A compound 'X' when treated with phthalic anhydride in presence of concentrated H$$_2$$SO$$_4$$ yields 'Y'. 'Y' is used as an acid/base indicator. 'X' and 'Y' are respectively

We are told that compound X, when treated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$, yields compound Y. Y is used as an acid/base indicator.

First, identify compound Y.

The reaction of a phenolic compound with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ (a Friedel-Crafts type condensation reaction) produces phenolphthalein, which is a well-known acid/base indicator.

Therefore, Y = Phenolphthalein.

First, identify compound X.

Phenolphthalein is synthesized by the condensation of two molecules of phenol with one molecule of phthalic anhydride in the presence of concentrated $$H_2SO_4$$:

$$2 \, C_6H_5OH + C_8H_4O_3 \xrightarrow{H_2SO_4} C_{20}H_{14}O_4 + H_2O$$

Therefore, X = Phenol.

Now, match with options.

Phenol is also known as carbolic acid. Looking at the options:

Option A: Anisole, methyl orange — Incorrect. Anisole is methyl phenyl ether, not phenol.

Option B: Salicylaldehyde, Phenolphthalein — Incorrect. Salicylaldehyde is not used to make phenolphthalein.

Option C: Toludine, Phenolphthalein — Incorrect. Toludine (methylaniline) cannot form phenolphthalein.

Option D: Carbolic acid, Phenolphthalein — Correct. Carbolic acid is phenol, and its reaction with phthalic anhydride gives phenolphthalein.

The correct answer is Option D.