What happens when 2-Bromobutane is made to react with alcoholic potassium hydroxide?
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Solution and Explanation
This involves a dehydrohalogenation reaction ($\beta$-elimination).
Step 2: Meaning
Alcoholic KOH acts as a strong base, removing a hydrogen atom from a $\beta$-carbon and the bromine atom from the $\alpha$-carbon, resulting in the formation of a double bond (alkene).
Step 3: Analysis
2-Bromobutane ($CH_3-CH_2-CH(Br)-CH_3$) has two different sets of $\beta$-hydrogens.
Elimination can yield either But-1-ene or But-2-ene. According to Zaitsev's rule,
the highly substituted alkene is thermodynamically more stable and is formed as the major product.
Elimination of the internal $\beta$-hydrogen forms But-2-ene, which has two alkyl groups attached to the double bond.
Step 4: Conclusion
Therefore, the reaction predominantly produces But-2-ene, with a minor amount of But-1-ene.
Final Answer: A dehydrohalogenation ($\beta$-elimination) reaction occurs. According to Zaitsev's rule, But-2-ene is formed as the major product because it is the more highly substituted, stable alkene. (A small amount of But-1-ene is also formed as a minor product).
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