CBSE Class 12 Chemistry Amines NEET JEE Questions Set 01

Single Correct Answer

Question 1. Which of the following reaction is an example of ammonolysis ? (JEE Main 2021)
(a) C₆H₅CH₂CN → C₆H₅CH₂CH₂NH₂
(b) C₆H₅COCl + C₆H₅NH₂ → C₆H₅CONHC₆H₅
(c) C₆H₅CH₂Cl + NH₃ → C₆H₅CH₂NH₂
(d) C₆H₅NH₂ → C₆H₅NH₃⁺Cl⁻
Answer:
Ans. (c)
Ammonolysis is the cleavage of the C–X bond of an alkyl (or benzyl) halide by NH₃: C₆H₅CH₂Cl + NH₃ → C₆H₅CH₂NH₂ + HCl.
(a) is reduction, (b) is benzoylation (acylation) and (d) is salt formation.

Teacher's Note:
1. In ammonolysis NH₃ acts as the nucleophile and replaces the halogen; it works for alkyl and benzyl halides, not for aryl halides.

Question 2. The reaction of propanamide with ethanolic sodium hydroxide and bromine will give (NEET 2020)
(a) methylamine
(b) propylamine
(c) aniline
(d) ethylamine
Answer:
Ans. (d)
Aliphatic or aromatic 1° amides on treatment with Br₂/NaOH undergo Hofmann bromamide reaction to yield 1° amines containing one carbon less than the 1° amides.
CH₃CH₂CONH₂ (Propanamide) + Br₂ + 4NaOH → CH₃CH₂NH₂ (Ethylamine) + Na₂CO₃ + 2NaBr + 2H₂O
Thus, option (d) is correct.

Teacher's Note:
1. Count carbons: the C=O carbon of the amide is lost as CO₃²⁻, so a C₃ amide gives a C₂ amine.

Question 3. Which of the following reaction does not involve Hofmann bromamide degradation ? (JEE Main 2021)
(a) C₆H₅CN ((i) KOH, H₂O; (ii) Br₂, NaOH) → C₆H₅NH₂
(b) C₆H₅CH₂COCH₃ ((i) Br₂, NaOH, H⁺; (ii) NH₃, Δ; (iii) LiAlH₄/H₂O) → C₆H₅CH₂NH₂
(c) C₆H₅COCl ((i) NH₃, NaOH; (ii) Br₂, NaOH) → C₆H₅NH₂
(d) C₆H₅CH₂CONH₂ (Br₂, NaOH) → C₆H₅CH₂NH₂
Answer:
Ans. (b)
Options (a), (c) and (d) undergo Hofmann degradation but option (b) does not.
In (b): C₆H₅CH₂COCH₃ + Br₂, NaOH, H⁺ → C₆H₅CH₂COOH (Haloform reaction, −CHBr₃); C₆H₅CH₂COOH + NH₃/Δ → C₆H₅CH₂CONH₂; C₆H₅CH₂CONH₂ + LiAlH₄/H₂O → C₆H₅CH₂CH₂NH₂
Here the amide is reduced by LiAlH₄ (no loss of carbon), so no Hofmann bromamide degradation is involved.

Teacher's Note:
1. Hofmann degradation always shortens the chain by one carbon; LiAlH₄ reduction of an amide keeps all the carbons.

Question 4. Which of the following reactions is appropriate for converting acetamide to methanamine ? (NEET 2017)
(a) Hofmann hypobromamide reaction
(b) Stephen's reaction
(c) Gabriel phthalimide synthesis
(d) Carbylamine reaction
Answer:
Ans. (a)
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Acetamide (C₂) gives methanamine (C₁), i.e., one carbon less: this is the Hofmann bromamide (hypobromamide) reaction.

Teacher's Note:
1. Amide → amine with one carbon less: think of Hofmann; amide → amine with the same carbons: think of LiAlH₄ reduction.

Question 5. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br₂ used per mole of amine produced are (JEE Main 2016)
(a) one mole of NaOH and one mole of Br₂
(b) four moles of NaOH and two moles of Br₂
(c) two moles of NaOH and two moles of Br₂
(d) four moles of NaOH and one mole of Br₂
Answer:
Ans. (d)
RCONH₂ + Br₂ + 4NaOH → R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Teacher's Note:
1. Balance it once and remember it: 1 Br₂ and 4 NaOH per mole of amide (and of amine).

Question 6. One of the following amides will not undergo Hofmann bromamide reaction :
(a) CH₃CONHCH₃
(b) CH₃CH₂CONH₂
(c) CH₃CONH₂
(d) C₆H₅CONH₂
Answer:
Ans. (a)
Only 1° amides undergo Hofmann bromamide reaction. CH₃CONHCH₃ being a 2° amide does not undergo this reaction.

Teacher's Note:
1. The nitrogen must carry two H atoms: one is replaced by Br and the second is removed by base before the alkyl/aryl group migrates.

Question 7. An organic compound A upon reacting with NH₃ gives B. On heating, B gives C. C in presence of KOH reacts with Br₂ to give CH₃CH₂NH₂. A is (AIPMT Mains 2011 ; JEE Main 2013)
(a) CH₃CH₂COOH
(b) CH₃COOH
(c) CH₃CH₂CH₂COOH
(d) CH₃–CH(CH₃)–COOH
Answer:
Ans. (a)
CH₃CH₂COOH (Propanoic acid, A) + NH₃ → CH₃CH₂COONH₄ (Amm. propanoate, B)
CH₃CH₂COONH₄ −H₂O (Δ) → CH₃CH₂CONH₂ (Propanamide, C)
CH₃CH₂CONH₂ + Br₂–KOH (Hofmann bromamide reaction) → CH₃CH₂NH₂ (Ethanamine)

Teacher's Note:
1. Work backwards: a C₂ amine from Hofmann means a C₃ amide, hence a C₃ acid.

Question 8. The following sequence of reactions on A gives
A = o-(H₂NCOCH₂)C₆H₄COOCH₃ (methyl 2-(carbamoylmethyl)benzoate)   1. Br₂/NaOH; 2. Heat → ?
(a) isochroman-1,3-dione (benzene fused to a six-membered cyclic anhydride, –CH₂–CO–O–CO–)
(b) isoquinoline-1,3(2H,4H)-dione (benzene fused to a six-membered cyclic imide, –CH₂–CO–NH–CO–)
(c) isoindolin-1-one (benzene fused to a five-membered lactam, –CH₂–NH–CO–)
(d) naphthalene-1,3(2H,4H)-dione (benzene fused to a six-membered ring, –CO–CH₂–CO–CH₂–)
Answer:
Ans. (c)
o-(H₂NCOCH₂)C₆H₄COOCH₃ (A) + Br₂/NaOH (Hofmann bromamide reaction) → o-(H₂NCH₂)C₆H₄COOCH₃
On heating, the –NH₂ group attacks the ester carbonyl intramolecularly: o-(H₂NCH₂)C₆H₄COOCH₃ −CH₃OH (Heat) → isoindolin-1-one (five-membered lactam).

Teacher's Note:
1. The CH₂CONH₂ group becomes CH₂NH₂ (one carbon lost), so the ring formed with the ester is five-membered, not six-membered.

Question 9. Which of the following compounds can be prepared in good yield by Gabriel phthalimide synthesis ? (JEE Main 2020)
(a) C₆H₅CH₂NH₂
(b) CH₃CH₂–NH–CH₃
(c) C₆H₅CH₂CONH₂
(d) C₆H₅NH₂
Answer:
Ans. (a)
Only 1° aliphatic and 1° aralkylamines (but not 1° aromatic amines) can be prepared in good yield by Gabriel phthalimide reaction. Thus, C₆H₅CH₂NH₂ being 1° aralkylamine can be prepared in good yield by Gabriel phthalimide reaction, i.e., option (a) is correct.

Teacher's Note:
1. Gabriel synthesis needs an S_N2 attack of phthalimide anion on R–X, so it fails for aryl halides and never gives 2° or 3° amines.

Question 10. Method by which aniline cannot be prepared is (RE-AIPMT 2015)
(a) degradation of benzamide with bromine in alkaline solution
(b) reduction of nitrobenzene with H₂/Pd in ethanol
(c) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution
(d) hydrolysis of phenylisocyanide with acidic solution
Answer:
Ans. (c)
Aniline cannot be prepared by Gabriel phthalimide method because chlorobenzene (i.e., aryl halide) does not undergo nucleophilic substitution reaction with potassium phthalimide in aqueous NaOH solution.

Teacher's Note:
1. The C–Cl bond of chlorobenzene has partial double bond character, so the phthalimide anion cannot displace Cl.

Question 11. The major product of the following reaction is
Phthalimide ((i) KOH; (ii) Br–C₆H₄–CH₂Cl (p)) → ? (IIT Paper I, 2011)
(a) N-(4-bromobenzyl)phthalimide, C₆H₄(CO)₂N–CH₂–C₆H₄–Br (p)
(b) N-[4-(chloromethyl)phenyl]phthalimide, C₆H₄(CO)₂N–C₆H₄–CH₂Cl (p)
(c) 3-[(4-bromobenzyl)oxy]-1H-isoindol-1-one (O–CH₂–C₆H₄–Br attached to the ring C=N carbon)
(d) 3-[4-(chloromethyl)phenoxy]-1H-isoindol-1-one (O–C₆H₄–CH₂Cl attached to the ring C=N carbon)
Answer:
Ans. (a)
Aralkyl halides are more reactive than aryl halides towards nucleophilic substitution reactions. Therefore, reaction occurs at the more reactive CH₂Cl instead of at Br as shown below :
Phthalimide + KOH −H₂O → potassium phthalimide (C₆H₄(CO)₂N⁻K⁺)
C₆H₄(CO)₂N⁻K⁺ + ClCH₂–C₆H₄–Br −KCl → C₆H₄(CO)₂N–CH₂–C₆H₄–Br

Teacher's Note:
1. The phthalimide anion reacts through nitrogen, and only at the benzylic carbon; the aryl C–Br bond stays untouched.

Question 12. The following reaction,
C₆H₅NH₂ + ClCOC₆H₅ (NaOH) → C₆H₅NHCOC₆H₅
is known by the name (RE-AIPMT 2015)
(a) Perkin's reaction
(b) Acetylation reaction
(c) Schotten-Baumann reaction
(d) Friedel-Craft's reaction
Answer:
Ans. (c)
Benzoylation of amines (or phenols) with benzoyl chloride in the presence of aqueous NaOH is called the Schotten-Baumann reaction. The base neutralises the HCl formed.

Teacher's Note:
1. It is benzoylation, not acetylation: the acyl group added is C₆H₅CO–.

Question 13. A (C₈H₆Cl₂O) (NH₃) → (C₈H₈ClNO) (Br₂/NaOH) → 4-chloro-3-methylaniline (CH₃ meta and Cl para to NH₂)
Consider the above reaction, compound A is (JEE Main 2022)
(a) 4-chloro-2-methylbenzoyl chloride
(b) 3-chloro-4-methylbenzoyl chloride
(c) 4-chloro-3-methylbenzoyl chloride
(d) 4-chloro-3-(chloromethyl)benzaldehyde
Answer:
Ans. (c)
Since compound A (C₈H₆Cl₂O) on treatment with NH₃ gives another compound I (C₈H₈ClNO) which is obtained by replacement of one Cl by NH₂ group. Therefore, compound A must be an acid chloride and compound (I) must be an amide. This compound (I) on treatment with Br₂/NaOH undergoes Hofmann bromamide reaction to give compound (II) with loss of CO. Therefore, position of NH₂ in the same as that of COCl group in compound (A).
4-Cl-3-CH₃C₆H₃COCl (A) + NH₃ −HCl → 4-Cl-3-CH₃C₆H₃CONH₂ (I) (Hofmann bromamide reaction) → 4-Cl-3-CH₃C₆H₃NH₂ (II)
Thus, option (c) is correct.

Teacher's Note:
1. In Hofmann degradation the NH₂ takes exactly the ring position of the old CONH₂ group, so the substitution pattern is unchanged.

Question 14. The reaction of an amide with bromine and KOH gives RNH₂ as the end product. Which one of the following is the intermediate product that is hydrolysed to give RNH₂ in this reaction ? (JEE Main 2022)
(a) RCONHBr
(b) R–NH–Br
(c) R–N=C=O
(d) RCONBr₂
Answer:
Ans. (c)
RCONH₂ + Br₂ + KOH → RCONHBr (N-bromoamide); RCONHBr + OH⁻ → RCON⁻Br → R–N=C=O (alkyl isocyanate) + Br⁻
R–N=C=O + 2OH⁻ → RNH₂ + CO₃²⁻. Thus, the isocyanate is the intermediate that gives the amine on hydrolysis.

Teacher's Note:
1. Remember the sequence amide → N-bromoamide → isocyanate → amine; the R group migrates from carbon to nitrogen in the second step.

Question 15. Butylamine (I), diethylamine (II) and N, N-dimethylethylamine (III) have the same molar mass. The increasing order of their boiling points is (JEE Main 2022)
(a) III < II < I
(b) I < II < III
(c) II < III < I
(d) II < I < III
Answer:
Ans. (a)
Extent of H-bonding increases as the number of H-atoms on the N-atom increases and hence boiling points increase in the order : 3° < 2° < 1°, i.e., III < II < I.

Teacher's Note:
1. For isomeric amines, b.p. falls 1° > 2° > 3°: a 3° amine has no N–H and cannot H-bond with itself.

Question 16. The correct order of basic strength of CH₃NH₂ (I), (CH₃)₂NH (II), (CH₃)₃N (III), C₆H₅CH₂NH₂ (IV) in gaseous phase is (AIIMS 2010)
(a) IV < III < II < I
(b) IV < III < I < II
(c) I < II < III < IV
(d) IV < I < II < III
Answer:
Ans. (d)
In the gaseous phase, basicity increases as the + I-effect of the alkyl groups increases, i.e., CH₃NH₂ (I) < (CH₃)₂NH < (CH₃)₃N (III). However, due to –I-effect of the C₆H₅– group, C₆H₅CH₂NH₂ (IV) is even a weaker base than CH₃NH₂ (I). Thus, the overall, basic character increases in the order :
IV < I < II < III.

Teacher's Note:
1. In the gas phase there is no solvation, so only the inductive effect counts and the 3° amine is the strongest base.

Question 17. The correct statement regarding the basicity of arylamines is (NEET Phase-I 2016)
(a) arylamines are generally more basic than alkylamines because of aryl group
(b) arylamines are generally more basic than alkylamines because the nitrogen atom in arylamines is sp-hybridised
(c) arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalised by interaction with the aromatic ring π-electron system
(d) arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalised by interaction with the aromatic ring π-electron system
Answer:
Ans. (c)
Due to nitrogen lone pair of electrons are delocalized by interaction with the aromatic ring π-electrons, the lone pair is less available for protonation. Hence arylamines are less basic than alkylamines.

Teacher's Note:
1. Draw the resonance structures of aniline: the lone pair moves into the ring at the o- and p-positions, which lowers basicity.

Question 18. The correct order of the basic strength of methyl substituted amines in aqueous solution is (NEET 2019)
(a) CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
(b) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
(c) (CH₃)₃N > CH₃NH₂ > (CH₃)₂NH
(d) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂
Answer:
Ans. (b)
In aqueous solution, basicity depends on +I effect, solvation (H-bonding) of the conjugate acid and steric hindrance together.
For methylamines the order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N.

Teacher's Note:
1. Learn both aqueous orders: methyl series 2° > 1° > 3°; ethyl series 2° > 3° > 1°.

Question 19. In the following compounds, the increasing order of basic strength will be (JEE Main 2019)
(a) (C₂H₅)₂NH > NH₃ > C₂H₅NH₂
(b) (C₂H₅)₂NH > C₂H₅NH₂ > NH₃
(c) C₂H₅NH₂ > NH₃ > (C₂H₅)₂NH
(d) NH₃ > C₂H₅NH₂ > (C₂H₅)₂NH
Answer:
Ans. (b)
Alkyl groups (+I effect) increase the electron density on N, and in aqueous solution the 2° amine is the strongest. Thus, basic strength: (C₂H₅)₂NH > C₂H₅NH₂ > NH₃.

Teacher's Note:
1. All aliphatic amines are stronger bases than NH₃; among ethylamines in water, the 2° amine is the strongest.

Question 20. The decreasing order of basicity of the following amines is
aniline (i), pyridine (ii), cyclohexylamine (iii), pyrrole (iv) (JEE Main 2020)
(a) (iii) > (i) > (ii) > (iv)
(b) (ii) > (iii) > (iv) > (i)
(c) (i) > (iii) > (iv) > (ii)
(d) (iii) > (ii) > (i) > (iv)
Answer:
Ans. (d)
In cyclohexylamine (iii), N is present in a sp³-orbital while in pyridine (ii), it is present in a sp²-orbital, therefore, cyclohexylamine (III) is more basic than pyridine (ii).
In aniline (i), the lone pair of electrons on N is delocalized over the benzene ring and hence is weakly available for protonation, therefore, it is a weaker base than even pyridine.
In pyrrole (iv), the lone pair of electrons on N is donated towards the formation of aromatic sextet and hence is unavailable for protonation. Therefore, it is such a weak base that it acts as an acid in presence of a strong base such as K metal.
Combining all the facts, the basicity of the four bases decreases in the order : (iii) > (ii) > (i) > (iv), i.e., option (d) is correct.

Teacher's Note:
1. Ask where the lone pair is: sp³ and free (strongest), sp² and free, delocalised into a ring, or part of the aromatic sextet (weakest).

Question 21. The increasing order of pK_b values of the following compounds is
I: p-CH₃O–C₆H₄–N(CH₃)₂; II: C₆H₅N(CH₃)₂; III: m-NC–C₆H₄–NHCH₃; IV: m-HO–C₆H₄–NHCH₃ (JEE Main 2020)
(a) II < IV < III < I
(b) I < II < IV < III
(c) II < I < III < IV
(d) I < II < III < IV
Answer:
Ans. (b)
p-Methoxy-N, N-dimethylaniline (I) is more basic than N, N-dimethylaniline (II). Since OH group at m-position acts as an electron-withdrawing group like CN group, therefore both anilines (III) and (IV) are less basic than even aniline (II). Since CN is a stronger electron-withdrawing group than OH group, therefore, aniline (IV) is more basic than aniline (III).
Combining both the trends, the overall basicity of the four anilines decreases in the order : I > II > IV > III.
Conversely, their pK_b values increase in the reverse order, i.e., I < II < IV < III, i.e., option (b) is correct.

Teacher's Note:
1. At the meta position only the –I effect of OH or OCH₃ operates, so they lower basicity there; at the para position +R wins.

Question 22. The increasing order of basicity of the following compounds is
pyridine (I), pyrrole (II), pyrrolidine (III), imidazole (IV) (JEE Main 2020)
(a) (II) < (I) < (IV) < (III)
(b) (IV) < (I) < (II) < (III)
(c) (II) < (I) < (III) < (IV)
(d) (I) < (II) < (III) < (IV)
Answer:
Ans. (a)
(i) In amine (III), N is sp³-hybridized while in all the remaining three (I, II and IV) amines it is sp²-hybridized. Since the less electronegative sp³-orbital can donate a pair of electrons more readily than a sp²-orbital, therefore, amine (III) is the strongest of the given four amines.
(ii) Since in amine (II), the lone pair of electrons on the N-atom is donated towards the aromatic sextet formation, therefore, it is not available for protonation and hence it is the weakest base.
(iii) Since the conjugate acid of the amine (IV) (imidazolium ion) is stabilized by two equivalent resonance structures but none of the resonance structures of pyridine (pyridinium ion) are equivalent, therefore, amine (IV) is more basic than amine (I).
From the above discussion, it follows that the basicity of the four amines increases in the order : II < I < IV < III, i.e., option (a) is correct.

Teacher's Note:
1. Imidazole is more basic than pyridine because its protonated form is a symmetrical cation stabilised by two equivalent resonance structures.

Question 23. The increasing order of pK_b for the following compounds will be
NH₂–CH=NH (A); 1,5,7-triazabicyclo[4.4.0]dec-5-ene (B, bicyclic compound with the group –N=C(–N<)–NH–); CH₃NHCH₃ (C) (JEE Main 2020)
(a) (B) < (C) < (A)
(b) (C) < (A) < (B)
(c) (A) < (B) < (C)
(d) (B) < (A) < (C)
Answer:
Ans. (d)
(B) has guanidine type structure, its conjugate acid is stabilized by four equivalent structures and hence is the most basic amine.
(A) has amidine type structure, its conjugate acid (NH₂–CH=NH₂⁺ ↔ ⁺NH₂=CH–NH₂) is stabilized by equivalent resonance structures and hence is less basic than guanidine type structure (B).
(C) is a 2° amine, its conjugate acid is only stabilized by weak H-bonding, therefore, it is the weakest base.
Thus, the basicity of these amines follows the order : B > A > C. Conversely, their pK_b values increase in the reverse order : B < A < C, i.e., option (d) is correct.

Teacher's Note:
1. The more the positive charge of the conjugate acid is spread over equivalent N atoms, the stronger the base: guanidines > amidines > simple amines.

Question 24. The basicity of the following amines
I: aniline; II: N,N-dimethylaniline; III: 2,4,6-trinitroaniline; IV: N,N-dimethyl-2,4,6-trinitroaniline
decreases in the order : (JEE Advanced 2020 modified)
(a) II > I > III > IV
(b) II > I > IV > III
(c) I > II > III > IV
(d) I > II > IV > III
Answer:
Ans. (b)
(i) Due to + I-effect of the two CH₃ groups, N, N-dimethylaniline (II) is more basic than aniline (I).
(ii) Due to strong – I-effect of the three NO₂ groups, amines (III and IV) are much less basic than even amine (II).
(iii) However, due to steric hindrance between the two CH₃ groups on N with the two NO₂ groups at o-positions, the p-orbital on N containing the lone pair of electrons is no longer parallel to the p-orbitals of the benzene ring. Consequently, resonance is prevented and hence the lone pair of electrons on N in amine (IV) is not contributed towards the benzene ring to the same extent as in amine (III). Therefore, the lone pair of electrons on N in amine (IV) is more easily available for protonation than in amine (III) and hence amine (IV) is a stronger base than amine (III).
Combining all the three trends, basicity of the four amines follows the order : II > I > IV > III.

Teacher's Note:
1. This is steric inhibition of resonance: bulky ortho groups twist the NMe₂ group out of the ring plane, so its lone pair is no longer delocalised.

Question 25. Which among the following is the strongest Bronsted base ? (JEE Main 2022)
(a) (CH₃CH₂)₃N
(b) (CH₃)₂NH
(c) N-methylpyrrolidine
(d) quinuclidine (1-azabicyclo[2.2.2]octane)
Answer:
Ans. (d)
Quinuclidine (d) is the strongest base because of maximum + I-effect and minimum steric hindrance due to rigid structure of the three alkyl groups which are projected away from the nitrogen lone pair.

Teacher's Note:
1. In a cage amine the alkyl groups are tied back, so the lone pair is fully exposed and the cation is easily solvated.

Question 26. Which among the following is the strongest conjugate acid ?
(a) C₆H₁₁NH₃⁺ (cyclohexylammonium)
(b) C₆H₅NH₃⁺
(c) C₆H₅CH₂NH₃⁺
(d) CH₃NH₃⁺
Answer:
Ans. (b)
A weak base has a strong conjugate acid. Since aniline is the weakest base, therefore, its conjugate acid is the strongest acid.

Teacher's Note:
1. Weakest base ↔ strongest conjugate acid: find the least basic amine first.

Question 27. The correct order of basicity of the following compounds is
1. CH₃CH₂NH₂   2. CH₃CH=NH   3. CH₃C(=NH)NH₂   4. H₂NC(=NH)NH₂ (West Bengal JEE 2016)
(a) 1 < 2 < 3 < 4
(b) 1 < 2 < 4 < 3
(c) 2 < 1 < 3 < 4
(d) 4 < 3 < 2 < 1
Answer:
Ans. (c)
The conjugate acid obtained by addition of a proton to amine 4 is stabilized by three (I–III) equivalent resonance structures while that obtained from amine 3 is stabilized by only two (IV–V) equivalent resonance structures.
H₂N–C(=NH)–NH₂ + H⁺ → [H₂N–C(=NH₂⁺)–NH₂ ↔ H₂N⁺=C(NH₂)–NH₂ ↔ H₂N–C(NH₂)=NH₂⁺] (I–III)
CH₃–C(=NH)–NH₂ + H⁺ → [CH₃–C(=NH₂⁺)–NH₂ ↔ CH₃–C(NH₂)=NH₂⁺] (IV–V)
Therefore, amine 4 is more basic than amine 3. In amine 2 the lone pair of electrons is present on a sp²-hybridized N while in amine 1 it is present on a sp³-hybridized N. Since a sp² hybridized N is more electronegative than a sp³ hybridized N, therefore, amine 2 is less willing to donate its lone pair than amine 1. In other words, amine 2 is less basic than amine 1. Combining both the trends, the basicity of the four amines, i.e., increases in the order : 2 < 1 < 3 < 4, i.e., option (c) is correct.

Teacher's Note:
1. Imines (sp² N) are weaker bases than amines, but amidines and guanidines are much stronger because their cations are resonance stabilised.

Question 28. The correct order of basicities of the following compounds is
1. CH₃C(=NH)NH₂   2. CH₃CH₂NH₂   3. (CH₃)₂NH   4. CH₃CONH₂
(a) 2 > 1 > 3 > 4
(b) 1 > 3 > 2 > 4
(c) 3 > 1 > 2 > 4
(d) 1 > 2 > 3 > 4
Answer:
Ans. (b)
CH₃–C(=NH)–NH₂ (1) + H⁺ → CH₃–C(=NH₂⁺)–NH₂ (i) ↔ CH₃–C(NH₂)=NH₂⁺ (ii, iii)
The conjugate acid (i) obtained by addition of a proton to 1 is stabilized by two equivalent resonance structures and hence compound (1) is the most basic. Further, 2° amines are more basic than 1° amines while amides are least basic due to delocalization of lone pair of electrons of N over the C = O group. Thus, the order : 1 > 3 > 2 > 4, i.e., option (b) is correct.

Teacher's Note:
1. Amides are almost neutral: the N lone pair is delocalised onto the C=O oxygen.

Question 29. The order of basicity among the following compounds is
I. CH₃C(=NH)NH₂ (acetamidine)   II. 4,5-dihydro-1H-imidazole (2-imidazoline)   III. imidazole   IV. H₂NC(=NH)NH₂ (guanidine) (JEE Advanced 2017)
(a) IV > II > III > I
(b) II > I > IV > III
(c) I > IV > III > II
(d) IV > I > II > III
Answer:
Ans. (d)
As explained in Ans. 30 amine (IV) is more basic than amine (I).
Now compound (III) is aromatic while compound (II) is not. Therefore, in compound (III) lone pair of electrons on N₁ are involved in aromatic sextet formation and hence not available for protonation. In other words, the basicity of compound (III) is only due to the lone pair of electrons on N₃. In contrast, in compound (II) the lone pair of electrons on both, N₁ and N₃ are available for protonation, therefore, compound (II) is more basic than compound (III).
Combining both the trends, the overall basicity of the four compounds follows the order :
IV > I > II > III, i.e., option (d) is correct.

Teacher's Note:
1. Aromaticity costs basicity: in imidazole one N lone pair is locked in the aromatic sextet.

Question 30. In pyrrole (ring atoms numbered N1, C2, C3, C4, C5) the electron density is maximum on (NEET Phase-II 2016)
(a) 2 and 3
(b) 3 and 4
(c) 2 and 4
(d) 2 and 5
Answer:
Ans. (d)
Pyrrole has two different types of hydrogens, viz., 2(or 5) and 3(or 4).
Obviously, electron density is high where electrophilic substitution occurs predominantly. To decide this, let us compare the stabilities of carbocations resulting from attack of the electrophile at C₂ (or C₅) or at C₃ (or C₄) positions.
Since carbocation resulting from electrophilic attack at C₂ or C₅ is stabilized by three resonance structures (I–III) while that resulting from attack at C₃ or C₄ is stabilized by only two resonance structures (IV–V), therefore, electron density is maximum at C₂ and C₅ positions, i.e., option (d) is correct.

Teacher's Note:
1. Electrophiles attack pyrrole at C-2 (α-position) because that intermediate cation has the most resonance structures.

Question 31. A compound with molecular mass 180 is acylated with CH₃COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is (JEE Main 2013)
(a) 6
(b) 2
(c) 5
(d) 4
Answer:
Ans. (c)
During acetylation, one H atom (at mass = 1 amu) of the NH₂ group is replaced by an acetyl group, i.e., CH₃CO (mol. mass = 43 amu).
–NH₂ + CH₃COCl → –NHCOCH₃ + HCl
In other words, acetylation of each NH₂ group increases the mass by 43 – 1 = 42 amu. Now since the mol. mass of the organic compound is 180 while that of the acetylated product is 390, therefore, increase in the mass due to acetylation = 390 – 180 = 210 amu.
∴ No. of –NH₂ groups = \( \frac{210}{42} \) = 5

Teacher's Note:
1. Each acetylation adds CH₃CO and removes one H, a net gain of 42 per NH₂ group.

Question 32. In the reaction shown below, the major product formed is/are
m-(H₂NCH₂)C₆H₄CONH₂ (Acetic anhydride, CH₂Cl₂) → product(s) (JEE Advanced 2014)
(a) m-(CH₃CONHCH₂)C₆H₄CONH₂ + CH₃COOH
(b) m-(H₂NCH₂)C₆H₄CONHCOCH₃ + CH₃COOH
(c) m-(CH₃CONHCH₂)C₆H₄CONHCOCH₃ + H₂O
(d) m-(CH₃COO^{− +}NH₃CH₂)C₆H₄CONHCOCH₃
Answer:
Ans. (a)
Due to delocalization of the lone pair of electrons on the N atom of the CONH₂, amides are much less nucleophilic than amines. Therefore, acetylation with acetic anhydride occurs only at the –NH₂ group, i.e., option (a) is correct.

Teacher's Note:
1. When an amine and an amide are both present, the amine nitrogen is the nucleophile; acetylation with (CH₃CO)₂O gives CH₃COOH as by-product.

One or More Correct Answers

Question 33. Consider the following compounds :
pyrrole (I), pyridine (II), aniline (III)
Which of the following statements are correct?
(a) I is more basic than II
(b) II is more basic than I and III
(c) III is more basic than II
(d) I is weakly acidic
Answer:
Ans. (b, d)
Due to contribution of the lone pair of electrons of N towards the aromatic sextet formation, pyrrole (I) is weakly acidic.
Pyridine (lone pair in an sp² orbital, not delocalised) is more basic than aniline (lone pair delocalised into the ring) and much more basic than pyrrole. Hence (b) and (d) are correct.

Teacher's Note:
1. Pyrrole N–H can lose H⁺ (to K metal or strong base) because the anion stays aromatic.

Passage-Based Questions

Question 34. Comprehension 1: All aliphatic amines are more basic than ammonia but due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength. The base strengthening effect of the electron-donating groups and base weakening effect of the electron-withdrawing groups is, however, more pronounced at p- than at m-position. However, due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing.
Arrange the following amines in decreasing order of their basic strength
aniline (I), o-chloroaniline (II), m-chloroaniline (III), p-chloroaniline (IV)
(a) I > II > III > IV
(b) I > IV > II > III
(c) II > I > IV > III
(d) I > IV > III > II
Answer:
Ans. (d)
Cl is an electron-withdrawing (–I) group, so all chloroanilines are weaker bases than aniline (I).
At the p-position the –I effect is weakest and is partly opposed by the +R effect of Cl, so IV is the strongest of the three; at the m-position only –I operates (III); the o-isomer is weakest due to ortho effect (II).
Thus, basic strength : I > IV > III > II.

Teacher's Note:
1. The –I effect falls off with distance (o > m > p), and the ortho effect makes o-substituted anilines the weakest.

Question 35. Comprehension 1: All aliphatic amines are more basic than ammonia but due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength. The base strengthening effect of the electron-donating groups and base weakening effect of the electron-withdrawing groups is, however, more pronounced at p- than at m-position. However, due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing.
Among the following, weakest base is
(a) C₆H₅CH₂NH₂
(b) C₆H₅CH₂NHCH₃
(c) O₂N–CH₂NH₂
(d) CH₃NHCHO
Answer:
Ans. (d)
Amides are weaker bases than amines.
In CH₃NHCHO the lone pair on N is delocalised over the C=O group (N–C=O ↔ N⁺=C–O⁻), so it is hardly available for protonation.

Teacher's Note:
1. An electron-withdrawing group on carbon weakens an amine, but a C=O directly on N (amide) weakens it far more.

Matrix Match Questions

Teacher's Note:
1. Use the molecular formula of each final product to check the match: count the N atoms first, then the O atoms.

Teacher's Note:
1. Match each compound to its functional-group test: 1° amine (carbylamine), terminal alkyne (Tollens'), ester (hydrolysis), 2° alcohol (Lucas, ~5 min).

Integer and Numerical Answer Questions

Question 38. How many isomeric amines have the molecular formula C₄H₁₁N ?
Answer:
Ans. 8
Eight amines are possible.
1° amines (4): CH₃CH₂CH₂CH₂NH₂, CH₃CH₂CH(NH₂)CH₃, (CH₃)₂CHCH₂NH₂, (CH₃)₃CNH₂
2° amines (3): CH₃NHCH₂CH₂CH₃, CH₃NHCH(CH₃)₂, CH₃CH₂NHCH₂CH₃
3° amine (1): (CH₃)₂NCH₂CH₃

Teacher's Note:
1. Count 1°, 2° and 3° amines separately; optical isomers of butan-2-amine are not counted as structural isomers here.

Question 39. The total number of optically active compounds from the following list is
C₂H₅NHCH₃, CH₃CH₂CH(CH₃)N(CH₃)C₂H₅, [Ph(CH₃)(C₂H₅)N⁺H]I⁻, [(CH₃)₂(C₂H₅)N⁺H] I⁻, C₆H₅(CH₃)(C₂H₅)N⁺– O⁻, (CH₃)(C₂H₅)NCH(CH₃)₂.
Answer:
Ans. 3
Three compounds are optically active. These are :
amine : CH₃CH₂C*H(CH₃)N(CH₃)C₂H₅ ;
quaternary ammonium salt : [Ph(CH₃)(C₂H₅)N⁺H]I⁻ and
amine N-oxide : C₆H₅(CH₃)(C₂H₅)N⁺–O⁻.

Teacher's Note:
1. A simple amine with three different groups on N is not resolvable because of rapid nitrogen inversion; N becomes a stable stereocentre when its four bonds are fixed (ammonium salt, N-oxide).

Question 40. How many of the following amines will undergo diazotization ?
tert-Butylamine, ethanamine, aniline, N-methylaniline, p-toluidine, m-chloroaniline, 2-phenylethanamine, o-anisidine, 2, 4, 6-tribromoaniline.
Answer:
Ans. 5
Five amines, i.e., aniline, p-toluidine, m-chloroaniline, o-anisidine and 2, 4, 6-tribromoaniline, being aromatic primary amines undergo diazotization.

Teacher's Note:
1. Here diazotization means forming a stable diazonium salt at 273–278 K: only 1° aromatic amines do this; 1° aliphatic amines give unstable salts that lose N₂.

Assertion–Reason Questions

Question 41. Statement-1: CuCl₂ gives a deep blue coloured solution with ethylamine.
Statement-2: Ethylamine molecules coordinate with cupric ions forming a blue coloured complex.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Ethylamine, like ammonia, acts as a Lewis base and coordinates with Cu²⁺ ions to form a deep blue complex, [Cu(C₂H₅NH₂)₄]²⁺. Hence Statement-2 explains Statement-1.

Teacher's Note:
1. Amines are ligands like NH₃: the lone pair on N is donated to the metal ion.

Question 42. Statement-1: m-Methoxyaniline (m-anisidine) is a weaker base than p-methoxyaniline (p-anisidine).
Statement-2: At m-position, methoxy group can exert only +R-effect but not –I-effect
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct statement-2. At m-position, methoxy group can exert only –I-effect but not +R-effect.

Teacher's Note:
1. Resonance effects reach only the ortho and para positions; at meta only the inductive effect acts.

Question 43. Statement-1: Aniline reacts with bromine water to form 2, 4, 6-tribromoaniline.
Statement-2: Aniline is resonance stabilized.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Correct explanation. The strong electron-donating effect of the NH₂ group activates p- and both the o-positions.

Teacher's Note:
1. A true reason must explain the observation: here it is the strong activation by –NH₂, not the resonance stability of aniline.

Question 44. Statement-1: Aniline on reaction with NaNO₂/HCl at 0°C followed by coupling with β-naphthol gives a dark blue coloured precipitate.
Statement-2: The colour of the compound formed in the reaction of aniline with NaNO₂/HCl at 0°C followed by coupling with β-naphthol is due to the extended conjugation. (IIT Paper II 2008)
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (d)
Correct statement-1. Aniline reacts with NaNO₂/HCl at 0°C followed by coupling with β-naphthol to give an orange red coloured precipitate.

Teacher's Note:
1. Remember the colour of the β-naphthol test for 1° aromatic amines: an orange-red dye.

Question 45. Statement-1: In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.
Statement-2: The group is migrated in Hofmann reaction to electron-deficient atom. (JEE Main 2022)
(a) Both Statement-1 and Statement-2 are correct.
(b) Both Statement-1 and Statement-2 are incorrect.
(c) Statement-1 is correct but Statement-2 is incorrect.
(d) Statement-1 is incorrect but Statement-2 is correct.
Answer:
Ans. (d)
Correct statement-1. Migration of alkyl/aryl group occurs from carbon to nitrogen.
Please note both alkyl and aryl groups migrate.

Teacher's Note:
1. Hofmann degradation works for aromatic amides too (benzamide gives aniline), so aryl groups also migrate.

Question 46. Statement-1: Nitration of aniline with a mixture of conc. HNO₃ and conc. H₂SO₄ gives a mixture of o-nitroaniline and p-nitroaniline.
Statement-2: o-Nitroaniline is the predominant product. (JEE Main 2021)
(a) Both Statement-1 and Statement-2 are correct.
(b) Both Statement-1 and Statement-2 are incorrect.
(c) Statement-1 is correct but Statement-2 is incorrect.
(d) Statement-1 is incorrect but Statement-2 is correct.
Answer:
Ans. (b)
Correct statement-1. Nitration of aniline into conc. HNO₃/conc. H₂SO₄ gives o, m- and p-nitroanilines.
Statement-2. p-Nitroaniline and m-nitroaniline are the predominant products.

Teacher's Note:
1. In the nitrating mixture most aniline is protonated to the meta-directing anilinium ion, so a lot of m-nitroaniline forms.