The following spectra for A and B correspond to two structural...

Question

The following spectra for A and B correspond to two structural isomers. The NMR singlet at δ1.16 in spectrum A disappears when the sample is shaken with D₂O. The singlet at δ0.6 ppm in the spectrum of B disappears on shaking with D₂O. Propose structures for these isomers, and show how your structures correspond to the spectra. Show what cleavage is responsible for the base peak at m/z 44 in the mass spectrum of A and the prominent peak at m/z 58 in the mass spectrum of B.
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Accepted Answer

All right. Hi everyone. So, for this question, um based on the data provided, we must propose the structures for the structural I swimmers A and B. So let's get into it. So, um part a here says that when the sample is shaken with D 20. For I see a. The single it at 1.5 parts per million disappears from the NMR spectrum. Similarly, the single IT at 3.3 PPm in struck in Iceland or B spectrum vanishes when shaken with D 20 for ice. Maybe the IR spectrum on the other hand, revealed two peaks at 3300 for A and only one peak at 3300 for B. So, before we, before we start talking about um let her be here in the text of the question, let's talk about the information that we're given in part A. Right? Because part A is interesting, it's telling us that we have protons in this molecule that are exchanging with our solvent, which is dude aerated water. Right? So if we have exchangeable protons, recall that amines can exchange very easily with a solvent which can create um which can cause rather the disappearance of certain peaks. Like the question describes in part A. So, if we have A means, Right, um we have a means that can be explained by the single it here at 1. parts per million for a. And also the single at 3.3 parts per million in B. So those are our means. And now, what's interesting is the information that we're given from the IR spectrum. So, I summer A has two peaks at 3300 inverse centimeters. Which recall for a second that if you have two peaks at inverse centimeters, that's where you would normally see. Ura means, right. And if we have two peaks, that means we have two hydrogen, which is characteristic of a primary amine. So we know that um We know that is a must have a primary amine. And similarly, if we have only one peak at 3300, then that means our amine only has one hydrogen. Therefore it must be secondary. So, um so I see A and B have a primary and secondary amine, respectively, which checks out if we look at the next part of the question, right? Because the mass spectrum shows a molecular ion peak of master charge 101 for both. As well as a prominent peak at M Z 72 for A and M Z 86 for B. Show which cleavage is responsible for these peaks. So, now, before we start talking about fragmentation, it would probably be a good idea to figure out the structure first. Right, So, for this question, right, recall that when you're talking about a mass spectrum, an odd number for your molecular ion peak corresponds to the presence of an odd number of um of nitrogen atoms in your molecule. So, because we do have a means for both customers, we know that we have one nitrogen for both. Right, corresponding to that. I mean, so with that being said, in order to figure out how many carbons you have, in your case you got to do a little bit of math. Right? So in order to figure out your structure here, or rather kind of the breakdown of how many of each item you have, I guess. Let's go ahead and do some math like I mentioned before. So we have a master charge peak of 101 for both um for both A and B. Right? So and we know also that we have one nitrogen atom. So if we subtract 14 from So 101 -14, you're going to get 87. That leaves you with a master charge of 87. Which you have to figure out how many carbons and hydrogen can fit there. So this is the part where you kind of have to manipulate um a little bit with your I guess you could say with your carbons and your hydrogen because you have to keep multiplying until you can fit enough carbons to explain your hydrogen, so to speak. So if you do that, you'll find out that you can fit six carbons with a massive within 87. So you're going to end up with a chemical formula of C six H 15. Sorry, guys C six H 15 N for both of these molecules. Right? Because notice how the question said their structural structural members, which must mean they have the same chemical formula. So this is going to be our structure, our molecular formula for both. Um Summers We have C six H 15 n. So now we can go ahead and talk about um talk about the structure right? In order to do that, we can start talking about our fragmentation for a second because here notice how we're given Were given the master charge of two relatively large fragments. Right? We have 72 for a and we have 86 for b in terms of master charge. But sometimes with these kinds of questions it might be easier for you to figure out the fragmentation of the smaller fragments in this case. So if I were to move um our molecular formula off to the side here to give myself some space we can start to talk about um a very quickly. So for I summer a it says that we have a master charge Or rather a fragment at a master charge of 72. So now If we subtract 72 from 101, We're going to end up with a master charge of 24, which if you recall is going to correspond actually to an ethyl group. So that's gonna correspond to an ethyl group. So we know that there must be an ethyl group on one side of our um in here. So Um # eight We discussed before was our primary amine. So on one side of our primary amine has to be an ethyl group. So if I draw an NH two here, this is our carbon binding are mean we must have an ethyl group on one side which leaves us with our remaining three carbons, So 123456. And that must be our structure for primary A with a fragmentation cleaning off this ethyl group here. Alright, so now we're making progress right now, we can go ahead and scroll down to give myself space to talk about part B. So now for part B let's do the same thing. Right? Let's subtract our master charge of 86 From our molecular ion peak of um of 101. So if we do 101 -86, we're going to end up with a master charge of 15. Which if you recall, is going to correspond to a method group. Right? Because one carbon is 12 mass units plus three hydrogen of one each, that's 15. So RB is our secondary, I mean and it's going to have a method group on one side. So if I draw our nitrogen here and five drawer and h right, because it's a secondary amine, we know we're going to have a method group on one side which is going to be fragmented to explain our master charge of 86. But now since our formula here is C six, H 15 N, we have to include our remaining five carbons, so we have And our method group is going to be six. So now, um these are going to be our answers for I, C, A and B. These are going to be your answers for it is A and B. And with that being said, if you made it to the end, thank you so much for watching and I hope you found this helpful.

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Isomerism

Mass Spectrometry

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