Question

The given reaction is called:
$Phenol{\text{ }}\frac{{conc.{H_2}S{O_4}}}{{NaN{O_2}}} > p - nitrosophenol$
${\text{A}}{\text{. Liebermann's nitroso reaction}} \\\ {\text{B}}{\text{. Lederer - manasse reaction}} \\\ {\text{C}}{\text{. Fries rearrangement}} \\\ {\text{D}}{\text{. Reimer - Tiemann reaction}} \\\$

Answer 1

Hint: In this reaction, the product gives a violet-blue color and this is an identification test for phenol group. This is the first step of a name reaction.

Complete answer:
Here, $HN{O_2}$ breaks as HO-NO so here, $N{O^ + }$ acts as electro-phile and attaches at the para-position because phenol has high electron density at its para-position.
On adding further reagents, the reaction proceeds with an attack by the more nucleophilic N.

In absence of sulphuric acid, that is, in a dilute medium, the color changes occur as shown:
In the presence of an acid the color changes occur as:

First, brown red color appears on warming phenol and conc. $H_2SO_4$.
This red then changes to blue then green.
On dilution, color changes to red and finally, to greenish blue or violet on treatment with alkali.
The test is given by only 2 degree amines (both aliphatic and aromatic). Two degree amine is converted into nitroso-amine by treating the amine with $HNO_2$. So, this is a detection test for secondary amines. Since, phenol is the starting substrate in this reaction, so this test is an indicator of phenol group as well.
So, the given reaction in the question is the first step of Libermann’s Nitroso reaction.

So the correct option is (A).

Note: Name reactions are really important in chemistry. The particular reagent of the reaction, substrate and the product all should be known correctly to do such questions.