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1H NMR:Spin-Splitting (N + 1) Rule
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1H NMR:Spin-Splitting (N + 1) Rule
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15. Analytical Techniques:IR, NMR, Mass Spect / 1H NMR:Spin-Splitting (N + 1) Rule / Problem 4
Problem 4
Explain why the cis and trans isomers of the given alkene show no coupling between the a and c, as well as between the b and c protons.
A
There is no coupling between a and c and between b and c protons because they are both too near to each other.
B
There is no coupling between a and c and between b and c protons because they are 6 and 4 pi bonds away from each other, respectively.
C
There is no coupling between a and c and between b and c protons because they are 4 and 6 sigma bonds away from each other, respectively.
D
There is no coupling between a and c and between b and c protons because they are 4 and 6 pi bonds away from each other, respectively.
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