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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    In the following sequence of reactions, \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{KOH(alc.)}(A)\xrightarrow{HBr}(B)\] \[\xrightarrow{KOH(aq.)}\,(C)\] The product C is:

    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: \[\underset{\text{Propyl}\,\,\text{bromide}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,\xrightarrow[KOH]{Alc.}\underset{\text{Propene}}{\mathop{C{{H}_{3}}CH==C{{H}_{2}}}}\,\] \[\xrightarrow{HBr}\overset{Br}{\mathop{\overset{|}{\mathop{C{{H}_{3}}CH\cdot C{{H}_{3}}}}\,}}\,\xrightarrow{KOH(aq.)}\] \[\underset{(C)\,\text{Propan-2-ol}}{\mathop{C{{H}_{3}}CH(OH)\cdot C{{H}_{3}}}}\,\]


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