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I formed this product (reason: alkene has 6 alpha hydrogens)

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However, it turns out, this is the answer:

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The reason why I suspect this is e1cb is because NO2 could stabilize the carboanion (conjugate base) formed after reaction with NaOCH3

I would also like to know if a substrate has two sites for E2 elimination to occur, how do I decide site which would be preferred. Is it the one with the most acidic hydrogen or the one which would lead to a more stable alkene?

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    $\begingroup$ Your thought carbanion formation is correct except I don't understand your point about "6 alpha hydrogens". $\endgroup$
    – user55119
    Jun 18 at 15:03
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    $\begingroup$ @MeetLalwani It is almost always the most acidic hydrogen that decides how elimination goes. $\endgroup$
    – Waylander
    Jun 18 at 15:25
  • $\begingroup$ @user55119 stability of alkene is decided by the number of alpha hydrogens right??? $\endgroup$ Jun 18 at 15:28
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    $\begingroup$ now I see what you are saying. For a given set of isomers the more carbon-substituted double bond increases "stability" via hyperconjugation with the attached hydrogens. That is, the heats of formation and combustion are lowered. But this is a thermodynamic issue. Your example is one of kinetics. What process occurs faster is at issue. $\endgroup$
    – user55119
    Jun 18 at 18:28
  • $\begingroup$ About the last point, it depends on the base and leaving group, for example if the base is strong, it will prefer the most acidic hydrogen. If the base is not so strong and say the leaving group is good, the stability of product would matter more. $\endgroup$ Jun 19 at 7:34

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