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JIPMER Jipmer Medical Solved Paper-2002

  • question_answer
    \[CH\equiv \equiv CH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{{{H}_{2}}O/H{{g}^{2+}}}X\xrightarrow{LiAl{{H}_{4}}}Y\xrightarrow{{{P}_{4}}/B{{r}_{2}}}Z,\] Here Z is:

    A)  ethylene bromide

    B)  ethanol

    C)  ethyl bromide

    D)  ethylidene bromide

    Correct Answer: C

    Solution :

    The reaction is as follows \[CH\equiv CH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{{{H}_{2}}O/H{{g}^{2+}}}C{{H}_{3}}-CHO\] \[\xrightarrow[(reduction)]{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{{{P}_{4}}/B{{r}_{2}}}\underset{Ethyl\,bromide}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,\]


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