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Three IR spectra are shown. Three possible chemical compoundsare also shown. Match each compound to its associatedIR spectrum. Label and identify two peaks that hel...

Question

Three IR spectra are shown. Three possible chemical compoundsare also shown. Match each compound to its associatedIR spectrum. Label and identify two peaks that helped youmake the association between structure and spectrum. The first oneis done as an example.

Three IR spectra are shown. Three possible chemical compounds are also shown. Match each compound to its associated IR spectrum. Label and identify two peaks that helped you make the association between structure and spectrum. The first one is done as an example.



Answers

The IR and mass spectra for three different compounds are shown below. Identify each compound.

So absorption is 824,720 suggests we have all the hide. 3000 absorption suggests we have hydrogen attached to SP three rods and SP to hybridize carbon. So therefore, from this analysis, we may determine to have two missile butor are behind which is drawn just below moving on to our second example here we have broadband at 3300 corresponding to an n H group. This is confirmed by the 1600 absorption. Secondly, there is an absorption at 1280 which is a me throw. Therefore we may have pro pro No a might present Well, we have that a might functional group moving on to our last example Here we have an absorption absence at 1600 meaning that there is no benzene ring. We don't have any meats are groups either are suggested by the absorption at 1380. Strong absorption at 1700 though, indicates the presence of a key tone. So our structure may in fact be set cloak Ito. Where are key tone is a carbon I'll that has to armed groups where neither off the ah group or protein protons on. We have six members Cyclical ring will the single bonds

This is the answer to Chapter 19. Problem number seven Fromthe Smith Organic chemistry. Textbook on this problem says, Explain how you could use I r spectroscopy to distinguish among the following three compounds. So we're talking about the three molecules that have drawn here. Um, and we're talking about how to use I r to distinguish between the three of these. Um and so basically, what we need to do is look for the signals that are going to be different for each molecule here. So they're all gonna have sp three hybridized carbon bound to hydrogen. So they're all gonna have i our signals around 28 52 to 3000 wave number. So we can safely ignore those signals. They're not gonna help us here. However, for this first molecule, um, our function are functionality here. Eyes Oops. Wanted that in green. So our functionality here is a car box, Silly Gassid, and, um, in the i r of a carb oxalic acid. We're going to have a peek around 17. 10 wave number. Um, and that's gonna be our carbon oxygen double bond. Ah, and then we're gonna have a second peak. Um, it probably Ah, wide broad peak somewhere between 2500 and 30. 500 wave number. Um, And that is gonna be due to the O h of a carb oxalic acid. So remember, that's ah, it's a little different than the O. H oven alcohol, but it's sort of around the same region. And so if we have this carb oxalic acid, we should see both of those signals. Um, so the middle molecule here is an esther. Um, And since it's an Esther, the, uh, carbon oxygen single bond, um isn't really gonna show up, But we will get that strong characteristic Carbonell Peak. Um, right around 1700. So we'll see one signal of interest here right around 1700. And that's how we would be able to tell that this is an Esther. Um, and so this third molecule, uh, the functionality of interest is gonna be the alcohol. Um, and so this is gonna be ah, broad peak between roughly 3200 and 3600 wave number. Um, and that is again gonna be due to this. Uh oh. H and so that's That's how we would tell. So if we see one peek on Lee between 32 3600 we would know that we have the third molecule. One peek on Lee. Um, around 1700. We know that we have the Esther. Ah, and if we see two peaks one and roughly each of those regions. So again, right around 1700 for the carbon Neil. Um, and then maybe maybe a little lower than the alcohol's 253,500 for the O. H of a carb oxalic acid. And so that's how we would sell that. We have the first molecule. Ah, and so that's Ah, that's the answer here. We just need to know that different functionalities air going to give us different I our signals. Ah, and that's the answer to Chapter 19. Problem number seven.

This is the answer to Chapter thirteen, problem number forty from the Smith Organic Chemistry textbook. And so this problem is giving us six pairs of molecules and asking us to identify what the differences would be in there. I are spectrums. And so for a we have a, um cycle Pantene Versace. Um, thiss out kind. And so the cycle Pantin was going to show carbon carbon double bond at roughly sixteen. Fifty and C S p to hybridized hydrogen apartment. Yes, Oh Spoto Iraq's carbon hydrogen bond. And so that's going to show between three thousand and thirty one fifty. Thie al Kind is obviously going to show the carbon carbon triple bond at twenty two fifty and the C S P H bond at thirty three hundred for be. The main difference in these molecules is that the first one, the carb oxalic acid, is going to have that O H. Bond. And remember in a car box Hill Gassid. It's hard to say exactly where that will show, but it will show above three thousand wave number. The second molecule for B does not have a no h bond in it, and so there should be no signal above three thousand, and so that makes that pretty straightforward to identify. Moving to C A C is also very straightforward. There should be the distinctive key tone signal in the first molecule at around seven hundred wave number. The second molecule and see should show carbon carbon double bond at about sixteen fifty two C S P to H bond again between three thousand and thirty one fifty. And the oh age bond on this O. H is not a carb oxalic acid as one and B wass. And so we know it should show between thirty, two hundred and thirty six hundred wave number, and it should be a nice broad peak for D. The second molecule and day has, ah, carbon double bonded to an oxygen. So again we should see that right around seventeen hundred on the first molecule, indeed, doesn't have a carbon double bounded to an oxygen. So again, that's very straightforward to identify. If we were given an I R. On knew that it had to be one of these two molecules, we could look right at seventeen hundred, and if there's a peek, it's the one on the right and If there's not a peak in seventeen hundred, it's the one on the left, so moving on e So when he this first al kind is symmetrical and so we're not going to see the carbon carbon triple bond signal because the molecule is symmetrical. But for the second molecule in a, we will see the carbon carbon triple bond. And so that should show at twenty two. Fifty in the C S. P. H bond should show it thirty three hundred for F again we're looking at you know how to distinguish these two molecules. And so the first molecule in F has a ness pee through. Pardon me. Jeez, The first molecule in F has an S p hybridized carbon bonded to a hydrogen which will show it thirty three hundred. And the second molecule in F does not have a nasty hybridized carbon down to ah ha Jin. So there will be no signal at thirty three hundred. And that's how you differentiate them on DH. This is, you know as faras experimental organic chemistry. This is one of the main uses of I R. If you know what you should have, what functional groups should be there. It's very straight for forward to look at an I R and see if the signals are where they're supposed to be. And so that's sort of what we use this for an industry, a lot of time. And so that is the answer to Chapter thirteen. Problem number four.

So here we have been absorb ins are 3300. So we have an uncle group present. We lack a peak at 1700 wave number. So we do not have a carbon ill. 3000 wave number corresponds to hide regions that are attached to SP three carbons. Therefore, we may deduce that our structure is built on one all so beats out meaning force. We have full carbons in our chain. 123 full on our first carbon. We haven't alcohol substitute which is O h being on to our second structure. Now we don't have an absorbent 3300. Therefore, we do not have an alcohol functional group. We do in fact, have a peek that is lesser than the expected average carbon on group. So therefore are benzene Ring may be responsible for this so we won't have me thout penal Keaton which is drawn out below the last example. Now we do not have an alcohol functional group we have absorb Ince's are 3000 corresponding to hydrogen to touch to SP three carbons we have absorb Ince's are 1700 suggesting a carbon I'll so therefore we may have four teeth out Cyclo hacks unknown where we have a key tone present in all six member ring where on off fools carbon We haven't a Thile substitution.


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