Two isomers 'A' and 'B' with molecular formula C$$_4$$H$$_8$$ give different products on oxidation with KMnO$$_4$$ in acidic medium. Isomer 'A' on reaction with KMnO$$_4$$/H$$^+$$ results in effervescence of a gas and gives ketone. The compound 'A' is

We are given two isomers A and B with molecular formula C$$_4$$H$$_8$$ that undergo oxidation with KMnO$$_4$$/H$$^+$$ to give different products. Isomer A produces effervescence of CO$$_2$$ and yields a ketone. The degree of unsaturation for C$$_4$$H$$_8$$ is $$\frac{2(4) + 2 - 8}{2} = 1$$, indicating either one double bond or one ring in each isomer. The possible alkene isomers of C$$_4$$H$$_8$$ include but-1-ene ($$CH_2=CH-CH_2-CH_3$$), but-2-ene (cis and trans, $$CH_3-CH=CH-CH_3$$), and 2-methylpropene [$$ (CH_3)_2C=CH_2$$].

Under KMnO$$_4$$/H$$^+$$ oxidative cleavage, a terminal $$=CH_2$$ group is oxidized to $$CO_2 + H_2O$$ (causing effervescence), whereas an internal >C= fragment is oxidized to a ketone. Considering 2-methylpropene [$$ (CH_3)_2C=CH_2$$], oxidation proceeds as:

$$ (CH_3)_2C=CH_2 \xrightarrow{KMnO_4/H^+} (CH_3)_2C=O + CO_2 \uparrow + H_2O $$

This reaction yields acetone [$$ (CH_3)_2C=O$$] and $$CO_2$$ gas, matching the effervescence and ketone formation observed for isomer A. By contrast, but-1-ene would give propanoic acid and $$CO_2$$ (an acid rather than a ketone), and but-2-ene (either cis or trans) would yield two moles of acetic acid with no $$CO_2$$ evolution. Therefore, isomer A must be 2-methylpropene.

The correct answer is Option D: 2-methylpropene.