I read that to test for the presence of phenol groups in an organic compound the aniline dye test can be performed. If the solution has phenol present it gives a red or orange precipitate to confirm the presence, however as a side note it was mentioned that this test is applicable for the detection of small quantities of phenols, which have a free para position. Why is it so?
1$\begingroup$ Umm, why don't you just read up about synthesis of azo compounds? That should make things clear. $\endgroup$– MithoronSep 19, 2019 at 22:21
First, the aniline is converted to a diazonium halide salt in the presence of acid and sodium nitrite, per this reference.
Then, the diazonium salt reacts with the phenol to give an azo compound, per this reference (see the section entitled Coupling reactions of diazonium ions
Note that the diazonium ion is shown reacting with the para position of the phenol. That is because the -OH group in phenol is an ortho,para-director (reference), meaning that it enhances the nucleophilicity of the ortho and para positions on the phenol ring. However, the para position is significantly more nucleophilic because of the relative lack of steric hindrance in comparison to the ortho positions, which are hindered by the proximity of the -OH group.
So, having a free para position on the phenol means that the phenol is especially reactive towards electrophiles, including the diazonium salt in this example, meaning that it will react faster and in higher yields - increasing the sensitivity of the aniline dye test.