Draw the major product of this reaction. Include stereochemistry if applicable. Ignore byproducts. Br (CH3)3COK heat Select to Draw
The Correct Answer and Explanation is:
Here is the major product of the reaction.
Now, let’s break down the reasoning behind the product formation. This reaction involves tert-butoxide (CH₃)₃COK, a strong, bulky base, and heat, suggesting an elimination mechanism rather than a substitution. Given that the substrate contains a bromine (Br) leaving group, it undergoes an E2 (bimolecular elimination) reaction, where elimination occurs in a single step.
Stepwise Explanation:
- Elimination Mechanism Selection:
- The presence of the strong, sterically hindered tert-butoxide favors E2 elimination over SN2 substitution.
- The reaction occurs in one step without a carbocation intermediate.
- Formation of the Alkene:
- The base abstracts a β-hydrogen (hydrogen on the carbon adjacent to the carbon bearing bromine).
- The electrons from the C-H bond shift to form a new π-bond (double bond) between adjacent carbons.
- Bromine leaves as Br⁻, completing the elimination process.
- Regioselectivity—Zaitsev vs. Hofmann:
- Normally, elimination follows Zaitsev’s rule, forming the most substituted alkene.
- However, due to the steric bulk of tert-butoxide, Hofmann elimination dominates, favoring the less substituted alkene.
- Stereochemistry:
- Since E2 requires antiperiplanar (coplanar but opposite) geometry, the specific hydrogen removed must be anti to bromine.
- This influences the resulting alkene’s stereochemistry, typically favoring the more stable E-isomer.
Thus, the major product is the least substituted alkene, as dictated by Hofmann elimination. The reaction proceeds efficiently due to the strong base and heat, ensuring rapid elimination. Let me know if you’d like further clarification!
