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Great! Right answer!Indeed, the three protons of the methyl group and the single proton of the carbonyl group (-CH3 und -CHO) give rise to two signals. (If you want to, you can check out the reasons for this on page20)The same reasoning holds for the two protons of the methylene group (-CH2-), since they also rotate freely, and since the two protons have mirror symmetry. They also give only one shared signal. ![]() C1 and C3 rotate for all concerns and purposes freely around the single bonds C1-C2 and C2-C3, respectively. Therefore both methylene protons experience the same average amount of shielding over time from the CH3- and the CHO- groups. To summarize: In systems with unrestricted movement all the protons of aComplications with regard to this rule only arise when the rotation around the single bond produces conformations that are related through a symmetry operations (e.g., when the CH2- group is next to an asymmetrical carbon atom). You can find further explanations of this on page 132!
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