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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        End product of the following reaction is:\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[red\,p]{C{{l}_{2}}}\] \[\xrightarrow[{}]{alcoholic\text{ }KOH}\]

    A)  \[C{{H}_{3}}\underset{\begin{smallmatrix}  | \\  OH \end{smallmatrix}}{\mathop{C}}\,HCOOH\]       

    B)  \[\underset{\begin{smallmatrix}  | \\  OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}COOH\]

    C)  \[C{{H}_{2}}=CHCOOH\]             

    D)  \[\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,\underset{\begin{smallmatrix}  | \\  OH \end{smallmatrix}}{\mathop{CH}}\,COOH\]

    Correct Answer: C

    Solution :

                    \[C{{H}_{3}}C{{H}_{2}}COOH\underset{(HVZ\text{ }reaction)}{\mathop{\xrightarrow[red\,P]{C{{l}_{2}}}}}\,C{{H}_{3}}\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C}}\,HCOOH\]\[\xrightarrow[(elimination)]{alcoholic\text{ }KOH}\underset{acrylic\,acid}{\mathop{C{{H}_{2}}=CHCOOH}}\,\]


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