1H NMR: Number of Signals

Concept: Concept: General Assumption for 1H NMR Signals

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Video Transcript

Let's discuss the first piece of information that we can derive from a proton NMR and that's the total number of signals.
On a typical proton NMR, there's going to be as many signals on the spectrum as there are unique non-equivalent types of protons. For that, we need to understand what's an equivalent or non-equivalent proton. Well, an equivalent proton is going to be a proton that has the same perspective on the molecule as another proton. If two protons are in pretty much the same place on a molecule, like for example, let's say the three protons that are attached to a methyl group. A methyl group usually has three H's on it, so all three of those would be said to have the same position on the molecule, so all three of them would be what we call equivalent.
For right now, how are we going to be able to determine if something's equivalent or non-equivalent? We're just going to go with a really easy rule which is that let's go ahead and assume that hydrogen is bound to the same atom or equivalent. So like I just said, the three hydrogens on a methyl group or on a carbon would be equivalent, so that would apply to other atoms as well, not just carbon.
In general, a rule that we can go by is that any type of symmetry is going to reduce the total number of signals. This is because if you have any planes of symmetry then you're by definition going to have some protons that are the same as other protons on the other side of the molecule. Symmetry is something you have to watch for when we're using this type of information.
What we're going to do is I'm going to go ahead and do practice problem (a) as a worked example and then I'll save the other three for you guys to do on your own. Let's just go ahead and read this question. It says how many different types of protons or signals are there on each molecule. Let's look at (a).
What we notice is that (a) has obviously a bunch of hydrogens on it that aren't drawn, but it has four different atoms. What I'm wondering is how many different signals do you think this is going to have. Now notice three of them are carbon, one of them is oxygen. Is there any plane of symmetry, etcetera. Those are the things we need to be thinking about.
Well, I'm just going to tell you right now. The answer is that there are going to be four different types of hydrogens here. Let's see why. The reason is because we could just start counting from the oxygen. Let's say that the oxygen is going to be letter A so the oxygen has a hydrogen that's attached to it. And there's no other hydrogen like that, so for sure that's one type of hydrogen. No other hydrogens on that molecule look like that one.
Now we have all the rest of these hydrogens. You might have through that we could group them all together since they're all on carbon, so maybe you're thinking you have one type of hydrogen with the O and the second type is on the carbons. But it turns out that no, there are actually more separated than that because, for example, the hydrogens that are attached to this carbon are closer to the oxygen than the hydrogens attached to this carbon. That means that theoretically, the red hydrogens, the two hydrogens that are on this red carbon, are going to be a little bit more deshielded than the blue ones because they're going to be closer to something electronegative. So I would actually expect that the red ones would be a little bit more downfield because remember those words downfield and up-field.
Anyway, because of the fact those two red hydrogens are attached to the same atom, we're going to say that's the second type. The two hydrogens attached to this atom are the third type and then finally the three hydrogens on this last carbon are the fourth. So in total, we get one, two, three, four different types of protons. So not so bad.
Another thing to note is you might have been thinking maybe there was symmetry here, but really this molecule isn't symmetrical. The way that it's drawn, the oxygen is on one side and then you've got this asymmetry that goes through the whole molecule so that's why every single atom needed its own peak or its own signal.
Now go ahead with that knowledge, try to do question (b) and then I'll go ahead and solve it. 

Concept: Example 1: Identifying Proton Signals

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All right guys, what was the answer for question (b). Three. Let's go ahead and check it out.
First of all, did you guys find any symmetry in this molecule? Actually yes, this is molecule with a plane of symmetry down the middle. That means that whatever conclusion I make let's say about this carbon over here, also to applies to the one across from it, meaning that if you are able to identify the amount of unique hydrogens on one side of that dotted line, the same exact thing applies to the other side, so you don't even count the ones on the other side.
For example, I noticed that this one is on a double bond, so I'm going to make this as hydrogen type A. Then I notice that these hydrogens are on an alkane, a regular sp3 hybridized carbon. That's going to be another type of hydrogen. Now I also notice that there's this carbon here. I'm wondering did you guys give that a signal or not. Actually, this carbon doesn't even count because that carbon doesn't have any hydrogens. Remember this is called proton NMR because it only responds to protons. Even though that is a unique position on the molecule, it doesn't have hydrogens so we don't count it.
Then finally we have over here, we have the hydrogens that are on that one. That's its own unique place.
We've got those three different signals. Now would we also have to draw the signals on the other side? No, because this one is also A. This one is also B. And this one is also C. That's what you do with a plane of symmetry. It means that any conclusions you have about one side, are going to be the same exact ones on the next side.
Awesome. So three different ones. Let's go ahead and move on to the next question. 

Concept: Example 2: Identifying Proton Signals

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And the answer for problem C was just one, okay? So, I know some of you guys got back because it turns out, first of all, is there any symmetry? actually yes, there's actually two planes of symmetry, there's a plane of symmetry here, I'll make that the red plane and there's also plane of symmetry here, I'll make that the blue plane, okay? Now, what does it means to have a plane of symmetry again? it means that basically whatever you figure out for one side of the mirror applies to the other, okay? In this case since I have two overlapping planes of symmetry that means all I need to do is figure out the number of unique positions in one quadrant and that will apply to all the other three, okay? So, all I need to figure out is how many different positions do I have in one quadrant, okay? And, what we notice is in that one quadrant we have only two different positions, we have the red position and we have the blue position, okay? So, my question is, which of these get signals, okay? Well, red. Notice red actually already has four bonds. So, red isn't going to count, red isn't going to get a signal. Now, blue is going to get a signal but notice that this blue is the same as this one, it's the same, oops, trying to use blue here, the same as this one this one this one, it's also the same as this, okay? Those are all the same, so the answers that you're just going to get one. So, I'm just going to put the letter A because there's nothing else, okay? Awesome next question.

Concept: Example 3: Identifying Proton Signals

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So, what was the answer that you got for d? Well, this was actually kind of a high number, I want to start off by just talking about symmetry, were you able to find any symmetry on this molecule? if you did you're kind of like deluding yourself because it actually isn't, this is an asymmetrical molecule.

Now, it would actually be easy to turn it into a symmetrical molecule if you didn't have this part, let's say this ethyl group didn't exist, that would be symmetrical because now we would have a dotted line down the middle and now, anything that I determined for one half applies to the other, but that's not the molecule I gave you, I gave you a molecule that the two alkanes are symmetrical but then one side has an ethyl on the other side doesn't, that means there's no symmetry, that literally means that every single atom needs its own signals. So, I'm going to get atom A, signal B, signal C, signal D, signal E, signal F, signal G, and signal H, okay? So, all of those are going to get their own unique signal, it's eight signals. Now, you might be wondering guys, I know a few of you guys are wondering, Johnny why did you give A and F asked different signals, they look just the same to me, okay? Why would you give them different signals? Well, notice, F is closer to the ethyl group and A is further from the ethyl group, so that means there isn't perfect symmetry here, right? If the ethyl group didn't exist then A and F would have been the same thing but since that ethyl group is there there's no symmetry, that means that A and A are their own unique Peaks, okay? So, hopefully that makes sense so far, let's move on to the next part.

Problem: How many types of electrically unique protons (peaks) are there in the following molecule?

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Problem: How many types of electrically unique protons (peaks) are there in the following molecule?  

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Problem: How many types of electrically unique protons (peaks) are there in the following molecule? 

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Problem: How many types of electrically unique protons (peaks) are there in the following molecule? 

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Problem: How many types of electrically unique protons (peaks) are there in the following molecule? 

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Problem: How many types of electrically unique protons (peaks) are there in the following molecule?

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Concept: Concept: Q-Test (Proton Relationships)

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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 going to be some exceptions where these protons will actually have different relationships depending on the original chirality of the molecule, this is where the exceptions begin and if you learn to accept the exceptions and just roll with them this is going to go a 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 hydrogens are going to have different relationships based on their original chirality and the test 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 3 hydrogens we take 1 hydrogen we cross it out and we replace it with a random letter in this case Q the only reason Q is just to be random so that we don't confuse it with an atom, OK? So, we're going to replace H with the symbol Q and we're going to analyze that carbon now that atom that the Q is attached to and I'm going to say did I just achieve a new chiral center by adding that Q there? So imagining that this H just randomly turned into some atom would this now be a new chiral center, OK? 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 chiral center because the Q is there? No guys, remember from organic chemistry 1 that chiral centers only exist when you have 4 completely different atoms all around the same carbon, OK? Or all around the same central atom and in this case why is this not a new chiral center? Well the Q is different this Bromo Ethyl group is different but I've got to 2 hydrogens, 2 of the same exact atom on that carbon So this obviously cannot be a chiral center, OK? So what do we do what do we conclude when your Q test does not generate a new chiral center, OK? Well this is going to be the relationships with your protons are going to be based on results like these from the Q test, OK? So if your Q test does not yield a new chiral center then these protons are always going of the same relationship that's called Homotopic, OK? These protons are always homotopic and because homotopic 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 can just assume that all the hydrogens on the carbon have the same peak it holds true in this case because we just said that they're homotopic so they're the same, perfect. And in general guys you don't want to use the Q test on stuff like CH3, right? Because notice that CH3 is always going to have 2 hydrogens left over after we use the Q test so at this point we don't need to be wasting time drawing Q tests for methyl groups, right? Because we just know that it's going to yield no chiral centers, perfect.

So now the next two relationships that we're going to discuss are going to be when we actually do find a new chiral center, OK? When it does create a new chiral center what's that relationship now? It's not homotopic anymore let's check it out, so this next category let's just go ahead and look at the picture first and see with different about this one notice that here instead of starting off with CH3 I'm starting off with a CH2 when I use the Q test on one of the Hs and get a Q instead, do I get a new chiral center? Yes Guys this is a chiral center this is a prime perfect example of a chiral center, why? Because notice that this carbon now has four completely different groups on it if we were labeling chiral centers we could label them as 1Q some kind of high priority atom, 2 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 4, OK? Now we're not going to be naming R and S here that's not important all I care about is that you decide is it a chiral center? Yes, it is, OK? 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 chiral center we need to analyze the chirality of the original molecule, we have to look at the original molecule and say did that original molecule have a chiral center? What do you guys think? So now I'm going to inspect this one and you guys are going to tell me is there chiral center on that molecule? Remember that in order to have a chiral center I need to have 4 different groups on an atom and absolutely not there are no chiral centers, this has 2 hydrogens, this has 2 hydrogens, the other one has 2 hydrogens and then the double bond can't be a chiral center so the answer here is that there is no original chiral centers so if you have a combination of 1 your Q test does yield a new chiral center, 2 you have no original chiral centers that is called in an Enantiotropy proton, OK? So if your professor asks you a specific question what's the relationship of these hydrogens? In this case you would say Enantiotopic whereas for the CH3 you would have said homotopic, makes sense? Now even though the relationships are the same, guess what? In proton NMR, protein NMR is not specific enough to detect the difference between Homotopic and Enantiotopic actually sees them both the same so that means that these protons even though they're technically little different they're still it's not what I want to use they're still equivalent, how crazy is that, OK? So that means that my assumption about how you could give every carbon on a molecule one peak because all the hydrogens are the same it still holds true for Enantiotopic because my proton NMR is going to look at both of these two H and even though they're Enantiotopic it's not going to tell the difference it's still going to say that they get one signal and that it, got it so far? So they still share a signal so you can see how my assumption works pretty well, right? Because it works for most cases, OK? Now I am going to throw in one side note some advanced forms of proton NMR can actually resolve the difference between an Enantiomers but that's only if you use a chiral 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 an Enantiotopic specific NMR we're not going to get into that, OK?

So then let's get into the last kind which is Diastrophic, well you might have already guessed what combination this is going to be Diastrophic happens when 1 your Q test does yield a new chiral center just like before but when you analyze your original molecule you did have one or more original chiral centers let's look at this example and see how it meets the criteria. Here once again I have this carbon I have two Hs I'm going to cross one out I'm going to replace it with Q I'm going to see did this just make a new Chiral center? Yea it did, I've got 1 2 3 4 different groups I put a little star there so I know I made a new chiral center, can we just assume this is Enantiotopic? We haven't answered the second question yet; the second question is how many chiral centers did I begin with? So for that I look at my original molecule did that original molecule have a chiral center? Absolutely guys, this one did because notice that this alcohol is chiral, OK? It has four different groups you just have to include that hydrogen that wasn't drawn, 4 different groups so now this is a situation we haven't seen before where now we have a Q test that is successful makes a chiral center with original chiral centers now these protons are going to have a Diastrophic relationship, OK? And Diastrophic is the only situation in which the protons now become non-equivalent, OK? And when they become nonequivalent that means that each proton is going to get its own signal, OK? So now when a proton NMR sees these two Hs it's not just going to give them both peak A, OK? one proton would get peak A and one proton would get signal B, OK? Signal A Signal B, isn't that interesting? Now they will be very close together they may almost be indistinguishable, OK? 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 his information for a specific question of what's the relationship between proton Ha and proton Hb in this case you would say Diastrophic based on the chirality of the molecule, OK? Is that making sense? Awesome guys so now we're going to move on to some practice problems and what I want you to do for all 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, OK? 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 chiral center, why? OK well if you think about it it's because the Q test only yields a result that changes your outcome if it yields a chiral center, OK? And if your molecule had an original chiral center if your molecule does not have an original chiral center then it's either going to be homotopic or Enantiotopic so you don't care remember that I'm just going to go back so you guys don't get confused here remember that if you don't have an original chiral center then you're either going to be homotopic, OK? Or you're going to Enantiotopic the only way that you can be Diastrophic is if you have a chiral center, OK? That's the only situation that you would actually give 2 different peaks so for this first molecule A we would analyze is there a chiral center? If there's no chiral center then you don't need to use the Q test because there's no chance of it being the Diastrophic so we're even going to write this so you can be extra clear when to use Q test, OK? And the answer is only when there's an original chiral center, OK? So only when there's an original chiral center you'll know you write that down, OK? I mean you'll use the Q test, OK? So in this molecule you have to analyze is there chiral center? If there is no chiral center then you would not use the Q test now I would go ahead and write that down because I'm actually going to erase that so that we'll have space to solve the questions, OK? So now go ahead and try to solve the first one and then I'll answer it for you.

Concept: Example 4: Identifying Proton Signals using Q-Test

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So, notice that this question is asking the same thing that we've already answered before, how many signals will each molecule possess in 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 going to 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 chiral centers, okay? Now, the logic behind that is this, if you don't already have a chiral Center then for sure it's going to be either homotopic or enantiotopic, right? If you don't have any chiral centers present from the get-go, okay? homotopic it or enantiotopic and remember that homotopic and enantiotopic both result exactly the same in proton NMR, they share signal, so that means the only time I have to worry about using the q test is if I already had one or more chiral centres 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 chiral centers present. Notice that you might think, this is a chiral Center but it has two of the same exact group on it, it's CH3, CH3. So, it's not a chiral Center, there's no chiral centers here. So, that means, I don't have to use the q test, I'm just going to 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 signal A, this is signal B, this is signal C, this is signal D and notice that after atom D we actually do have a plane of symmetry developing, where both of these are going to be equivalent E and 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 methyl groups are the same as 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 B, 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 B.

Concept: Example 5: Identifying Proton Signals using Q-Test

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So, did we have to use the q test on problem B? the answer is yes because I do have a chiral Center present from the beginning, okay? You might be wondering where is that chiral Center, it's right in the middle guys because notice that I have a methyl, I have a hydrogen, I have an ethyl on one side and I have an isopropyl on the other, so that is one chiral Center, okay? So, now does that mean that I have to use the q test on every single atom? actually no you only use the q test, you only use the cutest one you have one plus chiral centers and only on CH2's, okay? Because the fact that I told you guys CH3's are always formal topics no matter what because they don't make chiral centers and CH's, if you have one H 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 going to get its own peak A, okay? We can conclude that these two are going to get their own peak, B, 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 C, because it's a methyl group, okay? On top of that we've got this hydrogen, which is obviously unique because it's the only one on that chiral center so that must be hydrogen D, we've also got a hydrogen here, which since it's the only hydrogen there it must get its own peak because it's the only hydrogen that's in between two methyls 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? only on any CH2's that we have, do we have a CH2 present? yes, we have a CH2 right here, and on that CH2 we need to use the q test. So, I'm going to take an H, I'm going to take an H and I'm going to replace one of the H's with the q and now I'm going to ask myself, once I've done that did I just make a new chiral Center. So, what do you think? is that a new chiral Center now that I added a q? yes it is, because I've got one group, two groups, a methyl and then four, which is the rest of that junk, I can't even name it, it's a pretty big substituent.

So, that's definitely a new chiral Center. So, what does it mean,when the q test gives you a new chiral Center and you already had one chiral Center? what's the conclusion? it means that these hydrogen's, hydrogen 1, and hydrogen 2 are diastereotopic but most importantly these questions didn't what the relationship was, it says how many signals are we going to get. So, most importantly, what that means guys, this is the important part, what that means is that this H gets its own letter F, ugly F, let's do it again, this H gets its own letter F and this H gets its own letter G because remember that whenever you have diastereotopic 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 two diastereotopic protons get their own signal each, okay? So, that was complicated, if 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 going to answer for C, you're going to tell me, do you use the q test or not, so go ahead and try to figure it out.

Concept: Example 6: Identifying Proton Signals using Q-Test

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So did we have to use the q test on C? the answer's no, guys this molecule has no chiral centers, you might be thinking, but Johnny, isn't that chiral Center? no, because I have two of the same exact group on both sides. So, there's no chiral 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? You might be wondering, how in the world is that possible? one side obviously has the alcohol, one side obviously has the methyl, Johnny you're nuts, what's wrong with you? no, because guys remember that tetrahedral 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 OH is in the front or the OH is the back since wegend 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, you're splitting, let's say the alcohols in the front, you're splitting it right down the middle and you're also splitting the methyl right down the middle as 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 methyl is unique, that's going to be Type B, nothing else like that, we have a carbon here that were not going to count because it doesn't even have any H's and then we've got molecule C, we've got proton C and we got proton D, which are going to be mirrored on the other side because of symmetry, so this is also C and this is also D, so that means that this only had four signals, got it? Sool. So, let me know if that made sense, I'm sorry that was little bit tricky, just got to get practice with this, alright? So let's move on to the next part.

Problem: Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic. 

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Problem: Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic.

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Problem: Identify the indicated set of protons as unrelated, homotopic, enantiotopic, or diastereotopic.

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Concept: Concept: E/Z Diastereoisomerism

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Video Transcript

There's one more type of proton relationship that we should be aware of and that's called e and z diastereomerism, whew, big word, so how does that work? Well, this exists when your q test is used on a terminal double bond and it yields what we call a new trigonal Center. Now, this is a term we haven't been over in a long time but if you guys recall a trigonal Center is basically like a cis and trans isomer, okay? So, a trigonal Center would be like the relationship between a cis double bond versus a trans double bond as we call a trigonal Center. So, notice if I use the q test on a terminal double bond, let's say I scratch out this age replace it with an q. Notice that what I just did by putting the q at the bottom would be I just drew the cis version, right? cis, or if you're using a and z notation I would be you z, sis slash z, right? but who says I have to put it at the bottom, what if I had put the q at the top? Well, if I had put the q at the top then that would be trans, right? So, that means that the fact that I can make two different isomers depending on where I put the q means that this e-z relationship is possible, okay? So, these protons are always going to be diastereotopic, okay? We don't have to look for prior chiral centers anything like that, anytime you can make E or Z or cis and trans relationship with q, with the q test, always diastereotopic. Now, based on what you learned about diastereotopic before, do you think that these hydrogen's will have share a peak, share signal? or do you think they'll get different signals? it says it right there, they're going to be non equivalent and they're going to get different signals, okay?

Now, I know that I've gone a lot through the rules of how to figure this out but I just want to explain this really quickly intuitively because you might be struggling to think? Well, why would they be nonequivalent? Well, think about it like this, let's say you put this double bond, right? And we draw one of my H's there and we draw one of my H's there, okay? So, it's basically the same example that we have on the top, I just told you that because there's an E-Z relationship between these H's they get their own peak, but how does that make sense in terms of shielding. Remember, in terms of putting on the wool coat, how does that make sense? well, notice that this H is always going to be closer to that ethyl group because it's cis to it, this H is always going to be further from that ethyl group because it's trans to it, so that means that the red H and the green H will be shielded slightly differently, one of them is going to be a little bit more bundled up and one of those going to be a little less bundled up because of how far they are away or how close they are to this group over here. So, that means that they're going to have to each get their own peak they're, each, going to get their own signal, okay? So, in this example, specifically this would be HA and this would be HB, they would each get their own signal because of their unique position on the molecule, okay? So, that said, we have a few more practice problems, go ahead and try to figure out for A, if, you know, you think that these two hydrogens on the double bond deserve to have the same signal or different and then tell me the total number of signals you would get. Alright, so check it out.

Concept: Example 7: Determining Diastereoisomerism

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So, let's figure this one out, when you use the q test on these H's, what was your conclusion? guys they're actually exactly the same and they're going to share a signal the reason is because, let's say I replace the H, replace it with a q, do we get an easy relationship or a cis-trans relationship? no we don't, because notice that if I make the q on the right side then it's going to be next to an ethyl group however, if I put the q on the left side it's also next to an ethyl group, since this double bond is symmetrical, since it's symmetrical there is no cis and trans possible, okay? Since there's no cis and trans possible that means that they're going to share a peak, okay? So, these are actually homotopic, okay? They're are homotopic. So, let's go ahead and then just count out the peaks like normal, it would be that these are A, we've got this carbon? Well, I mean, we already counted that carbon and those H's, we've got this carbon, which we're going to scratch out because it doesn't have any H's, we've got this plane of symmetry so that means that this is B and that means that this is C and that's it, we've got three different ones, the other sides are also B and C. So, this would get three signals. Alright, try that with the next one and let me know what you get.

Concept: Example 8: Determining Diastereoisomerism

3m
Video Transcript

So, were the hydrogen's on this one homotopic or diastereotopic and after using the q test I've got H, I've got a H, I use q over here and what I notice after using the q test is it actually does make a difference work with the q, if I put the q at the top I get trans, if I put the q at the bottom I get cis. So, actually these hydrogen's are diastereotopic. So, I have a diastereotopic relationship and since they're diastereotopic that means I would, I would give each of them their own signal. So, I have HA and HB. Now, I just have to count up the rest, this would be C this would definitely be D. Now, what do you guys think about the last two? did you guys give them the same signal? Good, okay, I'm glad, I could tell a lot of you guys gave these the same signal because there's a plane of symmetry there, okay? So, both of them have the same.

Now, there is that one detractor out there, okay? And it's, okay? I'm glad you're thinking so hard but um, I know you're out there saying, Johnny but isn't this E, isn't this one closer to the double bond and isn't this one further away, right? So, shouldn't they actually get different signals, right? Shouldn't they be different, and the answer is actually no guys because the only way that you can have diastereotopic in this sense is on a double bond because there's no free rotation but notice that since these two methyl groups are on a single bond this thing can rotate as much as it wants, okay? So, that means that the E, let's say, this is E1, I'm going to say this is E2, right? E1 right now happens to be closest to the double bond but after it rotates then E2 can be just as close. So, actually there is no difference between E1 and E2, they can rotate, the only difference happens with HA and HB because they can't rotate, HB is always locked in the downwards position, HA is always locked in the outwards position. So, that's why they become diastereotopic in that sense, okay? I know the rest of you guys are like, Johnny, you're overcomplicating it but just letting you know you have to think about just the rules that I told you, don't over complicate it and don't try to think too hard about these because I'm giving you a pretty good set of rules already, okay? So, let's move on to the next topic.

1H NMR: Number of Signals Additional Practice Problems

Protons Ha and Hb in the molecule below are………….and therefore will give rise to ……………… signal(s)

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How many signals would you expect for the 1H-NMR and 13C-NRM spectra of the following compounds respectively? 

 

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Which compound below would give rise to 4 signals in the proton NMR spectrum and 6 signals in the carbon NMR spectrum? (Assume you can separate and see all peaks.)

A) I

B) II

C) III

D) IV

E) More than one of the above.

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How many 1HNMR signals would trans-1,2- dichlorocyclopropane give?

A) 1

B) 2

C) 3

D) 4

E) 5

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The 1HNMR spectrum of which of these compounds would consist of a triplet, singlet and quartet only?

A) 2-chloro- 4-methylpentane

B) 3-chloro- 2-methylpentane

C) 3-chloropentane

D) 1-chloro- 2,2-dimethylbutane

E) 3-chloro- 3-methylpentane

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For the following compound how many different signals would you see in the carbon NMR? (Assume that you can see them all.)

A) 3

B) 4

C) 5

D) 8

E) 9

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An unknown molecule X has 4 signals in the 1H NMR spectrum. Which of the following corresponds to molecule X?

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How many signals would you expect for vanillin?

A) 4

B) 5

C) 6

D) 7

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How many signals would you expect to find in each range of the 1H-NMR spectrum of the following compounds? Enter a number in each box. 

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How many signals would you expect to find in each range of the 1H-NMR spectrum of the following compounds? Enter a number in each box. 

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How many signals would you expect in the 1H NMR spectrum of the following compound?

a. 4 
b. 5 
c. 6 
d. 7 
e. 8

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In each molecule below, identify the number of unique 1H and 13C NMR signals you would expect to observe on a spectrum and write the number in the box provided. 

 

 

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How many signals do you expect to see in the 1H NMR spectrum of the following molecule?

a) 4

b) 5

c) 6

d) 7

e) 8

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For the structure below, label the types of protons using the a,b,c labeling protons. 

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Labeling: For each structure below, label the types of protons, using the a,b,c labeling protons. (This is only a labeling exercise.)

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Labeling: For each structure below, label the types of protons, using the a,b,c labeling protons. (This is only a labeling exercise.)

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Labeling: For each structure below, label the types of protons, using the a,b,c labeling protons. (This is only a labeling exercise.)

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Labeling: For each structure below, label the types of protons, using the a,b,c labeling protons. (This is only a labeling exercise.)

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How many 1H NMR signals would trans- 1,2-dichlorocyclopropane give?

 

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