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

  • question_answer
    When propionic acid is treated with aqueous sodium bicarbonate \[C{{O}_{2}}\] is liberated. The 'C' of \[C{{O}_{2}}\] comes from [IIT-JEE (Screening) 1999]

    A) Methyl group   

    B) Carboxylic acid group

    C) Methylene group           

    D) Bicarbonate

    Correct Answer: D

    Solution :

    \[\underset{\text{Propionic}\,\text{acid}}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH(aq)}}\,\,+\underset{\text{sod}\text{.}\,\text{bicarbonate}}{\mathop{NaHC{{O}_{3}}(aq)}}\,\to \]\[C{{H}_{3}}C{{H}_{2}}COONa+C{{O}_{2}}+{{H}_{2}}O\]


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