Ch. 15 - Analytical Techniques: IR, NMR, Mass SpectWorksheetSee all chapters
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Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins

Concept #1: Building Molecular Sentences

Transcript

So guys, now we're going to discuss what is basically the Holy Grail of our analytical technique section. That is the skill of structure determination.
At some point this semester, you may be asked to produce a structure from scratch. That means draw out a structure from scratch using nothing but a combination of molecular formula, NMRs, and IR spectra. That means literally all you have is a bunch of peaks and a bunch of spikes and stalactites and you're supposed to actually turn that into a carbon structure that is exactly the right structure.
When students see this type of question, they tend to freak out because this is a very complex skill. We're having to synthesize tons of information. We're having to get creative. And students just start drawing every structure they can think of.
Now for you guys, you guys are smart. You're already watching my videos. That means you're trying to get ahead. I'm going to tell you, that is the first way to lose points on the exam because you're going to run out of time. You're going to draw a bunch of structures that aren't correct and it's just not efficient. So we need to be much more strategic in how we draw these structures. That's why I always teach my students to build a strong molecular sentence before we even begin getting creative in drawing structures.
The way we build a molecular sentence is by gathering clues. We're going to gather as many clues as possible from that NMR, from that IR, from the molecular formula. We're going to put it all together in a very ordered way. Where almost going to basically create like a mini-essay on this molecule and then from that sentence we can then go ahead and propose structures that are actually relevant.
This isn't going to be easy guys. This is actually a skill that takes a lot of work and it's one of the harder things you may have to do this semester. But, I promise you that by using this strategy it's going to cut down on the learning curve big time.
So let's go ahead and just talk about the steps to build a strong molecular sentence. The very first step is to determine the IHD, which is a skill that we learned in organic chemistry one and that I have included in this section so you can review it. The IHD is just basically going to tell us about double bonds, rings triple bonds, etcetera.
Then we're going to use the NMR, the IR, splitting patterns and integrations, we're going to look at all of that for major clues. Now, specifically, I put all four of these things for a reason. The reason is because we tend to find extremely helpful clues with these four things.
For example, NMR, what if I have a chemical shift in my NMR that's like 9.1. I'm just giving you examples here. But there's a lot of different shifts that we learned. What if you saw that had an NMR shift of 9.1. What would you suspect about that molecule? Well, there's really only one functional group that results in the 9 to 10 range and that would be an aldehyde. So immediately I would be suspecting is there an aldehyde.
So now what if I look at my IR spectrum and there's also a peak there at 1710. Then would that confirm my suspicion that I have an aldehyde or would it deter me? It would confirm it because remember aldehydes have a carbonyl peak at 1710. Then that would kind of confirm the aldehyde suspicion.
Now, what if I look at my splitting patterns and I notice that in my NMR, I actually have a triplet and a quartet present. Well, that's one of the splitting patterns I told you guys to memorize and that's very indicative of an ethyl group. So already I know that I'm looking for aldehydes that have some kind of ethyl group on them. That's a big deal.
Now, what if I look at my integration. What kind of information do you get from integration? You get number of hydrogens. Why is that important? Well, what if I have that shift at 9.1 for my NMR, but it actually has an integration of 2H. What would that tell me? If my integration was 2H for a shift of 9.1 that tells me that I actually don't just have one aldehyde, I actually have two aldehydes. The reason is because every aldehyde only has one hydrogen that results in the 9 to 10 range.
If I have two hydrogens, that must mean that I have two aldehydes. So these are the kind of clues that we gather right away. And you have to get good at learning where to find those clues.
Now we've done that. Now we do something that's kind of like a Clutch Prep special. This is something that you're not going to see in your textbook, but sometimes it's helpful. And that's to do something that I call calculating the proton NMR peak or the proton NMR signal to carbon ratio. Basically, this is just a test of symmetry.
What I do is you look at the number of signals that you have and you put that over the number of carbons that you have. If that number, turns out to be less than ½, then that suggests it's a symmetrical compound. Whereas, if that number tends to be above ½, then it's probably going to be a pretty asymmetrical compound.
The logic behind here being that let's say that you have a molecule that's like C6H14. And you've got your proton NMR. It starts at 0 it ends at 13. And all you have is one peak. Let's just say. You have one peak. Well, what that's going to suggest to me is that. I only have one signal. So that's going to be the one in my fraction. And I have six carbons. That's going to tell me that a lot of these hydrogens are exactly the same as each other. Actually, they're all exactly the same if I'm only getting one signal.
The only way that they could all be the same is if the molecule is symmetrical. This must be a pretty symmetrical molecule if it's giving me a fraction that's so below ½, it's 1/6th instead of ½. Does that kind of make sense? Basically, my fraction is just a measure of how symmetrical is this molecule probably or how asymmetrical is it.
Because of the fact that I literally made this up. It's something that I've used for many years, but it's also not for sure. It's not like a tried and true method. What that means is that you can never just rule out a structure because of symmetry. I've had students that say, “Oh, but that molecule doesn't look symmetrical so that can't be it.” Don't do that. This is more of a hint than anything else. It should serve as guidance, but it shouldn't serve as your final cut.
For example, symmetry in straight chains can be very difficult to visualize. I don't want you guys to just go draw structures just based off of this rule. Just use it as a helping hint more than anything else. Also, just another point, it tends to be really helpful at the extremes and not very helpful in the middle.
If you take your fraction and it happens to be exactly ½. Let's say it happens to be 2/4. That's not very useful to me. 2/4 is ½ that could really be anything. When this rule becomes really helpful is when I have something either like a very, very low number, like 1/10 let's say. That would tell me it's extremely symmetrical. Or when I have a really, really high number, like 8/10. Then that would also tell me this is very asymmetrical and I would start to think of ways that I could arrange this in a very asymmetrical way.
Enough about that rule, the last thing is that you restate at the end, after you've gathered all these clues, after you've gathered your symmetry, you restate the number of proton NMR signals needed because you should only be drawing structures that actually have the number of signals needed. For example, if your proton NMR only has three signals, you should only draw structures that could yield three signals on a proton NMR. It's a waste to draw a structure that doesn't give that number.
At this point, this is when you get creative. Unfortunately, this is a word that many of you are scared of, but what I'm trying to do here is take the most creativity out of it as possible because I know that's the hard part is trying to really think about what could it be. Could it be this or that? I'm going to try to give you a system so that when you do get creative, there's not that much to think about. It's literally maybe you have two, three or four different drawings that are possible, but not more than that.
This is where you draw out everything that fits in the above criteria and then finally once you've drawn all the possible structures that could fit all of those clues, that molecular sentence that we built, you finally use a combination of shifts, integrations and splitting to confirm which structure is the right one.
I know that was a mouthful. What we want to do for this next example is we're finally going to get into structure termination. But I'm not going to actually give you guys this molecule yet. I just want you guys to learn how to built that strong molecular sentence ahead of time.
Basically, what we're doing here is I've given you a molecular formula. I've given you data from an IR and I've give you data from a proton NMR. These numbers, by the way, are the shifts. So the 2.2, 9.4. You guys should know what 4H and 2H are. Those are integrations. You should recognize what the IR is. These are all major clues. There's a ton of clues going on around here.
What I want to do is I'm going to go ahead and stop the video and I'm going to have you guys go step-by-step and I want you guys to figure out the IHD. I want you to figure out every clue possible that you can gather here. I've already given you a lot of hints. I want you to think about symmetry if that's important. And then finally, I want you to only draw structures that have two peaks in a proton NMR.
At that point, I'll kind of take over from there. But I just want you guys to build the molecular sentence and then only draw possible structures that fit that sentence and have two peaks in the proton NMR. Go ahead and take a stab at it. I'll be right back. 

Example #1: Building a molecular sentence

Transcript

Alright guys so I'm going to start off with the IHD, so the IHD can be calculated using the formula IHD equals (2N+2-H)/2, OK? Where N is equal to the number carbons so here it's N is equal to 4 that means IÕve got 10-H, now H is equal to the number of hydrogens or hydrogen equivalents I've got 6 hydrogens 2 oxygens the oxygens so it's literally (10-6)/2 that's going to give me 4/2 which is equal to 2 IHD, OK? Now if you guys recall from IHD that means that either I have 2 double bonds, 2 rings, a combination of that I could even have a triple bond so the IHD didn't tell me that much yet it just provided like a framework a basis for what I'm looking for. Alright so now it's time to actually gather the clues remember I told you you look at four different things for clue building, we're going to look at the NMR shifts, the IR, the splitting and the integrations so let's see what we can gather so let's just look at NMR first do you guys see anything suspicious about these shifts and the NMR? Absolutely we've got a 9.4 which happens to be in the range of 9 to 10 so immediately I'm suspecting aldehyde, OK? So I'm just going to write down my clues here and then I'll put it all together so I'm suspecting aldehyde, OK? What else I've got a shift of 2.2, 2.2 is actually in what range that's in our ZCH range right and remember that anything around 2 was kind of in the area of that it's either an H that's next to a carbon that basically has a carbonyl or that has a benzene, OK? So we also said you know is possible our would look like this, OK? So we're looking for something like that 2.2 is kind of distinctive it's right in that 2 range it's going to be one of these three things, OK? Cool so we've kind of suspected those shifts give us a little bit of information, does our IR tell us anything? Does it confirm any of this? Well definitely what it's looking like I have is a carbonyl and it's looking like I have a complex carbonyl, right? Because I've got both a peak in the 1700 range the carbonyl range but I've also got this peak at 2700 which is distinctive of aldehydes, OK? Now you guys might notice a discrepancy here which is that I told you guys that aldehydes have a wave number of 1710 and here I have 1720 and you are going to completely ignore that because it doesn't matter just so you guys know all the values I told you they could all change by a little bit, OK? Just really depends on the textbook, on the way your professor wants to ask it so for my purposes 1720 and 1710 are the exact same thing, OK? So this definitely confirms that I have an Aldehyde, OK? What my peak at 28950 tell me? Nothing, it just tells me that I have SP3CH bonds which hopefully recall me saying that every single molecule has that so that doesn't really help me, OK? So now I've got my NMR, I've got my IR clues do we see any splitting patterns that we learned to memorize that could give the structure away? Well I see a doublet and a triplet and actually that's not the splitting pattern I taught you, OK? I taught you about triplet quartet but that's not this so actually splitting patterns kind of didn't work you know. Finally integration do we see anything interesting with the integrations that might peak us into how many functional groups we have? And actually it looks like we just hit gold because notice that my aldehyde 9.4 actually has an integration of 2H that means that I must have two aldehyde presents in order to give me two 9.4 shifted hydrogens, OK? So this is actually becoming pretty great so now if I were to start off my sentence what I would say is that well actually really quick before I build the sentence I want to talk about this, we just gathered all these clues it looks like I have now 2 aldehydes, does this conclusion correlate with my IHD? Remember that my IHD said that I have 2 IHD so do we now know where those IHDs are coming from? Yes, we do, they're coming from both of the carbonyl groups on my dual aldehydes so that means that what I'm looking for so this is my sentence I'm going to then start building it, OK? So what I'm going to say is that I'm looking for a 4 carbon Dialdehyde, is there anything else that we can say? Do you have splitting patterns? No, that's really all we can say right now, it's acyclic could we say that it's acyclic? Because if it had a ring it would have an IHD of three, right? So we could even stick in parentheses acyclic, right? So it's an acyclic dialdehyde, now it's time to look at the symmetry thing so now we're done that's basically all we can say so far that's actually a lot though because there's not a whole lot of dialdehydes that you can draw with only four carbons so we're doing great.

Now, Let's look at symmetry now the way we did the symmetry thing was you take the number of HNMR signals which is 2 and put it over the number of carbon which is 4 that gives me a ratio or fraction simplifies down to 1/2 remember that 1/2 was really actually my cut off between symmetrical and Asymmetrical meaning that this symmetry trick is pretty much worthless right now, it's not going to help me because I you guys that symmetry really only hopes at the extremes it helps me if it's very symmetrical or very unsymmetrical but in this case it's since it's right down the middle I'm just going to ignore it, it didn't work out for me this time, OK? Finally now that I did symmetry we have to restate the number of protons or proton NMR peaks we need so we would say we need a 4-carbon acyclic dialdehyde with how many proton NMR peak or signals? 2, with 2 HNMR signals and guys this is exactly the kind of sentence that you need to start getting good at building this is the kind of sentence that if you can build.... IÕm sorry my head's in the way if you can build this sentence you are so far ahead of your classmates because your classmates are going to be struggling with the basics they don't even know where to begin, meanwhile you have like this beautiful little sentence here that perfectly captures what you're trying to draw, OK? So now this is the part where we get creative and we actually draw structures, OK? And I'm going to take over from here, Ok? And you're just going to see how I do it but what we find out is that this sentence is so good it's so strong that there's not that many structures we could draw so first of all let's start off with how many different 4 carbon chains can we make that are acyclic? Well 2, we could make a 4-carbon chain that is straight chain and I'm sorry guys are really running out of room because we're at the end of the page but I'll make it work and we can also have a branched 4 carbon chain like that, OK? So now could you also do like a 4-carbon ring like this? Would that work? No, because we said it was acyclic so we're going to take that out. Now on the first chain, how many places could we put 2 aldehydes? Well this is the easy part by definition aldehydes always have to be terminal, they always have to be at the end of the chain meaning that there's only one place I can put it which is here and here, OK? Let me just give you an example if I were to choose to try to put my aldehyde in the middle would that work? That's not poly-ketone so No the aldehyde by definition has to be on the edge so there we go, OK? Now with my other four carbon chain the branched one, would I be able to...Where can I put the aldehydes there? Yeah well, I could put that on the ends of these corners, OK? So that's another possibility so I've got 1 I've got 2 is there anywhere any other combination of atoms that I could put these aldehydes on? On the straight chain no and on the branch chain No so actually these are kind of are only options right now, there's nothing else we can do now it turns out we have to remember we need to only draw structures that are going to give us the right amount of signals that's what we always used to screen first so my question to you is let's look at the first compound let's says this is structure A and this is structure B would structure A yield 2 NMR signals? Would structure B yield 2 NMR signals? proton NMR? For A the answer is yes, it would yield only 2 because these hydrogens would be let's say peak A and this one would be peak B and then the same exact thing would be created on the other side so I would have B and A so I would only get two signals so that's a check mark. For structure B would I get 2 signals? Well I do have a plane of symmetry so I would get A, B but look I've also got this mess up here that's going to give me C so am I going to get two signals with this one? Nope this is automatically crossed out which means preprocess of elimination this has to be my correct structure, OK? But we're not, I'm letting you off that easy though because sometimes you're not going to get the answer that quick so what I want to do is I want to even though we know that it has to be a structure I want to use the rest of the information to confirm that it actually is that structure, for example would this actually have a doublet and a triplet, we have to analyze that so we can really be sure so A and I'm talking about proton A notice it's the aldehyde H how would that be split? How would proton A be split in you know with the N+1 rule? Well is it adjacent to any nonequivalent hydrogens? Yes, it's adjacent to 2 right here so that means that for proton A, N is equal to 2 so proton A should be a triplet that works so far let's look at proton B, is proton B next to any non-equivalent hydrogens? Well its next to two hydrogens here so are we going to split with those? Actually guys this is a really really tricky and good example the answer is if you go to the right of B you are not going to split here, even though there's two hydrogens why would that be so? Why am I telling you that if you go to the right there's 2 Hs there you're actually not going to count them towards N+1? Because guys remember in order to split not only do your hydrogens have to be adjacent which is right and left they also have to be nonequivalent, these hydrogens I have to the right are actually equivalent they're both called proton B you can't have equivalent protons splitting the same type of proton so even though I'm going to the right and even though I'm counting up these 2 protons here to the right those protons are the same as the ones that are being split so they don't split, OK? So that's the whole deal with adjacent and nonequivalent, if it has the same letter if it's the same type of hydrogen it's not going to split, let's go to the right if you go to I'm sorry I keeps mixing up my right and left my apologies I don't know if it's the same for you but that was to my right, OK? Now if I go to my left do I have any adjacent nonequivalent protons? Yes right here I've got a proton that has a different letter it's A and it's adjacent it's got a carbon right next to it but it's only one of them so that means my N equals to 1 for B which means that 1+1 would be a doublet. Does that make sense with the peaks or the splits that I actually saw with my proton NMR? Yes it does, ok I'm actually going to take myself out of the screen really quick so that we don't have to deal with my head being in the way, OK? So anyway triplet, doublet does that make sense? Yeah it does now notice that Proton A is the triplet and proton A should be the one with the shift of 9.4, so is my triplet the one with the 9.4 shift? Yes so that even makes sense that my triplet and my 9.4 are happening on the same proton, OK? Likewise does it make sense that my doublet would have a shift of 2.2? Well notice that Proton B is right next to carbonyl and if you're next to a carbonyl then you would be right around 2 in fact I told you guys I think 2.1 when we were doing our frequencies and our shifts to learn so that's exactly right. Now finally do the integrations make sense? Does it make sense that we have 4 Hs...I'm sorry let me start with 2 Hs, does it make sense that we have 2 Hs for my aldehydes? Yes, does it make sense that I have 4 Hs for my doublets the 2.2s? Well 1 2 3 4 it looks like it makes sense so guys everything makes sense here, this is confirmed to be the structure so I know this is a huge huge mess but I'm just going to draw it one more time for you guys just so you can have it clear in your notes, the answer was this now I know that a lot of you guys here you're doubting you're saying Johnny you know what I guess that makes sense but I lost you in 10 minutes ago and there's no way I could have done all of this? Guys I told you this is not going to be easy this is something that you have to practice and don't worry we're going to give you practice you can get better at it, the biggest deal that I'm trying to make here is always build your sentence first and if you can build a strong molecular sentence you've almost done all of the work, confirming is the easy part the hard part is really just making sure that you gather all your clues ahead of time, OK? So anyway guys I hope that section made sense and best of luck practicing with the problems let's go ahead and just wrap up this topic.

Practice: Propose a structure for the following compound that fits the following 1H NMR data: 

Formula: C3H8O2                                                  1H NMR: 3.36 δ (6H, singlet) 

                                                                                         4.57 δ (2H, singlet)

Practice: Propose a structure for the following compound that fits the following 1H NMR data: 

Formula: C2H4O2                                            1H NMR: 2.1 δ (singlet, 1.2 cm) 

                                                                                   11.5 δ (0.5 cm, D2O exchange)

Practice: Propose a structure for the following compound that fits the following 1H NMR data: 

Formula: C10H14                                      1H NMR: 1.2 ppm (6H, doublet) 

                                                                             2.3 ppm (3H, singlet) 

                                                                             2.9 ppm (1H, septet) 

                                                                             7.0 ppm (4H, doublet)

Practice: Propose a structure for the following compound, C7H12O2 with the given 13C NMR spectral data: 

Broadband decoupled 13C NMR: 19.1, 28.0, 70.5, 129.0, 129.8, 165.78 δ 

DEPT-90: 28.0, 129.8 δ 

DEPT-135: 19.1 δ (↑), 28.0 (↑) , 129.8 δ (↑) , 70.5 δ (↓)&129.0δ (↓)

Practice: Propose a structure for the following compound, C5H10O with the given 13C NMR spectral data: 

Fully Broadband decoupled 13C NMR and DEPT: 206.0 δ (↑); 55.0 δ (↑); 21.0 δ (↓)& 11.0 δ (↑).

Practice: Provide the structure of the unknown compound from the given information. 

Formula: C4H10O            IR: 3200-3600 cm-1           1H NMR: 0.9 ppm (6H, doublet) 

                                                                                               1.8 ppm (1H, nonatet) 

                                                                                               2.4 ppm (1H, singlet) 

                                                                                               3.3 ppm (2H, doublet)

Practice: Provide the structure of the unknown compound from the given information. 

Formula: C4H9N              IR: 2950 cm-1, 3400 cm-1           1H NMR: 1.0 ppm (4H, triplet) 

                                                                                                         2.1 ppm (4H, triplet) 

                                                                                                         3.2 ppm (1H, singlet)