Why is the alkyl halide below not capable of undergoing an e2 reaction upon treatment with sodium ethoxide?

See explanation below

Explanation:

You are not providing the alkyl halide, however, I found a question very similar to this, and I'm gonna answer it so you can know the answer of this kind of questions.

Picture 1 is the alkyl halide.

Picture 2 is how E2 reactions undergoes with a base such an sodium ethoxide, using easel projections. Picture 3 shows why is not possible to do it.

E2 reactions is a fast process, that takes only one step to do so, the base substract an electrophyle (Hydrogen) and the lone pair of electrons goes down to form a pi bond, and that, makes that the halide goes out of the molecule. However, the primary condition to do this, is that the C-H bond with the C-X bond, should always be antiperiplanar oriented. If they are not oriented in this direction, the reaction that it will take place instead would be an Sn2.

As you can see in picture 3 with the projections, Br and H are not antiperiplanar oriented, therefore, the base, could not substract the electrophyle, because of steric hindrance of the rest of the substituent in the molecule. in the opposite directions, the base can actually go and substract in an easier way the hydrogen, and then, the halide will go out.

Hope this helps

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.