Without worrying about the relative location of the signals (i...

Question

Without worrying about the relative location of the signals (i.e., the chemical shift) or the splitting patterns, draw a spectrum of the following molecule. Be sure to label each signal based on the set of equivalent hydrogens to which it corresponds.

Chemical structure with labeled hydrogen groups: a, b, c, d, e, indicating equivalent hydrogens for NMR analysis.

Accepted Answer

All right. Hi everyone. So this question is saying to draw a spectrum of the given compound without worrying about the chemical shifts and the splitting of the peaks remember to label each peak according to the set of equivalent protons it belongs to. And our starting material is Isobel pli. Now, I apologize for the formatting, but our task here is to draw specifically a proton NMR spectrum. So because the problem is requiring us to draw a proton NMR spectrum without concerning ourselves with chemical signals or splitting patterns. The only thing we have to discuss here is the number of signals that is Souto palate is going to produce. So once we determine the amount of signals, our only task would be to draw however many peaks in the resulting spectrum. So without specifying either the parts per million, the chemical shift and without specifying the multiplicity either. So we're only looking for the number of signals. So recall that the number of signals in a proton NMR spectrum corresponds to the number of non equivalent protons we have in the molecule itself. Now, what that means is that if the substitution of any group of hydrogens produces the same exact compound, then those hydrogens are considered equivalent. And where when they are equivalent, right, they correspond to the same chemical environment and would therefore correspond to the same exact signal. So multiple protons can actually correspond to the same signal because of this. And let's go ahead and showcase this by looking at our iso Beal pol here now is Souto palate. First of all has one ordinary carbon in our church beetle group. And because it has no protons, that ordinary carbon is not going to produce a signal. However, we do have three methyl groups attaching to the same exact ordinary carbon. And because of that right, substituting any of the nine protons across all three methyl groups would in fact result in the same compound and are therefore considered to be equivalent. So now if we focus our attention on the Isobel group here, that's on the right side, then right next to the oxygen, we've got one methylene carbon, right, because this carbon here has two protons associated to it. Now, because those two methylene protons are on the same carbon, they are also considered to be equivalent. So that's an additional signal. And in addition to that, we've got another tertiary carbon right next to the methylene carpet and that tertiary carbon has its own proton that corresponds to its own chemical environment. And last but not least attaching to the tertiary carbon on the very right side of the structure are two methyl groups right here. Now, because these methyl groups are attaching to the same carbon, substituting any of the six protons across both metal groups would also result in the same compound, which therefore means that these two methyl groups are equivalent to each other. So when we go ahead and re tally all this, right, all these circles that I've drawn in different colors, I've drawn four different colors, which means that I'm going to have a total of four signals because I have four sets of non equivalent protons in this molecule. So now let's go ahead and draw my spectrum. And again, we're not too picky about what the spectrum looks like because our only task is to make sure that there are four signals depicted. So I'm starting off with my baseline, right. And I'm going to go ahead and draw four peaks spaced out just enough. So you can distinct them from each other or distinguish them. So if I go ahead and I draw a redraw, actually the structure of Isobel palate here, then what I can do is I can match each signal to each set of protons going from left to right. So going from left to right, I will be labeling all of my signals A through D and in the structure of Isobel Pate itself, I'm going to go labeling by non equivalent or rather my signals, excuse me A through D going from left to right as Well, and right here is going to be our final answer because in this scenario, what we just did was successfully depict the fact that Isobel Pol should have four signals in the proton NMR spectrum. And the reason for that is because Isobel palate has four sets of protons that are distinct from each other. So they would result in four distinct signals and there you have it. So with that being said, thank you so very much for watching. And I hope you found this helpful.

Without worrying about the relative location of the signals (i.e., the chemical shift) or the splitting patterns, draw a spectrum of the following molecule. Be sure to label each signal based on the set of equivalent hydrogens to which it corresponds.

Key Concepts

Equivalent Hydrogens

NMR Spectroscopy

Chemical Shift

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