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?

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.

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.

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?