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In class we are going over intramolecular elimination. In this question I will provide two similar reactants that provide a different mixture of products.

The chugaev elimination gives a 50/50 mixture of methylcyclohexene isomers while selenoxide elimination yields only one major product, but I do not understand why this is happening.

From what I can see, there is only one syn B-hydrogen available in both reactions. My professor went on to explain the significance of comparing the two reactions, saying that selenoxide elimination does not use heat while chugaev elimination does, causing a ring flip in one but not the other, which is understandable to me.

My analysis/work on the problem at hand

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  • $\begingroup$ • The selenoxide reaction has a low activation energy proceeding via a syn-elimination giving only the less substituted alkene. The Chugaev requires temperatures of 200oC by a syn-elimination in acyclic systems. Although some syn-elimination may occur, the appearance of the trisubstituted alkene argues for an E1 process with this molecule. Formation of the trisubstituted alkene by a concerted syn-elimination would give rise to an E-methylcyclohexene (trans-cyclohexene) through a transition state high in energy. The foregoing implies kinetically controlled reactions and not equilibration. $\endgroup$ – user55119 Mar 22 at 0:08

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