CBSE Class 12 Chemistry Amines NEET JEE Questions Set 02

Single Correct Answer

Question 1. The product formed by the reaction of an aldehyde with a primary amine is (NEET Phase-I 2016)
(a) carboxylic acid
(b) aromatic acid
(c) Schiff's base
(d) ketone
Answer:
Ans. (c)
RCHO + R′NH₂ → RCH=NR′ + H₂O. The product, an imine, is called Schiff's base.

Teacher's Note:
1. A Schiff's base (imine, C=N–R) forms only with 1° amines because two H atoms on N are needed to lose water.

Question 2. Hofmann bromamide degradation of benzamide gives product A which upon heating with CHCl₃ and NaOH gives product B. The structure of A and B are (JEE Main 2021)
(a) A = C₆H₅NH₂, B = C₆H₅NC
(b) A = C₆H₅NH₂, B = o-H₂NC₆H₄CHO
(c) A = p-BrC₆H₄NH₂, B = 5-bromo-2-isocyanobenzaldehyde
(d) A = p-BrC₆H₄CONH₂, B = o-OHCC₆H₄CONH₂
Answer:
Ans. (a)
C₆H₅CONH₂ (Benzamide) + Br₂/KOH (Hofmann bromamide reaction) → C₆H₅NH₂ (A)
C₆H₅NH₂ + CHCl₃/NaOH (Carbylamine reaction) → C₆H₅NC (B)
Thus, option (a) is correct.

Teacher's Note:
1. Bromine is not introduced into the ring in Hofmann degradation; CHCl₃/base with a 1° amine gives the isocyanide (carbylamine test), not an aldehyde.

Question 3. Which of the following amine will give the carbylamine test ? (NEET 2020)
(a) C₆H₅NHCH₃
(b) C₆H₅N(CH₃)₂
(c) C₆H₅NHC₂H₅
(d) C₆H₅NH₂
Answer:
Ans. (d)
Only 1° aliphatic or aromatic amines give carbylamine test.
Thus, option (d) is correct.

Teacher's Note:
1. The carbylamine test distinguishes 1° amines from 2° and 3° amines; the foul smell of the isocyanide is the sign.

Question 4. The compound that would produce a nauseating smell/odour with a hot mixture of chloroform and ethanolic potassium hydroxide is (West Bengal JEE 2017)
(a) PhCONH₂
(b) PhNHCH₃
(c) PhNH₂
(d) PhOH
Answer:
Ans. (c)
1° Amines (C₆H₅NH₂) give bad smelling isocyanides.
C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O

Teacher's Note:
1. Amides and phenols do not respond; only an NH₂ group on carbon (1° amine) gives the isocyanide.

Question 5. Positive carbylamine test is shown by
(a) N,N-dimethylaniline
(b) triethylamine
(c) N-methylaniline
(d) p-methylbenzylamine
Answer:
Ans. (d)
Only 1° amines, i.e., p-methylbenzylamine, (p) CH₃–C₆H₄–CH₂NH₂, give carbylamine test.

Teacher's Note:
1. Classify each amine first: only a 1° amine (aliphatic, aralkyl or aromatic) gives the test.

Question 6. Compound A is converted to B on reaction with CHCl₃ and KOH. The compound B is toxic and can be decomposed by C. A, B, C respectively are : (JEE Main 2021)
(a) secondary amine, isonitrile compound, conc. NaOH
(b) primary amine, isonitrile compound, conc. HCl
(c) primary amine, nitrile compound, conc. HCl
(d) secondary amine, nitrile compound, conc. NaOH
Answer:
Ans. (b)
Only 1° amines undergo carbylamine reaction with CHCl₃ and KOH to form isonitriles which are toxic compounds. These are decomposed by conc. HCl to form a 1° amine and HCOOH.
R–NH₂ (1° Amine, A) + CHCl₃, KOH (Carbylamine reaction) → R–N⇌C (Isonitrile, B) (Toxic)
R–NC + 2H₂O (Conc. HCl) → R–NH₂ + HCOOH

Teacher's Note:
1. Isonitriles are hydrolysed by acids (not by alkalis) to the 1° amine and formic acid.

Question 7. An organic compound (C₃H₉N) (A) when treated with nitrous acid, gave an alcohol and N₂ gas was evolved. (A) on warming with CHCl₃ and caustic potash gave (C) which on reduction gave isopropylmethylamine. Predict the structure of (A). (AIPMT Main 2012)
(a) (CH₃)₂CH–NH₂
(b) CH₃CH₂–NH–CH₃
(c) CH₃–N(CH₃)–CH₃
(d) CH₃CH₂CH₂–NH₂
Answer:
Ans. (a)
Since compound A (C₃H₉N) reacts with HNO₂ to give alcohol and N₂ gas, therefore, it must be a primary aliphatic amine. Further, since 1° aliphatic amine (A) on warming with CHCl₃ and caustic potash gave compound (C) which on reduction gave isopropylmethylamine, therefore, 1° amine (A) must be isopropylamine, i.e., option (a) is correct.
(CH₃)₂CH–NH₂ (Isopropylamine, A) + HNO₂ → (CH₃)₂CHOH (Alcohol) + N₂
(CH₃)₂CH–NH₂ + CHCl₃/KOH, Δ (Carbylamine reaction) → (CH₃)₂CH–N⇌C (Isopropylcarbylamine, C)
(CH₃)₂CH–NC + Ni/H₂ (Reduction) → (CH₃)₂CH–NH–CH₃ (Isopropylmethylamine)

Teacher's Note:
1. N₂ with HNO₂ and a positive carbylamine test both point to a 1° amine; the isopropyl group survives unchanged on the nitrogen.

Question 8. The optically active compound (X), C₆H₅–C*H(CH₃)–NH₂, on treatment with NaNO₂/HCl gives
(a) 1° alcohol with retention of configuration
(b) 2° alcohol with inversion of configuration
(c) racemic mixture of 2° alcohols
(d) racemic mixture of 1° alcohols
Answer:
Ans. (c)
CH₃–C*H(C₆H₅)–NH₂ (+) or (−) + HONO → CH₃–C*H(C₆H₅)–N⁺≡N (unstable) −N₂ → CH₃–C⁺H–C₆H₅ (H₂O, −H⁺) → CH₃–CH(OH)–C₆H₅ (±)
Since carbocations are planar species, therefore, nucleophilic attack by H₂O gives a racemic mixture of 2° alcohols.

Teacher's Note:
1. A planar carbocation can be attacked from either face, so the chiral amine loses its optical activity.

Question 9. Predict the product :
C₆H₅–NHCH₃ + NaNO₂ + HCl → Product (AIPMT 2009)
(a) C₆H₅–N(CH₃)–NO₂
(b) o-ON–C₆H₄–NHCH₃ + p-ON–C₆H₄–NHCH₃
(c) C₆H₅–N(OH)–CH₃
(d) C₆H₅–N(CH₃)–N=O
Answer:
Ans. (d)
C₆H₅–N(CH₃)–H + NaNO₂/HCl → C₆H₅–N(CH₃)–N=O + H₂O
2° amines react with nitrous acid to give N-nitrosoamines (yellow oily compounds).

Teacher's Note:
1. 2° amines (aliphatic or aromatic) give N-nitroso compounds; ring nitrosation happens only with 3° aromatic amines.

Question 10. What product will form when N, N-dimethylaniline reacts with NaNO₂ and dilute HCl at low temperature ? (J & K CET 2018)
(a) p-Nitroso-N, N-dimethylaniline
(b) Methyl-n-hexylamine
(c) m-Benzenediazonium chloride
(d) N-Nitroso-N-methylaniline
Answer:
Ans. (a)
3° aromatic amines have no H on N; the nitrosonium ion (NO⁺) attacks the activated ring at the p-position. p-Nitroso-N, N-dimethylaniline is formed.
C₆H₅N(CH₃)₂ + HNO₂ → p-ON–C₆H₄–N(CH₃)₂ + H₂O

Teacher's Note:
1. The strongly activating N(CH₃)₂ group directs the weak electrophile NO⁺ to the para position.

Question 11. Identify A in the following reaction.
C₆H₅NH₂ (K₂Cr₂O₇) → A (JEE Main 2021)
(a) C₆H₅NO₂
(b) p-KO–C₆H₄–NH₂
(c) nitrocyclohexane
(d) p-benzoquinone
Answer:
Ans. (d)
Aniline is oxidised by K₂Cr₂O₇ to p-benzoquinone
C₆H₅NH₂ (Aniline) + K₂Cr₂O₇/H⁺ (Oxidation) → O=C₆H₄=O (p-Benzoquinone)

Teacher's Note:
1. Acidified dichromate oxidises aniline at the ring, giving p-benzoquinone, not nitrobenzene.

Question 12. With respect to the following reaction, consider the statements :
C₆H₅NH₂ (HNO₃, H₂SO₄, 288 K) → Products
I. o-Nitroaniline and p-nitroaniline are the predominant products
II. p-Nitroaniline and m-nitroaniline are the predominant products
III. HNO₃ acts as an acid.
IV. H₂SO₄ acts as an acid.
Choose the correct option. (JEE Main 2022)
(a) (I) and (III) are correct statements
(b) (I) and (IV) are correct statement
(c) (II) and (IV) are correct statements
(d) (II) and (III) are correct statement
Answer:
Ans. (c)
p-Nitroaniline (51%) and m-nitroaniline (47%) are the predominant products. Thus, statement (II) is correct. o-Nitroaniline is obtained only in 2% and hence statement (I) is wrong.
H₂SO₄ acts as an acid while HNO₃ acts as a base because H₂SO₄ acts on HNO₃ to generate the active electrophile, NO₂⁺ which brings about the nitration.
HO–NO₂ + H⁺ (from H₂SO₄) → H₂O⁺–NO₂ −H₂O → NO₂⁺
Thus, statements (II) and (IV) are correct, i.e., option (c) is correct.

Teacher's Note:
1. Remember the approximate yields for direct nitration of aniline: p 51%, m 47%, o 2%.

Question 13. Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali. (NEET 2021)
(a) CH₃CH₂–N(CH₃)–CH₂CH₃
(b) (CH₃CH₂)₂NH
(c) CH₃CH₂–NH–CH₃
(d) CH₃CH₂–NH₂
Answer:
Ans. (d)
The reaction of Hinsberg's reagent with 1° amines gives solid sulphonamides which dissolve in alkali due to the presence of an acidic hydrogen on nitrogen. For example,
CH₃CH₂–NH₂ (Ethylamine, 1°) + C₆H₅SO₂Cl (Benzenesulphonyl chloride, Hinsberg's reagent) −HCl → C₆H₅SO₂–NH–CH₂CH₃ (N-Ethylbenzenesulphonamide)
C₆H₅SO₂–NH–CH₂CH₃ + NaOH −H₂O → C₆H₅SO₂–N⁻Na⁺–CH₂CH₃ (Soluble in alkali)
Thus, option (d) is correct.

Teacher's Note:
1. Only the sulphonamide from a 1° amine still has an H on N, made acidic by the SO₂ group, so it dissolves in NaOH.

Question 14. An amine reacts with benzenesulphonyl chloride to form a white precipitate which is insoluble in aq. NaOH. The amine is
(a) (CH₃)₃N
(b) CH₃–NH–CH₂CH₃
(c) CH₃CH₂CH₂NH₂
(d) (CH₃)₂CHNH₂
Answer:
Ans. (b)
2° Amines react with benzenesulphonyl chloride to give benzenesulphonamides which do not dissolve in aq. NaOH due to the absence of acidic H on N in NaOH, i.e., option (b) is correct.

Teacher's Note:
1. 3° amines give no product at all with Hinsberg's reagent; an alkali-insoluble solid means a 2° amine.

Question 15. The compound that will react most readily with NaOH to form methanol is (IIT 2001 ; West Bengal JEE 2007)
(a) (CH₃)₄N⁺ I⁻
(b) CH₃OCH₃
(c) (CH₃)₃S⁺ I⁻
(d) (CH₃)₃CCl
Answer:
Ans. (a)
Due to greater electronegativity of N over S, +ve charge on N will make CH₃ group more electron deficient than +ve charge on S. Therefore, (CH₃)₄N⁺I⁻ will undergo nucleophilic substitution reaction more readily than (CH₃)₃S⁺I⁻.
HO⁻ Na⁺ + CH₃–N⁺(CH₃)₃I⁻ → CH₃OH + (CH₃)₃N + NaI

Teacher's Note:
1. (CH₃)₃N is a good leaving group, so OH⁻ can displace it from a methyl carbon by S_N2.

Question 16. Which of these will produce the highest yield in Friedel-Crafts reaction ? (JEE Main 2020)
(a) C₆H₅CONH₂
(b) C₆H₅NH₂
(c) C₆H₅Cl
(d) C₆H₅OH
Answer:
Ans. (c)
Aniline being a Lewis base reacts with Lewis acid AlCl₃ to form a salt
C₆H₅NH₂ + AlCl₃ → C₆H₅NH₂⁺AlCl₃⁻
Therefore, aniline does not undergo F.C. reaction. Similarly, phenol reacts with AlCl₃ to form a salt.
C₆H₅OH + AlCl₃ → C₆H₅OAlCl₂ + HCl
As a result, phenol gives a poor yield of F.C. reaction.
Likewise benzamide also forms a complex which does not allow F.C. reaction to occur. In contrast, chlorobenzene does not have any such problems and hence gives the highest yield in F.C. reaction, i.e., option (c) is correct.

Teacher's Note:
1. Any group with a lone pair that binds AlCl₃ (NH₂, OH, CONH₂) poisons the catalyst and deactivates the ring.

Question 17. Which of the following compounds will form the precipitate with aq. AgNO₃ solution most readily ? (JEE Main 2020)
(a) C₆H₅–N(CH₂CH₃)–CH₂Br
(b) N-(bromomethyl)piperidine, C₅H₁₀N–CH₂Br
(c) C₆H₅–N(CH₂CH₃)–CH₂CH₂Br
(d) CH₃CH₂–O–CH₂Br
Answer:
Ans. (b)
C₅H₁₀N–CH₂Br −Br⁻ → C₅H₁₀N–C⁺H₂ (I) ↔ C₅H₁₀N⁺=CH₂
CH₃CH₂–O–CH₂Br −Br⁻ → CH₃CH₂–O–C⁺H₂ (II) ↔ CH₃CH₂–O⁺=CH₂
Both carbocations I and II are stabilized by the lone pair of electrons on N and O respectively. Since N is less electronegative than O, therefore, N can donate the lone pair more readily than O and hence carbocation (I) is more stable than carbocation (II). In other words, compound under option (b) will readily give white ppt with AgNO₃ solution.
In (a), the lone pair on N is partly delocalised into the benzene ring, so it stabilises the carbocation less; in (c) the CH₂Br group is not directly attached to N.

Teacher's Note:
1. AgNO₃ reacts fastest with the halide that ionises to the most stable carbocation; an adjacent N lone pair gives an iminium ion.

Question 18. The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is (JEE Main 2021)
(a) Br₂/AlBr₃, HNO₃/H₂SO₄, Mg/ether, CO₂, H₃O⁺
(b) Br₂/AlBr₃, NaOH/H₃O⁺, HNO₃/H₂SO₄
(c) Br₂/AlBr₃, HNO₃/H₂SO₄, NaCN, H₃O⁺
(d) HNO₃/H₂SO₄, Br₂/AlBr₃, Mg/ether, CO₂, H₃O⁺
Answer:
Ans. (d)
C₆H₆ (Benzene) + HNO₃/H₂SO₄ (Nitration) → C₆H₅NO₂ (Br₂/AlBr₃, Nuclear bromination) → m-BrC₆H₄NO₂
m-BrC₆H₄NO₂ + Mg/ether (Grignard reagent) → m-BrMgC₆H₄NO₂ (CO₂/H₃O⁺, Carboxylation) → m-HOOC–C₆H₄–NO₂ (3-Nitrobenzoic acid)

Teacher's Note:
1. Introduce the meta-directing NO₂ group first so that bromine (later replaced by COOH) enters the meta position.

Question 19. The final major product of the following reaction is
3-methylaniline (m-toluidine) ((i) Ac₂O/pyridine; (ii) Br₂, FeCl₃; (iii) OH⁻/Δ) → ? (JEE Main 2020)
(a) 2-bromo-5-methylaniline
(b) 3-bromo-5-methylaniline
(c) 2-bromo-3-methylaniline
(d) 4-bromo-3-methylaniline
Answer:
Ans. (d)
m-CH₃C₆H₄NH₂ + Ac₂O/pyridine (Acetylation) → m-CH₃C₆H₄NHCOCH₃
(ii) Br₂, FeCl₃ (Bromination) → 4-bromo-3-methylacetanilide; (iii) OH⁻/Δ (Hydrolysis) → 4-bromo-3-methylaniline
Please note that due to steric hindrance at o-position, bromination preferentially occurs at p-position w.r.t. the bigger NHCOCH₃ group.
Thus, option (d) is correct.

Teacher's Note:
1. Acetylation tames the NH₂ group so that only monobromination occurs, and the bulky NHCOCH₃ group favours its para position.

Question 20. The correct stability order of the following diazonium salts is :
I. p-CH₃O–C₆H₄–N₂Cl   II. p-O₂N–C₆H₄–N₂Cl   III. C₆H₅–N₂Cl   IV. p-NC–C₆H₄–N₂Cl (JEE Main 2022)
(a) I > II > III > IV
(b) I > III > IV > II
(c) III > I > IV > II
(d) III > IV > II > I
Answer:
Ans. (b)
Electron donating groups push electrons towards positively charged nitrogen and thus stabilize the diazonium salts.
p-CH₃O–C₆H₄–N⁺≡NCl⁻ ↔ CH₃O⁺=C₆H₄=N–N⁻
In contrast, electron-withdrawing groups destabilize the diazonium salts. Since NO₂ is a stronger electron-withdrawing group than CN, II is the least stable. Thus, option (b) is correct.

Teacher's Note:
1. Electron donors at the para position spread the positive charge of the –N₂⁺ group and stabilise the salt.

Question 21. The reagent ‘R’ in the given sequence of chemical reactions is
2,4,6-tribromoaniline (NaNO₂, HCl, 0–5°C) → 2,4,6-tribromobenzenediazonium chloride (R) → 1,3,5-tribromobenzene (NEET 2021)
(a) CuCN/KCN
(b) H₂O
(c) CH₃CH₂OH
(d) HI
Answer:
Ans. (c)
2,4,6-Br₃C₆H₂NH₂ + NaNO₂, HCl (0–5°C, Diazotisation) → 2,4,6-Br₃C₆H₂N⁺≡NCl⁻ (Diazonium salt)
2,4,6-Br₃C₆H₂N₂⁺Cl⁻ + CH₃CH₂OH (R) (Reductive deamination) → 1,3,5-Br₃C₆H₃ + N₂ + CH₃CHO + HCl
The reagent R is CH₃CH₂OH, i.e., option (c) is correct.

Teacher's Note:
1. Ethanol (or H₃PO₂) replaces –N₂⁺ by H; this trick gives substitution patterns that direct substitution cannot.

Question 22. C₆H₅NH₂ (NaNO₂, HCl, 273–278 K) → [X] (‘A’) → C₆H₅OH (Major product) (JEE Main 2021)
(a) X = C₆H₅NO₂; A = H₂O/NaOH
(b) X = C₆H₅N₂⁺Cl⁻; A = H₂O/NaOH
(c) X = C₆H₅NO₂; A = H₂O/Δ
(d) X = C₆H₅N₂⁺Cl⁻; A = H₂O/Δ
Answer:
Ans. (d)
C₆H₅NH₂ + NaNO₂, HCl (273–278 K, Diazotisation) → C₆H₅N₂⁺Cl⁻ (H₂O, Δ, Hydrolysis) → C₆H₅OH (Phenol)
Diazonium salts decompose on heating in presence of H₂O. Thus, option (d) is correct.

Teacher's Note:
1. Warm water, not alkali, converts the diazonium salt into phenol; in alkali the phenoxide would couple with unreacted diazonium ion.

Question 23. The major product Z obtained in the following reaction scheme is
3-bromoaniline (NaNO₂/HCl, 273–278 K) → X (Cu₂Br₂) → Y (HNO₃, H₂SO₄) → Z (JEE Main 2020)
(a) 1,3-dibromo-5-nitrobenzene
(b) 1,3-dibromo-2-nitrobenzene
(c) 2,4-dibromo-1-nitrobenzene
(d) 1,3-dibromo-2,4-dinitrobenzene
Answer:
Ans. (c)
m-BrC₆H₄NH₂ + NaNO₂/HCl (273–278 K, Diazotisation) → m-BrC₆H₄N⁺≡NCl⁻ (X) (Cu₂Br₂, Sandmeyer reaction) → m-C₆H₄Br₂ (Y)
m-C₆H₄Br₂ + HNO₃/H₂SO₄ (Nitration) → 2,4-dibromonitrobenzene (Z)
Please note that due to steric hindrance, substitution never occurs in between the two groups. Thus, option (c) is correct.

Teacher's Note:
1. Both Br atoms direct o/p; C-4 is ortho to one Br and para to the other, while C-2 between them is too crowded.

Question 24. A given nitrogen containing aromatic compound ‘A’ reacts with Sn/HCl, followed by HNO₂ to give an unstable compound ‘B’. ‘B’, on treatment with phenol, forms a beautiful coloured compound ‘C’ with the molecular formula C₁₂H₁₀N₂O. The structure of compound ‘A’ is (NEET Phase-II 2016)
(a) C₆H₅NH₂
(b) C₆H₅NO₂
(c) C₆H₅CN
(d) C₆H₅CONH₂
Answer:
Ans. (b)
C₆H₅–NO₂ (Nitrobenzene, A) + Sn/HCl (Reduction) → C₆H₅–NH₂ (Aniline)
C₆H₅NH₂ + HNO₂ (Diazotisation) → C₆H₅–N⁺≡NCl⁻ (Benzenediazonium chloride, B)
C₆H₅N₂⁺Cl⁻ + C₆H₅OH (Coupling reaction) → C₆H₅–N=N–C₆H₄–OH (p-Hydroxyazobenzene, beautiful coloured, M.F. C₁₂H₁₀N₂O)

Teacher's Note:
1. A must first be reduced by Sn/HCl to an amine; aniline itself needs no reduction, so A is nitrobenzene.

Question 25. In the following reactions, the major product W is
C₆H₅NH₂ (NaNO₂, HCl, 0°C) → V (2-naphthol, NaOH) → W (JEE Advanced 2015)
(a) 1-(phenylazo)-2-naphthol (azo group at C-1, next to the OH at C-2)
(b) 4-(phenylazo)-2-naphthol
(c) 5-(phenylazo)-2-naphthol (azo group in the ring not carrying OH)
(d) 8-(phenylazo)-2-naphthol (azo group in the ring not carrying OH)
Answer:
Ans. (a)
Benzenediazonium chloride (V) couples with 2-naphthol in alkaline medium. Due to greater stabilization of the carbocation intermediate, coupling occurs at C₁.
C₆H₅N₂⁺Cl⁻ + 2-naphthol (NaOH) → 1-(phenylazo)-2-naphthol (W), an orange-red dye.

Teacher's Note:
1. In 2-naphthol the most reactive position is C-1 (the α-position next to OH); attack there keeps the second ring aromatic.

Question 26. Coupling of benzenediazonium chloride with 1-naphthol in alkaline medium will be (JEE Main 2019)
(a) 5-(phenylazo)-1-naphthol
(b) 4-(phenylazo)-1-naphthol
(c) 2-(phenylazo)-1-naphthol
(d) 3-(phenylazo)-1-naphthol
Answer:
Ans. (b)
The resonance structures of α-naphthol have –ve charge at C₂ and C₄
Because of steric hindrance at C₂, attack by diazo cation preferentially occurs at C₄
C₆H₅N⁺≡NCl⁻ + 1-naphthol (OH⁻) → 4-(phenylazo)-1-naphthol
Thus, option (b) is correct.

Teacher's Note:
1. Coupling goes para to the OH group whenever that position is free; 1-naphthol therefore couples at C-4.

Question 27. Consider the following reactions,
[P] ((i) NaNO₂/HCl, 0–5°C; (ii) β-Naphthol/NaOH) → Coloured solid
[P] (Br₂/H₂O) → C₇H₆NBr₃
The compound [P] is (JEE Main 2020)
(a) p-CH₃C₆H₄NH₂
(b) o-CH₃C₆H₄NH₂
(c) m-CH₃C₆H₄NHCH₃
(d) m-CH₃C₆H₄NH₂
Answer:
Ans. (d)
Since compound (P) on treatment with Br₂–H₂O gives a compound with MF C₇H₆NBr₃, therefore, [P] is a methylaniline with both its o- and p-position free since it gives a tribromo derivative. In other words, (P) is m-methylaniline or m-toluidine.
m-CH₃C₆H₄NH₂ (m-Toluidine) + Br₂–H₂O → 2,4,6-tribromo-3-methylaniline (M.F. C₇H₆NBr₃)
The coloured dye with β-naphthol shows that P is a 1° aromatic amine. Thus, option (d) is correct.

Teacher's Note:
1. Three Br atoms enter only if both ortho and the para positions of NH₂ are free; a meta substituent allows this.

Question 28. The product formed from the following reaction sequence is
C₆H₅CN ((i) LiAlH₄, H₂O; (ii) NaNO₂ + HCl; (iii) H₂O) → ? (NEET 2022)
(a) C₆H₅CH₂N₂⁺Cl⁻
(b) C₆H₅CH₂Cl
(c) C₆H₅CH₂OH
(d) C₆H₅CONH₂
Answer:
Ans. (c)
C₆H₅CN + (i) LiAlH₄, (ii) H₂O → C₆H₅CH₂NH₂ (NaNO₂/HCl) → [C₆H₅CH₂–N⁺≡NCl⁻] (Unstable)
[C₆H₅CH₂N₂⁺Cl⁻] + H₂O −N₂, −HCl → C₆H₅CH₂OH

Teacher's Note:
1. Benzylamine is an aralkyl (not aryl) amine, so its diazonium salt is unstable and gives benzyl alcohol with water.

Question 29. The product of the reaction
C₆H₅OH (phenol) + C₆H₅–N₂Cl (base) → is
(a) C₆H₅–N=N–C₆H₄–OH (p)
(b) C₆H₅–O–C₆H₅
(c) C₆H₅–C₆H₅
(d) C₆H₅–C₆H₄–OH (p)
Answer:
Ans. (a)
Benzenediazonium chloride couples with phenol in mildly alkaline medium at the p-position to give p-hydroxyazobenzene, an orange dye:
C₆H₅N₂⁺Cl⁻ + C₆H₅OH (OH⁻) → C₆H₅–N=N–C₆H₄–OH (p) + HCl

Teacher's Note:
1. In coupling reactions the N₂ group is retained; the diazonium ion is the electrophile and attacks the para position of phenol.

Question 30. The major product of the following reaction is
2′-aminobiphenyl-3-ol (OH at C-3 of one ring, NH₂ at C-2′ of the other) ((i) NaNO₂, HCl, 0°C; (ii) aq. NaOH) → ? (JEE Advanced 2017)
(a) sodium salt of the diazonium chloride (O⁻Na⁺ on one ring, –N₂Cl on the other)
(b) 2′-chlorobiphenyl-3-ol
(c) benzo[c]cinnolin-2-ol (tricyclic –N=N– bridge; OH para to the ring carbon bonded to N)
(d) benzo[c]cinnolin-4-ol (tricyclic –N=N– bridge; OH ortho to the ring carbon bonded to N)
Answer:
Ans. (c)
The diazonium salt undergoes intramolecular diazo coupling at the unhindered p-position w.r.t., –OH group to give product (c).
2′-aminobiphenyl-3-ol + NaNO₂, HCl 0°C → biphenyl-2-diazonium chloride (3′-OH) (NaOH, Intramolecular azo coupling) → benzo[c]cinnolin-2-ol

Teacher's Note:
1. Coupling needs an activated ring position; here the phenoxide ring couples para to O⁻, and that carbon is close to the diazonium group, closing a six-membered ring.

Question 31. In the following reaction, the product (A)
C₆H₅N⁺≡NCl⁻ + C₆H₅NH₂ (H⁺) → A (Yellow dye), is (AIPMT 2014)
(a) C₆H₅–N=N–NH–C₆H₅
(b) C₆H₅–N=N–C₆H₄–NH₂ (o)
(c) C₆H₅–N=N–C₆H₄–NH₂ (m)
(d) C₆H₅–N=N–C₆H₄–NH₂ (p)
Answer:
Ans. (d)
Benzenediazonium chloride couples with aniline in mildly acidic medium at the p-position of aniline to give p-aminoazobenzene, a yellow dye.
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ (H⁺) → C₆H₅–N=N–C₆H₄–NH₂ (p) + HCl

Teacher's Note:
1. Amines couple in weakly acidic medium and phenols in weakly alkaline medium; in both cases coupling is para.

Question 32. Three isomers A, B and C (mol. formula C₈H₁₁N) give the following results.
A and C (Diazotisation) → P + Q ((i) Hydrolysis; (ii) Oxidation (KMnO₄/H⁺)) → R (Product of A) + S (product of Q)
R has lower boiling point than S
B (C₆H₅SO₂Cl) → Alkali-insoluble product
A, B and C respectively are : (JEE Main 2020)
(a) o-H₂NC₆H₄CH₂CH₃, m-CH₃C₆H₄CH₂NH₂, m-H₂NC₆H₄CH₂CH₃
(b) p-H₂NC₆H₄CH₂CH₃, C₆H₅CH₂NHCH₃, o-H₂NC₆H₄CH₂CH₃
(c) m-H₂NC₆H₄CH₂CH₃, C₆H₅CH₂NHCH₃, o-H₂NC₆H₄CH₂CH₃
(d) o-H₂NC₆H₄CH₂CH₃, C₆H₅CH₂NHCH₃, p-H₂NC₆H₄CH₂CH₃
Answer:
Ans. (d)
Since isomers A and C undergo diazotisation to afford diazonium salts P and Q respectively which upon hydrolysis and subsequent oxidation give phenols R and S respectively. Since R has lower b.p. than S, therefore, A is the o-isomer and C is the p-isomer.
o-H₂NC₆H₄CH₂CH₃ (A) + p-H₂NC₆H₄CH₂CH₃ (C) (NaNO₂/HCl, 273–278 K) → P + Q (Hydrolysis) → o- and p-HOC₆H₄CH₂CH₃ (KMnO₄/H⁺) → o-HOC₆H₄COOH (R) + p-HOC₆H₄COOH (S)
R (salicylic acid) has intramolecular H-bonding and hence a lower b.p. than S, which has intermolecular H-bonding.
Since isomer B reacts with C₆H₅SO₂Cl to give benzenesulphonamide which is alkali insoluble, therefore, it must be 2° amine, i.e., C₆H₅CH₂NHCH₃ (B) (C₆H₅SO₂Cl/NaOH) → C₆H₅CH₂–N(CH₃)–SO₂C₆H₅ (Insoluble in alkali)
Thus, option (d) is correct.

Teacher's Note:
1. ortho-Hydroxy acids form intramolecular H-bonds and boil lower than their para isomers.

Question 33. Consider the following reaction,
C₆H₅–N(CH₃)₂ + Na^{+ −}O₃S–C₆H₄–N₂⁺Cl⁻ (p) (OH⁻) → ‘X’
The product ‘X’ is used (JEE Main 2020)
(a) in protein estimation as an alternative to ninhydrin
(b) in laboratory test for phenols
(c) as food grade colourant
(d) in acid-base titration as an indicator
Answer:
Ans. (d)
(CH₃)₂N–C₆H₅ + Na^{+ −}O₃S–C₆H₄–N₂⁺Cl⁻ (OH⁻, Coupling reaction) → (CH₃)₂N–C₆H₄–N=N–C₆H₄–SO₃Na (Methyl orange)
Methyl orange is used as an indicator in acid-base titration. Thus, option (d) is correct.

Teacher's Note:
1. Methyl orange is the coupling product of diazotised sulphanilic acid with N,N-dimethylaniline; it is red in acid and yellow in alkali.

One or More Correct Answers

Question 34. Examine the following structures for the anilinium ion and choose the correct statement from the ones given below :
I: C₆H₅–N⁺H₃   ↔   II: H₃N=C₆H₅⁺ (nitrogen joined to the ring by a double bond, positive charge on a ring carbon)
(a) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions.
(b) II is not an acceptable canonical structure because it is non aromatic
(c) II is not an acceptable canonical structure because the nitrogen has ten valence electrons
(d) II is an acceptable canonical structure.
Answer:
Ans. (a, c)
In II, the N atom already carrying four bonds (three N–H and one N–C) would need a fifth bond, i.e., ten valence electrons, which is not possible for a second period element. Further, II converts a stable ammonium ion into a less stable carbonium ion.
Hence II is not an acceptable canonical structure for the reasons given in (a) and (c).

Teacher's Note:
1. The anilinium ion has no lone pair on N, so it has no resonance with the ring; that is why –NH₃⁺ is meta-directing.

Question 35. The possible product(s) to be obtained from the reaction of cyclobutylamine with HNO₂ is/are (West Bengal JEE 2018)
(a) cyclobutanol
(b) cyclobutanone
(c) cyclopropylmethanol
(d) H₂C=CH₂
Answer:
Ans. (a, c)
Cyclobutylamine + HNO₂ → cyclobutyldiazonium ion −N₂ → cyclobutyl carbocation
Cyclobutyl carbocation + H₂O, −H⁺ → Cyclobutanol
Cyclobutyl carbocation (Ring contraction) → cyclopropylmethyl carbocation (H₂O, −H⁺) → Cyclopropylmethyl alcohol
Thus, options (a, c) are correct.

Teacher's Note:
1. 1° aliphatic amines give carbocations with HNO₂, so rearranged alcohols are expected along with the normal one.

Passage-Based Questions

Question 36. Comprehension 2: RCONH₂ is converted into RNH₂ by means of Hofmann bromamide degradation.
p-ClC₆H₄CONH₂ (i) → p-ClC₆H₄CONHBr (ii) → p-ClC₆H₄CON⁻–Br (iii) → p-ClC₆H₄–N=C=O (iv) → p-ClC₆H₄–NH–COO⁻ M⁺ (v) → p-ClC₆H₄–NH₂ (vi)
In this reaction, RCONHBr is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation is an intramolecular reaction. (IIT 2006)
How can the conversion (i) to (ii) be brought about ?
(a) KBr
(b) KBr + CH₃ONa
(c) KBr + KOH
(d) Br₂ + KOH
Answer:
Ans. (d)
The N-bromoamide (ii) is formed by the action of bromine in presence of alkali: RCONH₂ + Br₂ + KOH → RCONHBr + KBr + H₂O.
KBr cannot brominate the amide as Br⁻ is not an electrophile.

Teacher's Note:
1. Bromination of the N–H needs an electrophilic bromine source, i.e., Br₂ (as OBr⁻ in alkali).

Question 37. Comprehension 2: RCONH₂ is converted into RNH₂ by means of Hofmann bromamide degradation.
p-ClC₆H₄CONH₂ (i) → p-ClC₆H₄CONHBr (ii) → p-ClC₆H₄CON⁻–Br (iii) → p-ClC₆H₄–N=C=O (iv) → p-ClC₆H₄–NH–COO⁻ M⁺ (v) → p-ClC₆H₄–NH₂ (vi)
In this reaction, RCONHBr is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation is an intramolecular reaction. (IIT 2006)
Which is the rate-determining step in Hofmann bromamide degradation ?
(a) Formation of (i)
(b) Formation of (ii)
(c) Formation of (iii)
(d) Formation of (iv)
Answer:
Ans. (d)
The slowest step is the rearrangement of the N-bromoamide anion (iii), in which the aryl group migrates from carbon to nitrogen with loss of Br⁻, giving the isocyanate (iv).
This is why electron-donating groups on the phenyl ring (which help the migration) activate the reaction.

Teacher's Note:
1. The migration step to the electron-deficient nitrogen is the slow step; groups that make the aryl group more electron rich speed it up.

Question 38. Comprehension 2: RCONH₂ is converted into RNH₂ by means of Hofmann bromamide degradation.
p-ClC₆H₄CONH₂ (i) → p-ClC₆H₄CONHBr (ii) → p-ClC₆H₄CON⁻–Br (iii) → p-ClC₆H₄–N=C=O (iv) → p-ClC₆H₄–NH–COO⁻ M⁺ (v) → p-ClC₆H₄–NH₂ (vi)
In this reaction, RCONHBr is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation is an intramolecular reaction. (IIT 2006)
What are the constituent amines formed when the mixture of (i) and (ii) undergo Hofmann bromamide degradation ?
(i) m-D–C₆H₄–CONH₂   (ii) C₆H₅–CO¹⁵NH₂
(a) m-D–C₆H₄–NH₂, m-D–C₆H₄–¹⁵NH₂, C₆H₅–NH₂, C₆H₅–¹⁵NH₂
(b) m-D–C₆H₄–NH₂, C₆H₅–¹⁵NH₂
(c) m-D–C₆H₄–¹⁵NH₂, C₆H₅–¹⁵NH₂
(d) C₆H₅–¹⁵NHD, C₆H₅–NH₂
Answer:
Ans. (b)
Hofmann degradation is an intramolecular reaction: the migrating aryl group never leaves its own molecule, so no crossover products are formed.
Thus (i) gives only m-D–C₆H₄–NH₂ and (ii) gives only C₆H₅–¹⁵NH₂.

Teacher's Note:
1. Crossover (isotope-labelling) experiments prove a reaction is intramolecular when no mixed-label products appear.

Matrix Match Questions

Teacher's Note:
1. Aniline can be made by Hofmann degradation of benzamide but never by Gabriel synthesis.

Teacher's Note:
1. Lassaigne's test for N needs both C and N in the compound; AgCl is white, AgBr pale yellow and AgI yellow.

Integer and Numerical Answer Questions

Question 41. Total number of amines among the following which can be synthesized by Gabriel phthalimide reaction is
(CH₃)₂CHCH₂NH₂, CH₃CH₂NH₂, C₆H₅CH₂NH₂, C₆H₅NH₂, CH₃NHCH₂CH₃, C₆H₅CH₂NHCH₃, CH₃CH₂CH₂NHCH₃, (CH₃)₂NC₆H₅, C₆H₅–NHCH₃ (JEE Main 2021)
Answer:
Ans. 3
Three amines. Only 1° aliphatic and aralkylamines can be synthesized by Gabriel phthalimide reaction, i.e., (CH₃)₂CHCH₂NH₂, CH₃CH₂NH₂ and C₆H₅CH₂NH₂.

Teacher's Note:
1. Gabriel synthesis gives only 1° amines, and not aniline-type (aryl) amines.

Question 42. The diazonium salts of which of the following amines will form a coloured dye in reaction with β-Naphthol in NaOH.
C₆H₅–NH–CH₃, C₆H₅CH₂NH₂, C₆H₅NH₂, C₆H₅N(CH₃)₂. (JEE Main 2022)
Answer:
Ans. 1
One amine. The diazonium salts of only 1° aromatic amines form coloured dyes with β-naphthol, i.e., C₆H₅NH₂.

Teacher's Note:
1. Benzylamine is a 1° amine but not an aromatic amine (N is not on the ring), so its diazonium salt is unstable.

Question 43. In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is......... . The yield (%) corresponding to the product in each step is given in the parenthesis.
C₆H₅COCH₃ (NaOBr, H₃O⁺) → A (60%) (NH₃, Δ) → B (50%) (Br₂/KOH) → C (50%) (Br₂ (3 equiv.), AcOH) → D (100%) (JEE Advanced 2018)
Answer:
Ans. 495
C₆H₅COCH₃ (10 moles) ((i) NaOBr; (ii) H₃O⁺, Haloform reaction) → C₆H₅CO₂H (Benzoic acid, A) (60%) (NH₃, Δ) → C₆H₅CONH₂ (Benzamide, B) (50%) (Br₂/KOH, Hofmann-bromamide reaction) → C₆H₅NH₂ (Aniline, C) (50%) (Br₂ (3 equiv.), AcOH) → 2, 4, 6-Tribromoaniline (D) (100%)
Amount of D (in moles) = 10 × \( \frac{60}{100} \) × \( \frac{50}{100} \) × \( \frac{50}{100} \) × \( \frac{100}{100} \) = 1.5
Mol. mass of 2, 4, 6-tribromoaniline (C₆H₄NBr₃) = 6 × 12 + 4 × 1 + 1 × 14 + 3 × 80 = 330 amu
∴ Amount of D (in g) = 1.5 × 330 = 495

Teacher's Note:
1. Multiply the step yields as fractions to get the overall moles, then convert to grams with the molar mass of the final product.

Assertion–Reason Questions

Question 44. Assertion: Gabriel phthalimide reaction can be used to prepare aryl and alkyl amines.
Reason: Aryl halides have same reactivity as alkyl halides towards nucleophilic substitution reactions. (AIIMS 2015)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. Gabriel phthalimide reaction can be used to prepare only primary alkyl amines and not arylamines.
Correct reason. Alkyl halides undergo nucleophilic substitution reactions but aryl halides do not.

Teacher's Note:
1. Aryl halides are very unreactive towards S_N reactions, which is exactly why Gabriel synthesis cannot give aniline.

Question 45. Assertion: Anilinium chloride is more acidic than ammonium chloride.
Reason: Anilinium ion is resonance stabilized.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. Since ammonia is a stronger base than aniline, therefore, the conjugate acid of aniline, i.e., anilinium chloride is more acidic than conjugate acid of ammonia, i.e., ammonium chloride.

Teacher's Note:
1. It is aniline (the free base) that is resonance stabilised; the anilinium ion has no lone pair on N and no such resonance.

Question 46. Assertion: p-Toluidine is a stronger base than m-toluidine.
Reason: Methyl group from m-position exerts smaller electron-donating inductive (+I)-effect than from p-position. (AIIMS 2014)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. CH₃ group exerts greater +I-effect from m-position than p-position. However, it is the stronger hyperconjugation effect of the CH₃ group from p-position which increases the basic strength of p-toluidine w.r.t. m-toluidine.

Teacher's Note:
1. The inductive effect weakens with distance, so it is larger from meta; hyperconjugation operates only from o/p positions.

Question 47. Assertion: Pyrrole is more basic than pyridine.
Reason: In pyrrole, nitrogen is sp³-hybridized.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. Pyrrole is less basic than pyridine.
Correct reason. In pyrrole, nitrogen is sp² hybridized and the lone pair of electrons on N is involved in aromatic sextet formation.

Teacher's Note:
1. Protonating pyrrole on N would destroy its aromaticity, so pyrrole is an extremely weak base.

Question 48. Assertion: In benzimidazole (N-I: the =N– atom; N-II: the N–H atom of the five-membered ring), both the nitrogens (N-I and N-II) are basic.
Reason: Lone pair of electrons present on N-I are involved in delocalisation. (AIIMS 2009)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. Both the nitrogens are not basic.
Correct reason. Only the lone pair of electrons on N (II) are involved in delocalization.

Teacher's Note:
1. In imidazole-type rings, the N–H nitrogen gives its lone pair to the aromatic sextet; only the =N– nitrogen is basic.

Question 49. Assertion: N₃⁻ is a weaker base than NH₂⁻.
Reason: The lone pair of electrons on N atom in N₃⁻ is in a sp²-orbital while in NH₂⁻ it is in an sp³-orbital. (AIIMS 2010)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
An orbital with more s-character holds its electrons closer to the nucleus. The lone pair on N in N₃⁻ (in an orbital with more s-character) is less available for donation than the lone pair in the sp³-orbital of NH₂⁻. Hence N₃⁻ is a weaker base, and the reason explains the assertion.

Teacher's Note:
1. Basicity falls as s-character of the lone-pair orbital rises: sp³ > sp² > sp.

Question 50. Assertion: Aniline does not undergo Friedel-Crafts reaction.
Reason: Friedel-Crafts is an electrophilic substitution reaction. (AIIMS 2013)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. AlCl₃ forms a salt with aniline (C₆H₅NH₂⁺AlCl₃⁻) which deactivates the benzene ring thereby preventing F.C. reaction.

Teacher's Note:
1. Both statements are true, but the real cause is the Lewis acid–base reaction of the –NH₂ group with AlCl₃.