A) Propane
B) propene
C) Propyne
D) propanol
Correct Answer: B
Solution :
Alcoholic \[KOH\] is a dehydrohalogenation reagent, so when n-propyl bromide is treated with alcoholic \[KOH\], propene is obtained. \[\underset{n-propyl\,bromide}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,+alc\,\,KOH\xrightarrow{{}}\] \[C{{H}_{3}}CH=\underset{\text{Propene}}{\mathop{C{{H}_{2}}}}\,+HBr\]You need to login to perform this action.
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