An organic compound "A" on treatment with benzene sulphonyl chloride gives compound B. B is soluble

Q: Why doesn't a secondary amine form a soluble product?

When a secondary amine reacts with the Hinsberg reagent, it forms an N,N-dialkylbenzene sulphonamide. Because the nitrogen is bonded to two alkyl groups and the sulphonyl group, there are no hydrogen atoms left on the nitrogen. Without an acidic proton, it cannot react with NaOH to form a soluble salt.

Q: What happens if we react a tertiary amine with the Hinsberg reagent?

Tertiary amines do not have any hydrogen atoms attached to their nitrogen. Therefore, they cannot undergo the initial substitution reaction with benzene sulphonyl chloride to form a stable sulphonamide. They simply do not react under normal conditions.

Q: Why is the hydrogen in the primary amine's product considered acidic?

The hydrogen is attached to a nitrogen atom that is directly bonded to a highly electronegative sulphonyl ($-SO_2-$) group. This group exerts a strong electron-withdrawing effect ($-I$ and $-M$), pulling electron density away from the $N-H$ bond, making it easy for a base like $OH^-$ to abstract the proton.

Q: Can we use this test to separate a mixture of amines?

Absolutely! The Hinsberg test is not just for identification; it's a classic separation technique. The $1^\circ$ amine becomes a soluble salt in the aqueous layer, the $2^\circ$ amine forms an insoluble solid, and the $3^\circ$ amine remains unreacted in the organic layer.

Q: Is there any modern alternative to the Hinsberg test?

While the Hinsberg test is pedagogically excellent, modern laboratories often rely on spectroscopic techniques like NMR and IR spectroscopy, or mass spectrometry, to quickly and non-destructively identify the degree of substitution in amines.

Question

An organic compound "A" on treatment with benzene sulphonyl chloride gives compound $B$. $B$ is soluble in dil. NaOH solution. Compound $A$ is

Options:

(A) $C_6H_5-N-(CH_3)_2$

(B) $C_6H_5-NHCH_2CH_3$

(C) $C_6H_5-CH_2NHCH_3$

(D) $C_6H_5-	ext{CH}(	ext{CH}_3)-	ext{NH}_2$

Sigma Percentile · Accepted Answer

### The Magic of the Hinsberg Test

Welcome to a classic organic chemistry problem! We are dealing with the **Hinsberg test**, a brilliant chemical tool used to distinguish between primary ($1^\circ$), secondary ($2^\circ$), and tertiary ($3^\circ$) amines. The reagent used here is benzene sulphonyl chloride ($C_6H_5SO_2Cl$). Let's dive into the fascinating chemistry behind this test.

### The Reaction Mechanism

When a primary amine reacts with the Hinsberg reagent, the amine's nitrogen acts as a nucleophile. It attacks the electrophilic sulphur atom of the sulphonyl group, kicking out a chloride ion as a leaving group. This substitution reaction forms a product called **N-alkylbenzene sulphonamide**.

$$C_6H_5SO_2Cl + R-NH_2 ightarrow C_6H_5SO_2NHR + HCl$$

Notice a crucial structural detail: this product still has one hydrogen atom attached directly to the highly electronegative nitrogen atom.

### The Secret to Solubility

Now, here is where the magic happens! That remaining hydrogen on the nitrogen is sandwiched next to a powerful electron-withdrawing sulphonyl ($-SO_2-$) group. This group exerts a strong $-I$ and $-M$ effect, pulling electron density away from the $N-H$ bond. This makes the hydrogen **highly acidic**!

So, when we add a strong base like dilute sodium hydroxide ($NaOH$), it easily plucks off that acidic proton. This acid-base reaction forms a water-soluble sodium salt.

$$C_6H_5SO_2NHR + NaOH ightarrow C_6H_5SO_2N^\ominus R \quad Na^\oplus + H_2O$$

This solubility in aqueous alkali is the absolute hallmark of a primary amine!

### Identifying the Correct Compound

Our question states that compound $B$ (the sulphonamide) is soluble in dilute $NaOH$. This is a screaming neon sign telling us that compound $A$ must be a primary amine! Let's scan our options:

*   **(a)** $C_6H_5-N-(CH_3)_2$: The nitrogen is attached to three carbon groups. It's a **tertiary amine**. It won't even react with the Hinsberg reagent.
*   **(b)** $C_6H_5-NHCH_2CH_3$: The nitrogen is attached to two carbon groups. It's a **secondary amine**. It will react, but the product won't have an acidic hydrogen, so it will be insoluble in $NaOH$.
*   **(c)** $C_6H_5-CH_2NHCH_3$: Another **secondary amine**. Same story as (b).
*   **(d)** $C_6H_5-	ext{CH}(	ext{CH}_3)-	ext{NH}_2$: Look at the nitrogen! It is attached to only one carbon group. It's our **primary amine**!

So, without a shadow of a doubt, the organic compound $A$ is the primary amine given in option (d). It will react with the Hinsberg reagent to form a sulphonamide that happily dissolves in sodium hydroxide. Simple, elegant, and perfectly logical!

Animated Solution for Chemistry - Organic Chemistry: An organic compound "A" on treatment with benzene sulphonyl chloride gives compound $B$ . $B$ is soluble in dil. NaOH solution. Compound $A$ is

Select Answer:

Visualized Solution

\text{Hinsberg's Test}

\text{Reaction with } 1^\circ \text{ Amine}

\text{Acidic Hydrogen \& Solubility}

\text{Identifying the } 1^\circ \text{ Amine}

\text{Final Answer}

The Sigma Insight: Amines

The Magic of the Hinsberg Test

The Reaction Mechanism

The Secret to Solubility

Identifying the Correct Compound

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