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An $\mathrm{E2}$ (or $\mathrm{E1cb}$) mechanism requires antiperiplanar configuration. In case of restricted rotations, like in cycles, where rotation is not possible to achieve the antiperiplanar configuration, elimination occurs to favour the Hofmann product instead of the Zaitsev, if antiperiplanar configuration is possible at the less substituted β C.

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In menthyl chloride, antiperiplanar configuration is only seen with the red $\ce{H}$, so 2 menthene is the only product, although it is not the most substituted.

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In the molecule above, due to restricted rotation, the leaving group $\ce{Cl}$ and the base attacking the $\ce{H}$ will be on the same side. So will elimination be possible in this case?

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    $\begingroup$ An antiperiplanar while being the most efficient is not the only configuration from which elimination occurs. A synperiplanar configuration can give you an efficient elimination as well especially in the case of a double bond which prevents rotation and holds the eliminated H and leaving group fixed at the perfect positions for a syn elimination. It is all a question of orbital overlap: It is best in the case of an antiperiplanar configuration since the orbitals are perfectly aligned in this case, but in the synperiplanar configuration they are at least partly aligned and good overlap occurs. $\endgroup$ – Philipp Jun 11 '15 at 18:06
  • $\begingroup$ Elimination can occur from vinylic halides. This related question may be helpful. $\endgroup$ – ron Jun 11 '15 at 18:09
  • $\begingroup$ But if in an E2 mechanism, there is the possibility of both antiperiplanar and synperiplanar configurations, the anti is always preferred - despite the fact that it may be the less substituted product (in reference to the example of menthyl chloride in the question)? $\endgroup$ – Charles Jul 10 '15 at 5:05
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An E2 reaction will not occur with this reagent.

An E1 reaction, on the other hand, could do so (though quite slowly: E1 is an all around pathetic reaction mechanism, and this molecule does nothing special to stabilize it). The chlorine will leave and there will be a carbocation on the end of the alkene, whereupon the proton from the other side of the double can be extracted with a base in a second step, leaving you with an alkyne.

That said, if you want the alkyne, it would be infinitely infinitely easier to start with the cis isomer of the molecule you have and just run it through very strong base. E2 is far far better than E1 in terms of reaction rate and yield.

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