A) propane
B) propene
C) propyne
D) propanol
Correct Answer: B
Solution :
Alcoholic KOH is a dehydrohalogenating reagent, so when n-propyl bromide is treated with alcoholic KOH, propene is obtained. \[\underset{n-propyl\text{ }bromide}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,+alc\,\,KOC\xrightarrow{{}}\] \[\underset{Propene}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}+HBr}}\,\]
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