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Given the molecule,2-methyl-1-butene.

I can predict that there will be 4 signals present in the NMR spec. Problem is, the quiz says 5 is the correct answer. My question is, are alkene hydrogen not chemical-shift equivalent?

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    $\begingroup$ No, they are not. They have different relationships with the rest of the molecule. $\endgroup$
    – Zhe
    Sep 3 '18 at 13:28
  • $\begingroup$ Does that mean it shows up as two separate peaks? $\endgroup$ Sep 3 '18 at 14:22
  • $\begingroup$ Two 2J doublets, each further split up by allylic 4J couplings. They would have different reactivity in chemical reactions, so they obviously cannot be equivalent! CS and reactivity both are directly related to the local electron density distribution. $\endgroup$
    – Karl
    Sep 3 '18 at 17:10
  • $\begingroup$ Yes, the two alkene hydrogens are different. As to whether they are readily resolved may depend on the magnitude of the field of the NMR. $\endgroup$
    – user55119
    Sep 3 '18 at 18:42
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    $\begingroup$ @user55119 Ja. Typical setup for an AB spin system. $\endgroup$
    – Karl
    Sep 3 '18 at 20:25
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Since there isn't free rotation around the C=C bond in the alkene, the two hydrogens, Ha and Hb are experiencing different electronic environments. Whenever two hydrogen atoms are experiencing non-identical electronic environments, they are going to possess differing chemical shifts in proton NMR. If you heat this compound, and it doesn't decompose, I would expect the two signals from Ha and Hb to coalesce into one as the hindered rotation about the alkene C=C bond becomes unhindered.

https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_05%3A_Structure_Determination_II/5.2%3A_Chemical_equivalence

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  • $\begingroup$ Is the lack of free rotation due to double bonds having both Siqma and Pi bonds, essentially "locking" the hydrogens in place? $\endgroup$ Sep 13 '18 at 15:47

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