Organic Chemistry Basics 2 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
(2019 Main, 12 April I)
(a) nitrogen
(b) phosphorus
(c) fluorine
(d) sulphur
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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.
$$ \begin{gathered} \mathrm{H}{3} \mathrm{PO}{4}+12\left(\mathrm{NH}{4}\right){2} \mathrm{MoO}{4}+21 \mathrm{HNO}{3} \longrightarrow \ \left(\mathrm{NH}{4}\right){3} \mathrm{PO}{4} \cdot 12 \mathrm{MoO}{3} \downarrow+21 \mathrm{NH}{4} \mathrm{NO}{3}+12 \mathrm{H}_{2} \mathrm{O} \ \text { Yellow ppt. } \end{gathered} $$
