An unsaturated hydrocarbon X on ozonolysis gives A. Compound A when warmed with ammoniacal silver nitrate forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon X is:

The reaction requires identifying a hydrocarbon whose ozonolysis product gives a positive Tollens’ test.

Tollens’ reagent gives a positive test for aldehydes.

Among carboxylic acids, only $$\mathrm{HCOOH}$$ can also reduce Tollens’ reagent because it contains an aldehydic hydrogen.

For option $$\mathrm{(A)}$$, ozonolysis of $$\mathrm{2,3\text{-}Dimethyl\text{-}2\text{-}butene}$$ forms acetone, which is a ketone and does not respond to Tollens’ reagent.

For option $$\mathrm{(B)}$$, ozonolysis of isopropylidenecyclopropane forms acetone and cyclopropanone, both ketones, so Tollens’ test is negative.

For option $$\mathrm{(D)}$$, ozonolysis of $$\mathrm{2\text{-}Butyne}$$ forms acetic acid $$\mathrm{(CH_3COOH)}$$, which does not respond to Tollens’ reagent.

For option $$\mathrm{(C)}$$, ozonolysis of $$\mathrm{1\text{-}Butyne\ (HC\equiv C-CH_2-CH_3)}$$ produces propanoic acid and:

$$\mathrm{HCOOH}$$

Formic acid gives a positive Tollens’ test and forms the silver mirror.

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{HCOOH}$$

Thus, the correct option is C.