Answer any five questions of the following: (5×1=5) (a) N,N-diethyl-benzenesulphonamide is insoluble in alkali

Question

Q. Answer any five questions of the following: (5\times1=5)

(a) N,N-diethyl-benzenesulphonamide is insoluble in alkali. Give reason.

(b) Aniline does not undergo Friedel-Crafts reaction. Why?

(d) Write the chemical reaction involved in Gabriel phthalimide synthesis.

(e) How will you convert aniline to p-bromoaniline?

(f) Complete the following reaction :

C_6H_5\text{-}N_2^{+}Cl^{-} \xrightarrow[\text{(ii) } NaNO_2/Cu,\ \Delta]{\text{(i) } HBF_4}

(g) Write the structures of A and B in the following reaction :
Benzoic acid

C_6H_5COOH \xrightarrow[\Delta]{NH_3} A \xrightarrow{Br_2\ +\ NaOH} B

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Concept understanding — Hinsberg Test

The Hinsberg Test: Distinguishing Amines by Reactivity

Imagine you have three types of amines — primary ( $RNH_2$ ), secondary ( $R_2NH$ ), and tertiary ( $R_3N$ ) — and you need to tell them apart. They all smell fishy, they all are basic, but they differ in one crucial way: how many hydrogens are attached to the nitrogen.

A primary amine has two hydrogens on the nitrogen. A secondary amine has one. A tertiary amine has none. The Hinsberg reagent, benzenesulphonyl chloride ( $C_6H_5SO_2Cl$ ), exploits exactly this difference.

Note

The reagent is a sulfonyl chloride — think of it as a very reactive molecule that loves to swap its chlorine atom for a bond to nitrogen, but only if that nitrogen still has a hydrogen to give up during the reaction.

What Actually Happens

When you shake the amine with benzenesulphonyl chloride in the presence of aqueous alkali (usually KOH or NaOH), three different things happen:

Primary amine ( $RNH_2$ )

The nitrogen attacks the sulfur, kicking out chlorine. The product is a sulfonamide that still has one hydrogen on the nitrogen — $C_6H_5SO_2NHR$ . That hydrogen is acidic enough to be pulled off by the strong base, forming a soluble salt: $C_6H_5SO_2NR^-K^+$ . So the primary amine dissolves in the alkaline solution.

Secondary amine ( $R_2NH$ )

Same attack, but now the product is $C_6H_5SO_2NR_2$ — a sulfonamide with no hydrogen on the nitrogen. It cannot form a salt with base. It simply precipitates out as a solid.

Tertiary amine ( $R_3N$ )

No hydrogen on the nitrogen at all. The reaction cannot proceed to form a sulfonamide. The tertiary amine does not react — it either remains as an oily layer or, if it is a small molecule, may dissolve in the acid work-up later.

Watch out

A common mistake: students think tertiary amines do nothing at all. They do not form a sulfonamide, but they are still basic. In the alkaline medium they remain as free amines, often floating as an immiscible oil.

The Practical Workflow

Here is how you actually perform the test in the lab:

Take a small sample of the amine mixture in a test tube.
Add benzenesulphonyl chloride and a few mL of aqueous KOH.
Shake well. If the mixture warms up, cool it under the tap.
Observe:

Observation	What it means
Clear solution forms	Primary amine (soluble salt)
Solid precipitate forms	Secondary amine (insoluble sulfonamide)
Oily layer remains, no precipitate	Tertiary amine (no reaction)

To confirm the primary amine, you can acidify the clear solution with dilute HCl. The sulfonamide salt breaks down, and the free sulfonamide ( $C_6H_5SO_2NHR$ ) precipitates out — a white solid.

Tip

If you have a mixture, you can separate them: filter the secondary amine precipitate, acidify the filtrate to get the primary amine, and the tertiary amine stays as the oil layer that never reacted.

Why This Works — The Key Insight …

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This solution covers five selected short-answer questions from an organic chemistry exam, each explained with the core concept first — from the insolubility of N,N-diethylbenzenesulphonamide in alkali to the conversion of aniline to p-bromoaniline. The key idea in each case is the role of lone pair availability, steric hindrance, or directing effects.

(a) N,N-diethyl-benzenesulphonamide is insoluble in alkali. Give reason.

Concept: Sulphonamides are acidic when the nitrogen has at least one hydrogen attached — that hydrogen can be removed by a strong base like NaOH, making the compound soluble. But if both hydrogens are replaced by alkyl groups, the N–H bond is gone, and so is the acidity.

Benzenesulphonamide ( $C_6H_5SO_2NH_2$ ) has two N–H bonds. The strong electron-withdrawing effect of the sulphonyl group ( $-SO_2-$ ) makes these hydrogens weakly acidic. So it reacts with NaOH to form a water-soluble salt.
In N,N-diethyl-benzenesulphonamide, both hydrogens on nitrogen are replaced by ethyl groups. There is no N–H bond left.
Without an acidic hydrogen, the compound cannot donate a proton to base. It remains neutral and does not dissolve in alkali. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set DZ1 markMCQ
$Q. Which test is used to differentiate between primary, secondary and tertiary amines?$
(a) Brady test
(b) Tollen's test
(c) Hinsberg test
(d) Lucas test
›Reveal solutionSolution
The Hinsberg test (with benzenesulphonyl chloride, $C_6H_5SO_2Cl$ ) tells 1°, 2° and 3° amines apart, so option (c) is correct.

In the Hinsberg test the amine is treated with benzenesulphonyl chloride ( $C_6H_5SO_2Cl$ , Hinsberg's reagent) in the presence of KOH:

Primary amine → gives an N-substituted sulphonamide that has an acidic N–H; it dissolves in KOH (soluble product).

Secondary amine → gives an N,N-disubstituted sulphonamide with no N–H; it is insoluble in KOH. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Identify the Hinsberg reagent:$
(a) benzene ring - CONH2 (benzamide)
(b) benzene ring - O - COCl (phenyl chloroformate)
(c) benzene ring - SO2Cl (benzenesulfonyl chloride)
(d) CH3 - benzene ring - COCl (4-methylbenzoyl chloride)
›Reveal solutionSolution
Hinsberg's reagent is benzenesulfonyl chloride; its differing reaction behaviour with 1 degree, 2 degree, and 3 degree amines is the basis of the classic Hinsberg test.

Benzenesulfonyl chloride, C6H5-SO2Cl, reacts differently depending on the class of amine:

Primary amines form an N-substituted sulfonamide that has an acidic N-H, which dissolves in excess alkali (KOH) - giving a clear solution.

Secondary amines form an N,N-disubstituted sulfonamide with no acidic N-H, which is insoluble in alkali - it stays as a precipitate. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Hinsberg reagent is:$
(a) C6H5SO2Cl
(b) C6H5N2Cl
(c) C6H5NO
(d) C6H5SO3H
›Reveal solutionSolution

Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl, used to distinguish 1°, 2°, and 3° amines.

Hinsberg's reagent is benzenesulphonyl chloride (C₆H₅SO₂Cl). In the Hinsberg test, it reacts with a primary amine to give an N-alkylbenzenesulphonamide that is soluble in alkali (the remaining N–H is acidic due to the electron-withdrawing SO₂ group); with a secondary amine it gives a sulphonamide that is insoluble in alkali (no acidic N–H le …

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CBSE 2025Set D1 markMCQ
$Q. Which of the following is Hinsberg reagent?$
(a) Benzene sulphonyl chloride
(b) Benzene sulphonic acid
(c) Ethyl oxalate
(d) Acetyl chloride
›Reveal solutionSolution
Hinsberg's reagent = benzenesulphonyl chloride, C6H5SO2Cl.

Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl). It is used to distinguish the three classes of amines:

Primary amines form a sulphonamide that is soluble in alkali (has an acidic N-H). …
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CBSE 2025Set A1 mark
$Q. Write True or False: Hinsberg's reagent reacts with primary and secondary amines to form sulphonamides.$
›Reveal solutionSolution
Hinsberg's test uses benzenesulphonyl chloride to distinguish primary, secondary, and tertiary amines by whether/how a sulphonamide forms.

Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl). When treated with:

A primary amine: it forms an N-alkylbenzenesulphonamide (C6H5SO2NHR), which has an acidic N–H (due to the electron-withdrawing SO2 group) and so dissolves in alkali (KOH) to give a clear solution.

A secondary amine: it forms an N,N-dialkylbenzenesulphonamide (C6H5SO2NR2), which has no acidic hydrogen and stays insoluble in alkali. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Hinsberg's reagent is -$
(a) C6H5SO2Cl
(b) C6H5SO2Cl2
(c) C6H5SOCl2
(d) C6H5SO3Cl
›Reveal solutionSolution
Hinsberg's reagent is benzenesulfonyl chloride, C6H5SO2Cl.

In Hinsberg's test, an amine is treated with benzenesulfonyl chloride (Hinsberg's reagent) in the presence of aqueous KOH:

A primary amine forms an N-alkylbenzenesulfonamide with an acidic N-H, which dissolves in excess KOH (forms a soluble salt).

A secondary amine forms an N,N-dialkylbenzenesulfonamide with no acidic N-H, which is insoluble in KOH. …
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CBSE 2024Set A11 markMCQ
$Q. The reagents used to separate the mixture of methylamine and dimethylamine are :$
(a) $\text{CHCl}_3$ and $\text{HCl}$
(b) $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$ and $\text{KOH}$
(c) $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$ and $\text{HCl}$
(d) $\text{CHCl}_3$ and $\text{KOH}$
›Reveal solutionSolution
Hinsberg's reagent ( $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$ ) with $\text{KOH}$ separates a 1° from a 2° amine, so the answer is (b).

Methylamine is a primary amine and dimethylamine is a secondary amine. Hinsberg's reagent (benzenesulphonyl chloride, $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$ ) distinguishes them:

With methylamine (1°):

$\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + \text{H}_2\text{NCH}_3 \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{NHCH}_3 + \text{HCl}$

The product still has an N–H whose proton is acidic (activated by the $-\text{SO}_2-$ group), so it dissolves in $\text{KOH}$ to give a soluble salt.

With dimethylamine (2°):

$\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + \text{HN(CH}_3)_2 \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{N(CH}_3)_2 + \text{HCl}$ …
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CBSE 2024Set B1 mark
$Q. Match the pairs correctly. Column A item: "Hinsberg's reagent". Column B options:$
(a) C6H5SO2Cl
(b) Keratin
(c) C6H5NH2
(d) Rickets
(e) C6H5N2Cl
(f) Cobalt. Which option from Column B matches "Hinsberg's reagent"?
›Reveal solutionSolution
Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl), used in Hinsberg's test to distinguish primary, secondary and tertiary amines.

In the Hinsberg test, an amine is treated with benzenesulphonyl chloride (C6H5SO2Cl) in the presence of KOH:

1 degree amines give a sulphonamide that dissolves in excess KOH (the N-H is acidic enough to be deprotonated). …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Hinsberg's reagent which is used to test amines is$
(a) Benzene sulphonamide
(b) Benzene diazonium chloride
(c) Benzene sulphonyl chloride
(d) Acetanilide
›Reveal solutionSolution
Hinsberg's reagent is benzenesulphonyl chloride; it reacts differently with 1-degree, 2-degree, and 3-degree amines, which is the basis of the Hinsberg test used to distinguish them.

C6H5SO2Cl (benzenesulphonyl chloride) reacts with:

Primary amines -> forms an N-alkyl benzenesulphonamide that has an acidic N-H (soluble in alkali, since the sulphonyl group makes the remaining N-H acidic).

Secondary amines -> forms an N,N-dialkyl benzenesulphonamide with no N-H left (insoluble in alkali). …
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CBSE 2024Set ANNUAL1 mark
$Q. Name a test to distinguish between aniline and N-methylaniline.$
›Reveal solutionSolution
The carbylamine (isocyanide) test is specific to primary amines, so it cleanly distinguishes aniline (primary) from $N$ -methylaniline (secondary).

When a primary amine is warmed with chloroform and alcoholic potassium hydroxide, it is converted into an extremely foul-smelling alkyl/aryl isocyanide (carbylamine):

$C_6H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5NC\,(\text{phenyl isocyanide, offensive odour}) + 3KCl + 3H_2O$

This reaction requires an $-NH_2$ group with two replaceable hydrogens on nitrogen to form the isocyanide's $N \equiv C$ linkage — a mechanistic requirement that only primary amines satisfy.

Aniline ( $C_6H_5NH_2$ ) is a primary amine $\Rightarrow$ gives a positive carbylamine test (pungent, unpleasant smell of phenyl isocyanide). …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The primary, secondary and tertiary amines can be distinguished by -$
(a) Fehling's test
(b) Tollen's test
(c) Libermann test
(d) Hinsberg's test
›Reveal solutionSolution
Hinsberg's reagent (benzenesulphonyl chloride) reacts differently with 1°, 2°, and 3° amines, giving products with distinct solubility behaviour — this difference is the basis of the test.

Hinsberg's test: the amine is treated with benzenesulphonyl chloride (C $_6$ H $_5$ SO $_2$ Cl) in the presence of KOH:

1° amine: $RNH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR$ (N-alkylbenzenesulphonamide). This product still has an acidic N–H (due to the electron-withdrawing SO $_2$ group), so it dissolves in excess KOH to give a soluble salt; on acidification, the sulphonamide is regenerated as a precipitate.

2° amine: $R_2NH + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NR_2$ . No N–H remains, so this product is insoluble in KOH and separates out directly as an oily/solid precipitate. …
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CBSE 2023Set BZ1 markMCQ
$Q. Hinsberg reagent is:$
(a) Ethyl oxalate
(b) Trimethyl amine
(c) Benzene sulphonyl chloride
(d) Benzyl chloride
›Reveal solutionSolution

Hinsberg's reagent is benzenesulphonyl chloride, $C_6H_5SO_2Cl$ , used to distinguish primary, secondary and tertiary amines.

…

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Q.Answer any five questions of the following: (5×1=5)

(a) N,N-diethyl-benzenesulphonamide is insoluble in alkali. Give reason.

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