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Answer: Molecule b)
Reasoning:The two molecules differ only in the position of their double bond. Due to the linkage to the second ring only one sp2 hybridized carbon can also carry a proton. We should therefor find only one singulet in the expected spectral range for these protons (d = 5.3 ... 7.0).
What we actually find in the spectrum, however, are two doublets. This is in good agreement with the spectra we would expect to find for the molecule b). The signals of the two non-equivalent protons connected to sp2 hybridized carbons are split into doublets due to coupling between them.
(Compare the spectrum to the one for methylene protons in the first problem of this set).
On to the next problem! (We're almost done)