How would you expect the IR and ¹H NMR spectra for propanamide...

Question

How would you expect the IR and ¹H NMR spectra for propanamide and N,N-diethylpropanamide to differ?

Chemical structures of propanamide and N,N-diethylpropanamide, highlighting differences in amide groups.

Accepted Answer

All right. Hi, everyone. So for this question, lets go ahead and enumerate the differences between the Ir and proton en spectra of butter amide and those of NN di ethyl butter amide. So recall really quickly that both Ir and proton R are essential for structure determination in different ways. First, the reason why we rely on Ir spectroscopy is to identify what primary functional groups are present within a molecule. So things like car bals, things like hydroxy groups, et cetera, but the proton R spectra, the proton and a more spectra is essential for letting the snow how many no equivalent protons exist within the molecule. Now, that is essential to help us determine, for example, what functional groups might be close together as well as what types of carbons might be close together. Because the proton NMR spectra will also let us know the multiplicity which is dependent on how many adjacent protons there are relative to one specific one. But with all this in mind, right, we can go ahead and acknowledge here that both butter amide and NN di ethyl butter amide have a very simple or similar carbon chain. Excuse me, because they all have a four carbon chain alongside their car bono. So by understanding what makes them different, right, kind of ignoring this similarity or not ignoring. But by examining the differences, we're better able to distinguish which is which, right. So let's go ahead and start with Beter Aide. And when it comes to the most important ir bands to look for in beauty amide, one of the first ones we're going to check for is that of the carbo right now, recall that the carbo band in the IR spectrum tends to appear at around 1685 inverse centimeters or around 1700. But in addition, right, with bu aide, we can check for one more thing specifically and that is the band that represents the bond between nitrogen and hydrogen over here, right? And generally, we tend to see that at around 3300 inverse centimeters in an IR spectrum. But in addition to both of these, we can also check for the band representing the bond between our car bono carbon and nitrogen at around 1000 in centimeters. But when we go ahead and we examine the structure of NN Diho Beter amide, what we'll notice is that our hydrogen here, excuse me, our nitrogen is tertiary, right? And it's tertiary because there are three other carbons bound to it directly. Specifically, there are no protons attached to the nitrogen of this amide, which means that we would not see the corresponding band in the IR spectrum. So the band at around 3300 inverse centimeters that represents the bond between nitrogen and hydrogen would not be present in the IR spectra of nn die eo butter A mi. So that would be the difference between them. Bye. You can identify or differentiate between our two amides by checking for whether or not the NH band is present in the IR spectrum. So now that we've gone ahead and discussed the differences in the IR spectrum, the differences observed in the proton and R spectra are going to be similar, right? Because in the proton and more spectra of butter amide, we would go ahead and we would see a peak representing the protons directly bound to nitrogen and recall that those tend to occur in between around 6 to 8 parts per million. So those would be characteristic of the amino group of the amide. But NN diho butter amide has no such protons instead, we have these two EHO groups. So because we have the two ethel groups and not two protons, we would see two different peaks in our proton and Mr spectrum specifically, we would see additional peaks for the di ethyl groups in NN Diho Beter Aide. Now, because our ethel groups here are all groups connecting to a an electron atom like nitrogen. Those are going to appear approximately between 2 to 4 parts per million. And there we have our primary differences between these two compounds. So for the sake of organization, I'm actually going to go ahead and organize them a bit. I'm going to write them on the bottom of the screen here. So when examining the differences in both the IR and proton NMR spectra, the primary difference between both amides and the IR spectra is that there would be no nitrogen hydrogen band at a round 3300 inverse centimeters four nn Diy beer aut. And the reason for this is because the amide nitrogen has no protons, which means that we wouldn't see any in the IR spectrum. But for the proton and a more spectrum, well, for the proton and a more spectral, there would be no nitrogen hydrogen bands at around 6 to 8 parts per million four nn die ethel be a mind. And on the other hand, there would be additional NN diho peaks at around 2 to 4 parts per million four nn oops die eo be a mi, right? So you wouldn't see any bands representing protons bound to nitrogen, but you would see bands corresponding to all kill groups bound to an electron nitrogen and there you have it right. So the bottom line here for this question is that by examining the differences between two given compounds, you can identify or rather differentiate one from the other by examining what peaks may be missing or what other peaks may be found in one structure compared to another or one spectra compared to another. So if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

How would you expect the IR and ¹H NMR spectra for propanamide and N,N-diethylpropanamide to differ?

Key Concepts

Infrared (IR) Spectroscopy

¹H Nuclear Magnetic Resonance (NMR) Spectroscopy

Amide Functional Group

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