A) propan-2-ol
B) propan-1-ol
C) propyne
D) propene
Correct Answer: A
Solution :
Alkyl hallides dehydrohalogenates in presence of alcohalic alkali to give alkenes when added HBr. Hence, the reactions proceed as follows: \[\underset{\text{Propyl}\,\,\text{bromide}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,\xrightarrow[KOH]{Alc.}\underset{\text{Propene}}{\mathop{C{{H}_{3}}CH==C{{H}_{2}}}}\,\] \[\xrightarrow{HBr}\overset{Br}{\mathop{\overset{|}{\mathop{C{{H}_{3}}CH\cdot C{{H}_{3}}}}\,}}\,\xrightarrow{KOH(aq.)}\] \[\underset{(C)\,\text{Propan-2-ol}}{\mathop{C{{H}_{3}}CH(OH)\cdot C{{H}_{3}}}}\,\]You need to login to perform this action.
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