Question

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

• A. Methoxide gives more and t-butoxide gives less Hofmann product
• B. Methoxide gives more and t-butoxide gives less Saytzeff product
• C. Saytzeff product is the major product in both cases
• D. Hofmann product is major in both cases

Answer 1

Answer: (B)

Methoxide gives more and t-butoxide gives 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 favored over the least substituted alkene, known as the Hofmann 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, 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 favored.