Animated Solution for Chemistry - Organic Chemistry: The major product 'X' formed in the following reaction is

Welcome to a classic exploration of chemoselectivity in organic chemistry! In this problem, we are presented with a fascinating multifunctional molecule and asked to predict its fate when treated with a specific reagent. Let's break down the thought process step-by-step.

Imagine you are looking at the reactant molecule under a microscope. It's a six-membered ring, but it's far from simple. We can identify three distinct functional groups residing on this framework:

1. A Ketone Group: Located at the top of the ring ($C=O$). 2. An Ester Group: Attached as a side chain ($-C{H}_{2}COOC{H}_3$). 3. An Isolated Double Bond: A carbon-carbon double bond ($C=C$) situated within the ring, separated from the ketone by $s{p}^3$ hybridized carbons.

Our task is to determine how these three groups will react when exposed to our reagent: Sodium Borohydride (${\text{NaBH}}_4$) in methanol.

Sodium borohydride is a staple in the organic chemist's toolkit. It acts as a source of nucleophilic hydride ions ($H^{-}$). However, its defining characteristic is its mildness and high selectivity.

Unlike its aggressive cousin, Lithium Aluminum Hydride (${\text{LiAlH}}_4$), ${\text{NaBH}}_4$ is quite picky about what it attacks.

Aldehydes and Ketones: It readily attacks the highly electrophilic carbonyl carbon of aldehydes and ketones, reducing them to primary and secondary alcohols, respectively. Esters and Carboxylic Acids: The carbonyl carbon in an ester is less electrophilic because the adjacent oxygen atom donates electron density through resonance. ${\text{NaBH}}_4$ simply isn't strong enough to overcome this and leaves esters completely untouched. Isolated Double Bonds:* ${\text{NaBH}}_4$ is a nucleophile, and isolated $C=C$ double bonds are electron-rich. Therefore, they repel each other, and the double bond remains perfectly intact.

Now, let's apply these rules to our specific molecule:

The ketone at the top of the ring will be successfully reduced to a secondary alcohol ($-OH$). The ester side chain will ignore the ${\text{NaBH}}_4$ and remain exactly as it is. The isolated double bond* in the ring will also remain completely unaffected.

When we look at the options provided, we see that Option (b) perfectly reflects this selective transformation. The ketone has become an alcohol, while the ester and the double bond stand proud and unchanged.

Always remember: in organic synthesis, choosing the right tool for the job is just as important as knowing what the tool does!