H NMR/resonance structure

[tym]

---

Members don't see this ad.

Hey guys, here is an ochem question in TPR practice test 2. Im so confused. Hope someone can help 🙂

Amides have two common resonance structures because of the lone pair of electrons on the nitrogen. This allows partial double-bond character between the carbonyl carbon and the nitrogen, significantly restricting rotation. How many peaks would be exhibited in the 1H NMR spectrum at room temperature for ethanamide? A. 2 B.3 C.4 D. 5

Here is the explanation given by TPR:
One resonance will come from the CH3 group. The two hydrogen atoms on the nitrogen are different due to partial double-bond character between the carbonyl carbon and the nitrogen. Two more resonances will result from these two distinct protons. This gives a total of 3 resonances.

But I thought resonance is a 'false' theory that none of the structures actually exists in real life. Rather, the molecule exists as a hybridized intermediate between its resonance forms. So will 1H NMR be able to detect several peaks?

Thanks in advance! Any suggestion will be greatly appreciated!!

 

[Schenker]

I'm not sure if part of your confusion comes from mixing up the words "peak" and "resonance." In the TPR explanation you typed out,

One resonance will come from the CH3 group. The two hydrogen atoms on the nitrogen are different due to partial double-bond character between the carbonyl carbon and the nitrogen. Two more resonances will result from these two distinct protons. This gives a total of 3 resonances.

the bolded instances of "resonance" don't refer to resonance structures; they refer to the "peaks" you would see on an H-NMR spectrum and to which you would assign protons in orgo lab. (The reason TPR calls these peaks "resonances" is that the peaks represent resonant frequencies of hydrogen nuclei. Here, TPR's usage of the word "resonance" signifies something totally different from the theory of resonance that you talk about in your 4th paragraph.)

Sorry if that was already clear to you, but in case that wasn't, I will only use the word "resonance" when I'm talking about resonance structures.

The answer to the practice problem is that there are 3 peaks, so let's go over each of those three peaks. Remember that each peak represents a set of chemically identical protons.

The first peak will be from the 3 protons of the methyl (CH3) group. It's easy to see that these 3 protons are chemically identical, because there is free rotation about the bond between the methyl carbon and the amide carbon. That bond is a single bond.

On the other hand, can the C-N bond freely rotate? In other words, is the C-N bond a true single bond? No, because you can draw a resonance structure with a double bond, i.e. C=N. This means that the carbon-nitrogen bond has partial double bond character and will not be able to rotate freely. Thus, the hydrogen on the nitrogen that is 'cis' to the methyl group is chemically distinct from the hydrogen that is 'trans,' and each of these two hydrogens will have a distinct peak on the H-NMR spectrum. Hence, there are three peaks in total.

(That answers the question, but there is a bit more subtlety to the question that is worth looking into. Read on if you would like, but prioritize learning the basics.)

What's interesting is that if you raise the temperature, you can make it possible for the carbon-nitrogen bond to rotate 180 degrees and flip the hydrogens. (This has to do with overcoming the difference in bond energy of a C=N bond and a C-N bond.) In other words, raising the temperature enough would make the two amide protons chemically identical, and there would not be separate peaks for each hydrogen, i.e. at a high enough temperature there would be only 2 peaks for ethanamide. That's why the question specifies that the H-NMR spectrum is taken at room temperature.

I did a search just now, and someone else answered this question; have a look in case that explanation is more clear. http://forums.studentdoctor.net/archive/index.php/t-976538.html

 

[tym]

I'm not sure if part of your confusion comes from mixing up the words "peak" and "resonance." In the TPR explanation you typed out,



the bolded instances of "resonance" don't refer to resonance structures; they refer to the "peaks" you would see on an H-NMR spectrum and to which you would assign protons in orgo lab. (The reason TPR calls these peaks "resonances" is that the peaks represent resonant frequencies of hydrogen nuclei. Here, TPR's usage of the word "resonance" signifies something totally different from the theory of resonance that you talk about in your 4th paragraph.)

Sorry if that was already clear to you, but in case that wasn't, I will only use the word "resonance" when I'm talking about resonance structures.

The answer to the practice problem is that there are 3 peaks, so let's go over each of those three peaks. Remember that each peak represents a set of chemically identical protons.

The first peak will be from the 3 protons of the methyl (CH3) group. It's easy to see that these 3 protons are chemically identical, because there is free rotation about the bond between the methyl carbon and the amide carbon. That bond is a single bond.

On the other hand, can the C-N bond freely rotate? In other words, is the C-N bond a true single bond? No, because you can draw a resonance structure with a double bond, i.e. C=N. This means that the carbon-nitrogen bond has partial double bond character and will not be able to rotate freely. Thus, the hydrogen on the nitrogen that is 'cis' to the methyl group is chemically distinct from the hydrogen that is 'trans,' and each of these two hydrogens will have a distinct peak on the H-NMR spectrum. Hence, there are three peaks in total.

(That answers the question, but there is a bit more subtlety to the question that is worth looking into. Read on if you would like, but prioritize learning the basics.)

What's interesting is that if you raise the temperature, you can make it possible for the carbon-nitrogen bond to rotate 180 degrees and flip the hydrogens. (This has to do with overcoming the difference in bond energy of a C=N bond and a C-N bond.) In other words, raising the temperature enough would make the two amide protons chemically identical, and there would not be separate peaks for each hydrogen, i.e. at a high enough temperature there would be only 2 peaks for ethanamide. That's why the question specifies that the H-NMR spectrum is taken at room temperature.

I did a search just now, and someone else answered this question; have a look in case that explanation is more clear. http://forums.studentdoctor.net/archive/index.php/t-976538.html

wow thats so clear. Thank you so much Schenker!! I was stuck on that question for a really long time

 

[Czarcasm]

I'm not sure if part of your confusion comes from mixing up the words "peak" and "resonance." In the TPR explanation you typed out,



the bolded instances of "resonance" don't refer to resonance structures; they refer to the "peaks" you would see on an H-NMR spectrum and to which you would assign protons in orgo lab. (The reason TPR calls these peaks "resonances" is that the peaks represent resonant frequencies of hydrogen nuclei. Here, TPR's usage of the word "resonance" signifies something totally different from the theory of resonance that you talk about in your 4th paragraph.)

Sorry if that was already clear to you, but in case that wasn't, I will only use the word "resonance" when I'm talking about resonance structures.

The answer to the practice problem is that there are 3 peaks, so let's go over each of those three peaks. Remember that each peak represents a set of chemically identical protons.

The first peak will be from the 3 protons of the methyl (CH3) group. It's easy to see that these 3 protons are chemically identical, because there is free rotation about the bond between the methyl carbon and the amide carbon. That bond is a single bond.
On the other hand, can the C-N bond freely rotate? In other words, is the C-N bond a true single bond? No, because you can draw a resonance structure with a double bond, i.e. C=N. This means that the carbon-nitrogen bond has partial double bond character and will not be able to rotate freely. Thus, the hydrogen on the nitrogen that is 'cis' to the methyl group is chemically distinct from the hydrogen that is 'trans,' and each of these two hydrogens will have a distinct peak on the H-NMR spectrum. Hence, there are three peaks in total.

(That answers the question, but there is a bit more subtlety to the question that is worth looking into. Read on if you would like, but prioritize learning the basics.)

What's interesting is that if you raise the temperature, you can make it possible for the carbon-nitrogen bond to rotate 180 degrees and flip the hydrogens. (This has to do with overcoming the difference in bond energy of a C=N bond and a C-N bond.) In other words, raising the temperature enough would make the two amide protons chemically identical, and there would not be separate peaks for each hydrogen, i.e. at a high enough temperature there would be only 2 peaks for ethanamide. That's why the question specifies that the H-NMR spectrum is taken at room temperature.

I did a search just now, and someone else answered this question; have a look in case that explanation is more clear. http://forums.studentdoctor.net/archive/index.php/t-976538.html

I'm having a little trouble understanding this -- but from a 3D perspective, if one of N-H's hydrogens points as a wedge and the other as a dash, isn't there technically a mirror plane of symmetry in the molecule? Wouldn't that then indicate that both hydrogen's are equivalent regardless on whether or not rotation about the C-N bond is hindered?

 

[Czarcasm]

Being sp2, the nitrogen and it's attached H's are trigonal planar, and that means one of the H's must be closer in space to the carbonyl oxygen, and one farther away. The H's therefore experience different local electronic environments and different shielding from the NMR magnetic field, resulting in distinct chemical shift values in the spectrum.

Ahhh, I forgot that the hybridization changes from sp3 to sp2 so it changes the orientation of the hydrogen molecules. That makes sense now!

Thanks, and great explanation by the way.