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

  • question_answer
    Compound 'A' (molecular formula \[{{C}_{3}}{{H}_{8}}O\]) is treated with acidified potassium dichromate to form a product 'B' (molecular formula \[{{C}_{3}}{{H}_{6}}O\]). 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of \[{{H}_{2}}NCONHN{{H}_{2}}.HCl\] and sodium acetate gives a product 'C'. Identify the structure of 'C'?

    A) \[C{{H}_{3}}C{{H}_{2}}CH=NNHCON{{H}_{2}}\]

    B) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=NNHCON{{H}_{2}}\]

    C) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=NCONHN{{H}_{2}}\]

    D) \[C{{H}_{3}}C{{H}_{2}}CH=NCONHN{{H}_{2}}\]

    Correct Answer: A

    Solution :

    [a] \[\underset{A}{\mathop{{{C}_{3}}{{H}_{8}}O\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}}{{C}_{3}}{{H}_{6}}O}}\,\]  \[\underset{B\,(-CHO)}{\mathop{\xrightarrow{{{H}_{2}}NCONHN{{H}_{2}}}C}}\,\] Since B reduces Tollen's reagent, it indicates that it has an \[-CHO\] group, so it must be\[C{{H}_{3}}C{{H}_{2}}CHO\]. Hence \[\underset{[A]}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,\to \underset{[B]}{\mathop{C{{H}_{3}}C{{H}_{2}}CHO}}\,\xrightarrow{{{H}_{2}}NHCON{{H}_{2}}}\]\[\underset{[C]}{\mathop{C{{H}_{3}}C{{H}_{2}}CH}}\,=NNHCON{{H}_{2}}\]


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