Question

Which of the following compounds will give red colour on the Lassaigne test ?
a.) NaCNS
b.) $N{H_2}CSN{H_2}$ (thiourea)
c.) $N{H_2}CON{H_2}$ (urea)
d.) p -$N{H_2}{C_6}{H_4}S{O_3}H$ (p -aminobenzene sulphonic acid)

Answer 1

Hint : The presence of Nitrogen and Sulphur in the organic compound will result in reaction to form thiocyanate in the Lassaigne test. This product formed on heating with $FeS{O_4}$ or $FeC{l_3}$ gives out a complex which has red colour. Thus, the solution is turned red.

Complete p answer :
First, let us understand what the Lassaigne test is.
It is also called the Sodium fusion test. It is used in qualitative analysis for the determinations of halogens, nitrogen and sulphur elements.
Initially, a small amount of Sodium metal and organic compound to be tested is put in the fusion tube and heated to certain extent. Then rapidly this fusion tube is put in a container containing water. The tube burst leaving the contents out. This solution called sodium fusion extract is tested for the presence of various groups in the organic compound.
Halogen:
In case of presence of halides, sodium reacts with them giving sodium halide which further on reaction with silver nitrate will give coloured solutions. The reaction is as-
$Na + X \to NaX$
$NaX + AgN{O_3} \to AgX + NaN{O_3}$
Nitrogen:
 In case Nitrogen is present in the extract, then it will react with sodium giving sodium cyanide which will further react with ferrous sulphate giving sodium ferrocyanide as-
$Na + C + N \to NaCN$
$6NaCN + FeS{O_4} \to Na\left[ {Fe{{\left( {CN} \right)}_6}} \right] + N{a_2}S{O_4}$
Some ferric ions are generated during the process. These ferric ions and sodium ferrocyanide react giving ferric ferrocyanide which forms a ppt of Prussian blue colour.
 $Na\left[ {Fe{{\left( {CN} \right)}_6}} \right] + F{e^{3 + }} \to F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}$
Sulphur:
The sulphur will react with sodium giving sodium sulphide which is identified by sodium nitroprusside as-
$\begin{gathered}
  Na + S \to N{a_2}S \\
  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] \\
\end{gathered} $
This will have violet colour.
So, till now you would have been wondering that we read about both nitrogen and sulphur but still no red colour, why so.
This is because red colour is observed when nitrogen and sulphur both are present in the organic compound.
Both nitrogen and sulphur:
When both these elements are present, they will react with sodium to give sodium thiocyanate which will react with ferric ions from $FeS{O_4}$ or $FeC{l_3}$ to give complex having red colour.
$\begin{gathered}
  Na + C + N + S \to NaSCN \\
  F{e^{3 + }} + 2Na \to {\left[ {Fe\left( {SCN} \right)} \right]^{2 + }} \\
\end{gathered} $
So, now as we know the concept let us see the options.
When we see options then we observe that options a.), b.) and d.) contain both these elements in the organic compound. So, these three will give red colour.

So, the correct answer is options a.), b.) and d.).

Note :
In actuality, the elements like Halogens, nitrogen and sulphur are covalently bonded in the organic compound. Thus, the qualitative analysis is difficult to perform for such elements. Using Sodium fusion tests, they are converted into ionic forms using metal. On heating the organic compounds with sodium, we inorganic sodium salts such as sodium halide, sodium cyanide and sodium thiocyanate etc.