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NEET Chemistry Aldehydes, Ketones Question Bank Self Evaluation Test - Aldehydes, Ketons and Carboxylic Acids

  • question_answer
    A substance \[{{C}_{4}}{{H}_{10}}O\] yields on oxidation a compound, \[{{C}_{4}}{{H}_{8}}O\] which gives an oxime and a positive iodoform test. The original substance on treatment with cone. \[{{H}_{2}}S{{O}_{4}}\] gives \[{{C}_{2}}{{H}_{8}}\] The structure of the compound is

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

    B) \[C{{H}_{3}}CHOHC{{H}_{2}}C{{H}_{3}}\]

    C) \[{{\left( C{{H}_{3}} \right)}_{3}}COH\]

    D) \[C{{H}_{3}}C{{H}_{2}}-O-C{{H}_{2}}C{{H}_{3}}\]

    Correct Answer: B

    Solution :

    [b] \[{{C}_{4}}{{H}_{8}}\xrightarrow[(-{{H}_{2}}O)]{conc.\,{{H}_{2}}S{{O}_{4}}}{{C}_{4}}{{H}_{10}}O\xrightarrow{oxidation}\] \[{{C}_{4}}{{H}_{8}}O\left( R-COC{{H}_{3}} \right)\] Thus \[{{C}_{4}}{{H}_{8}}O\] should be \[C{{H}_{3}}C{{H}_{2}}COC{{H}_{3}}\] hence \[{{C}_{4}}{{H}_{10}}O\] should be \[C{{H}_{3}}C{{H}_{2}}CHOHC{{H}_{3}}\]


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