Compound X is an isomer of nitrotoluene. Oxidation of X with acidic...

Isomers of nitrotoluene are $o$-nitrotoluene, $m$-nitrotoluene, and $p$-nitrotoluene.

Oxidation of these isomers with acidic $\text{KMnO}_4$ yields $o$-nitrobenzoic acid, $m$-nitrobenzoic acid, and $p$-nitrobenzoic acid, respectively. Due to the ortho-effect (a combination of steric and electronic factors), ortho-substituted benzoic acids are generally stronger acids than their meta and para isomers. Thus, $o$-nitrobenzoic acid is the strongest acid among them.

Reduction of the isomers with $\text{Sn}/\text{HCl}$ yields $o$-toluidine, $m$-toluidine, and $p$-toluidine, respectively. Due to the ortho-effect (steric inhibition of solvation of the conjugate acid), ortho-substituted anilines are weaker bases than their meta and para isomers, as well as aniline itself. Thus, $o$-toluidine is the weakest base among them.

Therefore, Compound X must be $o$-nitrotoluene.

Answer: $o$-Nitrotoluene