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JEE Main & Advanced Chemistry Carboxylic Acids Question Bank Critical Thinking Carboxylic Acids

  • question_answer
    In a set of the given reactions, acetic acid yielded a product C \[C{{H}_{3}}COOH+PC{{l}_{5}}\to A\underset{anh.\,AlC{{l}_{3}}}{\mathop{\xrightarrow{{{C}_{6}}{{H}_{6}}}}}\,\]\[B\underset{ether}{\mathop{\xrightarrow{{{C}_{2}}{{H}_{5}}MgBr}}}\,C\].Product C would be                                         [CBSE PMT 2003]

    A)                 \[\overset{\overset{{{C}_{2}}{{H}_{5}}}{\mathop{|\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{3}}-C(OH){{C}_{6}}{{H}_{5}}}}\,\]     

    B)                 \[C{{H}_{3}}CH(OH){{C}_{2}}{{H}_{5}}\]

    C)                 \[C{{H}_{3}}CO{{C}_{6}}{{H}_{5}}\]         

    D)                 \[C{{H}_{3}}CH(OH){{C}_{6}}{{H}_{5}}\]

    Correct Answer: A

    Solution :

               \[C{{H}_{3}}COOH+PC{{l}_{5}}\to C{{H}_{3}}COCl\underset{\text{anh}.AlC{{l}_{3}}}{\mathop{\xrightarrow{{{C}_{6}}{{H}_{6}}}}}\,\]                                        \[C{{H}_{3}}CO{{C}_{6}}{{H}_{5}}\underset{\text{Ether}}{\mathop{\xrightarrow{{{C}_{2}}{{H}_{5}}MgBr}}}\,C{{H}_{3}}-\overset{_{\,|}^{{{C}_{2}}{{H}_{5}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C\,(OH)\,{{C}_{6}}{{H}_{5}}}}\,\]


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