13-50 (a), (b) and (c) The six isomers are drawn here. Below each structure is the number of proton signals and the number of carbon signals. (Note that splitting patterns would give even more clues in the proton NMR.) H C CH H CH 3 'rI" 3 H CH3 H H H _ l-. /' 13 ;L >=< >=< y� D IxH IxC 6 H H H3C H H3C CH3 H 3xH* 3xC 2xH 3xC 2xH 2xC 2xH 2xC *Rings always present challenges in stereochemistry. When viewed in three dimensions, it becomes apparent that the two hydrogens on a CHz are not equivalent: on each CHz, one H is cis to the methyl and one H is trans to the methyl. These are diastereotopic protons. A more correct answer to part (b) would be four types of protons; whether all four could be distinguished in the NMR is a harder question to answer. For the purpose of this problem, whether it is 3 or 4 types of H does not matter because either one, in combination with three types of carbon, will distinguish it from the other 5 structures.
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