An organic compound 'A' is oxidizod with Na_2O_2 followed by boiling with HNO_3 . The 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:

Practical Organic Chemistry PYQ — JEE Main 2019 · 12 Apr · Shift 1 · Q16 with Solution

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Correct answer: B

Identify the test being described

The compound is first oxidized with $Na_2O_2$ and then boiled with $HNO_3$ . After that, the solution is treated with ammonium molybdate, giving a yellow precipitate.
Meaning of the yellow precipitate with ammonium molybdate

Ammonium molybdate gives a canary yellow precipitate with phosphate ion in nitric acid medium.

The reaction corresponds to formation of ammonium phosphomolybdate.
Why oxidation with $Na_2O_2$ is done

If the organic compound contains phosphorus, oxidation converts it into phosphate:

$P \longrightarrow PO_4^{3-}$

On acidifying with $HNO_3$ , the phosphate remains in solution and then reacts with ammonium molybdate to form the yellow precipitate.
Check the options
- A: Fluorine — does not give this ammonium molybdate yellow precipitate.
- B: Phosphorus — correct, because phosphate gives yellow ammonium phosphomolybdate.
- C: Nitrogen — nitrogen is tested differently (Lassaigne's test, Prussian blue, etc.).
- D: Sulphur — sulphur is tested by sodium nitroprusside or lead acetate tests, not this one.
Conclusion

Therefore, the element present is phosphorus.

View written solutionFree

Correct answer: B

Identify the test being described

The compound is first oxidized with $Na_2O_2$ and then boiled with $HNO_3$ . After that, the solution is treated with ammonium molybdate, giving a yellow precipitate.
Meaning of the yellow precipitate with ammonium molybdate

Ammonium molybdate gives a canary yellow precipitate with phosphate ion in nitric acid medium.

The reaction corresponds to formation of ammonium phosphomolybdate.
Why oxidation with $Na_2O_2$ is done

If the organic compound contains phosphorus, oxidation converts it into phosphate:

$P \longrightarrow PO_4^{3-}$

On acidifying with $HNO_3$ , the phosphate remains in solution and then reacts with ammonium molybdate to form the yellow precipitate.
Check the options
- A: Fluorine — does not give this ammonium molybdate yellow precipitate.
- B: Phosphorus — correct, because phosphate gives yellow ammonium phosphomolybdate.
- C: Nitrogen — nitrogen is tested differently (Lassaigne's test, Prussian blue, etc.).
- D: Sulphur — sulphur is tested by sodium nitroprusside or lead acetate tests, not this one.
Conclusion

Therefore, the element present is phosphorus.