Give the structures of A and B in the following sequence of reactions: (a) CH₃COOH xrightarrowNH₃ A xrightarro

Question

Q. Give the structures of A and B in the following sequence of reactions:

(a)

CH_3COOH \xrightarrow{NH_3} A \xrightarrow{NaOBr} B

(b)

C_6H_5NO_2 \xrightarrow{HCl/Fe} A \xrightarrow[0-5\,°C]{NaNO_2,\ HCl} B

(c)

C_6H_5N_2Cl \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B

(OR)

(a) How will you distinguish between the following pairs of compounds:

(i) Aniline and Ethanamine

(ii) Aniline and N-methylaniline

(b) Arrange the following compounds in decreasing order of their boiling points: Butanol, Butanamine, Butane

CBSECBSE Class XII Board 2020Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Diazonium Salt Reactions

Diazonium Salt Reactions – A First Look

Imagine you have a benzene ring, and you want to attach a new group — say a chlorine, a bromine, a cyano group, or even a hydroxyl — directly onto the ring. The benzene ring is stubborn; it doesn't easily let go of its hydrogen atoms for simple substitution. But there is a clever trick: first convert the ring into a diazonium salt, a highly reactive intermediate that will let you swap in almost any group you want.

That is the core idea. A diazonium salt is a temporary, energetic handle on the benzene ring that you can then replace with a wide variety of substituents. It is one of the most powerful tools in aromatic synthesis.

What is a Diazonium Salt?

A diazonium salt has the general formula Ar–N₂⁺ X⁻, where Ar is an aryl group (like phenyl, C₆H₅–), N₂⁺ is a diazonium cation (two nitrogen atoms triple-bonded, with a positive charge on the terminal nitrogen), and X⁻ is a counterion like chloride, bromide, or hydrogensulfate.

The key structural feature: the –N₂⁺ group is attached directly to the benzene ring. This group is unstable — it wants to leave as N₂ gas. That instability is exactly what makes it useful: when the N₂ leaves, the ring is left with a highly reactive carbocation-like intermediate that can be attacked by a nucleophile.

Watch out

Diazonium salts are thermally unstable and can explode if dried. They are almost always prepared and used in cold solution (0–5 °C) without isolation.

How Do You Make One? (Diazotization)

You start with a primary aromatic amine (Ar–NH₂). Treat it with nitrous acid (HNO₂) at low temperature (0–5 °C). The reaction is:

\text{Ar–NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} \text{Ar–N}_2^+ \text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

The nitrous acid is generated in situ from sodium nitrite and a mineral acid. The amine gets converted into the diazonium salt almost instantly. You must keep the solution cold; if it warms up, the diazonium salt decomposes and you get phenol and nitrogen gas.

Two Major Classes of Reactions

Once you have the diazonium salt in solution, you can do two fundamentally different things with it:

1. Substitution Reactions (N₂ leaves)

Here the –N₂⁺ group is replaced by another group. The nitrogen gas bubbles away, and the ring gets a new substituent. This is called dediazoniation. The leaving group is N₂, which is extremely stable, so the reaction is thermodynamically driven.

The most common substitutions:

Reagent/Condition	Product	Name
CuCl / HCl, heat	Ar–Cl	Sandmeyer reaction
CuBr / HBr, heat	Ar–Br	Sandmeyer reaction
CuCN / KCN, heat	Ar–CN	Sandmeyer reaction
KI, heat	Ar–I	Direct substitution
H₂O, heat	Ar–OH	Hydrolysis
H₃PO₂ (hypophosphorous acid)	Ar–H	Reduction (replaces N₂ with H)
Cu₂O, Cu(NO₃)₂, H₂O	Ar–NO₂	Replacement with nitro group

Tip

The Sandmeyer reaction uses copper(I) halide or cyanide as a catalyst. The copper helps transfer the halide or cyanide to the ring. Without copper, the reaction is much slower or gives different products.

The mechanism for Sandmeyer: the diazonium salt accepts an electron from Cu⁺, forming an aryl radical, which then abstracts a halogen from CuX₂. The N₂ leaves as a gas.

2. Coupling Reactions (N₂ stays)

Here the diazonium salt keeps its N₂ group and attacks another aromatic ring (usually an activated one like phenol or aniline). The result is an azo compound with the general structure Ar–N=N–Ar'. These compounds are intensely coloured — many are used as dyes.

The reaction is an electrophilic aromatic substitution. The diazonium cation is a weak electrophile, so it only attacks rings that are strongly activated (with –OH, –NH₂, –NHR, –NR₂ groups). The coupling occurs at the para position if available; otherwise ortho.

Example: coupling with phenol in alkaline medium:

\text{C}_6\text{H}_5\text{–N}_2^+ \text{Cl}^- + \text{C}_6\text{H}_5\text{–OH} \xrightarrow{\text{NaOH, 0–5}^\circ\text{C}} \text{C}_6\text{H}_5\text{–N=N–C}_6\text{H}_4\text{–OH} \text{ (p-hydroxyazobenzene, orange dye)}

Important

Coupling requires the coupling component (phenol or aniline) to be in its reactive form: phenol is used in alkaline solution (phenoxide ion is more activating), aniline is used in slightly acidic or neutral solution (to avoid protonation of the amino group).

Why Are Diazonium Salts So Versatile? …

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): (a) A = acetamide, B = methanamine (Hofmann bromamide); (b) A = aniline, B = benzenediazonium chloride; (c) A = benzonitrile, B = benzoic acid (Sandmeyer + hydrolysis). Part (b): aniline vs ethanamine → azo-dye test; aniline vs N-methylaniline → carbylamine test; boiling points Butanol > Butanamine > Butane.

Part (a)

(a) Ammonolysis → Hofmann bromamide degradation

$CH_3COOH$ with $NH_3$ (then heat) gives acetamide (A). Acetamide with $NaOBr$ ( $Br_2$ + NaOH) undergoes Hofmann rearrangement — the amide loses its carbonyl carbon as $CO_3^{2-}$ and gives an amine with one fewer carbon: methanamine (B).

CH_3COOH \xrightarrow{NH_3} CH_3CONH_2\,(A) \xrightarrow{NaOBr} CH_3NH_2\,(B)

(b) Reduction → diazotisation

Nitrobenzene is reduced by $Fe/HCl$ to aniline (A). Aniline with $NaNO_2/HCl$ at 0–5 °C gives benzenediazonium chloride (B) (kept cold, or it decomposes).

C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2\,(A) \xrightarrow[0\text{–}5^\circ C]{NaNO_2,HCl} C_6H_5N_2^+Cl^-\,(B)

(c) Sandmeyer → nitrile hydrolysis

The diazonium salt with $CuCN$ (Sandmeyer) gives benzonitrile (A); acidic hydrolysis of the nitrile gives benzoic acid (B). …

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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 60 on this concept.

CBSE 2026Set V11 markMCQ
$Q. p-Hydroxyazobenzene is formed by the reaction of benzene diazonium chloride with phenol. It is$
(a) an electrophilic substitution reaction
(b) a nucleophilic substitution reaction
(c) a hydrogenation reaction
(d) a halogenation reaction
›Reveal solutionSolution

Formation of p-hydroxyazobenzene by coupling of benzene diazonium chloride with phenol is an electrophilic aromatic substitution.

Benzene diazonium chloride, $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ , reacts with phenol in mildly alkaline medium to give the orange azo dye p-hydroxyazobenzene:

$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \rightarrow p\text{-}\text{HO-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 + \text{HCl}$ …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Aqueous solution of benzene diazonium chloride in presence of dil. H2SO4 is heated to produce:$
(a) Aniline
(b) Benzene
(c) Chlorobenzene
(d) Phenol
›Reveal solutionSolution

Heating an aqueous benzenediazonium salt solution hydrolyses it to phenol, releasing nitrogen gas.

Benzenediazonium chloride, $C_6H_5-N_2^+Cl^-$ , is thermally unstable in aqueous acidic solution. When its aqueous solution is warmed (with dilute $H_2SO_4$ or simply with water), the diazonium group is displaced by a water molecule in a nucleophilic substitution, with loss of nitrogen gas:

$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl$

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write general formula of diazonium salt.$
›Reveal solutionSolution

Diazonium salts have the general formula R-N≡N+ X−, formed by diazotisation of a primary aromatic amine with NaNO2/HCl at 0-5°C.

…

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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
$Q. The reaction of ethylamine with chloroform in the alcoholic KOH gives:$
(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 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 2026Set ANNUAL1 markMCQ
$Q. The reagent X in the reaction C₆H₅NH₂ (aniline) —X / 273K\to C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride) is -$
(a)
(i) HNO₃
(b)
(ii) NaNO₂ and HCl
(c)
(iii) NaCl and HNO₃
(d)
(iv) NaNO₂ and H₂SO₄
›Reveal solutionSolution

The conversion of aniline to benzenediazonium chloride at 273 K (diazotisation) uses nitrous acid generated from $\text{NaNO}_2 + \text{HCl}$ . Correct option: (ii).

Concept. Diazotisation is the reaction of a primary aromatic amine with nitrous acid ( $\text{HNO}_2$ ) at low temperature (273–278 K) to form a diazonium salt. Nitrous acid is unstable, so it is prepared in situ by adding sodium nitrite ( $\text{NaNO}_2$ ) to a mineral acid.

Why HCl specifically (option ii not iv). The anion of the diazonium salt required here is chloride ( $\text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-}$ ), so HCl must be the acid supplying $\text{Cl}^{-}$ ; $\text{H}_2\text{SO}_4$ would give the sulphate salt.

Reaction. …

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CBSE 2025Set 56/4/11 markMCQ
$C_6H_5NH_2$
›Reveal solutionSolution
Aniline undergoes diazotization to form a benzenediazonium salt (X), which then reacts with ethanol to undergo reductive dediazoniation, replacing the diazonium group with hydrogen to yield benzene (Y). The correct option is (B).

This problem tests your understanding of two classic reactions in aromatic chemistry: diazotization and the replacement of a diazonium group. The key is to recognize that ethanol here acts as a reducing agent, not as a nucleophile.

Let’s walk through the sequence step by step.

Step 1: Diazotization of aniline Aniline ( $C_6H_5NH_2$ ) is treated with sodium nitrite ( $NaNO_2$ ) and hydrochloric acid ( $HCl$ ) at a low temperature ( $0-5^\circ C$ ). This is the standard condition for forming a diazonium salt. The reaction proceeds as:

$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
The product X is benzenediazonium chloride ( $C_6H_5N_2^+Cl^-$ ). This is a highly reactive intermediate, stable only in cold solution.

Watch out

A common mistake is to think that the diazonium salt itself is the final product Y. But the reaction sequence continues — X is just an intermediate.

Step 2: Reaction of the diazonium salt with ethanol When benzenediazonium chloride (X) is treated with ethanol ( $C_2H_5OH$ ), a reductive dediazoniation occurs. Ethanol acts as a reducing agent, donating a hydride ion ( $H^-$ ) or a hydrogen atom to replace the diazonium group. The overall transformation is: $C_6H_5N_2^+Cl^- + C_2H_5OH \longrightarrow C_6H_6 + N_2 + CH_3CHO + HCl$ …
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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 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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CBSE 2025Set ANNUAL1 markMCQ
$Q. When an aqueous solution of benzenediazonium chloride is boiled with water or steam distilled the product formed is:$
(a) Benzene
(b) Diphenyl
(c) Chlorobenzene
(d) Phenol.
›Reveal solutionSolution

Aqueous benzenediazonium chloride is thermally unstable; warming or steam-distilling it causes hydrolysis, replacing the $-N_2^+Cl^-$ group with $-OH$ and releasing nitrogen gas.

Benzenediazonium chloride ( $C_6H_5-N_2^+Cl^-$ ), when its aqueous solution is warmed/boiled (or steam distilled), undergoes hydrolysis. The diazonium group is displaced by a hydroxyl group from water, with loss of nitrogen gas and hydrochloric acid:

$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl$

…

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Q.Give the structures of A and B in the following sequence of reactions:

Part (a)

(a) Ammonolysis → Hofmann bromamide degradation

(b) Reduction → diazotisation

(c) Sandmeyer → nitrile hydrolysis

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