Gamma Girl

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

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?

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?

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.

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.

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

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

n-1

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.

Please open pdf.

Organic question.pdf

No

Elaborate, please.

Actually, I understand now. Thank-you again.

The sample is a molecule. MRI deals with magnets like in CAT scans.

The different spin states creates different energy differences for NMR due to different frequencies.

Why does a sample in an NMR spectrometer spin, but in an MRI the magnet spins?

Why does a sample in an NMR spectrometer spin, but in an MRI, the magnet spins?

I appreciate you, John

Why does a signal for Csp2-H still show up in symmetrical alkenes even though the C=C bond does not?