RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    A and B react with \[Na\] gives \[{{H}_{2}}\] gas and by reaction of both A and B formed ethyl acetate is formed then A and B is:

    A)  \[C{{H}_{3}}COOH,{{C}_{2}}{{H}_{5}}OH\]

    B)  \[C{{H}_{3}}COOH,C{{H}_{3}}OH\]         

    C)  \[{{C}_{3}}{{H}_{7}}COOH,{{C}_{3}}{{H}_{7}}OH\]

    D)  \[HCOOH,C{{H}_{3}}COOH\]

    Correct Answer: A

    Solution :

    \[C{{H}_{3}}COO\underset{(A)}{\mathop{H}}\,+Na\xrightarrow{{}}C{{H}_{3}}COONa+\frac{1}{2}{{H}_{2}}\] \[{{C}_{2}}{{H}_{5}}O\underset{(B)}{\mathop{H}}\,+Na\xrightarrow{{}}{{C}_{2}}{{H}_{5}}ONa+\frac{1}{2}{{H}_{2}}\] \[C{{H}_{3}}COO\underset{(A)}{\mathop{H}}\,+\underset{(B)}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\xrightarrow{{}}\]                                 \[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\]


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