Draw the ¹³C NMR spectrum you would expect to see f...

Question

Draw the ¹³C NMR spectrum you would expect to see for each of the molecules shown.

(b) Chemical structure of a bicyclo[3.1.0]hexane molecule, featuring a six-membered ring with a three-carbon bridge.

Accepted Answer

All right. Hi everyone. So for this question, let's go ahead and provide the estimated carbon 13 and a more spectrum of the following molecule. And the molecule in question is propan to iodine cyclo heine. So recall first and foremost, right, that the number of signals in a carbon 13 and a more spectrum is going to correspond to the number of so-called non equivalent carpets. And the reason we say non equivalent carbons is because recall that whatever a molecule has an element of symmetry, then multiple carbons are said to be in the same chemical environment because they have the same surroundings due to that element of symmetry. So because of that, right, multiple carbons can actually correspond to the same signal. So it's important to distinguish that carbon 13 won't tell you how many carbons there are in total, but it will tell you how many different kinds of carbons there are in the compound in question. So the reason I brought up symmetry is because this molecule right here actually does have an element of symmetry to it, right? Because I can draw an imaginary plane down the middle of this molecule vertically. And I will have two mirror images on either side of my divide. So because of that, right, I end up with a few different unique signals, right? And the number of signals will be less than the total number of carbons due to the symmetry. So my first signal actually let me go ahead and erase this divide to make it easier to see. But my first signal would be this carbon on the top here on the opposite side of my double bond. And that one, I'm going to label A. Now, after that, we have two carbons on either side of A that are actually equivalent due to the element of symmetry. So I'm labeling them both B and the same thing can be said for another set of two carbons next to both of the carbons that I labeled beat. So there we have C and each carbon of the double bond itself constitutes its own signal. So we've got C or D and E excuse me. And then last, but not least we have two methyl carbons bound to E which once again would correspond to the same signal because they are bound to the same carbon and the molecule is symmetrical. So both of them can be labeled F. So at this point, right, we can tell that we're going to have a total of six signals, right? We have six non equivalent carbons in the structure of the molecule given on the screen. So at this point, we can go ahead and draw out our NMR spectrum. And first, I'm going to start with our skill and recall that the parts per million scale, the chemical shift of a carbon 13 spectrum is actually larger than that of a proton and Mr spectrum. So I'm going to go ahead and count from approximately 140 parts for million. And I'm going to proceed downwards until I get to zero. And I'm going to do so in increments of 20 parts per million, just counting down here and here I've arrived at zero parts per million, but I'm going to go ahead and make smaller tick marks in between just to make the signals a little bit easier to scale. So at this point, I can go ahead and draw my actual peaks now that I've organized my scale. Now recall that the location of the signal, right. Their chemical shift is dependent on what other atoms they happen to be bound to, right? Because recalled proximity to electron atoms or electron rich areas will de shield the atom in question, right? It will de shield that signal, which means that that signal would appear more downfield or more to the left side of the spectrum itself. So the question then becomes which signal or signals will appear most downfield and the inch to that would be carbons D and E because carbons D and E are the alkene carbons, which would therefore appear more downfield than Alcaine carbons. However, right, carbon D is bound to carbon seat or carbon seat and carbon C are each bound to a carbon B whereas carbon E is only connecting to the methyl carbons labeled F. So because of that, right, carbon D will be slightly more downfield, then carbon eat. So alkene carbons tend to appear and I'm actually going to write that on the screen here. If I scroll down, recalled, alkene carbons tend to appear at approximately in between about 100 to 150 parts per million. So because the signal for carbon D is expected to be more downfield relative to that of E right, the signal for carbon D, I'm going to put at approximately 130 whereas carbon E, I'm going to go ahead and put next to it. And now that we're going ahead and talking about which carbons are downfield versus upfield, like I said previously, right, carbons F are methyl carbons, right? So they've got only three hydrogens connecting to it. So, in comparison to the other carbons in the structure of this molecule that are bound to other carbons, the one labeled F should actually be the most upfield or the most to the right of the spectrum itself. Because if I scrolled down, right, methyl carbons tend to appear in between zero and 35 parts per million. So very much upfield and for our structure, I'm going to go ahead and put them at about 20 parts per million. Now notice how there are two carbons labeled F. So therefore, the peak itself in the carbon 13 NMR spectrum should be taller relative to the signals for carbons DNE because there's only one carbon D and one carbon E. So if there's two carbons labeled F, the corresponding peak or signal F should be taller and now we can go ahead and group the remaining ones together are not together but put them in the spectrum. The carbons that we have yet to place in the spectrum are A B and C. So among those three options, right, the two carbons labeled C should be the most downfield of the three that we're discussing because they are the ones directly bound to the alkene carbon deep. So carbons A B and C are all examples of sp three hybridized carbons that are also secondary, right? So they generally appear and again, this is an approximation, but they generally appear between 15 and 55 parts per million. So it's important to make sure that my signals here do not fall beyond this range. So because signal C is expected to be the most downfield of the bunch, I'm going to go ahead and place it at around 30 and recall what we discussed previously about the peak height, right? Because signal C like signal F has two carbons associated to it. Now, right next to signal C should B signal B and signal B should be the same height as both C and F. Whereas signal A given that it's the farthest away from the double bond should be after, but before signal F, so I'm going to go ahead and place it in between. But I'm also going to make sure that they're the same height. A signal A is the same height as DNE because there is only one carbon A. And at this point, we've gone ahead and arrived at the final EMR spectrum, carbon 13 EMR spectrum. So by looking at the structure itself, right, due to the element of symmetry associated to it, we were able to tell that there were six non equivalent carbons in the structure of propan two iodine cyclo heine. And therefore, there were going to be six peaks in the overall carbon 13 spectrum. So according to what type of carbon they were, we could recall approximate ranges of chemical shifts to take into consideration where to place them. And then from there, the the height of each signal is dependent on how many carbons correspond to that signal. So with that being said, if you stuck around to the very end, thank you so much for watching. And I hope you found this helpful.

Draw the ¹³C NMR spectrum you would expect to see for each of the molecules shown.
(b)

Key Concepts

¹³C NMR Spectroscopy

Chemical Shift

Symmetry and Equivalent Carbons

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