Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is $\begin{array}{cc}{\text{CH}}_3& \text{O}\\ |& ||\\ {\text{CH}}_3-\text{CH}-{\text{CH}}_2-{\text{CH}}_2-\text{C}-\text{Cl}& \end{array}\underset{{\text{(ii) NaOH, Br}}_2}{\overset{{\text{(i) Alcoholic NH}}_3}{\to }}\underset{{\text{(iv) H}}_2\text{O}}{\overset{{\text{(iii) NaNO}}_2,\text{HCl}}{\to }}\text{Major product}$

A Journey of Transformations

Imagine you are an architect of molecules, tasked with transforming a specific starting material into a completely new functional group. In this problem, we are given an acid chloride, specifically 4-methylpentanoyl chloride, and we need to subject it to a sequence of three distinct chemical reagents.

Each reagent acts like a specialized tool, performing a precise operation on the molecule. To solve this, we must trace the journey of the molecule step-by-step, understanding the mechanism and the logic behind each transformation.

Ammonolysis of the Acid Chloride

The first reagent in our sequence is alcoholic ammonia ($N{H}_3$). When an acid chloride encounters ammonia, a classic nucleophilic acyl substitution takes place.

The nitrogen atom in ammonia possesses a lone pair of electrons, making it a fantastic nucleophile. It attacks the electrophilic carbonyl carbon of the acid chloride. This attack pushes the pi electrons of the $C=O$ double bond onto the oxygen atom, forming a tetrahedral intermediate.

However, oxygen is highly electronegative and wants to reform that stable double bond. As the pi bond reforms, it kicks out the best leaving group available—the chloride ion ($C{l}^{-}$). The result is the formation of an amide. In our specific case, 4-methylpentanoyl chloride is converted into 4-methylpentanamide.

The Hoffmann Bromamide Degradation

Now, we introduce the second set of reagents: sodium hydroxide ($NaOH$) and bromine ($B{r}_2$). Whenever you see an amide reacting with this specific combination, alarm bells should ring in your head! This is the legendary Hoffmann bromamide degradation.

This reaction is famous because it is a "step-down" reaction. It literally chops off the carbonyl carbon of the amide, releasing it as a carbonate ion ($C{O}_3^{2-}$).

The mechanism is fascinating: the base deprotonates the amide, and bromine reacts to form an N-bromoamide. Further deprotonation leads to a concerted rearrangement where the entire alkyl group migrates from the carbonyl carbon to the nitrogen atom, expelling the bromide ion to form an isocyanate intermediate.

Finally, the alkaline medium hydrolyzes the isocyanate, releasing carbon dioxide and leaving behind a primary amine. Because the carbonyl carbon was lost, our 4-methylpentanamide is degraded into a primary amine with one less carbon atom: 3-methylbutan-1-amine.

Diazotization and Hydrolysis

In the final phase, our newly formed primary aliphatic amine is treated with sodium nitrite ($NaN{O}_2$), hydrochloric acid ($HCl$), and water ($H_{2}O$).

The combination of $NaN{O}_2$ and $HCl$ generates nitrous acid ($HN{O}_2$) in situ, which further reacts to form the highly electrophilic nitrosonium ion ($N{O}^{+}$). The primary amine attacks this ion, and after a series of rapid proton transfers, an aliphatic diazonium salt is formed: $[R-N_2^{+}C{l}^{-}]$.

Here is the catch: unlike aromatic diazonium salts (which are stabilized by the resonance of the benzene ring at low temperatures), aliphatic diazonium salts are incredibly unstable. The $N_2$ group is arguably the best leaving group in all of organic chemistry. It immediately bubbles out of the solution as nitrogen gas, leaving behind a highly reactive carbocation.

In our molecule, the departure of nitrogen leaves a primary carbocation at the end of the chain. The abundant water molecules in the solution quickly act as nucleophiles, attacking the carbocation to form an alcohol.

Through this elegant sequence of ammonolysis, degradation, and diazotization, our starting acid chloride has been successfully transformed. The final product is 3-methylbutan-1-ol.

Comparing this to our given options, we find a perfect match. The correct answer is unequivocally option (c). By mastering the specific roles of these classic reagents, complex multi-step conversions become nothing more than a logical puzzle waiting to be solved!