Bouveault-Blanc reduction reaction involves:

We begin by recalling what the Bouveault-Blanc reduction is. This named reaction is classically carried out when an ester is treated with metallic sodium in the presence of an alcohol, most commonly ethanol. The overall transformation converts the ester functional group into the corresponding primary alcohols.

In other words, if we take a generic ester written as $$RCOOR'$$ and subject it to the reagents $$Na$$ (metallic sodium) and $$C_2H_5OH$$ (ethanol), the reaction proceeds as

$$RCOOR' \;+\; 4\,Na \;+\; 4\,C_2H_5OH \;\rightarrow\; RCH_2OH \;+\; R'OH \;+\; 4\,C_2H_5ONa.$$

So, the key points we must match with the options are:

1. The substrate must be an ester.

2. The reducing system must be $$Na/C_2H_5OH$$ (sodium in an alcoholic solvent).

Now we look at the given options one by one:

Option A mentions the reduction of an acyl halide with $$H_2/Pd$$. This is hydrogenation and is not the Bouveault-Blanc reduction.

Option B mentions the reduction of an anhydride with $$LiAlH_4$$. That is a lithium aluminium hydride reduction, again unrelated to Bouveault-Blanc.

Option C states the reduction of an ester with $$Na/C_2H_5OH$$. This matches exactly with the reaction we have just described.

Option D talks about reducing a carbonyl compound with $$Na/Hg$$ and $$HCl$$. That is the Clemmensen reduction, not Bouveault-Blanc.

Only Option C aligns perfectly with the definition and reagents of the Bouveault-Blanc reduction.

Hence, the correct answer is Option C.