Question

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Answer 1

Hint: The fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens is known as Lassaigne’s test and it is a general test for detection of nitrogen, sulphur and halogens. This test is also known as sodium fusion test and it is primarily used for qualitative analysis of various elements.

Complete step by step answer:
The elements such as halogens, nitrogen and sulphur are covalently bonded to the organic compounds. For their detection, we need to convert them to their ionic form for performing respective tests for their detection. As per principle, on fusion Na converts all elements present in an organic compound in ionic form. The ionic salts formed are extracted by boiling with distilled water and the process is known as sodium fusion extraction.
$$\begin{gathered} Na+C+N\to NaCN \\ 2Na+S\to N{a}_{2}SNa+X\to NaX\left(X=Cl,Br,orI\right) \end{gathered}$$
For testing nitrogen,
The extract is converted to sodium ferrocyanide by heating with ferrous sulphate and acidified with concentrated $H_{2}S{O}_4$. The presence Prussian blue precipitate of ferric ferrocyanide indicates the presence of nitrogen. The reactions occur,
$$6NaCN+FeS{O}_4\to N{a}_4\left[Fe{\left(CN\right)}_6\right]+N{a}_{2}S{O}_4$$
Here, the formula for Sodium ferrocyanide is $$N{a}_4\left[Fe{\left(CN\right)}_6\right]$$
$$N{a}_4\left[Fe{\left(CN\right)}_6\right]+F{e^3}^{+}\to F{e}_4{\left[Fe{\left(CN\right)}_6\right]}_3$$
Here, the formula for Ferric ferrocyanide is $$F{e}_4{\left[Fe{\left(CN\right)}_6\right]}_3$$
For testing Sulphur,
Sulphide ions are tested using sodium nitroprusside. The appearance of violet colour indicates presence of sulphur.
Below is the reaction that occurs
$$\begin{gathered} N{a}_{2}S+N{a}_2\left[Fe{\left(CN\right)}_{5}NO\right]\to N{a}_4\left[Fe{\left(CN\right)}_{5}NOS\right] \\ (Sodiumnitroprusside)(violetcolour) \end{gathered}$$
For testing halogens,
For the case of testing extract for the presence of halogens such as Cl, Br, I we will add silver nitrate solution after acidifying with dil. $HN{O}_3$ to it. A yellowish white precipitate sparingly soluble in ammonium hydroxide $(N{H}_{4}OH)$ indicates presence of Bromine(Br), yellowish precipitate insoluble in ammonium hydroxide $(N{H}_{4}OH)$ indicates presence of Iodine(I), White precipitate soluble in $(N{H}_{4}OH)$ indicates presence of Chlorine.

Note:
We only need to use sodium fusing with organic compounds as the resultant sodium compounds are soluble in water. If we discuss the use of potassium in place of sodium for fusion then the resultant compounds are insoluble in water and hence, the test is impossible to proceed.