Compound A has the formula C8H8. It reacts rapidly with acidic KMnO4 but reacts with only 1 equivalent of H2 over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, A reacts with 4 equivalents of H2, and hydrocarbon B, C8H16, is produced. The reaction of A with KMnO4 gives CO2 and a carboxylic acid C, C7H6O2. Required:
Draw the structure of compound B below.

C8H16 (Ethylcyclohexane).

Explanation:

From the given information:

Compound A is an alkene because it interacts with 1 unit of hydrogen across a palladium catalyst.

Also, we are given another hint that:

Compound A needs 4 equivalence of H2 to hydrogenate under circumstances that decrease aromatic rings, indicating that it is a phenyl substituted alkene.

Compound A with formula C8H8 reacts instantly with KMnO4 to produce CO2, as well as carboxylic acid, points out that Compound acts as a terminal alkene.

Therefore, we can opine that compound A is a terminal phenyl substituted alkene whose formula = C8H8 (Styrene)

The diagrammatic expression of the compound can be seen below.

However, in the presence of the palladium catalyst, the reduction of Compound A with 4 units of hydrogen produces Compound B: C8H16 (Ethylcyclohexane).