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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    An organic compound 'A' is oxidized with \[N{{a}_{2}}{{O}_{2}}\]followed by boiling with \[HN{{O}_{3}}.\]he resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is:                                                                                     [JEE Main Held on 12-4-2019 Morning]

    A) Sulphur                       

    B) Nitrogen

    C) Fluorine                      

    D) Phosphorus

    Correct Answer: D

    Solution :

    The phosphorus containing organic compound are detected by 'Lassaigne's test' by heated with an oxidizing agent (sodium peroxide) The phosphorus present in the compound in oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate.                                                       \[\underset{\left( Ammonium\text{ }molybdate \right)}{\mathop{\begin{matrix}    N{{a}_{3}}P{{O}_{4}}+3HN{{O}_{3}}\to {{H}_{3}}P{{O}_{4}}+3NaN{{O}_{3}}  \\    {{H}_{3}}P{{O}_{4}}+12{{(N{{H}_{4}})}_{2}}Mo{{O}_{4}}+21HN{{O}_{3}}\to   \\ \end{matrix}}}\,\] \[{{(N{{H}_{4}})}_{3}}P{{O}_{4}}.12Mo{{O}_{3}}\downarrow +21N{{H}_{4}}N{{O}_{3}}+12{{H}_{2}}O\] (Ammonium phosphomolybdate) (canary yellow precipitate)


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