Structure Determination without Mass Spect

Concept: Concept: Building Molecular Sentences

12m
Video 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. 

Concept: Example: Building a molecular sentence

17m

Concept: Practice 1: Proposing Structures from 1H NMR data

4m

Concept: Practice 2: Proposing Structures from 1H NMR data

3m

Concept: Practice 3: Proposing Structures from 1H NMR data

4m

Concept: Practice 4: Proposing Structures from 13C NMR data

9m

Concept: Practice 5: Proposing Structures from 13C NMR data

4m

Concept: Practice 6: Proposing Structures from multiple data

4m

Concept: Practice 7: Proposing Structures from multiple data

4m

Structure Determination without Mass Spect Additional Practice Problems

A compound with the molecular formula C7H14O2 is responsible for the fragrant smell of bananas. In either acidic or basic water, it decomposes to acetic acid and an alcohol. Given the NMR spectra below, draw the structure of this compound.

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What is the structure of the compound in the following 1H-NMR spectrum with the molecular formula C6H12 ? Relative integration is shown.

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Determine the likely structure for a compound A (C6H10O), which is found to decolorize bromine in carbon tetrachloride. Its spectral data is as follows:

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Deduce the structure of an unknown compound with molecular formula C5H12O using information given by its infrared spectrum.


Intensity (peak):  Frequency (cm–1):

m                         3300

m                         2900

m                         2800

m                         1465

m                         1450

m                         1375

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Draw the molecule with a molecular formula of C 5H8O2 described by the spectra below.

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A compound with a molecular formula C4H4N2 has the following 1H-NMR spectrum.

Show the structure that is consistent with the spectra in the labeled box. Show work for partial credit

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Given an organic molecule with the molecular formula C9H9N and the following IR and 1H NMR spectra, determine the structure of the molecule. 

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A compound with the formula C8H10O produces six signals in its 13C NMR spectrum. The 1H NMR spectrum is tabulated below. Provide an unambiguous structural formula for the compound from the data provided. 

1.19 ppm (3H triplet)

2.58 ppm (2H quartet)

4.65 ppm (1H broad singlet)

6.70 ppm (2H doublet)

7.05 ppm (2H doublet)

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A compound with the formula C7H14O has strong IR signals near 2900 and 1720 cm –1. The 1H NMR spectrum is tabulated below. Provide an unambiuous structural formula for the compound from the data provided.

1.07 ppm (6H doublet)

2.77 ppm (1H septet)

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Draw the structure of the compound that produces the 1H-NMR spectrum below. The molecular formula is in the upper left of the spectrum. Make sure to put your final answer in the box provided.

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Provide a structure for the compound with molecular formula C10H12O2 and with the following spectroscopic data.

IR: 1680 cm–1, 2750 cm–1, 2850 cm–1

1H NMR: 1.3δ (triplet, I = 3H), 3.6δ (quartet, I = 2H), 4.5δ (singlet, I = 2H), 7.3 (doublet, I = 2H), 7.7δ (doublet, I = 2H), 9.9δ (singlet, I = 1H)

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Identify the following carbonyls with molecular formula C4H8O from their  1H NMR.

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Predict the structure of the compound that gives rise to the following  1H NMR spectra. 

 

A) Molecular Formula: C4H8O

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An unknown compound has the molecular formula of C6H11O2Cl. The NMR spectrum of the compound in chloroform is presented below. The IR spectrum shows a strong peak at 1735 cm -1. What is the structure of the molecule? Draw your final solution below.

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