Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: The major products $A$ and $B$ for the following reactions are, respectively

This molecule features two key reactive sites: a ketone group and a primary alkyl iodide. The stereochemistry is also explicitly drawn, with the methyl group at carbon-4 pointing downwards.

Our goal is to trace the journey of this molecule through two distinct reaction conditions to identify the major products, $A$ and $B$.

The $S_{N}2$ Substitution The first set of reagents is $\text{KCN}$ in $\text{DMSO}$. This is a classic recipe for a nucleophilic substitution reaction.

$\text{DMSO}$ (Dimethyl sulfoxide) is a polar aprotic solvent. It is excellent at solvating cations like $K^{+}$, but it leaves the cyanide anion ($C{N}^{-}$) "naked" and highly reactive. Because the leaving group (iodide) is attached to a primary carbon, the conditions strongly favor an $S_{N}2$ mechanism.

The cyanide nucleophile attacks the primary carbon from the backside, displacing the iodide ion. The ketone group remains completely unaffected because cyanide is not a strong enough nucleophile to irreversibly attack the carbonyl carbon without an acid source to trap the cyanohydrin.

Thus, the iodine is replaced by a cyano group, yielding Product $A$: 3-methyl-5-oxohexanenitrile.

Catalytic Hydrogenation Next, Product $A$ is treated with hydrogen gas over a palladium catalyst ($H_2/Pd$). This is a standard method for catalytic hydrogenation.

While $H_2/Pd$ can reduce various functional groups, the carbon-nitrogen triple bond of a nitrile is highly susceptible to reduction under these conditions. The palladium catalyst facilitates the addition of hydrogen atoms across the pi bonds, fully reducing the nitrile group ($-C\equiv N$) to a primary amine ($-C{H}_{2}N{H}_2$).

Ketones can also be reduced by catalytic hydrogenation, but they typically require harsher conditions (higher pressure or temperature) compared to nitriles or alkenes. Given the options provided, it is clear that the ketone is meant to remain intact.

This selective reduction gives us our final Product $B$: 6-amino-4-methylhexan-2-one.

Option (c) perfectly matches the structures of both Product $A$ and Product $B$. Furthermore, it preserves the original stereochemical orientation of the reactant, with the methyl group pointing downwards. Therefore, option (c) is the correct answer.