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Write the fragmentation pattern for the following molecule and identify the peaks for each fragment based on the mass spectra_CH_CH-CH CH,OH1 ButanolNASS SPECTRUM1...

Question

Write the fragmentation pattern for the following molecule and identify the peaks for each fragment based on the mass spectra_CH_CH-CH CH,OH1 ButanolNASS SPECTRUM1m/z

Write the fragmentation pattern for the following molecule and identify the peaks for each fragment based on the mass spectra_ CH_CH-CH CH,OH 1 Butanol NASS SPECTRUM 1 m/z



Answers

Show the structures of the likely fragments you would expect in the mass spectra of the following molecules:

So here it was just predicting fragments that we might expect in all mass spectra off the following structures. So when it comes to identifying fragments, the material before fragments has a electron any positive charge where, when they and then split one molecule will have radical on one molecule will be the captain with a positive charge. So when we're looking at, fragments were looking to make synthetic cuts with ah starting materials. So, as you can see here, I've done this already where we can cut it several from brunch points in order to generate several different fragmented pieces for all three off the species that we were looking at.

So we were looking at Animal Spectra again. So the first example we have is the following structure. So what we would see is just one single because all of these protons are equivalent would have an integration of six. In the next example, we have the following structure. So we have three different peaks. But I wanted about 9.1 point 931.883 point 39 So we have the following Patton. What about the PK? One would be a triplet peak at just before two would be a multiple on the peak after three. There's another triplet. But this trip here will be the biggest because it has the most protons in the environment. Here we have an integration of three integration of two integration of two. So in the next example, we have C l h h c l. So here we have a locked structure confirmation the lock structure because of the double bond. So these protons would be seen as equivalent. We would just have one peak and again the same body. We have a lot structure, so just see one peak again for the e and the eczema so we can draw the sad I cinema where again ppm. These protons are in the same environment because the structure is confirmation. We locked due to the double bond. Then we have e. So we have a benzene ring followed by nitrogen, w and oxygen and oh minus. So we have to chemically equivalent also to equivalent matter hydrogen and one power. So we have three signals in total. And, uh, we have the following. So we have a symmetric molecules. We have two signals. One of these protons, this one and this one and then needs to will be in the same environment too. So in the next example, this is getting quite complicated now, So I'll just list out the signals rather than trying to draw a sketch just so it's easier to see what I am talking about. So here we have four types of different signals. One for each of the protons. 1234 This is because the molecules not symmetric anymore. Last one to look at, we have the following. So here we have a symmetry. So all of these protons are in the same environment. And so therefore, on the ppm spectra, we will just get singler because they're all considered equivalent

This is the answer to Chapter 13. Problem number 31 Fromthe Smith Organic chemistry textbook. Ah, And in this problem that we're asked for each compound assigned likely structures to the fragments at each m z value. Ah, and explain how each fragment is formed. Okay, so for a we have this molecule that have drawn here. Um, and we were told to account for peeks at M to Z 104 and 91. So, um, we're looking for peeks at 104 And at 91. Um, and so again, the way that I approached these is to, uh, see what the difference between our molecular ion peak and these peaks is. So the difference between 1 22 10 for 18. So right away. Um, since there's an alcohol here that tells me that's gonna be a dehydration, Um, since waterways 18. And the difference between 1 22 and 91 is 31 which to me is 18 plus. Ah. Um, well, okay, so actually, let me take that back. So, to me, that's 17 plus 14. Um, and from an earlier problem similar to this one, I explained that um, I always think of these in terms of subtraction. So, um, a method group is gonna be worth 15 and a methylene group is gonna be worth 14. Um, so this 31 tells me ah, that we're losing a methylene group and we're losing on alcohol. Um, so alcohol's 17 methylene is 14. Um, okay. And so with that in mind, I'm gonna go ahead and draw what I think is happening here. Um and so if we were to do a dehydration, um, so the dehydration is going to be minus water. As the name implies, I'm in. So our product here, it's going to look like this. Eso we'll have a double bond here as a result of the dehydration reaction. So ch ch two. Um and then we'll have the neutral radical. Um, okay. And so the mtz here, uh, does, in fact, equal 104 So that's where are one of four comes from, Um, and our other, um are are, uh, p get 91 is going to be in the form of this molecule. Um, and like I said, um, I'm thinking we're gonna cleave a methylene group and a no age group, So well, Cleveland, right here. And so that product is gonna be this C six h five c h each and there will be a carbon cat eye on right there. Um and so the mtz here is, in fact, 91. Okay, Um, yeah, and so I think that that's ah, good way to approach this type of problem. Um, you know, think about what your molecular ion is on then the molecular ions that you're asked to account for. What? The difference is between those ions and the molecular ion on and then sort of rationalized what groups you could cleave off, um, to account for that difference. So I'm gonna go ahead and apply the same logic to part B here. So our MTZ here is 86 were asked to account for four different ions this time. So 71 68 41 and 31. Okay, um and so right off the bat, uh, 71 is a difference of 15 from 86. And so this to me suggests losing a meth a group. Uh, if you remember from the previous page, a metal group equals 15. So that right there is cleavage of a metal group. The 68 is going to be dehydration, Right? Because it's a difference of 18 from 86 so this is gonna be minus water. 41 is 45 different from from 86. So how can we account for that? Um okay, so this is gonna be a difference of 45. Ah, the next one is gonna be a difference of 55. So these air obviously, um, fairly big cleavage is that we're making, um, since work leaving. Ah, about half and then more than half of the molecule in each case. Um, yeah. Okay. So in any event, well, we'll come back to those two. Okay? So the 1st 2 are very straightforward, right? So, um, we can start with the dehydration. So the dehydration is gonna be minus water again, as I've said, um, and so the product there is going to look like this siege to still have a double bond here. You started too high. S o C. H two. We still have our double bond when we moved is down a little bit. All right. Still have our double bond still have our metal group on. And so this will now be a C h. Okay. And then we have, ah, new double bond there. And so that's gonna be, uh, the dehydration product. Um, and of course, to really be correct, we need to put this in parentheses. Ah, Input. Uh, this neutral radical. OK, so that's that's our dehydration. So then we're gonna have three different cleavages, so make a trident type arrow here. Okay? Um and so our first cleavage now we already know is gonna be this. Uh, this mtz of 68 is our pardon me of 71 is just cleaning off a method group. So here, let me let me conclude this this is the MTZ equals 68. That's a dehydration. So this is gonna be the m to Z equals 71. So it's gonna look like this ch to see. See? Age too. See age too. Oh, each, um and we will have a corporal cat eye on there. Um, okay. And so this is going to be MTZ equals 71. Okay, so then we should think about where else we could cleave this molecule that's gonna get us our 41 our 31. Um, so there's not too many other spots, really, That we could cleave this. So I'm gonna go ahead and say, Well, let's think about, uh, cutting this bond and then this Bond. I mean, so if we cut that first bond there, the one right at the double bond, we're going to get something that looks like this ch to see ch three with a positive charge on this carbon on. And let's add that up and see what that's gonna look like. Um, so three carbons is going to be 36 5 Hydrogen is is going to be five. Morris, that's 41. Okay, so here's our mtz equals 41. And if you want to go back and rationalize it the way I started to at the beginning, what are we removing? We're removing two carbons for 24 five hydrogen cz for 29 then an oxygen for 16 MB, or for a total of 45. So this is gonna be minus ch to see h 20 H. Okay. Um and so I'm gonna go ahead out on a limb and say that, huh? I know that. Yeah. Okay, Um, no, that wouldn't work. Okay, so let's see. Okay, I see what's happening. So, um, if we were to cleave this second bond that I'm now going to arrow, um, if we were to Cleveland, But to put the carbo cat eye on, um, on this carbon that I'm gonna arrow in green, I have the carbo cat eye on Were there that would give us Ah, peak of 31 let me draw for you exactly what I mean. So if we were to have this ch two Ohh. With the cargo cat eye on on this carbon, that is gonna be a total of 12. Plus three is 15 plus 16 is 31. So that is where our MTZ 31 comes from. Okay? Um, yeah. And so that's Ah, that's the way to approach this problem. I like working backwards, as I started to do on the left over here. It certainly made the 1st 2 ions that we're looking for. Pretty apparent on dhe. Then too, get to the next to ions. We just had to think about what other bonds we could cleave. Um, And then when we do so, those become pretty straightforward as well. And yeah, So that's the answer to Chapter 13. Problem number 31

This is the answer to Chapter 12. Problem number four from the McMurray Organic chemistry textbook. This problem asks us to list the masses of the parent ion and of several fragments we might expect to find in the mass spectrum of the molecule that were given eso every drawn, that molecule. It's to metal to Penhall. Um and so we need to think about what types of fragmentation this molecule can undergo. But I guess the first thing that we should do is t count up. Um, all of the carbons hydrogen is and the oxygen in this molecule, um on Then use their am you values. So 12 for carbon, one for hydrogen, 16 for oxygen on. Add them all together. And that is going to tell us, Um, the mass of the parent I on here s of the M plus Peak is going to be 102. So we have six carbons, 14 hydrogen xenon, oxygen on. So when you do the math error, it adds up to 100 to, um okay, and then we need to think about possible fragmentation. Here s so this is an alcohol. That's the only functionality in this molecule. Um And so remember, from now this chapter of the textbook, that means that we can potentially see dehydration. Ah, and Alfa cleavage. And so, um, there are two possible Alfa cleavage pathways s so that the first pathway, which I will show in blue eyes gonna be this dehydration here. So the loss of water on and so how blue arrow that down. Then we can also get out for cleavage here, Uh, this carbon carbon bond or, um and probably more likely we could get Alfa Cleavage here at this carbon carbon bond. Okay, um, and so it's looked the Alfa cleavages first. Eso if we follow the red pathway if we break are the bond that I drew a line through in red What we would get, um, would be a methyl radical. So here's our radical. And remember, Since this isn't charged, we're not going to see it on the mass spectrum. Um, but what we will see is the rest of the molecule, since that's the charged fragment that is going to this. Okay, on this will you can't ions will have a positive charge there. Um, and the M two z for this fragment Eyes going the 87 to remember loss of a metal group, um, is loss of 15 atomic mass units. So our M plus people's +102 we've lost methyl group saw and two Z here is 87. Okay, um, the other alfa cleavage pathway that we can undergo. Ah, would be breaking the green bond there. Ah. And so when we break that, um, we're going to get a Pro Bowl radical. Um, again, there's no charge on this radical, so we're not actually going to see it. Um, And then the rest of the molecule, which we will see, I will look like this cake. And this is our cat eye on. Which is why it's the fragment that we see and the M two Z for this fragment. Eyes going to be 59. Okay. And then lastly, um, as I said, we can we can undergo Ah, dehydration. Ah. And so the product of that dehydration is going to this. Okay, um and so this, uh, will be the charge species here, Um, as well as being the radical, um, we'll also get water as the name dehydration implies. So we've lost water. I'm and in the MTZ for the charge. Species charged fragment eyes going to be 84 on and again. Remember that dehydration loss of water? Uh, you lose, um, 18 atomic mass units as a result of that. And so, with an M plus peak of +102 this fragment having lost 18 due to water, the MTZ will be 84. Okay. Ah, and that's the answer.


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