Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

When we look at our starting molecule, we are presented with a beautiful polyfunctional compound. It contains three distinct reactive sites that we need to evaluate:

1. A ketone group ($-C(=O)-$) 2. An N-methyl imine group ($-N=CH-$) 3. An isolated alkene ($-CH=CH-$)

The key to solving this problem lies in understanding the specific selectivity of our given reagent, sodium borohydride ($NaB{H}_4$).

Sodium borohydride is a mild and highly selective reducing agent. It functions by delivering a nucleophilic hydride ion ($H^{-}$) to electron-deficient (electrophilic) centers. Let's see how it interacts with each of our functional groups.

First, consider the ketone. The carbon-oxygen double bond is highly polar due to the electronegativity of oxygen. This leaves the carbonyl carbon with a partial positive charge, making it a prime target for the nucleophilic hydride ion. Consequently, $NaB{H}_4$ readily reduces the ketone into a secondary alcohol ($-CH(OH)-$ ).

Next, we evaluate the imine group. Similar to the carbonyl group, the carbon-nitrogen double bond is polar. The nitrogen atom pulls electron density away from the carbon, rendering it electrophilic. The hydride ion attacks this carbon, successfully reducing the imine to a secondary amine ($-NH-C{H}_2-$ ).

Finally, we must address the isolated alkene. Notice that the carbon-carbon double bond is separated from the ketone by a $-C{H}_2-$ group, meaning it is not conjugated. Because a $C=C$ double bond is non-polar and inherently electron-rich, it strongly repels the incoming nucleophilic hydride ion. Therefore, $NaB{H}_4$ cannot reduce isolated alkenes, and this group remains completely unaffected.

By combining these individual transformations, we can construct our final major product. The ketone has been reduced to an alcohol, the imine has been reduced to an amine, and the isolated alkene remains intact.

This specific combination of structural features perfectly matches Option (c).

A quick thought experiment: What if we had used catalytic hydrogenation ($H_2/Pd$) instead? That reagent would have easily reduced the isolated carbon-carbon double bond, leading to a completely different outcome. This highlights why mastering reagent selectivity is absolutely crucial in organic synthesis!