1H NMR:Q-Test

We just learned an assumption that stated that all of the hydrogen on a single atom are going to share one peak on Proton NMR. And while that assumption is very helpful for most cases, there are gonna be some exceptions where these protons will actually have different relationships, depending on the original Chire ality of the molecule. This is where the exceptions begin. And if you learn to accept the exceptions and just roll with, um, this is gonna go lot smoother for all of us. So let's dig right into Proton relationships and the Q test. So as I just mentioned, sometimes these hydrogen are gonna have different relationships based on their original Chire ality. And the tests that we use for that is called the Q test. So what the heck is the Q test? Well, the Q test is simply this. What we do is, let's just look at this example here we have three hydrogen, we take one hydrogen. We cross it out and we replace it with a random letter in this case. Q. The only reason we're using Q is just to be random so that we don't confuse it with an Adam Okay, so we're gonna replace H with the symbol que on. We're gonna analyze that carbon. Now that Adam that the queues attached to and I'm gonna say, Did I just achieve a new Kyrill center by adding that Q there, So imagining that this h just randomly turn into some Adam, would this now be a new Kyrill center? Okay, so that's what the Q test is. And now we're going to see how the Q test can yield some different results based on the original molecule. Let's just take this example into consideration. Do you think that that is a new Carl center? Because the queue is there? No guys remember from organic chemistry one that Cairo centers on Lee exist when you have four completely different atoms all around the same carbon, okay or all around the same central atom. And in this case, why is this not a new Carl center? Well, the Q is different. This brome Oh, Ethel group is different, but I've got to hydrogen is two of the same exact Adam on that carbon. So this obviously cannot be a Kyrill center. Okay, so what do we do? What we conclude when you're Q test does not generate a new carol center. Okay, well, this is going to be the relationship to your protons are gonna be based on results like these from the Q test. Okay, So if you're Q test does not yield a new carol center, then these protons air always can have the same relationship that's called Homo topic. Okay, thes protons, air always coma topic. And because home a topic means the same, that means they're considered equivalent. And that means that they actually do share a signal on proton NMR, meaning that the assumption that I taught you about how you could just assume that all the hydrogen on the carbon had the same peak. It holds true in this case because we just said that there homo topic. So they're the same perfect. And in general, guys, you don't have to use the Q test on stuff like ch three, right, because notice that CH three is always gonna have to hydrogen is left over after use the Q test. So at this point, we don't need to be wasting time drawing Q tests for metal groups, right, because we just know that it's gonna yield. No Kyrill centers. Perfect. So now the next to relationships that we're going to discuss are gonna be when we actually do find a new Carol center. Okay, when it does create a new Carl Center, what's that relationship now? It's not home a topic anymore. Let's check it out. So this next category let's just go ahead and look at the picture first and see what's different about this one. Notice that here, instead of starting off with CH three, I'm starting with the CH two. When I used the Q test on one of the h is and get a Q instead, do I get a new Carl center? Yes, Guys, this is a Kyrill center. This is a prime perfect example of a Carl Center. Why? Because noticed that this carbon now has four completely different groups on it. If we were labeling Kyle centers, we could label them as one cues, some some kind of high priority. Adam two. There's a double bond there. That double bond is different from a single bond, and then lastly, my hydrogen would be in the back. That's number four. Okay, Now we're not gonna be naming R and s here. That's not important. All I care about is that you decide. Is it a Cairo center? Yes, it is. Okay, so now is that enough to tell the relationship between those protons? No. We still need one more piece of information. Now that we know that it makes a carl center, we need to analyze the Cairo ality of the original molecule. We have to look at the original molecule and say, Did that original molecule have a Cairo center? What do you think? So now I'm gonna inspect this one and you guys are gonna tell me. Is there Carl sending on that molecule? Remember that in order to have a Carl center, I need to have four different groups on on an atom. And absolutely not. There are no carol centers. This has to hydrogen zones. This has to hydrogen. The other one has to hundreds. And then the dole bond can't be a Kyrill center. So theano, sir, here is that there's no Original Carol Center. So if you have that combination of won your cue test does yield a new carol center to you have no original Cairo centers that is called in an anti a topic. Proton. Okay, so if you're professor asks you a specific question, what's the relationship of these hydrogen? In this case, you would say in anti a topic, whereas for the ch three, you would have said home a topic makes sense now, even though their relationships are the same. Guess what? In proton NMR, Proton NMR is not specific enough to detect the difference between home a topic and an anti a topic. It actually sees them both the same. So that means that these protons, even though their technically a little different, they're still it's not what I wanted to use. They're still equivalent. How crazy is that? Okay, so that means that my assumption about how you could give every carbon on a molecule one peak because there are all the hydrogen of the same. It still holds true for an anti a topic because my proton NMR is gonna look at both of these two h is. And even though they're Ananta topic, it's not going to tell the difference. It's still going to say that they get one signal and that's it. Got it So far. So they still share signals. So you can see how my assumption works pretty well, right? Because it works for most cases. Okay, um, now I'm gonna throw in one side note, Okay? Some advanced forms of Proton NMR can actually resolve the difference between and anti MERS. But that's on Lee to use a Cairo solvent. And that's beyond the scope of this course. That would be something like advanced organic chemistry. Or if you're some kind of organic chemistry major, you may use in Ananta topic specific, and, um are we're not even going to get into that. Okay, so then let's get into the last kind, Which is dia stereotype topic. Well, you might have already guessed what combination this is gonna be. Dia stereotypic happens when one your Q test does yield a new carol center just like before. But when you analyze your original molecule, you did have one or more original Kyle centers. Let's look at this example and see how it meets the criteria here. Once again, I have this carbon I have to h is I'm gonna cross one out. I'm gonna replace it with Q I'm gonna see. Did this just make a new Carl center? Yeah, I did. I've got one, 234 different groups. I put a little star there, so I know I made a new carol center. Can we just assume this is in anti a topic? We haven't answered the second question yet. The second question is, how many Carl centers did I begin with? So for that, I look at my original molecule. Did that original molecule have a Carl center? Absolutely. Guys, this one did. Because notice that this alcohol is chi role. Okay, It has four different groups. You just have to include that hydrogen that wasn't drawn. Four different groups. So now this is a situation you haven't seen before. Where? Now we have a que test that is successful. Makes it Carl center with Original Kyle centers. Now, these protons, we're gonna have a diess stereotype topic relationship, okay? And die. Astrea topic is the Onley situation in which the protons now become non equivalent. Okay? And when they become non equivalent, that means that each proton is gonna get its own signal. Okay, so now when a proton NMR sees these two ages, it's not just gonna give them both peak A Okay, one proton would get Peek A and one proton would get signal. Be okay. Signal a signal be Isn't that interesting? Now they will be very close together. They might almost be indistinguishable. Okay, which is why some professors don't really care about this because it's very difficult to tell the difference between A and B. But that difference is still there. And your professor may want you to know this information for specific questions of what's the relationship between Proton h A and Proton H B. In this case, you would say die a stereotype. OPIC based on the core reality of the molecule. Okay, Is that making sense? Awesome, guys. So now we're gonna move on to some practice problems, and what I want you to do for all of these practice problems is tell me, First of all, I mean, you can use the assumption that I was using before, you know, if the molecules will, you know, every atom gets its own peak. But now the first question I need you to answer is, am I going to use the Q test? Okay, so for molecule A, the very first thing you should be thinking is am I going to use a Q test. And the way you know to use a Q test is if you had an original Cairo center. Why, Okay, well, if you think about it, it's because the Q test Onley yields a result that changes your outcome if it yields a Kyrill center. Okay, And if you're molecule Haddon Original Kyrill Center. If you're molecule does not have an original Kyrill center, then it's either gonna be home a topic or an into a topic. So you don't care. Remember that I'm just gonna go back so you guys don't get confused here. Remember that if you don't have original Cairo center, then you're either gonna be home a topic, okay? Or you're gonna be an anti a topic the Onley way that you could be dia stereotype. OPIC is if you have a carol center. Okay, that's the Onley situation, that you would actually give two different peaks. So for this first molecule a, we would analyze. Is there a Kyrill center? If there's no Carol center, then you don't need to use the Q test because there's no chance of it being dia stereotype. Ah, pick. So we'll just even right this instant, you guys could be extra clear when to use. Q test. Okay? And the answer is on Lee when there's an original Cairo center. Okay, so on Lee, when there's an original Cairo center, you'll you'll write that down. Okay, um, I mean, you're not You write that down. You'll use the Q test. Okay, so in this molecule you have to analyze is there Carl Center? If there's no Carl center, then you would not use the Q test. Now I would go ahead and write that down because I'm actually going to race. That's that will have space to solve the questions. Okay, so now go ahead and try to solve the first one and then I'll answer it for you.

so notice that this question is asking the same thing that we've already answered before. How Maney signals will each molecule possessing Proton NMR. But now that we know about the Q test, the very first thing we actually need to answer for all these problems is gonna be. Do I use the Q test or not? Okay, because you don't have to always use the Q test. So you might be wondering, Well, how do I know when I have to use it? Okay, so here's the hint we use the Q test when you already have one plus Cairo centers. Okay, Now, the logic behind that is this. If you don't already have a Cairo center, then for sure it's gonna be either home topic or an anti a topic, right? If you don't have any Carol Centers present from the get go, okay home a topic or an anti a topic, and remember that home a topic and an anti a topic. Both results exactly the same in Proton anymore. They share signal, so that means the only time I have tow worry about using the Q test is if I already had one or more. Kyrill centers, in which case it might change the answer. So my first question to you is for question A. Do we have to use the Q test? The answer is no. We don't because there's no Kyrill centers present. Notice that you might think this is a Carl Center, but it has two of the same exact group on it. It's ch three ch three. So it's not a Carl center. There's no Cairo centers here, so that means I don't have to use the Q test. I'm just gonna use the old rule that said that every single atom gets its own signal and watch for symmetry. So I would go ahead and I would say this is a signal A This is signal be This is signal. See, this is signal D and noticed that after Adam D, we actually do have a plane of symmetry developing where both of these are gonna be equivalent e e because after that you get a plane of symmetry. So we have symmetry on one part of the molecule, but not on the rest. That's still okay. It helps us to determine that these metal groups are the same azi each other. So the answer here would be five signals. Okay, so notice that it's going to become really important when you try to answer these questions that first to ask yourself, Do I need the Q test? Okay, so on this next problem problem be, that should be your first question. Do you need to use the Q test? If so, where do you use it when you use it? If not, then just go with the old rule we learned. So go ahead and solve question Be

So did we have to use the Q tests on problem be? The answer is yes, because I do have a Cairo center present from the beginning. OK, You know, if you're wondering, where is that Carroll Center? It's right in the middle, guys, because notice that I have a metal. I have ah, hydrogen. I haven't Ethel on one side, and I have an isopropyl on the other. So that is one Cairo center. Okay, so now does that mean that I have to use the Q test on every single Adam? Actually, no. You Onley use the Q test. You only use the Q test when you have one plus carol centers and on Lee on ch choose okay, because the fact that I told you guys ch threes are always home a topic no matter what, because they don't make Kyle centers. And C h is if you only have one age present, that kind of answers your question already because you only have one h so you don't have to worry about is it equivalent to another hydrogen? Okay, so let me just show you. So for example, we can already conclude that this is gonna get its own peek A. Okay, we can conclude that these two are going to get their own peak. That's be because their symmetry there. Okay, I'm trying to color code this for you guys. We can conclude that this is going to get its own peak, See, because it's a method group. Okay, On top of that, we've got this hydrogen, which is obviously unique because it's the only one on that carol center. So that must be hydrogen d. We've also got a hydrogen here which, since its the Onley hydrogen there, it must get its own peak because it's the Onley hydrogen that's in between two methods like that. So so far, we've been able to do all of this without the Q test. So where does the Q test really come into? Play Onley on any stage twos that we have? Do we have a CH to present? Yes, we have a CH two right here and on that ch two. We need to use the Q test. So I'm gonna take an h. I'm gonna taken age, and I'm going to replace one of the ages with Acute. And now I'm gonna ask myself once I've done that. Did I just make a new Karol center? So, what do you think? Is that a new Carl Center? Now that I added a cute Yes, it is because I've got, um, one group, two groups, um, Ethel and then four, which is the rest of that junk. I can't even name it. It's a pretty big substitue int, so that's definitely a new Kyrill center. So what does it mean when the Q test gives you a new Kyrill center and you already had one Kyle center? What's the conclusion? It means that these hydrogen hydrogen one in hydrogen to our diet hysteria topic. But most importantly, this question didn't ask me what the relationship was. It said How maney signals are we going to get so most importantly, what that means? Guys, this is This is the important part. What that means is that this H gets its own letter after it's ugly half suit again, this age gets its own letter F, and this age gets its own letter G because remember that whenever you have dia stereotype pick protons, they each get their own signal. So if you said six signals, that was the trick. Answer. The answer should actually be seven signals because of the fact that those to die Astrea topic protons get their own signal each. Okay, so if you think that was complicated you have any questions? Let me know. But that's the way you gotta approach these problems. Okay? That being said, what's the first thing you're gonna answer for, C? You're gonna tell me? Do you use the Q test or not? So go ahead and try to figure it out.

So did we have to use the Q test on C? The answer is no. Guys, this molecule has no Kyrill centers. You might be thinking, but Johnny, isn't that Karl Center? No, because I have two of the same exact group on both sides. So there's no Kyrill center here, meaning that there's no Q test. So that means that I'm literally just going to give every atom its own peak. Now, is there any symmetry on this molecule? That's the harder question. Unfortunately, yes, there's actually symmetry right down the middle. Okay, Now you might be wondering how in the world is that possible? One side obviously has the alcohol. One side obviously has the metal. Johnny, you're nuts. What's wrong with you? No, because guys, remember that Tetra hydro molecules don't really look like that. What they look more like is that you've got two groups to the side, and then you've got one group in the front and you've got one group in the back. Okay? Now, we don't know which one is which. We don't know if the H is in the front or the H is in the back. Since wedging dash wasn't given to us. But really, when you split this molecule down the middle, you're splitting it down that front and back molecule. So when you're splitting it down the middle, you actually have your splitting. Let's say the alcohol's in the front, you're splitting it right down the middle and you're also splitting the method right down the middle is well meaning that this actually is symmetrical. So how many different bonds would I have? Well, obviously that H is unique. That's going to be type A. Obviously, this method is unique. That's gonna be tight. Be nothing else is like that. We have a carbon here that we're not gonna count because it doesn't even have any ages. And then we've got molecules. See, we got proton, see? And we got Proton D, which are gonna be mirrored on the other side because of symmetry. So this is also see and this is also D. So that means that this only had four signals, got it cool. So let me know if that made sense. I'm sorry. That was a little bit tricky. Just got to get practice with this. All right, so let's move on to the next part