Question

In the sodium fusion test of organic compounds, the nitrogen of an organic compound is converted to:
(A) sodium cyanide
(B) sodium-amide
(C) sodium nitrite
(D) sodium nitrate

Answer 1

As the test is for the presence of nitrogen in an organic compound, traces of nitrogen would be present in the resulting compound. It is a poisonous compound.

Complete step by step solution:
The sodium fusion test is also known as Lassaigne’s Test popularly as it was developed by J. L. Lassaigne.
Nitrogen, Sulphur, and halogens present in organic compounds are detected by Lassaigne’s test. Here, a small piece of clean sodium metal is heated in a fusion tube with the organic compound. The principle is that, in doing so, Na converts all the elements present into ionic form. The variety of techniques has been described.
$Na+C+N\to NaCN$
$2Na+S\to N{{a}_{2}}S$
$Na+X\to NaX$ (X = Cl, Br, or I)
The formed ionic salts are extracted from the fused mass by boiling it with distilled water. This is called the sodium fusion extract.
When an organic compound is heated strongly with sodium, any halogen, nitrogen, and sulphur will be converted into organic sodium salts such as sodium halide (for halides), sodium cyanide (for nitrogen), sodium thiocyanate (for sulphur and nitrogen), the nitrogen is confirmed with ferrous sulphate.
Let us concentrate more on the test for nitrogen-
The extract is boiled with $FeS{{O}_{4}}$and acidified with concentrated ${{H}_{2}}S{{O}_{4}}$. The appearance of Persian blue colour indicates the presence of nitrogen.
The following reactions occur,
1. $F{{e}^{2+}}+6C{{N}^{-}}\to {{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$
2. $F{{e}^{2+}}+{{H}^{+}}\to F{{e}^{3+}}+{{e}^{-}}$
3. ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}+4F{{e}^{3+}}\to F{{e}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right].{{H}_{2}}O$
The acid helps in the oxidation of ferrous ions to ferric ions. The formation of ferriferous cyanide indicates the presence of nitrogen.
Therefore, option (A) sodium cyanide is the correct option.

Note: This test is not given by compounds containing only N but not C atoms. For example, $N{{H}_{2}}N{{H}_{2}}$ does not answer this test despite having N atom. This is because both N and C are required to form $C{{N}^{-}}$ ion.
This test is not given by diazonium salts as they decompose to give nitrogen gas on heating.