I am asked to explain why neither $\ce{E2}$ or $\ce{S_{N}}2$ products are formed when 3-chloro-2,2,4,4-tetramethylpentane is treated with NaOH.

I realize that since the reagent is a strong nucleophile & a strong base, it can only participate in either an $\ce{E2}$ reaction or in a combination of the $\ce{E2}$ and $\ce{{S_{N}}2}$ reactions, depending on the substrate. I think the substrate is secondary, so $\ce{E2}$ should predominate and $\ce{S_{N}2}$ should give a minor product.

The textbook solution says that since the substrate is tertiary, $\ce{{S_{N}}2}$ will not occur at an appreciable rate, and there are no β carbons, $\ce{E2}$ cannot occur.

My misunderstanding seems to stem from an inability to depict the molecule as is expected, so I am hoping someone can clarify whether or not my drawing and line of reasoning are correct.

enter image description here

  • $\begingroup$ Oh gosh, of course! Carbon can't have more than four bonds. $\endgroup$ – imaginov May 16 '15 at 16:50
  • $\begingroup$ You wrote, "there are no β carbons". There are two β carbons; there are no β hydrogens. Without a β hydrogen E2 elimination can not occur. $\endgroup$ – ron May 16 '15 at 17:17
  • $\begingroup$ @ron The textbook solution claims that there are no β carbons, this is why I was confused, as I thought there were two β carbons. I realize now that since the β carbons lack β hydrogens, an E2 mechanism cannot occur. Thanks! $\endgroup$ – imaginov May 16 '15 at 18:05

Remember that in a $S_{N}2$ reaction, the centre bearing the $\ce{Cl}$ atom would have to be attacked by $\ce{OH-}$ from the back side. This, however, is prevented by the by two bulky tert.-butyl substituents tetramethylchloropentane

As far as the alternative $E2$ pathway is concerned: There are no $\alpha$ hydrogen atoms that could be abstracted by $\ce{OH-}$.

I've uploaded a CML file of the structure to pastebin. I suggest to get yourself a copy of Avogadro and inspect the geometry on your own PC.

  • $\begingroup$ Thanks for putting things in 'perspective'. I am still unsure as to why the E2 reaction cannot occur, however. Am I correct in my understanding that the substrate is secondary, since the two tert-butyl substituents branching off of the alpha-carbon have been identified? $\endgroup$ – imaginov May 16 '15 at 16:44
  • 2
    $\begingroup$ The substrate is secondary as you have said. There are no $\beta$ hydrogens and so E2 cannot occur because there is no $\beta$ hydrogen to be abstracted by the hydroxide ion. $\endgroup$ – bon May 16 '15 at 17:43

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