CBSE Class 12 Chemistry Amines Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Why do amines behave as nucleophiles?
Answer:
Due to the presence of a lone pair of electrons on nitrogen atom, amines behave as nucleophiles.

Teacher's Note:
1. The same lone pair makes amines Lewis bases; nucleophilicity and basicity both rise with electron density on N.

Question 2. What is the role of pyridine in the acylation reaction of amines? [NCERT Exemplar]
Answer:
Pyridine and other bases are used to remove the side product, i.e., HCl from the reaction mixture.

Teacher's Note:
1. Removing HCl shifts the equilibrium to the right and stops HCl from protonating the unreacted amine.

Question 3. What is the role of HNO₃ in the nitrating mixture used for nitration of benzene? [NCERT Exemplar]
Answer:
HNO₃ acts as a base in the nitrating mixture and provides the electrophile NO₂⁺.

Teacher's Note:
1. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻: the stronger acid H₂SO₄ protonates HNO₃.

Question 4. The conversion of primary aromatic amines into diazonium salts is known as __________ . [CBSE (AI) 2014]
Answer:
Diazotisation.

Teacher's Note:
1. Spell it as one word and remember the conditions: NaNO₂ + HCl at 273–278 K.

Question 5. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt. [CBSE Delhi 2008]
Answer:
C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride) + KI →(Δ) C₆H₅I (Iodobenzene) + KCl + N₂

Teacher's Note:
1. No copper salt is needed for iodine – simply warming with KI works, unlike the Sandmeyer reactions for Cl and Br.

Question 6. Why is benzene diazonium chloride not stored and is used immediately after its preparation? [NCERT Exemplar]
Answer:
Benzene diazonium chloride is very unstable.

Teacher's Note:
1. In the dry state it can even explode, so it is always prepared in solution in the cold and used at once.

Question 7. What are the reactions involved in the reductive removal of nitro group from an aromatic compound?
Answer:
C₆H₅—NO₂ (Nitrobenzene) →[Sn/HCl, Reduction] C₆H₅—NH₂ (Aniline) →[NaNO₂/HCl, 273–278 K (Diazotisation)] C₆H₅—N⁺≡NCl⁻ (Benzenediazonium chloride) →[Aq. H₃PO₂ (Reduction)] C₆H₆ (Benzene)

Teacher's Note:
1. Replacing –N₂⁺ by H with H₃PO₂ (or ethanol) is the standard trick to remove an amino or nitro group after it has done its directing job.

Teacher's Note:
1. Link each name to its key reagent: NH₃ (ammonolysis), phthalimide/KOH (Gabriel), Br₂/NaOH (Hoffmann), CHCl₃/KOH (carbylamine).

Question 9. Write IUPAC name of the following compound:
(CH₃)₂N–CH₂CH₃ [CBSE Delhi Set-3 2017]
Answer:
N,N-Dimethylethanamine.

Teacher's Note:
1. Both methyl groups are on nitrogen, so each needs its own locant N: N,N-dimethyl.

Question 10. Write IUPAC name of the following compound:
CH₃–CH₂–CH₂–CH₂–N(CH₃)₂ [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
N,N-Dimethylbutan-1-amine

Teacher's Note:
1. The four-carbon chain is the parent and the –N carbon is C-1, so the name ends in butan-1-amine.

Question 11. Write IUPAC name of the following compound:
A benzene ring carrying –OH at C-1 and –NO₂ groups at the two adjacent positions C-2 and C-3. [CBSE Comptt. Delhi Set-1, 2, 3 2017]
Answer:
2,3-Dinitrophenol

Teacher's Note:
1. When –OH is present it is the principal group, so the ring is named as phenol with the OH carbon as C-1.

Question 12. Write IUPAC name of the given compound:
CH₃–NH–CH₂–CH(CH₃)–CH₃ [CBSE OD Set-2 2017]
Answer:
N,2-Dimethylpropan-1-amine

Teacher's Note:
1. Identical prefixes are combined even if one is on N and one on carbon: N,2-dimethyl, not N-methyl-2-methyl.

Question 13. Give the IUPAC name of H₂N–CH₂–CH₂–CH=CH₂. [CBSE Comptt. Delhi 2013]
Answer:
But-3-en-1-amine.

Teacher's Note:
1. The amino group gets the lowest locant (C-1); the double bond then falls at C-3.

Short Answer Questions (2 Marks)

Question 14. (i) Arrange the following in decreasing order of solubility in water:
(CH₃)₃N, (CH₃)₂NH, CH₃NH₂
(ii) Arrange the following in decreasing order of the basic character:
C₆H₅NH₂, (CH₃)₃N, C₂H₅NH₂
Answer:
(i) CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
(ii) (CH₃)₃N > C₂H₅NH₂ > C₆H₅NH₂
C₆H₅NH₂ is least basic due to presence of −I group i.e., −C₆H₅ group that withdraws unshared pair of electrons present on N-atom of –NH₂ group due to which it does not undergo protonation easily; whereas (CH₃)₃N is most basic due to strong +I effect of –CH₃ groups which release their bonding electrons towards N-atom and it undergoes protonation easily. Hence, the decreasing order of the basic character is (CH₃)₃N > C₂H₅NH₂ > C₆H₅NH₂ (gas phase).
In aqueous solution, where solvation and steric effects also act, the order is C₂H₅NH₂ > (CH₃)₃N > C₆H₅NH₂ (pK_b: 3.29, 4.22, 9.38).

Teacher's Note:
1. Aniline is the weakest base in either phase; only the order of the two aliphatic amines depends on whether the question means gas phase or water.

Question 15. (i) Arrange the following in increasing order of pK_b values:
C₆H₅CH₂NH₂, C₆H₅NHCH₃, C₆H₅NH₂
(ii) Arrange the following in decreasing order of solubility in water:
(C₂H₅)₂NH, C₂H₅NH₂, C₆H₅NH₂
Answer:
(i) C₆H₅CH₂NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
Greater the basic nature of amine, lesser will be the pK_b value of amine. Since, C₆H₅CH₂NH₂ is most basic, so, it possesses least pK_b value, while, C₆H₅NH₂ is least basic, so, it possesses highest pK_b value. Hence, the increasing order of pK_b values is C₆H₅CH₂NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂.
(ii) C₂H₅NH₂ > (C₂H₅)₂NH > C₆H₅NH₂
(C₂H₅)₂NH and C₂H₅NH₂ are aliphatic amines and C₆H₅NH₂ is aromatic amine. Lower aliphatic amines can form hydrogen bonds with water molecules. Therefore, such amines are soluble in water. When number of alkyl groups (hydrophobic) increases then molar mass of amines increases. This usually results in a decrease in its solubility in water. Hence, C₂H₅NH₂ is more soluble in water than (C₂H₅)₂NH. Whereas, aromatic amines are insoluble in water because of large hydrocarbon part (hydrophobic) which retards the formation of hydrogen bonding.

Teacher's Note:
1. Benzylamine behaves like an aliphatic amine because its –NH₂ is on a CH₂, not on the ring, so its lone pair is not delocalised.

Question 16. (i) Write IUPAC name of the following compound: (CH₃CH₂)₂NCH₃
(ii) Write IUPAC name of the following compound: CH₃NHCH(CH₃)₂
Answer:
(i) N-Ethyl-N-methylethanamine.
(ii) N-Methylpropan-2-amine.

Teacher's Note:
1. Choose the largest alkyl group on nitrogen as the parent alkanamine and list the other groups alphabetically with the prefix N-.

Question 17. (i) Write IUPAC name of the following compound: (CH₃)₂N–CH₂CH₃
(ii) Complete the following reaction equation:
C₆H₅N₂Cl + H₃PO₂ + H₂O → ..................
Answer:
(i) N,N-Dimethylethanamine
(ii) ArN₂⁺Cl⁻ + H₃PO₂ + H₂O → ArH (Benzene) + N₂ + H₃PO₃ + HCl (where Ar is C₆H₅)

Teacher's Note:
1. H₃PO₂ (hypophosphorous acid) replaces the diazo group by H; remember all four products for full marks.

Question 18. Write the chemical equations involved in the following reactions:
(i) Hoffmann-bromamide degradation reaction,
(ii) Carbylamine reaction. [CBSE Outside Delhi 2016]
Answer:
(i) Ar/R–CONH₂ + Br₂ + 4NaOH → Ar/R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
When an amide is treated with bromine in aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. Example:
CH₃CH₂CONH₂ (Propanamide) + Br₂ + 4NaOH → CH₃CH₂NH₂ (Ethanamine) + Na₂CO₃ + 2NaBr + 2H₂O
(ii) Ar/R–NH₂ + CHCl₃ + 3KOH \( \xrightarrow{\Delta} \) Ar/R–NC + 3KCl + 3H₂O
(where R = alkyl group, Ar = aryl group)
It is used as a test for detection of primary amines. When aliphatic or aromatic primary amines are heated with chloroform and alcoholic potassium hydroxide, carbylamines or isocyanides having foul smell are formed. Secondary and tertiary amines do not respond to this test.

Teacher's Note:
1. Examiners deduct marks for incomplete equations: write all by-products (Na₂CO₃, NaBr, H₂O; KCl, H₂O) and mention alcoholic KOH.

Question 19. Give reasons:
(i) Electrophilic substitution in aromatic amines takes place more readily than benzene.
(ii) CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.
Answer:
(i) –NH₂ group of aromatic amines strongly activates the aromatic ring through delocalisation of the lone pair of electrons of the N-atom over the aromatic ring. Due to the strong activating effect of the –NH₂ group, aromatic amines undergo electrophilic substitution reactions readily than benzene.
(ii) Due to resonance, the lone pair of electrons on the nitrogen atom in CH₃CONH₂ is delocalised over the keto group:
CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺
As a result, electron density on the N-atom in CH₃CONH₂ decreases. On the other hand, in C₂H₅NH₂, due to +I effect of the ethyl group, the electron density on the N-atom increases consequently, CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.

Teacher's Note:
1. Amides are almost neutral: whenever the N lone pair is in resonance with C=O, it is not free to accept a proton.

Question 20. Give reasons:
(i) Identify ‘A’ and ‘B’:
Aniline \( \xrightarrow[\text{273–278 K}]{\text{NaNO}_{2}\text{/HCl}} \) A \( \xrightarrow{\text{H}_{2}\text{O/H}^{+}} \) B
(ii) Aniline does not undergo Friedel Crafts reaction.
Answer:
(i) C₆H₅NH₂ (Aniline) \( \xrightarrow[\text{273–278 K}]{\text{NaNO}_{2}\text{/HCl}} \) C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride, A) \( \xrightarrow{\text{H}_{2}\text{O/H}^{+}} \) C₆H₅OH (Phenol, B)
(ii) A Friedel–Crafts reaction is carried out in the presence of anhydrous AlCl₃. But AlCl₃ used as catalyst is acidic in nature i.e., Lewis acid whereas aniline is a strong Lewis base. Thus, aniline reacts with AlCl₃ to form a salt.
C₆H₅NH₂ + AlCl₃ → C₆H₅NH₂⁺AlCl₃⁻ (Salt)
Due to the positive charge on the N–atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo Friedel–Crafts reaction.

Teacher's Note:
1. Diazotisation needs 273–278 K because the diazonium salt decomposes at higher temperature; warming it with water is exactly how phenol is made.

Question 21. Why does acetylation of –NH₂ group of aniline reduce its activating effect?
Answer:
Direct nitration of aniline is not possible on account of –NH₂ group present. However, nitration can be carried out after protecting the –NH₂ group by acetylation to give acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines.
The acetyl group being electron withdrawing attracts the lone pair of electrons of the N-atom towards carbonyl group.
As a result, the activation effect of –NH₂ group is reduced, that is, that lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of –NHCOCH₃ group is less than that of –NH₂ group.

Teacher's Note:
1. Even after acetylation the group remains ortho/para directing; it is only a weaker activator, which is what allows monosubstitution.

Question 22. Give reasons:
(i) Aniline is a weaker base than cyclohexylamine.
(ii) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Answer:
(i) Cyclohexylamine is more basic than aniline because aniline is a resonance hybrid of various resonance structures. As a result, in aniline the electron donating capacity of nitrogen for protonation is considerably decreased.
(ii) Ammonolysis of alkyl halides does not give single amine but gives a mixture of primary, secondary and tertiary amines.
C₂H₅I \( \xrightarrow[\text{–HI}]{\text{NH}_{3}\text{/343 K}} \) C₂H₅NH₂ (1°) \( \xrightarrow[\text{–HI}]{\text{C}_{2}\text{H}_{5}\text{I}} \) (C₂H₅)₂NH (2°) \( \xrightarrow[\text{–HI}]{\text{C}_{2}\text{H}_{5}\text{I}} \) (C₂H₅)₃N (3°) \( \xrightarrow{\text{C}_{2}\text{H}_{5}\text{I}} \) [(C₂H₅)₄N⁺]I⁻ (Quaternary ammonium salt)

Teacher's Note:
1. Each amine formed is itself a nucleophile (and often a better one than NH₃), so it keeps reacting with the alkyl halide; a large excess of NH₃ only favours the 1° amine.

Teacher's Note:
1. For Hinsberg’s test always write the solubility in alkali; that observation, not just the product, is what separates 1° from 2° amines.

Question 24. Complete the following reactions:
(a) Cyclobutanecarbonitrile (C₄H₇–CN) \( \xrightarrow{\text{H}_{2}\text{/Ni}} \)
(b) C₆H₅CH₂–NH₂ + CHCl₃ \( \xrightarrow{\text{Ethanolic KOH}} \)
Answer:
(a) C₄H₇–CN \( \xrightarrow{\text{H}_{2}\text{/Ni}} \) C₄H₇–CH₂NH₂ (cyclobutylmethanamine)
(b) C₆H₅CH₂–NH₂ + CHCl₃ + 3KOH \( \xrightarrow{\text{Ethanolic KOH}} \) C₆H₅CH₂NC + 3KCl + 3H₂O

Teacher's Note:
1. Catalytic reduction of a nitrile adds one CH₂ between the ring and the –NH₂, so the amine has one carbon more than the original alkyl group.

Question 25. How do you convert the following:
(a) N-phenylethanamide to p-bromoaniline
(b) Benzoic acid to aniline [CBSE Outside Delhi 2019]
Answer:
(a) C₆H₅NHCOCH₃ \( \xrightarrow[\text{CH}_{3}\text{COOH}]{\text{Br}_{2}} \) p-BrC₆H₄NHCOCH₃ \( \xrightarrow{\text{OH}^{-}\text{ or H}^{+}} \) p-BrC₆H₄NH₂ (p-bromoaniline)
(b) C₆H₅COOH + NH₃ \( \xrightarrow[\text{–H}_{2}\text{O}]{\Delta} \) C₆H₅CONH₂ \( \xrightarrow{\text{Br}_{2}\text{ + KOH}} \) C₆H₅NH₂
(or any other suitable method)
Benzoic acid to aniline can also be done as: C₆H₅COOH (Benzoic acid) \( \xrightarrow{\text{SOCl}_{2}} \) C₆H₅COCl (Benzoyl chloride) \( \xrightarrow{\text{NH}_{3}} \) C₆H₅CONH₂ (Benzamide) \( \xrightarrow{\text{Br}_{2}\text{/NaOH}} \) C₆H₅NH₂ (Aniline)

Teacher's Note:
1. Benzoic acid to aniline loses one carbon, which is the signal to use Hoffmann bromamide degradation through benzamide.

Question 26. Identify A and B in each of the following processes:
(i) CH₃CH₂Cl →[NaCN] A →[Reduction, Ni/H₂] B
(ii) C₆H₅NH₂ →[NaNO₂/HCl] A →[C₆H₅NH₂, OH⁻] B [CBSE (AI) 2010]
Answer:
(i) CH₃CH₂Cl →[NaCN] CH₃CH₂CN (Ethyl cyanide, A) →[Reduction, Ni/H₂] CH₃CH₂CH₂NH₂ (n-Propyl amine, B)
(ii) C₆H₅NH₂ →[NaNO₂/HCl] C₆H₅N⁺≡NCl⁻ (Benzene diazonium chloride, A) →[C₆H₅NH₂, OH⁻] C₆H₅N=N—C₆H₄—NH₂(p) (p-aminoazobenzene, B)

Teacher's Note:
1. Count carbons in (i): the CN group adds one carbon, so a 2-carbon halide gives a 3-carbon amine.

Short Answer Questions (3 Marks)

Question 27. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer:
Formula of the compound ‘C’ indicates it to be an amine. Since it is obtained by the reaction of Br₂ and KOH with the compound ‘B’ so compound ‘B’ can be an amide. It is also indicated because ‘B’ is obtained from compound ‘A’ by reaction with ammonia following by heating. So compound ‘A’ could be an aromatic acid. Formula of compound ‘C’ shows that it is aniline, then ‘B’ is benzamide and compound ‘A’ is benzoic acid. The sequence of reactions can be written as follows:
C₆H₅COOH (A) NH₃
⟶
Δ C₆H₅CONH₂ (B) Br₂ / KOH
⟶ C₆H₅NH₂ (C)
A = benzoic acid (benzenecarboxylic acid), B = benzamide (benzenecarboxamide), C = aniline (benzenamine).

Teacher's Note:
1. C₆H₇N is C₆H₅NH₂; working backwards one carbon is added at each Hoffmann step you reverse.

Question 28. Account for the fact that although N, N-dimethyl aniline is only slightly more basic than aniline, 2, 6-dimethyl N, N-dimethyl aniline is much more basic than 2, 6-dimethyl aniline.
Answer:
Extended π bonding between the amino nitrogen and the ring requires the σ bonds on N atom to become coplanar with the ring and its ortho bonds.
Presence of bulky substituents in the ortho positions (2, 6-positions) sterically hinder the attainment of this geometry (coplanarity) and thus interferes with the base–weakening extended π bonding.
Because of this effect, (called steric inhibition of resonance) 2, 6-dimethyl N, N-dimethyl aniline is much more basic than 2, 6-dimethyl aniline.
In N,N-dimethylaniline (H atoms at the ortho positions) the –N(CH₃)₂ group can lie in the plane of the ring, giving the resonance form with C=N⁺(CH₃)₂; in 2,6-dimethyl-N,N-dimethylaniline the ortho CH₃ groups twist the –N(CH₃)₂ group out of the plane, so its lone pair stays on nitrogen.

Teacher's Note:
1. Steric inhibition of resonance: when a group is pushed out of the ring plane, its lone pair can no longer overlap with the ring π system.

Question 29. Give reasons:
(i) Acetylation of aniline reduces its activation effect.
(ii) CH₃NH₂ is more basic than C₆H₅NH₂.
(iii) Although –NH₂ is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.
Answer:
(i) After acetylation of aniline, acetanilide is formed in which due to the presence of –CO–CH₃ group having −I effect, electron density on N-atom decreases and hence, activation effect of aniline gets reduced.
(ii) In C₆H₅NH₂ the lone pair of electrons on nitrogen is delocalised over the benzene ring and is less available for protonation, while in CH₃NH₂ the +I effect of the CH₃ group increases the electron density on nitrogen. Hence CH₃NH₂ is more basic than C₆H₅NH₂.
(iii) Nitration is carried out with conc. HNO₃ in the presence of conc. H₂SO₄. In the presence of these acids, the –NH₂ group of aniline gets protonated and is converted into –NH₃⁺ group. This positively charged group acts as a strong electron withdrawing group and meta-directing group. Hence, the incoming electrophile goes to m-position.

Teacher's Note:
1. In acetanilide the nitrogen lone pair is also delocalised onto the C=O oxygen, so less of it is available to the ring.

Question 30. Give reasons:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) Aromatic primary amines cannot be prepared by Gabriel’s phthalimide synthesis.
(iii) Aliphatic amines are stronger base than ammonia. [CBSE Delhi 2020]
Answer:
(i) Aniline does not undergo Friedel-Crafts reaction because aniline being a Lewis base forms a complex with AlCl₃ which is a Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution. Therefore the reaction is not possible.
C₆H₅NH₂ + AlCl₃ → C₆H₅NH₂⁺AlCl₃⁻
(ii) Aromatic primary amines can not be prepared by Gabriel’s phthalimide synthesis because haloarenes have to react with potassium phthalimide and they are least reactive. So, the bond cleavage does not take place.
(iii) Aliphatic amines are stronger base than ammonia because the alkyl group in aliphatic amines has +I effect. So, the alkyl group tends to increase the electron density on the nitrogen atom whereas the electron releasing tendency of amines becomes more than that of ammonia.

Teacher's Note:
1. For (ii) mention the partial double bond character of the aryl C–X bond: that is why haloarenes resist the S_N2 attack by the phthalimide anion.

Question 31. Arrange the following compounds as directed:
(i) In increasing order of solubility in water: (CH₃)₂NH, CH₃NH₂, C₆H₅NH₂
(ii) In decreasing order of basic strength in aqueous solution: (CH₃)₃N, (CH₃)₂NH, CH₃NH₂
(iii) In increasing order of boiling point: (C₂H₅)₂NH, (C₂H₅)₃N, C₂H₅NH₂ [CBSE Outside Delhi 2020]
Answer:
(i) Increasing order of solubility in water: C₆H₅NH₂ < (CH₃)₂NH < CH₃NH₂
CH₃NH₂ and (CH₃)₂NH are aliphatic amines. Aliphatic amines are soluble in water because they can form hydrogen bonds with water. When number of alkyl group (hydrophobic) increases, molar mass also increases which tends to decrease in solubility in water. So, CH₃NH₂ is more soluble in water than (CH₃)₂NH. Whereas, C₆H₅NH₂ is an aromatic amine which is insoluble in water due to more hydrophobic part which renders the formation of H-bond.
(ii) Decreasing order of basic strength in aqueous solution: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
Greater the +I effect, greater will be the basic nature of amine in aqueous solution. However, tertiary amine also exhibits steric hindrance due to the presence of three bulky (–CH₃) groups and its cation is least stabilised by hydration, due to which its basic strength is lower than secondary and primary amine.
(iii) Increasing order of boiling point: (C₂H₅)₃N < (C₂H₅)₂NH < C₂H₅NH₂
Greater the extent of hydrogen bonding, greater will be the boiling point of amine. Primary amines have highest boiling point due to the presence of two hydrogen atoms on N-atoms as more H-bonding takes place in primary amines.

Teacher's Note:
1. For methyl amines in water the order is 2° > 1° > 3°; for ethyl amines it is 2° > 3° > 1°. Learn both series.

Question 32. An aromatic compound ‘A’ on heating with Br₂ and KOH forms a compound ‘B’ of molecular formula C₆H₇N which on reacting with CHCl₃ and alcoholic KOH produces a foul smelling compound ‘C’. Write the structure and IUPAC names of compound A, B and C. [CBSE Delhi 2019]
Answer:
A = C₆H₅CONH₂, Benzamide; B = C₆H₅NH₂, Aniline; C = C₆H₅NC, Phenylisocyanide / Benzene isonitrile (isocyanobenzene)
C₆H₅CONH₂ (Compound A, benzamide) \( \xrightarrow[\text{in KOH}]{\text{Br}_{2}} \) C₆H₅NH₂ (Compound B, C₆H₇N, aniline) \( \xrightarrow[\text{alcoholic KOH}]{\text{CHCl}_{3}} \) C₆H₅N⁺≡C⁻ (Compound C, isocyanobenzene)

Teacher's Note:
1. Work backwards from C₆H₇N: it is aniline, and Br₂/KOH (Hoffmann) means A is the amide with one carbon more.

Question 33. Write the structures of compounds A, B and C in the following reactions:
(a) CH₃–COOH \( \xrightarrow{\text{NH}_{3}\text{/}\Delta} \) A \( \xrightarrow{\text{Br}_{2}\text{/KOH (aq)}} \) B \( \xrightarrow{\text{CHCl}_{3}\text{ + alc. KOH}} \) C
(b) C₆H₅N₂⁺BF₄⁻ \( \xrightarrow[\Delta]{\text{NaNO}_{2}\text{/Cu}} \) A \( \xrightarrow{\text{Fe/HCl}} \) B \( \xrightarrow{\text{CH}_{3}\text{COCl/pyridine}} \) C [CBSE Outside Delhi 2017]
Answer:
(a) A = CH₃CONH₂; B = CH₃NH₂; C = CH₃NC
CH₃COOH \( \xrightarrow{\text{NH}_{3}\text{/}\Delta} \) CH₃CONH₂ (Acetamide, A) \( \xrightarrow{\text{Br}_{2}\text{/KOH(aq)}} \) CH₃NH₂ (Methylamine, B) \( \xrightarrow{\text{CHCl}_{3}\text{ + alc. KOH}} \) CH₃–NC (Methyl isocyanide, C)
(b) A = C₆H₅NO₂; B = C₆H₅NH₂; C = C₆H₅NHCOCH₃
C₆H₅N₂⁺BF₄⁻ \( \xrightarrow[\Delta]{\text{NaNO}_{2}\text{/Cu}} \) C₆H₅NO₂ (Nitrobenzene, A) \( \xrightarrow{\text{Fe/HCl}} \) C₆H₅NH₂ (Aniline, B) \( \xrightarrow{\text{CH}_{3}\text{COCl/Pyridine}} \) C₆H₅NHCOCH₃ (N-phenylethanamide, C)

Teacher's Note:
1. Fluoroborate diazonium salt with NaNO₂/Cu is the standard way to introduce –NO₂ through the diazonium route.

Question 34. Write the structures of A, B and C in the following:
(a) C₆H₅–CONH₂ \( \xrightarrow{\text{Br}_{2}\text{/aq KOH}} \) A \( \xrightarrow[\text{0–5°C}]{\text{NaNO}_{2}\text{ + HCl}} \) B \( \xrightarrow{\text{KI}} \) C.
(b) CH₃–Cl \( \xrightarrow{\text{KCN}} \) A \( \xrightarrow{\text{LiAlH}_{4}} \) B \( \xrightarrow{\text{CHCl}_{3}\text{ + alc. KOH}} \) C. [CBSE Outside Delhi 2016]
Answer:
(a) A = C₆H₅NH₂, B = C₆H₅N₂⁺Cl⁻, C = C₆H₅I
C₆H₅–CONH₂ \( \xrightarrow{\text{Br}_{2}\text{/KOH}} \) C₆H₅NH₂ [A] \( \xrightarrow{\text{NaNO}_{2}\text{ + HCl}} \) C₆H₅N₂⁺Cl⁻ [B] \( \xrightarrow{\text{KI}} \) C₆H₅I [C]
(b) A = CH₃CN, B = CH₃CH₂NH₂, C = CH₃CH₂NC
CH₃Cl \( \xrightarrow{\text{KCN}} \) CH₃CN [A] \( \xrightarrow{\text{LiAlH}_{4}} \) CH₃CH₂NH₂ [B] \( \xrightarrow[\Delta]{\text{CHCl}_{3}\text{ + KOH}} \) CH₃CH₂NC [C]

Teacher's Note:
1. Track carbon count: Hoffmann (a) removes one carbon, while the KCN route (b) adds one, so B in (b) is ethylamine, not methylamine.

Question 35. Give reasons for the following:
(i) Aniline does not undergo Friedel-Crafts reaction,
(ii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
(iii) Primary amines have higher boiling point than tertiary amines. [CBSE Outside Delhi 2016]
Answer:
(i) Aniline is a Lewis base while AlCl₃ is Lewis acid. They combine to form a salt.
C₆H₅NH₂ + AlCl₃ → C₆H₅NH₂⁺AlCl₃⁻ (Salt). Due to the positive charge on the N–atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo Friedel–Crafts reaction.
(ii) Due to combined +I and solvation effects.
(CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. +I effect in alkyl group results in increasing the ease of donation of lone pair electron. Amine accepts a proton and form cation which will be stabilised in water by solvation. Higher the solvation by hydrogen bonding, higher will be the basic strength. Therefore, with increase in methyl group, hydrogen bonding and stabilisation by solvation decreases. This net effect results in decrease of basic strength from secondary to tertiary amine.
(iii) Due to presence of H-bonding in primary amines.
In tertiary amines, there are no H-atoms whereas in primary amines, two H-atoms are present on nitrogen. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding (···H–N(R)–H···H–N(R)–H···). As a result, extra energy is required to separate the molecules of primary amine. Therefore, primary amines have higher boiling point than tertiary amines.

Teacher's Note:
1. For (ii) compare the cations: R₂NH₂⁺ has two N–H bonds to H-bond with water, R₃NH⁺ only one.

Question 36. Arrange the following in increasing order of property specified:
(i) Aniline, ethanamine, N-ethylethanamine (solubility in water)
(ii) Ethanoic acid, ethanamine, ethanol (boiling point)
(iii) Methanamine, N,N-dimethylmethanamine and N-methylmethanamine (basic strength in aqueous phase)
Answer:
(i) Aniline < N-ethylethanamine < Ethanamine
(ii) Ethanamine < ethanol < ethanoic acid
(iii) N,N-dimethylmethanamine < methanamine < N-methylmethanamine

Teacher's Note:
1. Boiling point follows H-bond strength: O–H bonds beat N–H bonds, and carboxylic acids form dimers, so they boil highest.

Question 37. (i) Give a chemical test to distinguish between N-methylethanamine and N,N-dimethylethanamine.
(ii) Write the reaction for catalytic reduction of nitrobenzene followed by reaction of product so formed with bromine water.
(iii) Out of butan-1-ol and butan-1-amine, which will be more soluble in water and why? [CBSE SQP 2020-21]
Answer:
(i) N-methylethanamine is a secondary amine. When it reacts with benzene sulphonyl chloride, it forms N-ethyl-N-methylbenzenesulphonamide (insoluble in alkali) and N,N-dimethylethanamine is a tertiary amine; it does not react with benzenesulphonyl chloride.
(ii) C₆H₅NO₂ \( \xrightarrow{\text{H}_{2}\text{/Ni}} \) C₆H₅NH₂ \( \xrightarrow{\text{Br}_{2}\text{/H}_{2}\text{O}} \) 2,4,6-tribromoaniline (white ppt.)
(iii) Butan-1-ol.
Alcohol forms stronger hydrogen bonds with water than formed by amine due to higher electronegativity of O in alcohol than N in amine.

Teacher's Note:
1. Hinsberg’s test separates 2° from 3° amines; the carbylamine test cannot, since neither is a primary amine.

Question 38. How will you convert the following:
(i) Nitrobenzene into aniline,
(ii) Ethanoic acid into methanamine,
(iii) Aniline to N-phenylethanamide.
(Write the chemical equations involved.)
Answer:
(i) C₆H₅NO₂ \( \xrightarrow{\text{Sn/HCl}} \) C₆H₅NH₂
(ii) CH₃COOH \( \xrightarrow[\Delta]{\text{NH}_{3}} \) CH₃CONH₂ \( \xrightarrow{\text{Br}_{2}\text{ + NaOH}} \) CH₃NH₂
(iii) C₆H₅NH₂ \( \xrightarrow{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) C₆H₅NHCOCH₃ (N-Phenylethanamide)

Teacher's Note:
1. Ethanoic acid (2 C) to methanamine (1 C) is a step-down; it always goes through the amide and Hoffmann bromamide degradation.

Long Answer Questions (5 Marks)

Question 39. Write the structure of the reagents/organic compounds A to F in the following sequence of reactions:
A HNO₃ (Conc.)
⟶
H₂SO₄ (Conc.) B Sn / HCl
⟶
Heat C CHCl₃ / KOH
⟶
Heat D H₂ / Pt
⟶
H₂SO₄ (Conc.) E F
⟶ C₆H₅N(CH₃)COCH₃ + CH₃CO₂H
Answer:
Structure of reagents/organic compounds:
A = Benzene, C₆H₆
B = Nitrobenzene, C₆H₅–NO₂
C = Aniline, C₆H₅–NH₂
D = Phenyl isocyanide, C₆H₅–N≡C (C₆H₅–NC)
E = Methyl phenyl amine, C₆H₅–NH–CH₃
F = Acetic anhydride, (CH₃CO)₂O

Teacher's Note:
1. The by-product CH₃CO₂H tells you the acylating agent F is acetic anhydride, not acetyl chloride (which would give HCl).

Question 40. Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(i) In alcohols, the hydrogen atom is attached to more electronegative oxygen atom whereas nitrogen of amines is less electronegative. After the loss of H⁺ ion, the negative charge is more easily accommodated on oxygen than in case of nitrogen in amines. Hence, amines have lesser tendency to lose H⁺ ions, so they are less acidic than alcohols.
R – NH₂ → R – NH⁻ + H⁺
R – O – H → R – O⁻ + H⁺
(ii) Primary amines (R – NH₂) have two hydrogen atoms on nitrogen which can undergo intermolecular hydrogen bonding whereas no such hydrogen bonding is present in tertiary amines (R3N). So primary amines boil at a higher temperature than tertiary amines.
(iii) In aromatic amines, the lone-pair of electrons on nitrogen atom is involved in resonance with the benzene ring as shown below for aniline.
C₆H₅–NH₂ ↔ three structures with C=NH₂⁺ and a negative charge at the ortho, para and other ortho carbon ↔ Kekulé structure
It shows that this pair of electrons is less available for protonation. In case of aliphatic amines electron releasing alkyl groups increase electron density on nitrogen atom. So, aliphatic amines are stronger bases than aromatic amines.

Teacher's Note:
1. Aniline has five resonance structures; the three charge-separated ones place negative charge only at ortho and para carbons.

Question 41. (a) Write the structures of the main products of the following reactions:
(i) Aniline \( \xrightarrow[\text{Pyridine}]{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \)
(ii) C₆H₅SO₂Cl \( \xrightarrow{\text{(CH}_{3}\text{)}_{2}\text{NH}} \)
(iii) Benzonitrile, C₆H₅CN \( \xrightarrow{\text{LiAlH}_{4}} \)
(b) Give a simple chemical test to distinguish between aniline and N,N-dimethylaniline.
(c) Arrange the following in the increasing order of their pK_b values:
C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃
Answer:
(a) (i) C₆H₅NHCOCH₃
C₆H₅NH₂ \( \xrightarrow[\text{Pyridine}]{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) C₆H₅NHCOCH₃ (Acetanilide) + CH₃COOH
(ii) C₆H₅SO₂N(CH₃)₂
C₆H₅SO₂Cl \( \xrightarrow{\text{(CH}_{3}\text{)}_{2}\text{NH}} \) C₆H₅SO₂N(CH₃)₂ (N,N-dimethylbenzenesulphonamide) + HCl
(iii) C₆H₅CH₂NH₂
C₆H₅C≡N \( \xrightarrow{\text{LiAlH}_{4}} \) C₆H₅CH₂NH₂ (Benzylamine)
(b) Add chloroform in the presence of KOH and heat, then, aniline gives a offensive smell while N,N-dimethylaniline does not. (or any other correct test)
Carbylamine test: C₆H₅NH₂ (Aniline) + CHCl₃ + 3KOH \( \xrightarrow{\Delta} \) C₆H₅NC (Phenyl isocyanide) + 3KCl + 3H₂O
In this reaction, aniline gives a offensive smell due to formation of phenyl isocyanide.
C₆H₅N(CH₃)₂ + CHCl₃ + 3KOH \( \xrightarrow{\Delta} \) No reaction
N,N-dimethylaniline does not give carbylamine test as no hydrogen is present on nitrogen atom.
(c) C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
Greater the basic strength of amine, lesser will be the pK_b value. Alkyl amines are more basic than arylamines due to presence of +I effect of alkyl group.

Teacher's Note:
1. pK_b runs opposite to basic strength, so an “increasing pK_b” list starts with the strongest base.

Question 42. Write the structure of A, B, C, D and E in the following reactions:
C₆H₅NO₂ \( \xrightarrow{\text{Sn/HCl}} \) A \( \xrightarrow[\text{Pyridine}]{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) B \( \xrightarrow[\text{288 K}]{\text{HNO}_{3}\text{ + H}_{2}\text{SO}_{4}} \) C \( \xrightarrow{\text{OH}^{-}\text{ or H}^{+}} \) D
A \( \xrightarrow{\text{H}_{2}\text{SO}_{4}} \) E
Answer:
A = C₆H₅NH₂ (Aniline)
B = C₆H₅NHCOCH₃ (Acetanilide)
C = p-O₂N–C₆H₄–NHCOCH₃ (p-Nitroacetanilide)
D = p-O₂N–C₆H₄–NH₂ (p-Nitroaniline)
E = p-H₂N–C₆H₄–SO₃H (Sulphanilic acid)

Teacher's Note:
1. Acetylation before nitration protects –NH₂ from oxidation and protonation, so the nitro group enters mainly para; hydrolysis then gives back the free amine.