Animated Solution for Chemistry - Organic Chemistry: Three isomers $A,B$ and $C$ (molecular formula ${\text{C}}_8{\text{H}}_{11}\text{N}$) give the following results : $A\text{ and }C\stackrel{\text{Diazotisation}}{\to }P+Q\underset{{\text{(ii) Oxidation (KMnO}}_4{\text{+H}}^{+})}{\overset{\text{(i) Hydrolysis}}{\to }}R\text{ (product of }A)+S\text{ (product of }C)$ $R\text{ has lower boiling point than }S.$ $B\stackrel{{\text{C}}_6{\text{H}}_5{\text{SO}}_2\text{Cl}}{\to }\text{alkali-insoluble product.}$ $A,B\text{ and }C$, respectively are

Organic chemistry problems that weave together molecular formulas, reaction mechanisms, and physical properties are like intricate detective stories. In this problem, we are given three isomers—$A$, $B$, and $C$—all sharing the molecular formula ${\text{C}}_8{\text{H}}_{11}\text{N}$. Our mission is to deduce their exact structures based on a series of chemical tests and physical clues. Let's break down the evidence step by step.

The first step in any structural elucidation is calculating the Degree of Unsaturation (DU), also known as the Double Bond Equivalent (DBE). The formula is:

Plugging in our values for ${\text{C}}_8{\text{H}}_{11}\text{N}$:

A DU of 4 is a massive structural hint. It almost universally points to the presence of a benzene ring (which accounts for three double bonds and one ring). Therefore, we can confidently assume that isomers $A$, $B$, and $C$ are aromatic compounds.

The problem states that isomers $A$ and $C$ undergo diazotisation to form products $P$ and $Q$. This is a highly specific reaction. Among amines, only primary aromatic amines (where the $-{\text{NH}}_2$ group is directly attached to the benzene ring) undergo diazotisation with ${\text{NaNO}}_2/\text{HCl}$ at low temperatures to form stable diazonium salts.

This tells us that $A$ and $C$ must have the structure $\text{Ar}-{\text{NH}}_2$. Since the total formula is ${\text{C}}_8{\text{H}}_{11}\text{N}$ and the aniline core accounts for ${\text{C}}_6{\text{H}}_7\text{N}$, the remaining fragment is an ethyl group ($-{\text{C}}_2{\text{H}}_5$). Thus, $A$ and $C$ are isomers of ethylaniline.

Let's follow the chemical transformations for $A$ and $C$:

1. Diazotisation: The $-{\text{NH}}_2$ group is converted into a diazonium group ($-{\text{N}}_2^{+}{\text{Cl}}^{-}$). 2. Hydrolysis: Heating the diazonium salt with water replaces the excellent leaving group ($-{\text{N}}_2^{+}$) with a hydroxyl group ($-\text{OH}$), yielding an ethylphenol. 3. Oxidation: The ethylphenols are then subjected to vigorous oxidation using acidic potassium permanganate (${\text{KMnO}}_4/{\text{H}}^{+}$). This reagent is notorious for cleaving any alkyl side chain possessing benzylic hydrogens, oxidizing it entirely into a carboxylic acid group ($-\text{COOH}$).

Consequently, the final products $R$ and $S$ are isomers of hydroxybenzoic acid.

We are given a crucial physical property: $R$ has a lower boiling point than $S$.

Why would two isomers of hydroxybenzoic acid have different boiling points? The answer lies in hydrogen bonding.

- Ortho-hydroxybenzoic acid (Salicylic acid): The $-\text{OH}$ and $-\text{COOH}$ groups are adjacent. This proximity allows them to form intramolecular hydrogen bonds (bonding within the same molecule). Because the molecule is "busy" bonding with itself, it has less capacity to form intermolecular bonds with neighboring molecules. Weaker intermolecular forces mean a lower boiling point. - Para-hydroxybenzoic acid: The groups are on opposite sides of the ring. They cannot reach each other to form intramolecular bonds. Instead, they form extensive intermolecular hydrogen bonds, creating a strong, sticky network of molecules. Stronger intermolecular forces mean a higher boiling point.

Since $R$ has the lower boiling point, $R$ must be the ortho isomer. Tracing this back, isomer $A$ must be ortho-ethylaniline. Consequently, $S$ is the para isomer, meaning isomer $C$ is para-ethylaniline.

Finally, we turn our attention to isomer $B$. It reacts with benzenesulfonyl chloride (Hinsberg's reagent) to yield an alkali-insoluble product.

The Hinsberg test is the gold standard for distinguishing amine classes: - Primary amines form sulfonamides with an acidic proton on the nitrogen, making them soluble in aqueous alkali. - Secondary amines form sulfonamides lacking an acidic proton, making them insoluble in aqueous alkali. - Tertiary amines generally do not react to form stable sulfonamides.

Since $B$ forms an alkali-insoluble product, it must be a secondary amine. Looking at our options, N-ethylaniline is a secondary amine that perfectly matches the molecular formula ${\text{C}}_8{\text{H}}_{11}\text{N}$.

Piecing it all together: - $A$ is ortho-ethylaniline. - $B$ is N-ethylaniline. - $C$ is para-ethylaniline.

This perfectly aligns with option (a). By systematically applying our knowledge of reaction mechanisms and physical chemistry, we've successfully cracked the case!