Question

During sodium extract preparation for lassaigne’s test both N and S present in an organic compound change to:
A) ${\text{NaCN}}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}$
B) ${\text{NaN}}{{\text{H}}_2}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$
C) ${\text{NaSCN}}$
D) ${\text{NaN}}{{\text{O}}_3}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}$

Answer 1

Lassaigne’s test is used to determine the foreign elements in the presence of organic compounds like carbon and nitrogen. These foreign elements are nitrogen, sulfur and halogen.

Complete answer:
Lassaigne’s test is also known as sodium fusion test. The halogen, Nitrogen and Sulfur are covalently bonded to the organic compound and in order to detect them the element first need to converted into ionic form and this is extracted in the aqueous solution and further easily be detected by simple chemical test and the extract is called sodium fusion extract or Lassaigne’s extract. In this reaction first sodium fuses with carbon and nitrogen and forms sodium cyanide and further it reacts with sulphur to form sodium thiocyanate.
The reaction is-
${\text{Na + C + N + S }}\xrightarrow{{fuse}}{\text{NaSCN }}\left( {{\text{Sodium Thiocyanate}}} \right)$

${\text{NaCN}}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}$ is not possible because only one product will be formed by fusion , first ${\text{Na}}$ reacts with ${\text{N}}$ and ${\text{C}}$ to form ${\text{NaCN}}$ and then it again fuses with the Sulfur to form ${\text{NaSCN}}$ . In the same way other two products like${\text{NaN}}{{\text{H}}_2}$, ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$, ${\text{NaN}}{{\text{O}}_3}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}$are not possible. Therefore, the correct answer is ${\text{NaSCN}}$

Hence the correct answer is option ‘C’.

Note: In Lassaigne’s test for nitrogen, Prussian blue colour is obtained due to formation of ferriferous cyanide complex. Violet colour indicates presence of sulphur. White precipitate indicates the presence of chlorine.