Draw the peak that would correspond to the hydrogens at C₃ in ...

Question

Draw the peak that would correspond to the hydrogens at C₃ in the molecule shown. Be sure to indicate the integration and place it at an appropriate chemical shift.

Molecule structure with numbered carbon atoms 1 to 5, showing an ester functional group between C3 and C4.

Accepted Answer

All right. Hi, everyone. So for this question, let's go ahead and draw the expected proton NMR signal for the indicated hydrogen show the integration and chemical shift. For. So here we have two bits of information that we have to recall, right. But here the compound in question is going to be NN di metal Han amid. So let's go ahead and talk about what the integration in chemical shift actually refer to. Right, starting off with the integration. Now the integration represents the number of equivalent hydrogens that correspond to one specific signal because recall that equivalent protons don't split each other's signal and therefore correspond to just one signal. Now, the chemical shift, right, which is expressed in parts for million reflects the location of that specific signal in a proton more spectrum and the location, right, the chemical shift depends entirely on what type of proton we happen to be looking at. So let's consider that right. But first, let's go ahead and consider the chemical shift of the signal for the indicated hydrogens. So here, right, both of the indicated hydrogens are attached to the same carbon, which therefore means that they're going to correspond to the same exact signal. So in terms of the integration, the integration should be equal to two because there are two equivalent protons corresponding to our given signal. Now, in terms of the chemical shift, right, the carbon that's connecting these two hydrogens together is right next to our nitrogen here, which is relatively electron negative. So because of this right protons that are connected to carbons that are connected to nitrogen tend to appear at approximately 2.8 parts per million. Because in this case, right, the nitrogen is going to remove electron density from these protons, which will therefore expose them more to the NMR effects, right, which means that these signals are relatively DS shielded. So they appear more down field or more to the left of the spectrum given that they are all kilt protons. So our value for the parts per million, right, for the chemical shift is going to be approximately 2.8. And in this case, the integration is going to be equal to two because there are only two equivalent protons that will correspond to this one signal. But recall that when it comes to a proton and A R signal, we have to worry about the multiplicity, right, we have to worry about the splitting pattern of any given signal. So in that scenario, right, we can go ahead and use the N plus one rule because recall that the N plus one rule can be used when all neighboring protons are equivalent to each other. So if we consider the hydrogens that are indicated, right? And we also consider what non equivalent protons may be less than three sigma bonds away. Then we draw her attention to this methyl group here or this methyl carbon, right, because this methyl carbon has three protons, right, three hydrogens that are adjacent to the hydrogens that are indicated in red, which means that these three hydrogens would correspond to the three neighbors that are going to split the signal that we're trying to draw. So going back to the N plus one rule N is the number of neighbors. So that's three plus one, which gives me a sum of four. Now, because the sum is four, right, that's going to mean that the signal for the protons indicated in red is going to appear in the spectrum as a quartet. So the one signal is going to split into four smaller peaks. And according to Pascal's triangle, the individual split peaks are going to have a ratio of 1 to 3 to 3 to 1. So with all this in mind, right, we can go ahead and actually draw our signal. So the first thing I'll do is set up my X axis with my given chemical shift in the setter. So here I'm putting my 2.8 parts per million in the center. And from here I can go ahead and draw my signal. So when it comes to drawing the split peaks right after you figured out your multiplicity ensure that there is an element of symmetry to it, right, the the inner split peaks should be larger than the outer split peaks. And I'm ensuring to keep my ratio approximately 1 to 3 to 3 to 1. So now in order to reflect my integration, I'll be drawing the integration symbol to the left and a value of two and there you have it right here is my signal in the proton and a more spectrum taking into consideration not only the chemical shift, but also the integration and the splitting pattern. So with that being said, thank you so very much for watching and I hope you found this helpful.

Draw the peak that would correspond to the hydrogens at C₃ in the molecule shown. Be sure to indicate the integration and place it at an appropriate chemical shift.

Key Concepts

Chemical Shift

Integration in NMR

NMR Peak Shape

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