Animated Solution for Chemistry - Organic Chemistry: An amine on reaction with benzene sulphonyl chloride produces a compound insoluble in alkaline solution. This amine can be prepared by ammonolysis of ethyl chloride. The correct structure of amine is

Imagine you are a chemical detective, and you've just been handed a mystery vial containing an unknown amine. Your job is to deduce its exact molecular structure using only two clues provided by its chemical behavior. This is exactly what this classic JEE problem asks us to do. Let's break down the clues one by one and solve the mystery.

The Hinsberg Test

The first piece of evidence is that our mystery amine reacts with benzene sulphonyl chloride to produce a compound that is insoluble in alkaline solution.

Benzene sulphonyl chloride is famously known as the Hinsberg Reagent. It is the ultimate tool for distinguishing between primary ($1^{\circ }$), secondary ($2^{\circ }$), and tertiary ($3^{\circ }$) amines. Here is how the chemistry unfolds:

1. Primary Amines: When a $1^{\circ }$ amine reacts with the Hinsberg reagent, it forms an $N$-alkylbenzene sulphonamide. Because the nitrogen atom still has one hydrogen atom attached to it, and this hydrogen is highly acidic (due to the strong electron-withdrawing nature of the sulphonyl group), the resulting compound easily dissolves in an aqueous alkali (like $\mathrm{NaOH}$) to form a soluble salt. 2. Secondary Amines: When a $2^{\circ }$ amine reacts, it forms an $N,N$-dialkylbenzene sulphonamide. Notice the difference? There are no hydrogen atoms left on the nitrogen! Without an acidic hydrogen, this compound cannot react with an alkali, making it insoluble. 3. Tertiary Amines: These do not react with the Hinsberg reagent at all under normal conditions because they lack a replaceable hydrogen atom on the nitrogen.

Since our product is insoluble in alkali, we can definitively conclude that our mystery compound is a secondary ($2^{\circ }$) amine.

The Ammonolysis Origin

The second clue tells us about the amine's origin story: it can be prepared by the ammonolysis of ethyl chloride.

Ammonolysis is a nucleophilic substitution reaction where ammonia (${\mathrm{NH}}_3$) or an amine acts as a nucleophile, attacking an alkyl halide and replacing the halogen atom. If we start with ethyl chloride (${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}$), the nucleophilic attack will attach an ethyl group to the nitrogen atom.

Even if the reaction proceeds further to form secondary or tertiary amines, every alkyl group added from this specific alkyl halide will be an ethyl group. Therefore, any amine prepared this way must contain at least one ethyl group (${-\mathrm{CH}}_2{\mathrm{CH}}_3$) in its structure.

Now that we have our two structural constraints—it must be a secondary amine AND it must contain an ethyl group—let's interrogate our options:

(a) $\mathrm{Ph}-\mathrm{NH}-{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3$: This is $N$-propylaniline. It is indeed a secondary amine, but look at its substituents: a phenyl group and a propyl group. There is no ethyl group to be found. This cannot be our molecule. (b) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$: This is ethanamine. It certainly has an ethyl group, but it is a primary ($1^{\circ }$) amine. If we subjected it to the Hinsberg test, the product would dissolve in alkali. Incorrect. (c) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NHCH}}_3$: This is $N$-methylpropan-1-amine. It is a secondary amine, but its alkyl groups are methyl and propyl. Again, the crucial ethyl group is missing. (d) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NHCH}}_2{\mathrm{CH}}_3$: This is $N$-ethylpropan-1-amine. Let's check the criteria. Is it a secondary amine? Yes. Does it contain an ethyl group? Yes, right there on the right side of the nitrogen.

Conclusion: Option (d) is the only structure that perfectly aligns with both chemical clues. The mystery is solved!