Question

Lassaigne’s test for the detection of nitrogen will fail in the case of:
A. $\text{N}{{\text{H}}_{2}}\text{CON}{{\text{H}}_{2}}$
B. $\text{N}{{\text{H}}_{2}}\text{CON}{{\text{H}}_{2}}.\text{HCl}$
C. $\text{N}{{\text{H}}_{2}}\text{N}{{\text{H}}_{2}}$
D. ${{\text{C}}_{6}}{{\text{H}}_{5}}\text{NHN}{{\text{H}}_{2}}.2\text{HCl}$

Answer 1

Hint: Lassaigne’s extract is prepared to convert the covalent compound into ionic form. This is done by fusing the organic compound with sodium metal. Carbon elements should also be present in the organic compound for giving sodium cyanide to test the presence of nitrogen.

Complete step by step solution:
Lassaigne’s test is the general test for the detection of halogens, nitrogen and sulphur in an organic compound. In order to detect them, these have to be converted into their ionic forms.
Test for Nitrogen-
The presence of nitrogen in the organic compound is detected by fusing organic compounds with sodium metal to give sodium cyanide (NaCN) soluble in water. This is converted into sodium ferrocyanide by the addition of sufficient quantities of ferrous sulphate. Ferric ions generated during the process react with ferrocyanide to form Prussian blue precipitate of ferric ferrocyanide.
$\text{Na+C+N}\to \text{NaCN}$
$\text{6NaCN+ FeS}{{\text{O}}_{4}}\to \text{N}{{\text{a}}_{4}}[\text{Fe(CN}{{\text{)}}_{6}}]+\text{N}{{\text{a}}_{2}}\text{S}{{\text{O}}_{4}}$
$\text{N}{{\text{a}}_{4}}[\text{Fe(CN}{{\text{)}}_{6}}]+\text{F}{{\text{e}}^{3+}}\to \text{F}{{\text{e}}_{4}}{{[\text{Fe(CN}{{\text{)}}_{6}}]}_{3}}$
From the above discussion, we saw that there is a presence of carbon atoms in the organic compound and there is a formation of a complex of cyanide ions as a ligand. Therefore, the compound which does not have carbon atoms in it, will surely fail the nitrogen test done by Lassaigne’s extract.
Therefore, the correct option is (c) $\text{N}{{\text{H}}_{2}}\text{N}{{\text{H}}_{2}}$.

Note: We should also know that when nitrogen and sulphur both are present in an organic compound, sodium fusion will convert it into sodium thiocyanate which then reacts with $\text{F}{{\text{e}}^{3+}}$ to form blood colour complex${{[\text{Fe(SCN) }\!\!]\!\!\text{ }}^{2+}}$.