But-2-ene on reaction with alkaline KMnO$$_4$$ at elevated temperature followed by acidification will give:

We are given the alkene but-2-ene, whose condensed structural formula is $$\mathrm{CH_3-CH=CH-CH_3}$$. In an oxidation reaction with hot, alkaline potassium permanganate ($$\mathrm{KMnO_4 + OH^-}$$), followed by acidification (addition of $$\mathrm{H^+}$$), the double bond undergoes oxidative cleavage.

First, recall the general rule (state the formula):

$$\text{Alkene}\;(\mathrm{R_1-CH=CH-R_2})\; +\;[\mathrm{O}]\;\xrightarrow{\text{hot } \mathrm{KMnO_4}/\mathrm{OH^-}} \; \text{cleaves to}\; \mathrm{R_1-COO^-} + \mathrm{R_2-COO^-}$$

After this oxidative cleavage step, the basic medium gives carboxylate ions. Subsequent acidification converts each carboxylate ion into its corresponding carboxylic acid:

$$\mathrm{R-COO^- + H^+ \;\longrightarrow\; R-COOH}$$

Now we apply the rule to but-2-ene. Identify $$R_1$$ and $$R_2$$ on either side of the double bond:

We have $$\mathrm{CH_3-CH=CH-CH_3}$$, so $$R_1 = \mathrm{CH_3}$$ attached to the left carbon and $$R_2 = \mathrm{CH_3}$$ attached to the right carbon. Both sides are identical.

Performing oxidative cleavage:

$$ \mathrm{CH_3-CH=CH-CH_3} \;+\;[\mathrm{O}] \;\xrightarrow{\text{hot } \mathrm{KMnO_4}/\mathrm{OH^-}} \; \underbrace{\mathrm{CH_3-COO^-}}_{\text{acetate ion}} \;+\; \underbrace{\mathrm{CH_3-COO^-}}_{\text{another acetate ion}} $$

Next we acidify (add $$\mathrm{H^+}$$):

$$ \mathrm{CH_3-COO^- + H^+ \;\longrightarrow\; CH_3-COOH} $$

The same step occurs for the second acetate ion. So we finally obtain two molecules of acetic acid:

$$ \mathrm{CH_3-COOH + CH_3-COOH} $$

No aldehyde or diol is produced because the oxidative cleavage under these strong conditions fully oxidises each alkene carbon to the carboxylic acid level.

Therefore, the product set is “2 molecules of $$\mathrm{CH_3COOH}$$”, which matches Option A.

Hence, the correct answer is Option A.