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Manipal Medical Manipal Medical Solved Paper-2015

  • question_answer
    An unknown alkyl halide (X) reacts with alcoholic KOH and produces a hydrocarbon\[({{C}_{4}}{{H}_{8}})\]as the major product. Ozonolysis of this hydrocarbon affords one mole of propanaldehyde   and   one   mole   of formaldehyde.  Suggest which organic compound among the following has the correct structure of the above alkyl halide (X)?

    A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]

    B) \[C{{H}_{3}}CH(Br)C{{H}_{2}}C{{H}_{3}}\]

    C) \[C{{H}_{3}}CH(Br)CH(Br)C{{H}_{3}}\]

    D) \[Br{{(C{{H}_{2}})}_{4}}Br\] 

    Correct Answer: A

    Solution :

    As the ozonolysis gives propanaldehyde and formaldehyde thus the hydrocarbon is \[\underset{1-butene}{\mathop{C{{H}_{3}}C{{H}_{2}}CH}}\,=C{{H}_{2}}({{C}_{4}}{{H}_{8}})\] 1-butene can be obtained by \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-Br\xrightarrow[-HBr]{Alc.\,KOH}\]                                               \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}\]


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