|Jul7-07, 02:00 PM||#1|
Proton NMR - Spin-Spin Splitting and Multiplicity
1. What will be the multiplicity due to spin-spin splitting of the highlighted protons in the molecule: (C6H5)-CH2-CH2-CH2-OH?
2. Relevant equations
3. 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 text book, 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.
|Jul9-07, 09:25 AM||#2|
The assumption you are asked to make... is this first order coupling? Generally first order coupling is assumed for [tex]\Delta v\ /J > 20[/tex], where [tex]\Delta v\ [/tex] is the frequency difference from the centroid of the multiplets. Sometimes first order coupling is apparent where [tex]\Delta v\ /J =10[/tex]. 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.
|Jul9-07, 06:08 PM||#3|
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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