The most appropriate reagent for conversion of $$\text{C}_2\text{H}_5\text{CN}$$ into $$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$ is:

We have the starting compound $$\text{C}_2\text{H}_5\text{CN}$$, which is also written as $$\text{CH}_3\text{CH}_2\text{CN}$$. The functional group present is a nitrile $$(-\,\text{C}≡\text{N})$$. Our target molecule is $$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$, that is, a primary amine containing three carbon atoms.

To go from a nitrile $$\text{R-C}≡\text{N}$$ to a primary amine $$\text{R-CH}_2\text{NH}_2$$, the standard method is complete reduction of the triple bond. The commonly used laboratory reagent for this purpose is lithium aluminium hydride, $$\text{LiAlH}_4$$.

First, recall the general reduction formula:

$$\text{R-C}≡\text{N} \overset{\text{LiAlH}_4}{\underset{\text{ether}}{\rightarrow}} \text{R-CH}_2\text{NH}_2$$

This tells us that every nitrile carbon receives two hydride ions from $$\text{LiAlH}_4$$, converting the $$C≡N$$ triple bond first to an imine intermediate and finally to the saturated $$\text{CH}_2\text{NH}_2$$ group. No carbon atoms are lost or gained in the process.

Applying the same transformation to our substrate:

$$\text{CH}_3\text{CH}_2\text{C}≡\text{N} \overset{\text{LiAlH}_4}{\underset{\text{ether}}{\rightarrow}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$

Thus the reagent that achieves the desired conversion is clearly $$\text{LiAlH}_4$$.

Why the other options fail:

• $$\text{NaBH}_4$$ is a milder hydride donor; it reduces aldehydes and ketones but does not normally touch nitriles.
• $$\text{CaH}_2$$ is not a typical hydride reagent for organic reductions; it is mainly a drying agent.
• $$\text{Na(CN)BH}_3$$ (sodium cyanoborohydride) is even milder than $$\text{NaBH}_4$$ and is used for reductive amination, not for reducing nitriles.

So, only $$\text{LiAlH}_4$$ can reduce $$\text{C}_2\text{H}_5\text{CN}$$ all the way to $$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$.

Hence, the correct answer is Option C.