized if you guys recall. That means that it has three groups around it or should have three different types of bonds. A great example being the Clutch logo, benzene, as I put at the top. Notice that each of these carbons would have just three groups. It would have a hydrogen, a carbon, and a carbon. This hydrogen because it's an alkene, because it's sp2 hybridized, is going to result higher. Usually, alkanes result around 2,900 and alkenes result around 3,100. About 200 wave numbers higher.
Then finally, alkynes are going to result the highest. Now alkynes are triple bonds. But there is kind of a caveat to this which is that it's only terminal alkynes. Terminal alkynes will result here. Why? Because if you remember, this is what an alkyne looks like. We already talked about the wave number for that bond. Where does the triple bond result? The triple bond should result at 2,200. The only time you're going to get an additional absorption is when you have an H present. That only happens when you have a terminal alkyne. If your alkyne is in the middle, you only get 2,200, but if your alkyne is on the end of a chain, so it has an H, then you get a 2,200 plus a 3,300. In that sense, this is also a complex functional group, not a complex carbonyl, but a complex functional group.
We're almost to the end. We're almost finished with our hydrogens. We've got amines and we've got alcohols. Notice that they actually have overlapping ranges. My alcohols are a little bit higher, but they tend to overlap. These are mostly going to be distinguished by their shape which we will go into, but, in general, alcohols are more broad. As you can even see here as I drew. The range is very large with an alcohol. In fact, it looks like a parabola. It takes up almost the entire 3,000 area. Whereas amines tend to have much smaller peaks that are in the same range, but they definitely don't look like an alcohol, so you kind of assume that it's an amine.
That's pretty much the end of that. That was a mouthful. For the rest of today's – anything you're watching in terms of IR, feel free to use this as an open book exam. Feel free to use this as a reference because my biggest goal for you is not to memorize this right now, but to get familiar with it so that once you're working with it the values start coming back to you. Awesome. Let's move on to some practice problems really quick.
In these practice problems, what I'm going to have you guys do is I'm' going to have you guys look at each of these molecules and I don't need anything about shapes, I just need to figure out how many different absorptions would I see for these molecules. Would I see one? Would I see two? Would I see four? And what would be the values of those absorptions based on the values that I taught you above? Go ahead and look at the first problem and try to predict how many different values am I going to get there and what would be the actual numbers of those values.
Go ahead and take a second to do that and then I'll go ahead and answer the question. 

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


IR Spect: Frequencies Additional Practice Problems

A compound C5H10O gave the following spectral data:

1H NMR spectrum                              IR spectrum

doublet, δ 1.10                                  strong peak

singlet, δ 2.10                                   near 1720 cm    -1

septet, δ 2.50

Which is a reasonable structure for the compound?

Watch Solution

Below is the IR graph of a compound with the MF C4H8O2. Identify and label peaks a, b and c in the spectrum that are consistent with the molecular formula given. After, say whether the molecule contains a carboxylic acid? Briefly explain your answer as to why or why not.

Watch Solution

Identify the functional groups that are indicated in the following IR spectra.

Watch Solution

Which of the following compounds gives an infrared spectrum with peaks at 3000-3500 cm-1 and ~1750 cm-1?

a. 1          b. 2        c. 3       d. 4

Watch Solution

For the following reaction, which of the following change(s) in the IR spectrum is consistent with conversion of the reactant to the product?

a. absorption at 2150 cm -1 should disappear 
b. absorption at 3300 cm -1 and 2150 cm -1 should disappear 
c. absorption at 2250 cm -1 should disappear, a new absorption at 3300 cm -1 should appear 
d. absorption at 1650 cm -1 should disappear, a new absorption at 3300 cm -1 should appear
e. absorption at 3300 cm -1 and 2150 cm -1 should disappear, a new absorption at 1720 cm -1 should appear

Watch Solution

Which of the following compounds will show a broad absorption around 3300 cm-1 and a sharp absorption at 1650 cm-1?

a. I 
b. II 
c. III 
d. IV 
e. V

Watch Solution

Which one of the following compounds is consistent with the following IR spectrum?

a. I        b. II            c. III         d. IV          e. V

Watch Solution

Rank IR absorption of the indicated bonds in decreasing (highest to lowest) order of wavenumber.

e. none of these

Watch Solution

Which compound would be expected to show intense IR absorption at 1746 cm-1?


Watch Solution

A saturated compound (IHD = 0) that contains oxygen and shows strong broadened IR signal above 3000 cm –1 is likely 

a) ketone

b) an alcohol

c) a carboxylic acid

d) Any of the above

e) None of the above

Watch Solution

Which compound will likely NOT show a significant IR signal between 4000 & 2500 cm –1 ?

a) CH2Cl2

b) Cyclohexane

c) 1-Hexene

d) CCl4

e) All of the above are likely to show a significant IR signal between 4000 & 2500 cm  –1 .

Watch Solution

Which compound will likely show a significant IR signal between 2000 and 1500 cm –1 ?

a) Hexane

b) 1-Hexyne



e) None of the above

Watch Solution

A compound contains oxygen but has no IR signals above 3000 cm –1 or between 2000 and 1500 cm –1. This is consistent with the compound being

a) an ether.

b) an alcohol.

c) an aldehyde.

d) a ketone.

e) none of the above.

Watch Solution

In what region of an IR spectrum will the triple bond stretch of 1-heptyne be detected?

a) 4000-2500 cm –1

b) 2500-2000 cm –1

c) 2000-1500 cm –1

d) 1500-400 cm –1

e) None of the above

Watch Solution

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

Watch Solution

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?



Watch Solution

Rank the indicated bonds in decreasing (highest to lowest) order of wavenumber.

a) None of these

b) III > I > II

c) II > I > III

d) I > II > III

e) III > II > I

Watch Solution

Concentrated alcohols show a _____ absorption in the region of 3200-3600cm  –1, due to _____. 

a. sharp, hydrogen bonding

b. broad, hydrogen bonding

c. sharp, polarity

d. broad, polarity

Watch Solution

How would you distinguish the formula by IR spectroscopy? Please list the major difference in the spectra and provide frequencies.

Watch Solution

Which function group is missing in the following IR spectrum?

a) OH           b) sp  2C         c) sp 3C         d) C=O        e) all are present

Watch Solution

Which C=O stretch will occur at the highest frequency (cm  −1) in the IR spectrum?

Watch Solution

Rank the indicated bonds in decreasing (highest to lowest) order of wavenumber for IR absorbance.





e) none of these

Watch Solution

Which of the following bonds would produce the strongest IR absorption?

a) C=O

b) C–O

c) C≡C

d) C=N

e) spC–H

Watch Solution

For the following reaction, which of the following is consistent with a comparison of the IR spectra of the starting material vs. the product?

a) Absorption at 2250 cm –1 should disappear, new absorption at 2600-2800 cm –1 and 1720 cm –1 should appear.

b) Absorption at 2250 cm –1 should disappear and absorption around 3400 cm –1 should appear.

c) Absorption at 3200-3400 cm –1 and 1720 cm –1 should appear.

d) Absorption at 2250 cm –1 should disappear.

e) None of these.

Watch Solution

Match the following structures with the IR spectra shown below.


Watch Solution