An unknown alkene 'X' on vigorous oxidation with acidic KMnO_4...

Vigorous oxidation of alkenes with acidic $\text{KMnO}_4$ cleaves the $\text{C=C}$ double bond.

The products are propan-2-one ($\text{CH}_3\text{-CO-CH}_3$) and ethanoic acid ($\text{CH}_3\text{COOH}$).

Propan-2-one is a ketone, which indicates that one of the doubly bonded carbons was disubstituted, specifically a $\text{=C(CH}_3)_2$ group.

Ethanoic acid is a carboxylic acid, which indicates that the other doubly bonded carbon had one hydrogen atom and one methyl group, specifically a $\text{=CH-CH}_3$ group.

Joining these two fragments at the double bond gives the structure of the original alkene 'X': $\text{(CH}_3)_2\text{C=CH-CH}_3$.

The IUPAC name of this compound is 2-methylbut-2-ene.

Answer: 2-Methylbut-2-ene