Question

Consider the dehydrohalogenation of $2$$-bromo-$2$-methylbutane$ with ethoxide and $tert-butoxide$. Which of the following statements is true?

• A. Saytzeff product is major product in both cases.
• B. Hofmann product is major product in both cases.
• C. Ethoxide gives more, $t-butoxide$ give less Hofmann product.
• D. Ethoxide gives more, $t-butoxide$ give less Saytzeff product.

Answer 1

The correct option is D Ethoxide gives more, $t-butoxide$ give less Saytzeff product.

In $E_2$ elimination reactions, a base abstracts a proton that is beta to a halide. The removal of the proton and the loss of the leaving group occur in a single, concerted step to form a new double bond. When a small, unhindered base ethoxide is used for an $E_2$ elimination, the Saytzeff product is typically favoured over the least substituted alkene, known as the Hoffmann product.

For example, treating $2-bromo-2-methylbutane$ with sodium ethoxide in ethanol produces the Saytzeff product with moderate selectivity. According to Saytzeff's rule, during dehydration, a more substituted alkene is formed as a major product, since greater the substitution of double bond greater is the stability of the alkene.

Due to steric interactions, a bulky base such as potassium $t-butoxide$, cannot readily abstract the proton but a less sterically hindered proton is preferentially abstracted instead. As a result, the Hofmann product is typically favoured.