Protons in the NMR

[Gamma Girl]

Please open pdf.

Organic question.pdf

[Fuzzwood]

Please show what work you put into it already.

[Gamma Girl]

PLEASE DO NOT TORTURE ME FURTHER.

I know for structure II (n+1)(m+1) rule for 6 peaks.

 

For structure I, D makes it invisible on the left. I know t-butyl makes 1 peak from previous experiences but why again the 9 H integration? So n+1= 2 (again why isn't 9+1=10)

 

For structure III, the symmetry of the molecule is inflicting a pain. I can tell you easily it has 2 NMR signal, but the splitting is a demon of another sort.

[BabcockHall]

What were you taught with respect to the number of bonds that connect two atoms that show coupling?

[Gamma Girl]

n-1

[hypervalent_iodine]

PLEASE DO NOT TORTURE ME FURTHER.

I know for structure II (n+1)(m+1) rule for 6 peaks.

 

For structure I, D makes it invisible on the left. I know t-butyl makes 1 peak from previous experiences but why again the 9 H integration? So n+1= 2 (again why isn't 9+1=10)

 

For structure III, the symmetry of the molecule is inflicting a pain. I can tell you easily it has 2 NMR signal, but the splitting is a demon of another sort.

 

I think you are misunderstanding the question, or I am misunderstanding your issue. It doesn't matter how many peaks the t butyl gives. The question is asking about the splitting of the circled protons only.

 

Side note: we are not here to do your homework for you, so being rude to those kind enough to offer assistance will do you no favours.

[BabcockHall]

n-1

 

Gamma Girl,

 

Your answer does not make any sense. What is n? Also it seems to me that you may be confusing rules that govern splitting with rules that govern the integration of peaks.

 

With respect to the question that I asked, the answer is that hydrogen atoms that are connected by two or three bonds show coupling that is typically easily seen. Hydrogen atoms connected by four or more bonds couple with each other with such small coupling constants that it is often difficult or impossible to detect. Unless your instructor told you otherwise, it is probably a safe bet that coupling between hydrogen atoms that are separated by four or more bonds can be ignored.

[Gamma Girl]

CASE IS CLOSED. PROBLEM SOLVED BY WATCHING YOU TUBE AND KNOWBEE.

 

STRUCTURE 1 - 2 PEAKS USING (n+1) [(0 peaks +1) from the left, (0+1) from the right because Deuterium is invisible]

STRUCTURE 2- 6 PEAKS USING (n+1)(m+1)

STRUCTURE 3- 1 PEAK OBSERVED DUE TO THE PLANE OF SYMMETRY

 

n= proton neighbors attached to adjacent carbons

 

[hypervalent_iodine]

I'm glad you figured it out, but your working for structure one is off. The term n represents the total number of neighbouring protons. You do not do the calculation twice. By your logic, that would give n(total) = 0 and therefore only 1 peak, which has to be wrong based on the answers provided and given that structure two is split into 6 peaks.

 

In fact, the coupling that gives the doublet in structure one is due to the other proton attached to the same carbon. These CH2 hydrogens appear to be something known as diastereotopic hydrogens. This occurs when the rotation around one or more of the single bonds is effectively locked in place, providing different electronic environments for the two protons on the one carbon. This is a well known phenomenon in benzylic systems, for example.

[Gamma Girl]

Hypervalent_iodine, I see your point. My thinking for 1 is incorrect. What is the correct answer for structure 1?

[hypervalent_iodine]

Given my above post, what do you think would be the correct answer?

[Gamma Girl]

I'm glad you figured it out, but your working for structure one is off. The term n represents the total number of neighbouring protons. You do not do the calculation twice. By your logic, that would give n(total) = 0 and therefore only 1 peak, which has to be wrong based on the answers provided and given that structure two is split into 6 peaks.

 

In fact, the coupling that gives the doublet in structure one is due to the other proton attached to the same carbon. These CH2 hydrogens appear to be something known as diastereotopic hydrogens. This occurs when the rotation around one or more of the single bonds is effectively locked in place, providing different electronic environments for the two protons on the one carbon. This is a well known phenomenon in benzylic systems, for example.

Case reopened.

 

Structure 1= n is 0 because no neighbors, n +1= 0+1=1 The circled hydrogen has 1 peak.

Structure 2= (n+1)(m+1)= (2+1)(1+1)=6 The circled hydrogens observe 6 peaks.

Structure 3=n is 2, has 2 hydrogens as neighbors. (n+1) =(2+1)= 3 The circled hydrogen sees 3 peaks.

 

None of the above are a choice in the pdf. Do you know how to approach this question? It is literally keeping me up nights.

[hypervalent_iodine]

Okay, there appears to be some miscomprehension going on, or you simply didn't read my post. I didn't mention this previously because I thought it would give you the chance to figure it out for yourself, but it appears I was wrong. I didn't say your answer was incorrect, I said your working out was, even if it gave you the right answer. Moreover, I did in fact explicitly state the answer for structure one in my post.

 

Please reread my post carefully, and look up what diastereotopic hydrogens are.

[Gamma Girl]

Okay, there appears to be some miscomprehension going on, or you simply didn't read my post. I didn't mention this previously because I thought it would give you the chance to figure it out for yourself, but it appears I was wrong. I didn't say your answer was incorrect, I said your working out was, even if it gave you the right answer. Moreover, I did in fact explicitly state the answer for structure one in my post.

 

Please reread my post carefully, and look up what diastereotopic hydrogens are.

Diastereotopic hydrogens form stereoisomers that are not mirror images when the H is replaced by another atom like Cl. Concerning structure 1, a study group influenced me to change the answer from 2 to 1 peak because there are no neighbors and also hence structure 3. I felt they had a point, but it has something to do with diastereotopic hydrogens according to you. There is a blurb on that in this chapter, but the teacher skipped that term this time.

[hypervalent_iodine]

Your study group is wrong.

 

Diastereotopic protons will have different chemical shifts due to their different environments. The two hydrogens in the CH2 of structure one are diastereotopic. If they have different chemical shifts, do you think they would couple to each other or not? What would n equal in this case?

[Gamma Girl]

Diastereotopic hydrogens form stereoisomers that are not mirror images when the H is replaced by another atom like Cl. Concerning structure 1, a study group influenced me to change the answer from 2 to 1 peak because there are no neighbors and also hence structure 3. I felt they had a point, but it has something to do with diastereotopic hydrogens according to you. There is a blurb on that in this chapter, but the teacher skipped that term this time.

addendum: doublet from structure 1 is from t-butyl

Your study group is wrong.

 

Diastereotopic protons will have different chemical shifts due to their different environments. The two hydrogens in the CH2 of structure one are diastereotopic. If they have different chemical shifts, do you thnk they would couple to each other or not? What would n equal in this case?

No, the H's would not couple. n is 1 due to the t-butyl because the distance is 2 bonds.

Explain structure 3 to me, please.

[Gamma Girl]

Hypervalent,

I reread what you wrote in the daylight hours. The doublet is due to the other hydrogen attached and has nothing to do with the t-butyl group. Where can I see more examples like this? All we talk about is neighbors when analyzing these problems. I see the stereogenic center, and the protons in the CH2 are diastereotopic and have different environments, but had no idea that the doublet came from the other H.

I am grateful. Can you explain structure 3?

[hypervalent_iodine]

There are plenty of examples of this. Benzylic CH2's are great examples. You will sometimes see these called AB or ABX systems, but I wouldn't get too bogged down in the details of that as it can get very confusing. It's enough to know that diastereotopic hydrogens will quite often have different chemical shifts and therefore couple to one another.

Structure three you seemed fine with? Your explanation was correct in your previous post.

[Gamma Girl]

H H How many Peaks does the red Hydrogen observe?

I I

ClH2C--C--C--CH2Cl

I I

H H

 

Answer: 3 peaks

Dilemma: I thought 5 peaks since the environment is similar. How does symmetry play a role here?

[Fuzzwood]

Symmetry plays a role so that the 4 hydrogens closest to the symmetry plane are the same, ergo they would show up as a singlet if the outer 4 protons were not there.

[Gamma Girl]

The answer key says the red hydrogen sees 3 peaks. How is that?

[hypervalent_iodine]

Because that hydrogen is electronically equivalent to the other hydrogen attached to it and to the CH2 to the left. It won't couple to the CH2 side on the left due to this, but it will couple to the other CH2. This means n = 2, so you get a triplet.

[Fuzzwood]

The opposing CH₂Cl site is too far away to really influence coupling, if that would confuse you otherwise. If it would be right next to it (1,3 dichloropropane instead of the current 1,4 dichlorobutane), it would cause a 5-fold splitting.

[BabcockHall]

addendum: doublet from structure 1 is from t-butyl

No, the H's would not couple. n is 1 due to the t-butyl because the distance is 2 bonds.

Explain structure 3 to me, please.

 

Between the hydrogen atoms of the t-butyl group and the hydrogen in question there are four bonds, not two bonds. When there are four bonds, the coupling constant is usually small and may be undetectable.