(a) (i) In a chemistry practical class, the teacher gave his students an amine 'X' having molecular formula C₂

Question

Q. (a)

(i) In a chemistry practical class, the teacher gave his students an amine 'X' having molecular formula

C_2H_7N

, and asked the students to identify the type of amine. One of the students, Neeta, observed that it reacts with

C_6H_5SO_2Cl

, to give a compound which dissolves in NaOH solution. Can you help Neeta to identify the compound 'X' ?

(ii) Arrange the following in the increasing order of their

pK_b

value in aqueous phase :

C_6H_5NH_2

(CH_3)_2NH

NH_3

CH_3NH_2

(CH_3)_3N

(iii) Aniline on nitration gives considerable amount of meta product along with ortho and para products. Why ?

(iv) Convert aniline to : (I) p-bromoaniline (II) phenol

(OR)

(b)

(i) Arun heated a mixture of ethylamine and

CHCl_3

with ethanolic KOH, which forms a foul smelling gas. Write the chemical equation involved.

(ii) Identify A and B in the following reactions : A

\xrightarrow[\text{ethanol}]{H_2/Pd}

Aniline (

C_6H_5NH_2

, drawn ring bearing

-NH_2

)

\xleftarrow{Br_2/NaOH}

B (both arrows point TOWARDS aniline: A is converted to aniline by

H_2/Pd

in ethanol; B is converted to aniline by

Br_2/NaOH

)

(iii) Convert aniline to : (I) benzene (II) sulphanilic acid

CBSECBSE Class XII Board 2025Subjective· 5mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Carbylamine Reaction

The Carbylamine Reaction: A Stink Test for Primary Amines

Imagine you are a chemist in a lab with several unlabelled bottles of amines. You need to quickly tell which ones are primary amines (where the nitrogen is attached to exactly one carbon) and which are secondary or tertiary. You could run a complicated spectrum, or you could just make something that smells terrible. That is the carbylamine reaction.

The reaction is simple in practice: you take your suspected amine, add a little chloroform ( $CHCl_3$ ) and a strong base like alcoholic potassium hydroxide ( $KOH$ ), and warm the mixture. If a primary amine is present, you will produce an isocyanide (also called a carbylamine). Isocyanides have a famously repulsive, penetrating odour — often described as "foul" or "putrid". If you smell that, you have a primary amine.

Watch out

Common Mistake

Secondary and tertiary amines do not give this reaction. The carbylamine test is specific to primary amines. Do not confuse it with other amine tests like the Hinsberg test, which distinguishes all three classes.

The Precise Statement

Carbylamine Reaction: A primary amine ( $RNH_2$ ) reacts with chloroform ( $CHCl_3$ ) in the presence of a strong base (usually alcoholic $KOH$ ) to form an isocyanide ( $RNC$ ) and potassium chloride ( $KCl$ ) and water ( $H_2O$ ).

The general equation is:

RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O

The product $RNC$ is the isocyanide — the source of the foul smell.

Why Does It Happen? (The Mechanism)

The reaction proceeds through a dichlorocarbene intermediate. This is the key concept that makes the reaction work.

Generation of Dichlorocarbene: The strong base ( $KOH$ ) deprotonates chloroform, which then loses a chloride ion to form a highly reactive species called dichlorocarbene ( $:CCl_2$ ). This carbene is electron-deficient and desperately wants to react.

CHCl_3 + KOH \rightarrow :CCl_2 + KCl + H_2O

Attack by the Amine: The lone pair on the nitrogen of the primary amine attacks the electron-deficient carbon of the dichlorocarbene. This forms an unstable intermediate.
Elimination: The intermediate undergoes a series of dehydrohalogenation steps (loss of $HCl$ ) driven by the base, ultimately yielding the isocyanide ( $RNC$ ). …

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Part (a): X = ethylamine (1° amine, Hinsberg product soluble in NaOH); $pK_b$ order $(CH_3)_2NH<CH_3NH_2<(CH_3)_3N<NH_3<C_6H_5NH_2$ ; meta product from $-NH_3^+$ ; conversions to p-bromoaniline and phenol.

Part (b): Carbylamine gives ethyl isocyanide; A = nitrobenzene, B = benzamide; aniline $\rightarrow$ benzene ( $H_3PO_2$ ) and $\rightarrow$ sulphanilic acid (conc. $H_2SO_4$ ).

Part (a)

(i) Molecular formula $C_2H_7N$ corresponds to $CH_3CH_2NH_2$ , ethylamine, a primary amine. In the Hinsberg test, a primary amine reacts with benzenesulphonyl chloride:

C_2H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHC_2H_5 + HCl

This N-ethylbenzenesulphonamide has an acidic hydrogen on nitrogen (attached to strongly electron-withdrawing $SO_2$ ), so it dissolves in NaOH. This confirms X is a primary amine — ethylamine.

(ii) In aqueous medium basicity results from a balance of inductive (+I), steric and solvation effects. The observed order of basicity is $(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3>C_6H_5NH_2$ (aniline is weakest as its lone pair is delocalised into the ring). Therefore increasing $pK_b$ (weaker base = larger $pK_b$ ):

(CH_3)_2NH < CH_3NH_2 < (CH_3)_3N < NH_3 < C_6H_5NH_2

(iii) Nitration is done in a strongly acidic mixture. Aniline is largely protonated to the anilinium ion ( $C_6H_5\overset{+}{N}H_3$ ). The $-\overset{+}{N}H_3$ group is electron-withdrawing, hence deactivating and meta-directing, so a considerable proportion of the meta isomer is obtained along with o- and p-products (from the small amount of unprotonated aniline).

(iv)(I) Aniline $\rightarrow$ p-bromoaniline: direct bromination gives 2,4,6-tribromoaniline, so the $-NH_2$ is first protected:

C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3 \xrightarrow{Br_2} p\text{-Br-}C_6H_4NHCOCH_3 \xrightarrow{H_3O^+} p\text{-bromoaniline}

(iv)(II) Aniline $\rightarrow$ phenol: …

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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 40 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 mark
$Q. Write chemical equation for conversion from methylamine into methylisocyanide.$
›Reveal solutionSolution

Primary amines react with chloroform and alcoholic KOH to give a foul-smelling isocyanide - this is the carbylamine reaction, a test unique to primary amines.

…

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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
$Q. The reaction of ethylamine with chloroform in the alcoholic KOH gives:$
(a) C2H5NC
(b) C2H5CN
(c) CH3CN
(d) CH3NC
›Reveal solutionSolution

This is the carbylamine (isocyanide) test — a primary amine + chloroform + alcoholic KOH gives a foul-smelling alkyl isocyanide.

When a primary amine reacts with chloroform in the presence of alcoholic KOH, the carbylamine reaction occurs:

$C_2H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_2H_5NC + 3KCl + 3H_2O$

…

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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 2026Set ANNUAL1 mark
$C_6H_5NH_2 + CHCl_3 + KOH(alc.) \rightarrow ?$
›Reveal solutionSolution

This is the carbylamine (isocyanide) test, specific to primary amines: aniline reacts with chloroform and alcoholic KOH to give the foul-smelling phenyl isocyanide.

Reaction

$C_6H_5-NH_2 + CHCl_3 + 3KOH\ (alc.) \rightarrow C_6H_5-\overset{-}{N}\equiv \overset{+}{C} + 3KCl + 3H_2O$

Alcoholic KOH generates dichlorocarbene ( $:CCl_2$ ) from chloroform; this electrophilic carbene inserts into the N–H of the primary aromatic amine, and after loss of the remaining chlorines (as KCl, with KOH supplying the base) the isocyanide (carbylamine, $-NC$ ) functional group is formed. This reaction is diagnostic for …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The compound obtained by heating a mixture of 1° amine and chloroform with ethanolic KOH is -$
(a) Alkyl cyanide
(b) Amide
(c) Alkyl isocyanide
(d) None of these
›Reveal solutionSolution

This is the carbylamine (isocyanide) test for primary amines.

A primary amine, on heating with chloroform and ethanolic KOH, undergoes the carbylamine reaction to form a foul-smelling alkyl isocyanide:

$RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O$

…

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CBSE 2025Set D1 markMCQ
$Q. Reaction of primary amine with chloroform in the presence of alcoholic KOH is called$
(a) Hydrolysis
(b) Reduction
(c) Wurtz reaction
(d) Carbylamine reaction
›Reveal solutionSolution

Primary amine + CHCl3 + alcoholic KOH -> isocyanide (carbylamine reaction), a test for 1deg amines.

When a primary amine (aliphatic or aromatic) is warmed with chloroform and alcoholic potassium hydroxide, it forms an alkyl or aryl isocyanide (carbylamine), which has an extremely offensive smell:

R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O

…

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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 2025Set ANNUAL1 markMCQ
$Q. The test which is used to distinguish between primary amine and secondary amine is$
(a) Lucas test
(b) Carbylamine test
(c) Fehling's test
(d) Tollen's test
›Reveal solutionSolution

The carbylamine (isocyanide) test is positive only for primary amines, so it cleanly distinguishes them from secondary (and tertiary) amines, which give no reaction.

$\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{R-NC (foul smell)} + 3\text{KCl} + 3\text{H}_2\text{O}$

…

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Q.(a)

Part (a)

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