Under second-order conditions (strong base nucleophile), SN2and E2 reactions may occur simultaneously and compete with each other. Show what products might be expected from the reaction of 2-bromo-3-methylbutane (a moderately hindered 2° alkyl halide) with sodium ethoxide.
Here we are going to write the reaction of 2-bromo-3-methylbutane (a moderately hindered 2° alkyl halide) with sodium ethoxide with dominant product formation.
With a strong base, either the SN2 or the E2 (second order elimination) reaction is possible. Here, the reaction is as follows:
The E2 reaction requires abstraction of a proton on a carbon atom next to the carbon bearing the halogens. Thus in 2-bromo-3-methylbutane , there are two or more possibilities, mixtures of products may result. It follows the Zaitsev’s rule where more substituted alkene is the major product.
On the other hand,
Since, sodium ethoxide is a strong base as well as strong nucleophile. In the E2 reaction just shown, ethoxide reacts as a base rather than as a nucleophile. Most strong nucleophiles are also strong bases, and elimination commonly results when a strong base nucleophile is used with a poor substrate such as a 3° or hindered 2° alkyl halide. Thus the SN2 mechanism is blocked because the secondary alkyl halide is too hindered.