Question

DRAW THE MAJOR ELIMINATION PRODUCT FORMED IN THE FOLLOWING REACTION.

Answer 1

Concepts and reason

This problem is based on the concept of nucleophilic substitution reactions and elimination reactions.

Elimination reactions involve the abstraction of proton in order to form alkene. Elimination reactions are categorized in two categories which are and elimination reactions.

Fundamentals

Theelimination reaction involves the formation of carbocation and then proton abstraction takes place whereas elimination reaction does not involve an intermediate formation. The base abstracts the proton from the reactant in order to give an alkene.

The structure of the reactants in given below:

The mechanism of the chemical reaction involved is given below:

Ans:

The structure of the major product is given below:

Answer 2

Secondary alkyl halides can undergo both E1 and E2 eliminations. The combination of methoxide (a strong base) and DMSO (an aprotic polar solvent) creates the conditions for an E2 reaction. In E2 elimination, the base attacks a β‑proton relative to the halide group, and the C–H$\text{C}–\text{H}$ and C–Br$\text{C}–\text{Br}$ bonds break simultaneously with the formation of a C=C$\text{C}=\text{C}$ π bond.

Typically, E2 reactions follow Zaitsev's rule, and the major product is the most substituted alkene. However, if the alkyl halide has a double bond or a benzene ring positioned to produce a conjugated product, Zaitsev's rule does not apply. Conjugation has a bigger stabilizing effect than alkyl substituents, so the most stable product will not be the most substituted alkene, but the alkene with conjugated double bonds.

1. major product (Conjugation makes this alkene more stable than the more substituted alkene.) 

   

2. minor product (Although this alkene is more substituted, the new double bond is not conjugated.)