An organic compound 'A' is oxidized with Na2O2 followed by boiling with HNO3. 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 :

Question

An organic compound 'A' is oxidized with Na2O2 followed by boiling with HNO3. 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 : (A) Sulphur (B) Nitrogen (C) Fluorine (D) Phosphorus

Tardigrade Admin · Accepted Answer

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.

Na X_{3} PO X_{4} + 3 HNO X_{3} H X_{3} PO X_{4} + 3 NaNO X_{3}

H X_{3} PO X_{4} + (Ammonium molybdate) 12 (NH X_{4}) X_{2} MoO X_{4} + 21 HNO X_{3} (NH X_{4}) X_{3} PO X_{4} \cdot 12 MoO X_{3} ↓ + 21 NH X_{4} NO X_{3} + 12 H X_{2} O

(Ammonium phosphomolybdate)
(canary yellow precipitate)

Q. An organic compound ′A′ is oxidized with NaX2​OX2​ followed by boiling with HNOX3​. 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 :

Sulphur

Nitrogen

Fluorine

Phosphorus

Q. An organic compound $^{'} A^{'}$ is oxidized with $Na X_{2} O X_{2}$ followed by boiling with $HNO X_{3}$ . 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 :