|Ch. 1 - A Review of General Chemistry||4hrs & 48mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 19mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 25mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 32mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete|
|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|Purpose of Analytical Techniques||6 mins||0 completed|
|Infrared Spectroscopy||16 mins||0 completed|
|Infrared Spectroscopy Table||32 mins||0 completed|
|IR Spect: Drawing Spectra||41 mins||0 completed|
|IR Spect: Extra Practice||27 mins||0 completed|
|NMR Spectroscopy||10 mins||0 completed|
|1H NMR: Number of Signals||27 mins||0 completed|
|1H NMR: Q-Test||28 mins||0 completed|
|1H NMR: E/Z Diastereoisomerism||8 mins||0 completed|
|H NMR Table||24 mins||0 completed|
|1H NMR: Spin-Splitting (N + 1) Rule||25 mins||0 completed|
|1H NMR: Spin-Splitting Simple Tree Diagrams||11 mins||0 completed|
|1H NMR: Spin-Splitting Complex Tree Diagrams||8 mins||0 completed|
|1H NMR: Spin-Splitting Patterns||8 mins||0 completed|
|NMR Integration||18 mins||0 completed|
|NMR Practice||14 mins||0 completed|
|Carbon NMR||7 mins||0 completed|
|Structure Determination without Mass Spect||58 mins||0 completed|
|Mass Spectrometry||12 mins||0 completed|
|Mass Spect: Fragmentation||29 mins||0 completed|
|Mass Spect: Isotopes||32 mins||0 completed|
Concept #1: 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 you guys 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's two. There's a C-C bond and there's an sp3 CH bond. How many of these do we actually draw? Just the second one. We're only going to draw the sp3 CH.
Just so you guys know, sp3 CH, we know one thing about them, they result around 2,900. But what is their shape? Their shape is called choppy. 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 because, typically, there's so many of these signals, of these absorptions in a molecule, because think about it, there's hydrogens everywhere. They're all going to overlap with each other and make kind of a clustered mess around 2,900. Instead of giving it a specific shape, we just call it choppy.
Now you're wondering what is this going to look like, so 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 down. What I'm going to do is I'm really just going to neglect – as you can see this graph – my 1,500 starts here. I'm never going to draw anything significant below there. I'm just going to draw a line. I'm literally just making this part up. I'm not going to do anything until I get to right below 3,000. This is my 3,000 mark. Right before there, I'm going to start to draw something. What I'm going to draw is a choppy peak. When I get to 3,000, I abruptly stop and I continue.
Now notice what I just did there. It's got 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 half way 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. Notice that alkenes had more that we had to worry about because we still have that C-C, 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 sp2 CH. That's in addition to the rest of the hydrocarbon that still has sp3 CH. 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? Well, these are all going to have their own peaks that you need to know. So 1,600 – 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 a double bond region that's going to be 1,600 and what is that going to look like? That is going to be weak and sharp. The peak at 1,600 is expected to be weak and sharp.
Then we would expect both of these to be choppy. We know their ranges. We talked about how sp3 is 2,900. We also discussed how sp2 is 3,100. Basically, we're going to have a lot of choppiness everywhere from between 2,900 and 3,100. 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 about1,500, but wait. This time when I cross 1,500, I immediately have to draw a weak, sharp peak. That's going to represent my C-C double bond. This is my C-C double bond. That's my weak, sharp peak.
Now, once again, nothing really happens for a while. I'm bored. But I get to 2,900 and I have to start paying attention because now what I'm going to have is I'm going to use the 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.
Usually, if this was 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's these sp2 hydrogens, that means that some of this choppiness is going to continue past the 3,000 mark. 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. 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. Let's identify, once again, like we've always done all the different bonds here.
We've got, once again, C,C, who cares. I've also got my sp3 CH. By the way, heads up, you're always going to be drawing that sp3 CH. Get used to it. These are around to stay. Because remember that alkanes are the backbone of all organic molecules so what are the chances that there's no alkane component. Very little. You're almost always going to draw an sp3 CH absorption.
We're X'ing out this one. This one we're definitely going to draw. But what else do we have? Now we have a C triple bond C that we're adding and we've also got an sp CH 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 2,900. We know it's going to be choppy. But what do these other guys look like? C-C triple bond if you had to guess what it's going to be, triple bond region. That's going to be 2,200. 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 2,200 weak sharp.
Then we've got the sp CH which, because of what I told you guys about hybridization and the wave numbers, this one's actually going to result the highest at 3,300 centimeters. This one's actually going to be strong and sharp.
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. That can't be choppy. It's going to be just one lone sharp peak at 3,300.
Let's go ahead and give this a whirl. I'm going to start off. I completely ignore 1,500. By the way, just so you know, I've been ignoring it, but it might as well look like this. There might be stuff going on, but the whole point is that I don't care. It could have spikes, it could not have spikes. It doesn't really matter.
The point is that nothing is going to really happen until I get to 2,200. When I get to 2,200, I'll give my weak, sharp peak. 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 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. In this case, because there's kind of a gap between 2,900 and 3,300, 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 and sharp peak coming from my alkyne. This is my sp3 CH that's choppy, but now this is my strong and sharp sp hydrogen peak. Notice that there's actually a separation here.
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 H's there. Then what would happen? 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.
Concept #2: 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 one we want to focus on is really the granddaddy of all absorptions and that's alcohol. Alcohols, as we said, are one of the broadest absorptions that there is. They absorb from 3,200 all the way to 3,600 in their absorption. 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 alkane 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 CO single bond, which I'm excluding as well. I don't have to worry about anything else, I'm just going to draw those two absorptions.
Let's go ahead. I've got my fingerprint, who cares. Now I move all the way to 2,900 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. I expect different shapes coming up at this point. You got your choppy sp3 peaks. But what's going to happen? Well when we get to 3,200, it gets real. This alcohol is going to be huge, it's going to do something like this. Boom.
Now in this spectrum, I actually went past 3,600, so I probably drew it a little bit too big, but it's not a big deal. The whole point is 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 3,200 and 3,600? Remember our triple bonds? Our 3,300 alkyne. So imagine that you have an alcohol, but you also have a terminal alkyne that has a sharp peak around 3,300, 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. A primary amine has how many hydrogens? Two. That means that we would expect it to have two absorptions. In general, these absorptions are going to be weaker and sharper than alcohol.
If you guys recall, the range of amines was somewhere between 3,300 and 3,500. That means that amines could completely get covered by an alcohol if it was present. 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 can tell difference is by the shape because it's weaker and it's sharper.
Now you might be wondering why does it have these two different absorptions and I'm just going to tell you quickly this is beyond the scope of this course. You don't necessarily need to know this, but basically, there's two types of vibrations that are going to be happening with your amine. We talked about how there's stretching, wagging, rocking and all of those are under the umbrella of vibrations. When you have those two 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 two 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 3,300. The asymmetrical stretch is going to be a little higher. It's going to be around 3,400. That's why when you have two 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. We get to 2,900 and we draw our choppy sp3. So much fun. Then we get to around 3,300 and this is where we draw our amine peaks. We're going to get a weaker, sharper peak around 3,300 and then another peak around 3,400 and that's so ugly I'm just going to do it again. I'm sorry. That looks like slime or something. Let's just try this one more time. That's a lot more like what I was trying to draw. You've got this double peak where the lower peak, at the 3,300, has to do with the symmetrical stretching and the higher peak is with the asymmetrical stretching. Does that make sense so far? Awesome.
Note that another bond that we didn't discuss was our C single bond N bond, but recall that that would just go in your fingerprint so we don't have to draw it. Excellent.
Why is this symmetrical and asymmetrical thing important? Because look at secondary amines. Secondary amines only have one hydrogen. That means that when this thing is stretching it's only going to have one type of stretch possible which is symmetrical because there's no asymmetry here. There's no possibility of hydrogens going in different directions. Since it only has one H, we expect it to have just one absorption.
What I haven't told you is what is going to be the range of that absorption. Because it's symmetrical would you expect it to be lower, around 3,300, or higher, around 3,400? What do you guys think? Take a stab at it. You can use the above notes. Yeah, it's going to be a 3,300 because there's no reason for it to be boosted to that higher frequency. Hopefully, that difference is making sense. One hydrogen, one type of stretch, one signal.
We're just going to go ahead and draw this again. We've got our sp3 over here. Then when we get to 3,300, we just have one lone, weak, sharp peak around 3,300. As you can tell, these peaks, no matter what type of amine you have, they look far different from alcohol and they should never be confused with alcohol. In fact, if you confuse anything with alcohol, I'm going to be personally disappointed in you because alcohol is probably the most easy to recognize absorption there is. You can't really confuse anything with it.
Let me know if you have any questions. Oh, wait. One more thing. Don't leave yet.
We did primary amine. We did secondary amine. Why is tertiary amine not on this chart? Did I just run out of room? Is that the next thing we're going to talk about? Not necessarily. It's because notice that a tertiary amine, if I were just to draw it down here, tertiary amine, how many hydrogens does it have? No hydrogens. A tertiary amine, by definition, is that you have an N with only R groups around it. So are you going to see an amine peak with a tertiary amine? No, you're not. So this is actually an amine that would not result in the functional group region because we don't even have a hydrogen present. That's why you're only going to see your double peaks for primary and your single peaks for secondary and you're going to see zero peaks for tertiary.
All right. That's it for this topic. Let's move on.
Concept #3: 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, OK? So now recall that my definition of a simple carbonyl was just a carbonyl that's only going to result in one absorption, OK? It's a carbonyl you only have to worry about one time it doesn't come back to haunt you a second time, OK? And we're not going to cover all of them here but you know we've got 2 good examples we're going to cover ketones and esters and I'm going to explain why they're simple carbonyls, well we already know that we should be getting used to the peaks there being our SP3CH bonds, right? But we know that we have a C double bond O, OK? So now remember that C double bond Os 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, right? So that means that hopefully you guys remember these ranges that this is going to result at 1710, OK? Now Carbonyls are probably the second most distinctive or easy to recognize absorption in IR Spec because they're extremely strong and sharp and that's going to apply to all carbonyls not just the ketones not just the R that's going to apply 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 to be very very sharp, OK? So let's go ahead and draw this guy I'm in a 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, OK? Then I go ahead and I do nothing until I get to 2900 and I complete my chops, OK? Now this one is 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, OK? So there's no their bonds here that 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 you have to worry about there's no bonds to hydrogen, extra bonds or anything like that so we're done, OK?
So now what's going to be the difference the biggest difference between drawing a ketone and drawing an ester? Well let's look at the different types of bond we have and ester once again we have our SP3CH so we're also 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 of the O resulting at 1750 or around 1750, Ok? So I'm going expect 1750, OK? Now has anything changed in the shape? Is it going to be a different looking carbonyl that I'm just going to go like 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, OK? So 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, OK? Now if you look at the units that I have on this X axis 40 is almost an imperceivably difference with how low or 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, OK? But we do have this extra bond notice that in an ester we have a CO that's created, where O result, C single bond O? Guys this is 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, OK? And then finally have my SP3 blah blah blah I drew both ugly perfect so 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 said 1750 if it actually stated the number like right here then you would know oh this is an ester or if it stated 1710 then you think oh this is an aldehyde or a ketone get what I'm saying? So the numbers are really important here when it comes to these little tiny differences in wavenumber, a more specific number is probably necessary for you to be able to tell a difference, OK? So now you guys understand what a simple carbonyl is let's move on to more complicated systems so we're going to this topic let's move on.
Concept #4: Drawing Complex Carbonyls
So now we know how to draw simple Carbonyls we have to take the next logical step and move on to complex carbonyls so complex carbonyls remember are those carbonyls that have 2 absorptions, OK? So complex carbonyls are going to be those molecules that you're going to be kicking yourself about later after the exam because you drew the first absorption you forgot to draw the second, OK? Or maybe you didn't know what the second absorption was about so that's what we're going to go over now because these are a little bit more complicated, OK? So, Aldehyde are great example of a complex carbonyl because once again we've always got our SP3CH which you know you we'll have that taken care of, we have a CO double bond which in this case it's an Aldehyde so where would I expect that to result? I know you've got this, That's the R in corn Aldehyde ketones that's going to be 17 about 1710 again, OK? And we know the shape doesn't change its strong and sharp but notice that now we also have this other bond that you have to worry about...Just a second so we have this other bond you have to worry about which is CH but it's specifically Aldehyde C H specifically aldehyde and this one's tricky this one has its own set of peaks that we need to know and actually it turns out that Aldehyde the CH happens to have a double peak, OK? So, what we're going to see is double peak or double absorption, OK? At 2700 and 2800, OK? So usually I mean what I told you guys earlier I said 27 because that where it's going to be centered around but typically this shows up as a double absorption, OK? Now sometimes that 2800 peak is going to kind of blend into the 2900 peak that you get from Alkanes, right? So sometimes the real only thing that you're going to see is this 2700, OK? Because the 28 is kind of close to the 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 so let's go ahead and draw this we get to 1710 and we draw our very aggressive absorption that's our carbonyls that's going to be our C double bond 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 it out like this I'm going to draw that like a peak here and a peak here but notice that this second one is kind of got to blend in to the choppiness that happens around 3000, OK? 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, OK? 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, OK? But technically both of those peaks came due to the Aldehyde, OK? And then finally we have all of our SP3CH over here, cool makes sense so far? So complex carbonyl there's a little bit more that we have to worry about, OK?
Now let's move on to carboxylic acid, carboxylic acid is also a complex carbonyl so we've got our SP3CH, OK? We've got our C double bond 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 get what did I tell you guys about OH bonds previously? I told that these are very broad peaks that would be around 3200 to 3600, OK? 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 carbonyl so this OH is actually going to have a very different range it's actually going to have a very.... ItÕs first of all it's going to be very strong and it's going to be very broad in fact I'm even going to put the word here very because this is the most broad peak that there is, OK? And it's going to start at around 2500 and sometimes it ends around 3000 which can happen sometimes it can even go all the way up to about 3300, OK? So, this thing is a mess it kind of imagine that it's like you took the alcohol peak and you just shifted it down and stretched it out and that what you're going to get, now think about it do you know any other peaks that result between 2500 and 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, OK? So, we're going 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, OK? Then when I get to 2500 I start drawing this really incredibly broad peak that kind of looks like this, OK? So, it's going to just start off it's not quite as deep as strong as an alcohol but it definitely is broad so we got this thing going on but it's not going to end here remember that I have these 2900 peaks to draw, alright? So, 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, OK? So, what you wind up getting is kind of like an overlap of peaks that is very common with carboxylic acid and sometimes it clouds the view of your choppier Alkanes and alkenes and stuff like that which makes it kind of problematic, OK? So that would be another example of the complex carbonyl since I had to worry about my C double bond O but I also had to worry about specifically the OH that was involved in the carbonyl and then once again I always had the SP3CH as before, OK? So now you guys got it, you got simple and you got complex carbonyls so let's move on to the next topic.
Concept #5: 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, and these functional groups are all going to be, what we call concealed functional groups because they don't show up in IR. So, let's go ahead and talk about these and why exactly that happens and what they look like. So, these groups are as follows, we've got alkyl halides, ethers and tertiary amines, these groups even though they really seem what 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, okay? They have no absorptions in the functional group region, okay? Why is that? Well, because remember, that in order to have a 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, I'm just going to put numbers here. So, you need, one, a double bond, two, or you need a triple bond, three, or you need some kind of bond to hydrogen okay? Well, it's okay. So, now we look at these three and we wonder is what kind of bonds do they have, what we're going to notice is that they all have sp3 CH bonds, okay? So, all of them have that, you really pretty much every single molecule in the world is going to have that, okay? But then besides that what else do they have? Well, this one has C-X, this one is C-O, this one has C-N, 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, so 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, okay? 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 thing that, 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, okay? So, that means that these three functional groups are going to be in perceivable from alkanes. So, you're going to need to use other clues to figure out if you have them, okay?
So, 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 that's, I don't want to do that, it could have been cyclo butane or something, you know, it could have just been an alkane but it happens to be, you know, it happens to be an alkyl halide or happens to be an ether you're never going to tell an IR, okay? So, that's another limitation of IR, that IR doesn't do a great job of picking out these different function groups that don't show up in the functional group region, okay? Now, I mean a quick note that some professors might say, oh that you can use the fingerprint region to tell if you have these molecules or not, and the truth is it's complicated, the truth is is not as straightforward as they might say, and most courses in organic chemistry one and two are not going to cover that. So, 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. Alright, so hope that made sense guys, let me know if you have any questions, let's move on to the next topic.
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