A) hydrogenation
B) halogenation
C) dehydro halogenation
D) none of these
Correct Answer: C
Solution :
\[C{{H}_{3}}-\underset{Br}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[(-HBr)]{alc.\,KOH}\] \[\underset{Butene-1}{\mathop{C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{3}}}}\,\] 2-bromobutane on reaction with ale. \[KOH\] eliminates \[HBr\]. Therefore, this reaction is known as dehydrohalogenation.
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