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.

(c) Chemical structure of acetophenone, showing a benzene ring attached to a carbonyl group.

Accepted Answer

All right. Hi, everyone. So for this question, let's draw the estimated carbon 13 and a more spectrum of the molecule shown below which is Benz aldehyde. So recall first and foremost, right, that when it comes to a carbon 13 and a more spectrum, the number of signals that you see is going to be equal to the number of non equivalent carbons present within the molecule itself. Now, it's important to recognize that we have to look at the number of non equivalent carbons within the molecule, right? Because recall that when a structure has a plain of symmetry that results in less non equivalent carbons than you might expect to see. Otherwise take for example, Benz aldehyde, right. Benz aldehyde has an internal plain of symmetry in the benzene ray itself. So therefore, right, we have a total of five non equivalent carbons even though we don't have just five carbons in the molecule itself. So carbon A can be the carbon of my aldehyde and then I can go ahead and label the four non equivalent carbons of my ring. So given the plaintiff symmetry, right, I end up with a total of five non equivalent carbons and a total of five signals in my carbon 13 NMR spectrum. So then the question becomes, where do I put them in my spectrum? Right. So let's start with carbon A. Now recall that because carbon A is the carbonyl carbon of the aldehyde, it is going to be the most downfield due to the electron negativity of the oxygen atom. And recall that aldehyde carboy carbons generally appear between approximately 1 90 to 200 parts per million in the carbon 13 spectrum. That is the approximate chemical shift. Now because carbon B is directly adjacent to the carbon, it's going to appear more downfield due to the presence of the neighboring carbon group. So carbon B is going to appear up next. And now we have to consider our remaining three carbons, right CD and E no recall really quickly that the carbon of benz aldehyde is what we consider an electron withdrawing group or an ewg no ewgs decrease the electron density at both the PAA and the Ortho positions of the ring itself. So therefore, right, carbon C and E are going to appear more downfield than carbon D. However, carbon E being the the paras position is going to appear most downfield because it is most electron deficient. And we're going to follow follow that up with carbon C which is the the orthos position, carbon, the para and then carbon D is going to be the most upfield. So because carbons B through D are all going to be part of the ring here, recall that aero carbons generally appear between approximately 110 to 170 parts per million. So what that means is that carbons B through D or rather their peaks have to appear within that range, but we have to make sure that they are a reasonable distance apart so that we can differentiate between each distinct peak, right? So with that being said, we can go ahead and draw the spectrum itself and withdrawing the spectrum, I'm going to go ahead and proceed with my scale with my chemical shifts. So here I'm going to start from 200 and I'm going to go all the way down to zero in increments of 20 parts per million. So here I'm just adding tick marks and labeling them accordingly. And now I've gotten down to zero parts for a million. However, I'm going to add unlabeled tick marks in between so that I can better estimate where to put my peaks. So now let's go ahead and start with the most downfield peak after I draw my baseline of the carbon 13 NMR spectrum itself. Now the peak that's going to appear most downfield is going to be the car bottle of my aldehyde, which I'm going to go ahead and place at around parts per million. No recall that in a carbon 13 and a more spectrum you don't necessarily have to worry about splinting, right? Because all peaks appear as singlets. So that's why I'm going to go ahead and just draw one line for each P. And then from here, I can go ahead and draw my next peak, which is going to be Carbon B in the rink. So I'm going to go ahead and draw that one just before 140 parts per million. And then I'm going to use, I'm going to go ahead and use the peak for Carbon B as kind of a guide, right? The peaks for carbon's EC and D have to appear a little bit more upfield, but they have to remain within their respective range. So I'm going to go ahead and draw carbon E right next to the peak for carbon peak. So here I have carbon A on the very left followed by B followed by E. So now because carbon C and D are relatively similar, right? They're going to be very, very close together in the carbon 13 and Mr spectrum. So I'm going to draw the peak for carbon C first given that it is closer to the carbon. So here is the peak for carbon C and then I'm drawing the peak for carbon D in very close proximity to the one for carbon C and there you have it. Now, what I want to point out here is the fact that I made carbon C and D or rather I made their peaks slightly taller than the ones for A B and E. And that's because the height of a peak in a carbon 13 NMR spectrum to a certain extent can tell you how many equivalent carbons there are in that molecule. So due to the element of symmetry in Benz aldehyde, there are actually two equivalent carbons for both PC and D. So because there are more of those types of carbons, the peak is going to appear slightly taller in the carbon 13 NMR spectrum, whereas A B and E each have one carbon, so their peaks are going to appear a little bit shorter. But with that being said, here is the final spectrum and there you have it. So if you stuck around to the end, thank you very much for watching. And the main takeaway, right? When it comes to constructing an NMR spectrum or carbon 13 NMR spectrum, your first step is to figure out the number of non equivalent carbons and then place them in the carbon 13 NMR spectrum depending on their approximate chemical shifts. So, thank you very 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.
(c)

Key Concepts

13C NMR Spectroscopy

Chemical Shift

Symmetry and Equivalent Carbons

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