Which one of the following phenols does not give colour when condensed with phthalic anhydride in presence of conc. H$$_2$$SO$$_4$$?

When a phenol is heated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ the first step is an electrophilic acylation at a position that is ortho (or, more rarely, para) to the -OH group. After two such substitutions and intramolecular ring closure, a phthalein dye (e.g. phenolphthalein, fluorescein, etc.) is obtained. The reaction therefore has two essential requirements:

1. The aromatic ring must be activated toward electrophilic substitution (-OH is an ortho/para-directing activator).
2. At least one of the positions ortho or para to the -OH must be unsubstituted so that the acylation can occur.

Let us check the four phenols given in the question.

Phenol itself has both ortho and the para positions free; it readily forms the coloured dye phenolphthalein.

p-Cresol has both ortho positions free; it also condenses with phthalic anhydride to give a coloured p-tolyl-phthalein.

Resorcinol (1,3-dihydroxybenzene) is even more activated because it contains two -OH groups. It reacts smoothly, giving fluorescein (intense yellow-green fluorescence).

2,4,6-Trinitrophenol (picric acid) carries three -NO2 groups at all ortho and para positions relative to the -OH group. Two consequences follow:
(i) Every required entry position is already occupied; sterically the ring offers no vacant ortho/para site for the incoming acylium ion.
(ii) The strongly -M and -I nitro groups make the ring highly deactivated toward electrophilic substitution.
Because of these two factors, picric acid cannot undergo the initial acylation step, so no phthalein framework is formed and, hence, no colour is produced.

Therefore the only phenol that fails to give a coloured product with phthalic anhydride and conc. $$H_2SO_4$$ is the one in Option D, 2,4,6-trinitrophenol (picric acid).

Option D which is: 2,4,6-trinitrophenol (picric acid)