Animated Solution for Chemistry - Organic Chemistry: Reagent(s) which can be used to bring about the following transformation is(are)

Look closely at the reactant molecule provided in the problem. It is a beautifully complex structure that doesn't just have one, but four distinct functional groups. On the left side, we have a carboxylic acid ($-COOH$) and a bulky tert-butyl ester ($-COOR$). Moving towards the center, there is a highly strained three-membered epoxide ring. Finally, on the far right side, we find an aldehyde group ($-CHO$).

Now, let's observe the product. The transformation is incredibly specific. The carboxylic acid, the ester, and the fragile epoxide ring are exactly the same—they have been left completely intact. However, the aldehyde group has been successfully reduced to a primary alcohol ($-C{H}_{2}OH$). This observation is our biggest clue: we need a reducing agent that is highly selective.

Let's evaluate our options. Option A suggests using Lithium Aluminum Hydride ($LiAl{H}_4$). While $LiAl{H}_4$ is a fantastic and very strong reducing agent, it lacks the finesse required here. It will definitely reduce the aldehyde, but it will also aggressively attack and reduce both the ester and the carboxylic acid into primary alcohols. Therefore, it cannot be our answer.

Option B offers Borane ($B{H}_3$) in THF. Borane is famous in organic chemistry for its exceptional ability to chemoselectively reduce carboxylic acids over other carbonyl compounds. If we use $B{H}_3$, it will fail to maintain the required selectivity and will convert our precious carboxylic acid group into an alcohol.

Option D suggests Raney Nickel with Hydrogen ($H_2$). Catalytic hydrogenation is great for reducing aldehydes and alkenes, but it comes with a severe drawback for this specific molecule. The conditions can easily cause hydrogenolysis, which would cleave the strained epoxide ring open. So, this option is also rejected.

This leaves us with Option C: Sodium Borohydride ($NaB{H}_4$). $NaB{H}_4$ is renowned for being a mild and highly selective reducing agent. Its specialty lies in the fact that it is not nucleophilic enough to attack esters, carboxylic acids, or amides under standard conditions. Furthermore, it leaves epoxide rings completely untouched.

$NaB{H}_4$ strictly targets the more electrophilic carbonyl carbons found in aldehydes and ketones. It is the absolute perfect match for this transformation, flawlessly reducing the $-CHO$ group to $-C{H}_{2}OH$ while preserving the rest of the molecule's complex architecture.

It is clear that for this specific transformation, $NaB{H}_4$ is the only correct choice. This problem perfectly encapsulates the beauty of JEE Advanced chemistry—it doesn't just test if you know what a reagent does, but it rigorously tests your deep understanding of its chemoselectivity and limitations.