Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. $$H_2SO_4$$ followed by treatment with NaOH?

The reaction described here is the formation of a phenolphthalein-type dye. When a phenolic compound is heated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ (which acts as a dehydrating agent), a condensation reaction occurs. The resulting product, when dissolved in NaOH, gives a characteristic pink colour.

For this condensation to happen, the phenol must have a free $$-OH$$ group attached directly to the aromatic ring. The electrophilic substitution during the condensation takes place at the ortho or para positions relative to the $$-OH$$ group.

Now, 4-methylphenol (p-cresol) has the structure where the $$-OH$$ group is at position 1 and the $$-CH_3$$ group is at position 4. The para position (position 4) is occupied by $$-CH_3$$, but both ortho positions (positions 2 and 6) are free. The condensation with phthalic anhydride requires two available positions on the ring, and the ortho positions serve this purpose.

Among the other options, option 2 contains an alcoholic $$-OH$$ (not directly on the ring) which cannot undergo this phenol-specific condensation. Options 3 and 4 contain dihydroxy compounds (resorcinol-type), which would give fluorescein (a green fluorescent dye) rather than the pink phenolphthalein-type dye.

Hence, 4-methylphenol gives the pink colour with phthalic anhydride in concentrated $$H_2SO_4$$ followed by NaOH treatment.

Hence, the correct answer is Option 1.