An organic compound 'A' C$$_4$$H$$_8$$ on treatment with KMnO$$_4$$/H$$^+$$ yields compound 'B' C$$_3$$H$$_6$$O. Compound 'A' also yields compound 'B' on ozonolysis. Compound 'A' is:

We are told that the hydrocarbon ‘A’ has molecular formula $$C_4H_8$$. The general formula $$C_nH_{2n}$$ corresponds either to an alkene (one C=C bond) or to a cycloalkane (all C-C single bonds inside a ring). So ‘A’ must be either an alkene or a four-membered ring.

After hot acidic permanganate oxidation (written as $$KMnO_4/H^+$$) the only organic product obtained is ‘B’ with formula $$C_3H_6O$$. Hot $$KMnO_4$$ completely cleaves a C=C bond and converts each carbon of that double bond as follows:

• “Internal” carbon (no attached H) → Ketone.
• “Secondary” carbon (one attached H) → Carboxylic acid.
• “Terminal” vinyl carbon $$( =CH_2)$$ → $$CO_2$$.

Because only one organic fragment remains (three carbons long) and one carbon has been lost, the cleavage must have destroyed a terminal $$=CH_2$$ group, giving $$CO_2$$, while the other carbon of the double bond (which had no hydrogens) became the three-carbon ketone $$C_3H_6O$$.

Let us write the unknown alkene in the general form $$R_2C=CH_2$$. Applying the above rule:

$$$\displaystyle R_2C=CH_2 \;\xrightarrow[\text{hot}]{KMnO_4/H^+}\; R_2C=O\;+\;CO_2$$$

The ketone obtained has formula $$C_3H_6O$$. For a ketone the carbonyl carbon is already counted, so its alkyl parts must supply three carbons in total. The only way to supply three carbons with the pattern $$R_2C=O$$ is to make both $$R$$ groups methyl, because

$$$ (CH_3)_2C=O\;\text{ is acetone (propan-2-one), formula }C_3H_6O.$$$

Therefore

$$$ R = CH_3 \quad\Longrightarrow\quad A = (CH_3)_2C=CH_2 $$$

This structure is named 2-methylpropene.

Now we verify the second fact, ozonolysis. Ozonolysis also cleaves the C=C bond but under reductive work-up (Zn/H2O) stops at carbonyl compounds:

$$$ (CH_3)_2C=CH_2 \;\xrightarrow[\;Zn/H_2O\;]{O_3}\; (CH_3)_2C=O \;+\; H_2C=O $$$

The products are acetone $$(C_3H_6O)$$ and formaldehyde $$(CH_2O)$$. Thus the same compound ‘B’ (acetone) is again formed, satisfying the condition in the statement.

Any of the other options fails:

• $$\text{But-2-ene}\;(CH_3CH=CHCH_3)$$ would give two molecules of ethanal $$(C_2H_4O).$$
• $$\text{Cyclobutane}$$ and $$\text{1-methylcyclopropane}$$ contain no C=C bond; their oxidative cleavage patterns cannot furnish a single $$C_3H_6O$$ fragment while losing only one carbon.

Hence, the only structure consistent with all the data is 2-methylpropene.

Hence, the correct answer is Option A.