JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
આ MCQ મોડ્યુલ આના પર આધારિત છે: Chemical Reactions
ટોપિક 17 / 25
Chemical Reactions
🎓 Class 12
Chemistry
CBSE
Theory
Ch 9 – Amines
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Chemical Reactions
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Chemical Reactions of Amines
Part 3 dealt with the first reaction of Section 9.6 — basicity. The remaining six reactions fall into two groups. Reactions 2 to 6 happen at the nitrogen atom and are driven by its lone pair acting as a nucleophile. Reaction 7 happens on the benzene ring of an arylamine and is driven by the same lone pair pushing electron density into the ring. Three of these reactions — carbylamine, nitrous acid and Hinsberg — double as laboratory tests that distinguish 1°, 2° and 3° amines.
9.6 Chemical Reactions (continued)
2. Alkylation
Amines undergo alkylation on reaction with alkyl halides. The nitrogen lone pair attacks the carbon bearing the halogen in a nucleophilic substitution, and as we saw in Part 2 this generates the familiar cascade to secondary, tertiary and finally quaternary ammonium salts.
R–NH₂ —[R′X]→ R–NH–R′ —[R′X]→ R–N(R′)₂ —[R′X]→ R–N⁺(R′)₃ X⁻
3. Acylation
Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution. This reaction is known as acylation. It can be regarded as replacement of a hydrogen atom of the –NH₂ or >N–H group by an acyl group; the products are amides.
CH₃COCl + H₂N–C₂H₅ —[pyridine]→ CH₃CO–NH–C₂H₅ + HCl
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH
Why pyridine is added. The reaction is carried out in the presence of a base stronger than the amine, such as pyridine, which removes the HCl formed and shifts the equilibrium to the right-hand side. Without it, the HCl would simply protonate the starting amine and remove it from play.
Benzoylation. Amines also react with benzoyl chloride (C₆H₅COCl). This reaction is known as benzoylation.
CH₃NH₂ + C₆H₅COCl → CH₃NHCOC₆H₅ + HCl
methanamine benzoyl chloride N-methylbenzamide
methanamine benzoyl chloride N-methylbenzamide
And with carboxylic acids? Amines and carboxylic acids form salts at room temperature rather than amides. An amide requires the more reactive acid chloride or anhydride, or heating the salt strongly to drive off water.
4. Carbylamine reaction (isocyanide test)
Aliphatic and aromatic primary amines only, on heating with chloroform and ethanolic potassium hydroxide, form isocyanides (carbylamines), which are foul-smelling substances. Secondary and tertiary amines do not show this reaction. It is therefore used as a test for primary amines.
R–NH₂ + CHCl₃ + 3KOH(alc.) —[heat]→ R–NC + 3KCl + 3H₂O
C₆H₅NH₂ + CHCl₃ + 3KOH(alc.) —[heat]→ C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O
C₆H₅NH₂ + CHCl₃ + 3KOH(alc.) —[heat]→ C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O
5. Reaction with nitrous acid
The three classes of amines react differently with nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite (NaNO₂ + HCl).
(a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which, being unstable, liberate nitrogen gas quantitatively and give alcohols. This quantitative evolution of nitrogen is used in the estimation of amino acids and proteins.
R–NH₂ + HNO₂ → [R–N₂⁺Cl⁻] —[H₂O]→ R–OH + N₂↑ + HCl
(b) Aromatic amines react with nitrous acid at low temperature (273–278 K) to form diazonium salts — a very important class of compounds used for the synthesis of a variety of aromatic compounds, taken up fully in Part 5.
C₆H₅NH₂ + NaNO₂ + 2HCl —[273–278 K]→ C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(c) Secondary and tertiary amines react with nitrous acid in a different manner, and do not give the brisk quantitative evolution of nitrogen seen with primary aliphatic amines.
The diagnostic contrast. A primary aliphatic amine gives brisk effervescence of N₂ even in the cold and leaves an alcohol. A primary aromatic amine gives a stable, ice-cold solution of a diazonium salt with no gas evolution — which is precisely why arenediazonium salts can be isolated and used synthetically while alkyldiazonium salts cannot.
6. Reaction with arylsulphonyl chloride — the Hinsberg test
Benzenesulphonyl chloride (C₆H₅SO₂Cl), known as Hinsberg's reagent, reacts with primary and secondary amines to form sulphonamides.
(a) With a primary amine the reaction yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in this sulphonamide is strongly acidic, because of the strong electron-withdrawing sulphonyl group. Hence the product is soluble in alkali.
C₆H₅SO₂Cl + H₂N–C₂H₅ → C₆H₅SO₂–NH–C₂H₅ + HCl (soluble in KOH)
(b) With a secondary amine, N,N-diethylbenzenesulphonamide is formed. Since it contains no hydrogen attached to nitrogen, it is not acidic and hence insoluble in alkali.
C₆H₅SO₂Cl + HN(C₂H₅)₂ → C₆H₅SO₂–N(C₂H₅)₂ + HCl (insoluble in KOH)
(c) Tertiary amines do not react with benzenesulphonyl chloride at all.
Why the test works. This property of the three classes reacting differently is used both for distinguishing primary, secondary and tertiary amines and for separating a mixture of amines. These days benzenesulphonyl chloride is commonly replaced by p-toluenesulphonyl chloride.
7. Electrophilic substitution in aromatic amines
Aniline is a resonance hybrid of five structures. Where is the electron density greatest in these structures? At the ortho and para positions relative to the –NH₂ group. Thus the –NH₂ group is ortho- and para-directing and a powerful activating group.
(a) Bromination
Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline.
C₆H₅NH₂ + 3Br₂(aq) → 2,4,6-Br₃C₆H₂NH₂↓ (white ppt) + 3HBr
The main problem encountered in electrophilic substitution of aromatic amines is their very high reactivity — substitution tends to occur at all the ortho and para positions at once. To prepare a monosubstituted aniline derivative, the activating effect of the –NH₂ group must be controlled. This is done by protecting the –NH₂ group by acetylation with acetic anhydride, carrying out the desired substitution, and then hydrolysing the substituted amide back to the substituted amine.
How protection works. The lone pair of electrons on the nitrogen of acetanilide interacts with the carbonyl oxygen atom by resonance. Hence the lone pair is less available for donation to the benzene ring. The activating effect of the –NHCOCH₃ group is therefore less than that of the –NH₂ group — strong enough to keep the ring ortho/para-directing, but weak enough to stop at monosubstitution.
(b) Nitration
Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion, which is meta-directing. That is why, besides the ortho and para derivatives, a significant amount of the meta derivative is also formed.
However, by protecting the –NH₂ group by acetylation with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative obtained as the major product.
(c) Sulphonation
Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate, which on heating with sulphuric acid at 453–473 K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product.
Aniline does not undergo the Friedel–Crafts reaction (alkylation or acylation). Aluminium chloride, the Lewis acid catalyst, forms a salt with the nitrogen lone pair. The nitrogen thereby acquires a positive charge and acts as a strong deactivating group, shutting down further electrophilic attack on the ring.
🧪 Activity 9.4 — Identifying three unknown aminesL4 Analyse
Three unlabelled bottles contain ethanamine (1°), N-ethylethanamine (2°) and N,N-diethylethanamine (3°). This activity walks through the reasoning a chemist actually uses, applying two independent tests and checking that they agree.
Predict: Which single reagent would let you identify all three in one experiment? Write your answer before proceeding.
- Test 1 — carbylamine. To a portion of each, add chloroform and ethanolic KOH and warm gently in a fume cupboard. Record which produces an extremely offensive smell.
- Test 2 — Hinsberg. To a fresh portion of each, add benzenesulphonyl chloride, shake, then add aqueous KOH. Record which gives a clear solution, which gives an insoluble solid, and which gives no reaction at all.
- Tabulate both sets of results side by side.
- Check consistency: does the bottle positive in Test 1 match the bottle giving the alkali-soluble product in Test 2?
The Hinsberg test alone identifies all three; the carbylamine test confirms the primary amine.
Ethanamine (1°): foul isocyanide smell in Test 1 (positive carbylamine). In Test 2 it gives C₆H₅SO₂NHC₂H₅, whose N–H is made acidic by the electron-withdrawing sulphonyl group, so it dissolves in KOH.
N-Ethylethanamine (2°): no smell in Test 1. In Test 2 it gives C₆H₅SO₂N(C₂H₅)₂, which has no N–H, is therefore not acidic, and remains as an insoluble solid in KOH.
N,N-Diethylethanamine (3°): no smell in Test 1 and no reaction at all in Test 2, because it has no N–H hydrogen to be substituted; the amine layer simply remains.
The unifying idea: both tests are really counting N–H hydrogens. The carbylamine reaction needs two, so only primary amines respond. The Hinsberg reagent needs at least one, so primary and secondary amines respond but tertiary amines do not — and whether the product keeps an acidic N–H afterwards separates primary from secondary. Safety note: isocyanides are extremely foul-smelling and toxic, so the carbylamine test must always be done in a fume cupboard.
Intext questions
Intext 9.6 — Exhaustive alkylation of aniline
Aniline with an excess of methyl iodide in the presence of sodium carbonate solution undergoes successive methylation. C₆H₅NH₂ —[CH₃I]→ C₆H₅NHCH₃ —[CH₃I]→ C₆H₅N(CH₃)₂ —[CH₃I]→ C₆H₅N⁺(CH₃)₃ I⁻, N,N,N-trimethylanilinium iodide. The sodium carbonate neutralises the HI liberated at each stage, driving the reaction forward. The final product is a quaternary ammonium salt, so it has no lone pair and is no longer basic.
Intext 9.7 — Aniline with benzoyl chloride
C₆H₅NH₂ + C₆H₅COCl → C₆H₅CONHC₆H₅ + HCl. The product is N-phenylbenzamide (benzanilide). This is benzoylation — the Schotten–Baumann type acylation of an amine.
Intext 9.8 — Isomers of C₃H₉N and which liberate N₂ with nitrous acid
Four isomers: CH₃CH₂CH₂NH₂ (propan-1-amine, 1°); (CH₃)₂CHNH₂ (propan-2-amine, 1°); CH₃CH₂NHCH₃ (N-methylethanamine, 2°); (CH₃)₃N (N,N-dimethylmethanamine, 3°).
Which liberate nitrogen gas? Only the primary aliphatic amines, because they alone form the unstable aliphatic diazonium salt that decomposes quantitatively to an alcohol and N₂. So propan-1-amine and propan-2-amine evolve nitrogen; the secondary and tertiary amines do not.
Competency-Based Questions
A forensic chemist receives a sample seized as an unknown organic base, molecular formula C₇H₉N. Three tests are run. Test 1: with CHCl₃ and ethanolic KOH, warming produces an intensely offensive odour. Test 2: with benzenesulphonyl chloride followed by aqueous KOH, a clear solution is obtained. Test 3: with bromine water at room temperature, a white precipitate forms immediately. A second sample, also C₇H₉N, is negative in Tests 1 and 3 but gives an alkali-insoluble solid in Test 2.
1. What do Tests 1 and 2 together establish about the first sample? L2 Understand
Both point to a primary amine. The offensive odour in Test 1 is the carbylamine (isocyanide) reaction, which only primary amines give. In Test 2 the sulphonamide formed retains an N–H that is made strongly acidic by the sulphonyl group, so it dissolves in alkali — again characteristic of a primary amine.
2. What additional information does Test 3 provide, and what is the likely identity of the first sample? L4 Analyse
An immediate white precipitate with bromine water at room temperature is characteristic of a highly activated aromatic ring bearing a free –NH₂ group — the ring undergoes rapid tribromination. This tells us the compound is an arylamine, not merely an aliphatic one. Combined with the primary classification and formula C₇H₉N, the sample is a methylaniline (toluidine), for example 2-methylaniline or 4-methylaniline, which would give a tribromo derivative as an insoluble white solid. (Benzylamine, C₆H₅CH₂NH₂, is also C₇H₉N and primary, but its ring is not activated, so it would not give the instant precipitate — Test 3 is what rules it out.)
3. Identify the class of the second sample and name a compound of formula C₇H₉N that fits. L3 Apply
Negative carbylamine plus an alkali-insoluble sulphonamide means a secondary amine: it has exactly one N–H, so it reacts with Hinsberg's reagent, but the product C₆H₅SO₂NR₂ has no N–H left and is therefore not acidic. A compound of formula C₇H₉N fitting this is N-methylaniline, C₆H₅NHCH₃. Its negative bromine-water test is consistent, since the ring is less activated than in a free arylamine and no instant precipitate forms under those conditions.
4. The chemist needs to prepare pure p-nitroaniline from aniline. Explain why direct nitration fails and outline the correct three-step route. L4 Analyse
Direct nitration fails for two reasons. First, the nitrating mixture is a strong oxidising agent and aniline is easily oxidised, giving tarry oxidation products. Second, in the strongly acidic medium aniline is protonated to the anilinium ion, and –N⁺H₃ is meta-directing, so a substantial amount of m-nitroaniline is formed alongside the ortho and para products. Correct route: (1) protect — treat aniline with acetic anhydride to give acetanilide; (2) nitrate — the –NHCOCH₃ group is still ortho/para-directing but much less activating, so nitration gives predominantly p-nitroacetanilide; (3) deprotect — hydrolyse the amide with acid or alkali to give p-nitroaniline.
5. A student attempts to acetylate benzene using CH₃COCl and anhydrous AlCl₃ and succeeds, then tries the identical conditions on aniline and recovers only unreacted starting material. Explain the failure. L5 Evaluate
Aniline does not undergo the Friedel–Crafts reaction. Anhydrous AlCl₃ is a strong Lewis acid and the nitrogen of aniline is a good Lewis base, so the two combine to form a salt (an acid–base adduct) at nitrogen. Once complexed, the nitrogen carries a positive charge and becomes a strong deactivating, meta-directing group instead of the powerful activator it normally is. The ring is thereby deactivated towards the electrophile, and in addition the catalyst has been consumed by complexation rather than being available to generate the acylium ion. The student's observation is therefore exactly what theory predicts. If an acetyl group on the ring is genuinely required, the nitrogen must first be converted to a non-basic derivative — or the desired substituent must be introduced via a diazonium salt, as Part 5 will show.
Assertion–Reason Questions
For each pair choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
Assertion (A): The sulphonamide obtained from a primary amine dissolves in aqueous KOH, whereas that from a secondary amine does not.
Reason (R): The product from a primary amine retains an N–H hydrogen that is rendered acidic by the strongly electron-withdrawing sulphonyl group.
Answer: A. C₆H₅SO₂NHR keeps one acidic N–H and forms a soluble potassium salt, while C₆H₅SO₂NR₂ has none and stays insoluble. This is the basis of the Hinsberg test.
Assertion (A): Although the amino group is ortho- and para-directing, nitration of aniline gives a substantial amount of m-nitroaniline.
Reason (R): In the strongly acidic nitrating medium aniline is protonated to the anilinium ion, which is meta-directing.
Answer: A. It is the protonated form, –N⁺H₃, that directs meta. The free amine present in equilibrium still gives ortho and para products, which is why a mixture results.
Assertion (A): The carbylamine reaction is used to detect tertiary amines.
Reason (R): Isocyanides formed in the reaction have an extremely unpleasant smell.
Answer: D. The assertion is false — the carbylamine reaction detects primary amines only, since two hydrogen atoms on nitrogen are required; secondary and tertiary amines do not respond. The reason is a true statement about isocyanides and is indeed what makes the test so easy to detect, but it does not support the false assertion.
Coming next. Part 5 takes up Sections 9.7 to 9.10 — diazotisation, the physical properties of benzenediazonium chloride, the Sandmeyer, Gattermann and Balz–Schiemann reactions, azo coupling, and why diazonium salts are the key to aromatic compounds that direct substitution cannot reach.
Frequently Asked Questions
What is the carbylamine reaction and which amines give it?
Aliphatic and aromatic primary amines, heated with chloroform and ethanolic potassium hydroxide, form isocyanides (carbylamines) which have an extremely offensive smell. Secondary and tertiary amines do not react, so the test is specific for primary amines. The equation is R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O.
How does the Hinsberg test distinguish primary, secondary and tertiary amines?
Benzenesulphonyl chloride reacts with a primary amine to give C₆H₅SO₂NHR, whose N–H is made strongly acidic by the sulphonyl group, so the product dissolves in alkali. With a secondary amine it gives C₆H₅SO₂NR₂, which has no N–H, is not acidic and stays insoluble in alkali. Tertiary amines have no N–H at all and do not react. These days p-toluenesulphonyl chloride is often used in place of benzenesulphonyl chloride.
Why must the amino group of aniline be protected before nitration?
Direct nitration of aniline gives tarry oxidation products because the nitrating mixture oxidises the easily oxidised amine. In addition, the strongly acidic medium protonates aniline to the anilinium ion, and the –N⁺H₃ group is meta-directing, so a significant amount of the meta derivative forms. Acetylating the –NH₂ to –NHCOCH₃ moderates the activation while keeping the group ortho and para directing, so the para product predominates and can be hydrolysed back to the amine.
Why does aniline not undergo the Friedel-Crafts reaction?
Anhydrous aluminium chloride is a Lewis acid and the nitrogen lone pair of aniline is a Lewis base, so they combine to form a salt. The nitrogen then carries a positive charge and acts as a strong deactivating group, shutting down electrophilic attack on the ring. The catalyst is also consumed by this complexation rather than generating the electrophile.
How do the three classes of amines react differently with nitrous acid?
Primary aliphatic amines form unstable aliphatic diazonium salts that decompose at once, liberating nitrogen quantitatively and giving alcohols — a reaction used in estimating amino acids and proteins. Primary aromatic amines react at 273 to 278 K to give relatively stable arenediazonium salts. Secondary and tertiary amines react in a different manner and do not give the brisk quantitative evolution of nitrogen.
Why is pyridine added during the acylation of amines?
Acylation liberates HCl, which would protonate the starting amine and remove it from the reaction as its unreactive ammonium salt. Pyridine is a base stronger than the amine being acylated, so it mops up the HCl as it forms and shifts the equilibrium towards the amide product.
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JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
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