The E2 reaction involves the simultaneous removal of a leaving group and a hydrogen atom from adjacent carbon atoms to form a double bond. It occurs through a concerted mechanism, and its rate depends on the concentrations of the substrate and the base/nucleophile. Factors such as the strength of the base, stability of the leaving group, steric hindrance, and solvent effects influence the reaction.
When a bulky base reacts with a secondary alkyl halide, the
reaction can proceed through E2 elimination. The bulky base abstracts a proton
from the beta carbon adjacent to the halide, causing the leaving group to
depart simultaneously. This results in the formation of a double bond. However,
steric hindrance around the reacting carbon atoms may affect the rate and
selectivity of the E2 reaction.
When a secondary alkyl halide reacts with a bulky base, the
product outcome can vary in terms of regioselectivity, resulting in either the
Zaitsev or Hofmann product. Zaitsev product is typically favored in most E2
reactions, there are cases where the Hofmann product can be the major or
exclusive product, especially when steric hindrance is significant. The actual
product distribution will depend on the specific reaction conditions and the
substrate used.
In this reaction, the tert-butoxide acts as a base and
abstracts a proton from the beta carbon (adjacent to the bromine) in
2-bromobutane. At the same time, the bromine atom serves as the leaving group.
This leads to the formation of a double bond, resulting in the production of
2-butene. Additionally, tert-butyl bromide is generated as a byproduct.
The use of tert-butoxide as a bulky base can influence the
regioselectivity of the reaction. The steric hindrance provided by the bulky
tert-butoxide can hinder the approach to the more substituted beta carbon. As a
result, the reaction may exhibit some degree of Hofmann product selectivity,
leading to the formation of the less substituted alkene in some cases.
ANSWER (b)
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