Animated Solution for Chemistry - Amines: For the reaction sequence given below, the correct statement(s) is(are)

From Tetralin to Primary Amines

Welcome to a beautiful journey through organic synthesis! This problem is a masterpiece that tests your ability to connect different chapters of organic chemistry. We start with a hydrocarbon, aggressively oxidize it, and then seamlessly transition into one of the most famous name reactions for synthesizing primary amines. Let's break it down step by step.

Our starting material is tetralin (1,2,3,4-tetrahydronaphthalene). Imagine a molecule where a stable, aromatic benzene ring is fused to a saturated, aliphatic cyclohexane ring.

When we treat tetralin with hot, acidic potassium permanganate (${\text{KMnO}}_4,{\text{H}}^{+},\Delta$), we are unleashing a very aggressive oxidizing agent. The benzene ring is incredibly stable and resists oxidation, but the benzylic carbons (the carbons of the saturated ring directly attached to the benzene ring) are highly susceptible.

The ${\text{KMnO}}_4$ attacks these benzylic positions, completely cleaving the saturated ring. The carbon-carbon bonds break, and the benzylic carbons are fully oxidized to carboxylic acid groups. The result is phthalic acid (benzene-1,2-dicarboxylic acid).

Next, we introduce ammonia (${\text{NH}}_3$) and apply strong heat. Initially, the acid-base reaction forms ammonium phthalate. However, the strong heating ($\Delta$) drives off two molecules of water ($-2{\text{H}}_2\text{O}$).

This dehydration forces the molecule to undergo an intramolecular cyclization, forming a stable five-membered ring containing a nitrogen atom flanked by two carbonyl groups. This intermediate is phthalimide, which is our compound X.

If we look at the structure of phthalimide (${\text{C}}_8{\text{H}}_5{\text{NO}}_2$), it clearly contains two oxygen atoms. Keep this in mind as we evaluate the options!

Now, the magic happens. The sequence of reagents—(1) $\Delta$, (2) Ethanolic KOH, (3) $\text{R}-\text{Br}$—is the unmistakable signature of the Gabriel Phthalimide Synthesis.

First, the ethanolic KOH acts as a base. The N-H proton of phthalimide is unusually acidic because the resulting negative charge on the nitrogen is highly resonance-stabilized by the two adjacent carbonyl groups. The KOH abstracts this proton, forming the potassium phthalimide salt.

This phthalimide anion is a fantastic nucleophile. It attacks the alkyl halide ($\text{R}-\text{Br}$) via an ${\text{S}}_{\text{N}}2$ mechanism, displacing the bromide ion. The alkyl group $\text{R}$ attaches to the nitrogen, yielding N-alkylphthalimide, which is our compound Y.

Notice that N-alkylphthalimide still retains the two carbonyl oxygen atoms. Therefore, both X and Y are oxygen-containing compounds. Option (A) is absolutely correct!

Furthermore, because Y is an imide and not a primary amine, it will not respond to the carbylamine test (heating with ${\text{CHCl}}_3$ and KOH). It will not form a foul-smelling isocyanide. Thus, Option (B) is incorrect.

The final step is the hydrolysis of the N-alkylphthalimide (Y) using aqueous sodium hydroxide ($\text{NaOH}$). The strong base attacks the carbonyl carbons, cleaving the imide bonds.

This releases the aromatic portion of the molecule as a water-soluble salt, sodium phthalate. More importantly, it liberates the nitrogen atom along with its attached alkyl group and two new hydrogen atoms, forming a primary aliphatic amine ($\text{R}-{\text{NH}}_2$). This is our compound Z.

Because the $\text{R}$ group came from the alkyl halide ($\text{R}-\text{Br}$), and the Gabriel synthesis is strictly used for aliphatic amines (aryl halides do not readily undergo ${\text{S}}_{\text{N}}2$), Z is definitely an aliphatic primary amine, not an aromatic one. Therefore, Option (D) is incorrect.

Finally, does Z react with Hinsberg's reagent? Yes! Hinsberg's reagent is benzenesulfonyl chloride. Primary amines react with it to form N-alkylbenzenesulfonamides. Because these sulfonamides still have one acidic proton attached to the nitrogen, they are soluble in aqueous alkali. This confirms that Option (C) is correct.

By carefully mapping the reaction sequence, we've successfully navigated from a simple hydrocarbon to a pure primary amine, proving that statements (A) and (C) are the correct answers.