How might you use ¹³C NMR spectroscopy to different...

Question

How might you use ¹³C NMR spectroscopy to differentiate between the possible ortho, meta, and para products of the electrophilic aromatic substitution reaction shown?

Chemical reaction: toluene with Cl2 and AlCl3, arrow pointing to a question mark, indicating unknown product.

Accepted Answer

All right. Hi, everyone. So this question is asking how would you differentiate between the Ortho and para products of the following reaction using carbon 13 NMR only. And here we have a bro reaction of Tolu to produce one bromo, two methylbenzethon. So here we've got the Ortho product in which the methyl groups and bromine are right next to each other on the benzene ring. And we have the per product where the methyl group and the bro are on opposite carbons on the benzene ring. So recall really quickly that the beauty of carbon 13 and Mr is that it gives us insight into how many different kinds of carbons there are in a given compound because the number of signals in a carbon 13 NMR spectrum corresponds to the number of non equivalents carbons present in the molecule itself. Now, it's important that I mention non equivalent carbons because of the effect that symmetry has on a carbon 13 NMR spectrum and also a proton in our our spectrum. Because when a molecule has an element of symmetry to it, at least one, then multiple carbons can be said to be in the same chemical environment because they would be surrounded by similar types of carbons. So in the case of a molecule that's symmetrical, you would actually see less signals than there are carbons in the molecule itself. So if we're using carbon 13 NMR to differentiate between our or Ortho and our para products, then what that means is that the number of signals must be different, the Ortho and the para products must have a different number of signals to them, which is how we would be able to differentiate and that would be accurate, right? That's the point. And we can tell by considering the structures of our Ortho and para products themselves because notice how the Ortho product has no symmetry to it. Whereas the per product does because if we consider our Ortho product, first of all right, like I said previously, there is no element of symmetry to the Ortho product itself. So therefore, the six carbons of the benzene ring and the methyl carbon would all be non equivalent, which means that we would actually go ahead and see a total of seven different signals in its carbon 13 NMR spectrum because all seven carbons are not equivalent to each other. Now, on the other hand, right, our pair product does have an element of symmetry to it because we can imagine a line being drawn vertically across the benzene rink because that separates our pair product into two mirror halves, right, two mirror images of the other So when we imagine that the number of signals in the carbon 13 naar spectrum is reduced because for example, multiple carbons within the benzene ring would actually be equivalent to each other. Because first of all right, our methyl carbon here would correspond to its own signal because it has its own chemical environment. As would the carbon in the benzene ring that is attached to the methyl carbon. In this case, carbon four of the bending ring. No here, right. I've labeled carbon four of the para product as signal B and Signal B, carbon B is connected to two others. And those two others would correspond to the same signal because once again, right, this molecule is symmetrical. So these two carbons would be equivalent. And that's why I'm going to label both of them c now each carbon labeled C is bound to yet another one within the benzene ring. And once again, right, same chemical environment because they have the same neighbors, right? They're bound to the same type of carbon. So our next signal is labeled D and at the last but not least, right, carbon, one of the benzene ring bound to bromine gets its own distinct signal because it's the only carbon connecting to Bromy. So our para product has five signals associated to it. And there lies the answer to this question right. The way in which we would differentiate between the Ortho and the para products is by counting the number of signals in their carbon 13 NMR spectrums, right? Because whereas the Ortho product, the Ortho product will cause seven signals while the per product will cause five signals. And the reason for that difference, right, the reason why the para product has less signals in the Ortho product is because the para product has an element of symmetry to it. And that element of symmetry means that multiple carbons might correspond to the same chemical environment and would therefore correspond to the same signal and there you have it. So with that being said, thank you so much for watching. And I hope you found this helpful.

How might you use ¹³C NMR spectroscopy to differentiate between the possible ortho, meta, and para products of the electrophilic aromatic substitution reaction shown?

Key Concepts

Electrophilic Aromatic Substitution

¹³C NMR Spectroscopy

Chemical Shift and Splitting Patterns

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