Proton NMR - Spin-Spin Splitting and Multiplicity

AI Thread Summary
The discussion focuses on determining the multiplicity due to spin-spin splitting for the protons in the molecule (C6H5)-CH2-CH2-CH2-OH. The highlighted CH2 group is surrounded by non-equivalent protons, complicating the multiplicity calculation. Participants debate whether to assume first-order coupling, which typically applies when the frequency difference over the coupling constant is greater than 20. The spin-spin coupling constants for methylene groups are noted to be around 5-7 Hz, influencing the expected multiplicity. Ultimately, clarity on whether to assume first-order coupling is essential for accurately determining the multiplicity.
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1. What will be the multiplicity due to spin-spin splitting of the highlighted protons in the molecule: (C6H5)-CH2-CH2-CH2-OH?



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The Attempt at a Solution


I know that the ones surrounding the highlighted CH2 group are not equivalent. If they were, the multiplicity would be 5. When I make a branching diagram, I get 6 to a multiplet. In the textbook, there was an example of a similar case where you could "assume" that they were equal, even though they were not. How do I know if I can "assume" in this case? My answer choices are a) doublet b) triplet c) quartet d) quintet e) hextet. I'm stuck between d) a quintet, if I assume that they are equal and e) the minimum, probably, but couldn't a multiplet of more than 6 form also? Any help? Thank you.
 
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The assumption you are asked to make... is this first order coupling? Generally first order coupling is assumed for \Delta v\ /J > 20, where \Delta v\ is the frequency difference from the centroid of the multiplets. Sometimes first order coupling is apparent where \Delta v\ /J =10. The spin-spin coupling constants for methylene-methylene are usually about 5-7 Hz.

Remember that J coupling is independent of magnetic field strength whereas the absorption frequency is a function of field strength.
 
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Thank you for your help. The problem never said that I could assume it was first-order coupling, but my professor also cleared up the situation. Thanks again.
 
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