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Answer: Molecule a)
Reasoning:
In this case you can use the coupling between the different protons (as
identified by identical coupling constants in the multiplets) as a clue
for the determination of the proper structure. The proton(s) with a
signal at d = 0.81 ppm
(Methyl protons) are coupled to the ones with a chemical shift of d = 4.19 ppm (Methine proton).
Based on the number of bonds between the respective groups in the two
candidates, such a coupling can only occur in molecule A,
since there the two groups of protons are seperated by only 3 s-bonds (Doublet and quartet, J
= 6,4 Hz)
The signal at d = 7.4
ppm is a composite of the signals of the phenyl protons as well as the
amide proton of the NH-CO group (Irel = 6!).
And now for something completely
different!
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