In the following sequence of reactions:
Toluene $$\xrightarrow{KMnO_4}$$ A $$\xrightarrow{SOCl_2}$$ B $$\xrightarrow[ BaSO_4]{H_2/Pd}$$ C
The Product C is:

We start with toluene, whose molecular formula is $$C_6H_5CH_3$$.

First, the reagent potassium permanganate $$KMnO_4$$ is a strong oxidising agent. A well-known result in organic chemistry is that $$KMnO_4$$ oxidises any alkyl side chain attached to an aromatic ring completely to the carboxylic acid, no matter how long the side chain is. Symbolically,

$$C_6H_5CH_3 + 3[O] \;\;{\longrightarrow}\;\; C_6H_5COOH + H_2O.$$

So after the first step we obtain compound A, benzoic acid $$C_6H_5COOH$$.

Next, benzoic acid is treated with thionyl chloride $$SOCl_2$$. Before writing the equation, recall the general reaction:

Carboxylic acid $$RCOOH$$ + $$SOCl_2$$ $$\rightarrow$$ Acyl chloride $$RCOCl$$ + $$SO_2$$ $$\uparrow$$ + $$HCl$$ $$\uparrow$$.

Applying this to benzoic acid, we have

$$C_6H_5COOH + SOCl_2 \;\;{\longrightarrow}\;\; C_6H_5COCl + SO_2 + HCl.$$

Thus compound B is benzoyl chloride $$C_6H_5COCl$$.

Finally, compound B is subjected to $$H_2/Pd$$ in the presence of $$BaSO_4$$. This specific combination is called the Rosenmund catalyst. The reaction it brings about is known as the Rosenmund reduction. State the rule first:

Rosenmund reduction: An acyl chloride $$RCOCl$$, when hydrogenated over $$Pd$$ poisoned with $$BaSO_4$$, is reduced to the corresponding aldehyde $$RCHO$$.

Applying the rule to benzoyl chloride,

$$C_6H_5COCl + H_2 \;\;{\longrightarrow[\;Pd/BaSO_4\;]}\;\; C_6H_5CHO + HCl.$$

Hence compound C is benzaldehyde $$C_6H_5CHO$$.

Looking back at the options, benzaldehyde corresponds to Option A.

Hence, the correct answer is Option A.