Organic Chemistry Basics Result Question 2
2. An organic compound $A$ is oxidised with $\mathrm{Na} _2 \mathrm{O} _2$ followed by boiling with $\mathrm{HNO} _3$. 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
(a) nitrogen
(b) phosphorus
(c) fluorine
(d) sulphur
(2019 Main, 12 April I)
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Answer:
Correct Answer: 2. ( b )
Solution:
- Organic compound ’ $A$ ’ contain phosphorus as it gives positive test with ammonium molybdate. Phosphorus present in organic compound ’ $A$ ’ get oxidised with $\mathrm{Na}_2 \mathrm{O}_2$ and form $\mathrm{Na}_3 \mathrm{PO}_4$. $$ 2 \mathrm{P}+5 \mathrm{Na}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{Na}_3 \mathrm{PO}_4+2 \mathrm{Na}_2 \mathrm{O} $$
Compound Sodium phosphate $\mathrm{Na}_3 \mathrm{PO}_4$ in presence of $\mathrm{HNO}_3$ form $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{NaNO}_3$. $$ \mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3 $$
Upon cooling, a few drops of ammonium molybdate solution are added. A yellow ppt. confirms the presence of phosphorus in the organic compound. $$ \begin{gathered} \mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \longrightarrow \ \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 \downarrow+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O} \ \end{gathered} $$
Yellow ppt
BETA
