Predicting the Product of a Substitution/Elimination Reaction 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 &amp; 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.
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 


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.