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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    \[{{N}_{2}}\]gas is liberated when\[[HCl+NaN{{O}_{2}}]\]reacts with the following compounds; \[A.C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]      \[B.\,\,urea\]                     \[C.C{{H}_{3}}CON{{H}_{2}}\]                     \[D.{{C}_{6}}{{H}_{5}}N{{H}_{2}}\] The answer is

    A)  A, B, C                 

    B)                         B, C, D

    C)                         A, C, D                 

    D)                         A, B, D

    Correct Answer: A

    Solution :

    \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}\]                   \[C{{H}_{3}}C{{H}_{2}}OH+{{N}_{2}}+{{H}_{2}}O+NaCl\] \[N{{H}_{2}}CON{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}\]                                 \[2{{N}_{2}}+{{H}_{2}}O+C{{O}_{2}}+NaCl\] \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}\]                      \[C{{H}_{3}}COOH+{{N}_{2}}+{{H}_{2}}O+NaCl\] \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}\]                                 \[{{C}_{6}}{{H}_{5}}N_{2}^{+}C{{l}^{-}}+{{H}_{2}}O+NaCl\]


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