The dehydrohalogenation of 2-bromobutane with alcoholic KOH gives:
Correct answer: C. 2-butene as the major product
- A. Only 2-butene
- B. Only 1-butene
- C. 2-butene as the major product
- D. 1-butene as the major product
Explanation
In the dehydrohalogenation of 2-bromobutane with alcoholic KOH, the elimination reaction follows Zaitsev's rule, where the more substituted alkene is more stable and formed preferentially. In this case, 2-butene is more substituted than 1-butene, making it the major product. The third carbon's hydrogen is removed to form 2-butene, which is more stable. Options suggesting only one product or 1-butene as the major product are incorrect due to the stability of 2-butene.
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Hydrocarbons are studied through the structures and reactions of alkanes, alkenes, alkynes and aromatic compounds. Coverage includes free radical substitution in alkanes, electrophilic addition to carbon-carbon multiple bonds, benzene substitution, aromatic stability and the difference between addition reactions of alkenes and substitution reactions of benzene.
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