JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • question_answer
    In the reaction\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[{}]{SOC{{l}_{2}}}A\xrightarrow[{}]{N{{H}_{3}}(excess)}\]\[B\xrightarrow[{}]{B{{r}_{2}}/KOH}C\]Product C is

    A)  \[C{{H}_{3}}COC{{H}_{3}}\]                       

    B)  \[{{C}_{6}}{{H}_{5}}CHO\]

    C)  \[{{C}_{6}}{{H}_{5}}COC{{H}_{2}}C{{H}_{3}}\]    

    D)  \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}CHO\]

    Correct Answer: B

    Solution :

                    \[\underset{Propanoic\text{ }acid}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH}}\,\xrightarrow[-S{{O}_{2}}.-HCl]{SOC{{l}_{2}}}\underset{Propanoyl\text{ }chloride\text{ }(A)}{\mathop{C{{H}_{3}}C{{H}_{2}}COCl}}\,\] \[\xrightarrow[-N{{H}_{4}}Cl]{N{{H}_{3}}(excess)}\underset{Propionamide\,(B)}{\mathop{C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}}}\,\xrightarrow[{}]{B{{r}_{2}}/KOH}\]       \[\underset{ethylamine\text{ }(C)}{\mathop{C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}}}\,\]


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