CBSE Class 12 Chemistry Amines Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Direct nitration of aniline is not carried out. Explain.
Answer:
(H₂SO₄ + HNO₃) easily oxidise aniline into tarry complex products due to high e⁻ density on the benzene ring of aniline.

Teacher's Note:
1. Also, in acid aniline is protonated and gives a lot of meta product; acetylating it first avoids both problems.

Question 2. Among the following compounds, which will react with CH₃–CO–CH₃ to give a product containing >C=N–?
(i) C₆H₅NH₂   (ii) (CH₃)₃N   (iii) C₆H₅NHC₆H₅   (iv) C₆H₅NHNH₂
Answer:
C₆H₅NH₂ and C₆H₅NHNH₂.

Teacher's Note:
1. Only compounds with an –NH₂ group (two H on N) can lose water with C=O to form a C=N bond.

Question 3. How will you give an expression for K_b to indicate the basic strength of an amine?
Answer:
RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻
\( K_b = \frac{[\mathrm{RNH_3^{+}}][\mathrm{OH^{-}}]}{[\mathrm{RNH_2}]} \)

Teacher's Note:
1. Larger K_b (smaller pK_b) means a stronger base.

Question 4. What happens when aniline is treated with bromine water?
Answer:
C₆H₅NH₂ + 3Br₂ → 2,4,6-tribromoaniline + 3HBr
A white precipitate of 2,4,6-tribromoaniline is formed.

Teacher's Note:
1. The –NH₂ group activates all three ortho and para positions, so monobromination is not possible without first acetylating aniline.

Question 5. Write a chemical equation to illustrate ammonolysis.
Answer:
For alcohols:
C₂H₅OH NH₃
⟶
Al₂O₃/620 K C₂H₅NH₂ C₂H₅OH
⟶
−H₂O (C₂H₅)₂NH C₂H₅OH
⟶ (C₂H₅)₃N
For alkyl halides:
C₂H₅I NH₃/373 K
⟶
−HI C₂H₅NH₂ C₂H₅I
⟶
−HI (C₂H₅)₂NH C₂H₅I
⟶
−HI (C₂H₅)₃N C₂H₅I
⟶ [(C₂H₅)₄N]⁺I⁻

Teacher's Note:
1. Ammonolysis gives a mixture of 1°, 2°, 3° amines and the quaternary salt; a large excess of NH₃ favours the primary amine.

Question 6. Write the structure of p-toluidine.

Answer:
p-Toluidine is 4-methylaniline, p-CH₃C₆H₄NH₂:

Teacher's Note:
1. ‘Toluidine’ = methyl + aniline; p means the two groups are at positions 1 and 4.

Question 7. Prepare/convert nitrobenzene into aniline.
Answer:
C₆H₅NO₂ H₂/Ni
⟶
Reduction C₆H₅NH₂

Teacher's Note:
1. Sn/HCl or Fe/HCl can also be used; Fe/HCl is preferred industrially because the FeCl₂ formed is hydrolysed, releasing HCl.

Question 8. Convert C₆H₅COOH to C₆H₅NH₂.
Answer:
C₆H₅COOH NH₃(aq)
⟶
Δ C₆H₅CONH₂ Br₂/KOH
⟶
Δ C₆H₅NH₂

Teacher's Note:
1. Hoffmann bromamide degradation removes the C of the –CONH₂ group, so the amine has one carbon atom less.

Question 9. Write the isomerism exhibited by different amines.
Answer:
Chain, position, metamerism, functional.

Teacher's Note:
1. Example of functional isomers: CH₃CH₂NH₂ (1°) and (CH₃)₂NH (2°) have the same formula C₂H₇N.

Question 10. Predict the product of the following reaction : CH₃NO₂ + C₆H₅CHO \( \xrightarrow{\mathrm{KOH\ (alc.)}} \)
Answer:
Due to the strong electron-withdrawing effect of the NO₂ group, the H-atom of CH₃NO₂ are weakly acidic. Therefore, in presence of KOH (alc.), it forms a carbanion (I) which adds to >C=O group of C₆H₅CHO to form aldol (II) which subsequently undergoes dehydration to form β-nitrostyrene.
CH₃NO₂ (Nitromethane) \( \xrightarrow[\mathrm{-H_2O}]{\mathrm{OH^-}} \) ⁻:CH₂–NO₂ (Carbanion, I) \( \xrightarrow[\mathrm{(Nucleophilic\ addition)}]{\mathrm{C_6H_5CH=O}} \) C₆H₅–CH(OH)–CH₂NO₂ (II) \( \xrightarrow[\mathrm{(Dehydration)}]{\mathrm{-H_2O}} \) C₆H₅CH=CHNO₂
1-Nitro-2-phenylethene (β-Nitrostyrene)

Teacher's Note:
1. This is an aldol-type (Henry) condensation: the α-H next to NO₂ behaves like the α-H next to C=O.

Question 11. What is Baker-Mulliken's test?
Answer:
Both aliphatic and aromatic nitro compounds on reduction in neutral medium, i.e., Zn dust and NH₄Cl solution, give the corresponding hydroxylamines.
C₆H₅NO₂ (Nitrobenzene) \( \xrightarrow[\mathrm{\Delta}]{\mathrm{Zn/NH_4Cl,\ H_2O}} \) C₆H₅NHOH (Phenylhydroxylamine)
These hydroxylamines when warmed with Tollens' reagent are easily oxidised to the corresponding nitroso compounds and thus reduce Tollens' reagent to metallic silver
C₆H₅NHOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → C₆H₅N=O (Nitrosobenzene) + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O
This reaction is called Baker Mulliken's test and is used as a test for detection of nitro group in organic compounds.

Teacher's Note:
1. Remember the medium: neutral reduction (Zn/NH₄Cl) stops at the hydroxylamine, while acidic reduction (Sn/HCl) goes all the way to the amine.

Question 12. What are amines? (J & K Board 2012)
Answer:
The mono-, di or tri-alkyl, aryl or aralkyl derivatives of ammonia are called amines. For example,
CH₃–NH₂ (Methylamine), C₆H₅NHC₆H₅ (Diphenylamine), (C₆H₅CH₂)₃N (Tribenzylamine)

Teacher's Note:
1. Think of an amine as NH₃ in which one, two or all three H-atoms have been replaced by carbon groups.

Question 13. What are primary amines? Give one example. (Pb. Board 2012)
Answer:
1° Amines are formed when one H-atom of ammonia is replaced by an alkyl, aryl or aralkyl group. For example,
CH₃CH₂NH₂ (Ethylamine), C₆H₅NH₂ (Aniline), C₆H₅CH₂NH₂ (Benzylamine)

Teacher's Note:
1. For amines, 1°, 2°, 3° refers to the number of carbon groups on nitrogen, not to the type of carbon carrying the N (unlike alcohols).

Short Answer Questions (2 Marks)

Question 14. Draw the structures of the following compounds:
(i) N-isopropylaniline   (ii) t-butylamine
Answer:
(i) C₆H₅–NH–CH(CH₃)₂
(ii) (CH₃)₃C–NH₂, i.e. CH₃–C(CH₃)₂–NH₂

Teacher's Note:
1. ‘N-isopropyl’ means the isopropyl group is on nitrogen, not on the ring.

Question 15. Why is C₆H₅N⁺(CH₃)₃OH⁻ a stronger base than NH₄OH?
Answer:
C₆H₅N⁺(CH₃)₃OH⁻ is a quaternary ammonium hydroxide. It is completely ionised in water, giving a high concentration of OH⁻ ions, like NaOH or KOH.
NH₄OH is only feebly ionised (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻), so it gives few OH⁻ ions and is a weak base.

Teacher's Note:
1. Quaternary ammonium hydroxides are strong bases because the cation has no N–H proton to give back to OH⁻.

Question 16. Explain the K_b order: Et₂NH > Et₃N > EtNH₂ in aqueous solution.
Answer:
Basicity of amines in aqueous solution depends upon:
(i) +I effect of the alkyl groups.
(ii) Extent of hydrogen bonding with H₂O (solvation of the ammonium ion).
(iii) Steric effects of the alkyl groups.
In Et₂NH the combined effect of +I and solvation is the greatest. Et₃N has the largest +I effect but its ion is poorly solvated and more crowded, while EtNH₂ has the smallest +I effect. Hence Et₂NH > Et₃N > EtNH₂.

Teacher's Note:
1. For ethylamines in water the order is 2° > 3° > 1°, but for methylamines it is 2° > 1° > 3° – learn both.

Question 17. Distinguish between 1°, 2° and 3° amines by the HNO₂ test.
Answer:
1° amines give N₂ gas. 2° amines give a yellow oily compound. 3° amines form water-soluble salts.
CH₃CH₂NH₂ + HNO₂ → C₂H₅OH + N₂ + H₂O
R₂NH + HNO₂ → R₂N–N=O + H₂O
(CH₃CH₂)₃N + HNO₂ → [(CH₃CH₂)₃NH]⁺NO₂⁻ (soluble in water)

Teacher's Note:
1. The yellow oil from a 2° amine is an N-nitrosamine; the 1° aliphatic amine is recognised by brisk effervescence of N₂.

Question 18. A compound ‘A’ having molecular formula C₃H₇ON reacts with Br₂ in presence of NaOH to give compound ‘B’. This compound ‘B’ reacts with HNO₂ to form alcohol and N₂ gas. Identify compounds ‘A’ and ‘B’ and write the reactions involved.
Answer:
‘A’ is CH₃CH₂CONH₂ (propanamide) and ‘B’ is CH₃CH₂NH₂ (ethylamine).
CH₃CH₂CONH₂ Br₂
⟶
NaOH CH₃CH₂NH₂ (B) HNO₂
⟶ C₂H₅OH + N₂ + H₂O

Teacher's Note:
1. Br₂/NaOH on a compound with formula C_nH_2n+1ON points straight to an amide undergoing Hoffmann degradation.

Question 19. Write chemical equations for the following conversions:
(i) CH₃–CH₂–Cl → CH₃–CH₂–CH₂–NH₂
(ii) C₆H₅–CH₂–Cl → C₆H₅CH₂CH₂–NH₂
Answer:
(i) With NaCN and reduction, i.e., C₂H₅Cl NaCN
⟶
Ethanol C₂H₅–C≡N reduction
⟶ CH₃–CH₂–CH₂–NH₂
(ii) C₆H₅CH₂Cl NaCN
⟶
Ethanol C₆H₅CH₂–C≡N H₂/Ni
⟶ C₆H₅CH₂CH₂–NH₂

Teacher's Note:
1. Going through the nitrile adds one carbon atom – the opposite of Hoffmann degradation.

Question 20. Account for:
(i) Amino group in aniline is o- and p-directing in aromatic electrophilic substitution reactions, yet aniline on nitration gives a substantial amount of m-nitroaniline.
(ii) Aniline does not undergo Friedel–Crafts reaction.
Answer:
(i) It is because aniline is protonated to form the anilinium cation, in which the –NH₃⁺ group is meta-directing.
(ii) It is because aniline is basic and forms an adduct (salt) with AlCl₃, so the electrophile cannot be generated and the ring is deactivated.

Teacher's Note:
1. In nitration of aniline, about 47% of the product is the meta isomer because of the anilinium ion.

Question 21. Account for the following:
(i) Electrophilic substitution in aromatic amines takes place more readily than in benzene.
(ii) Nitro compounds have higher boiling points than hydrocarbons having almost the same molecular mass.
Answer:
(i) –NH₂ is an electron-releasing group (ERG); it increases the electron density on the ring, so electrophilic substitution takes place faster.
(ii) Nitro compounds are more polar than hydrocarbons, therefore they have stronger van der Waals forces (dipole–dipole attraction).

Teacher's Note:
1. The –NH₂ group releases electrons by resonance (+R), which raises electron density mainly at the ortho and para positions.

Question 22. Write short notes on:
(i) Coupling reaction
(ii) Ammonolysis
Answer:
(i) Coupling reaction: C₆H₅N₂⁺Cl⁻ + H–C₆H₄–NH₂ mildly acidic
⟶
(pH 4–5) C₆H₅–N=N–C₆H₄–NH₂ (p-aminoazobenzene) + HCl
(ii) Ammonolysis: R–X + NH₃ → R–NH₂ + HX
RNH₂ + R–X → R₂NH + HX
R₂NH + R–X → R₃N + HX
R₃N + R–X → [R₄N]⁺X⁻

Teacher's Note:
1. Coupling with amines is done in mildly acidic medium and with phenols in mildly alkaline medium; the –N=N– group is retained.

Question 23. Prepare a pure sample of a 1° amine from a 1° alkyl halide.
Answer:
By Gabriel phthalimide reaction:
Phthalimide KOH
⟶ potassium phthalimide (C₆H₄(CO)₂N⁻K⁺) C₂H₅Cl
⟶ N-ethylphthalimide (C₆H₄(CO)₂N–C₂H₅)
N-ethylphthalimide H₂O/H⁺
⟶ C₂H₅NH₂ + phthalic acid (C₆H₄(COOH)₂)

Teacher's Note:
1. Gabriel synthesis gives only the primary amine, with no 2°/3° amine contamination as in ammonolysis.

Question 24. What happens when an alkyl halide reacts with AgNO₂ and the product is reduced?
Answer:
R–X + AgNO₂  
⟶
−AgX R–NO₂ Sn/HCl
⟶
[H] R–NH₂
A nitroalkane is formed, which on reduction gives a primary amine.

Teacher's Note:
1. AgNO₂ gives mainly the nitroalkane (attack through N), while KNO₂ gives mainly the alkyl nitrite.

Question 25. Write the structures of the products in each case:
(i) CH₃CH₂NH₂ (CH₃CO)₂O
⟶
Δ
(ii) CH₃CONHC₆H₅ Br₂/Fe
⟶
Answer:
(i) CH₃CH₂NHCOCH₃, CH₃COOH
(ii) o-BrC₆H₄NHCOCH₃ and p-BrC₆H₄NHCOCH₃ (2-bromoacetanilide and 4-bromoacetanilide; the para isomer is the major product)

Teacher's Note:
1. The –NHCOCH₃ group is still ortho/para directing but less activating than –NH₂, so only monobromination occurs.

Question 26. Toluene is more easily nitrated than benzene. Explain.
Answer:
Nitration is an electrophilic substitution reaction and thus occurs more readily at a site where the electron-density is more. Now in toluene, there is electron-donating CH₃ group. In other words, the electron-density in the toluene ring is much more than in the benzene ring. As a result, toluene is nitrated more easily than benzene.
Further since hyperconjugative effect increases the electron-density at o- and p-positions, relative to m-positions, therefore, nitration of toluene gives a mixture of o- and p-nitrotoluenes.

Teacher's Note:
1. Electron-donating groups activate the ring and direct o/p; mention both points (rate and orientation) for full marks.

Question 27. How is classification of amines different from that of alcohols ?
Answer:
Classification of amines is made on the basis of the nature of nitrogen atom (whether it has one, two or no hydrogen atom) to which the alkyl groups are attached. The classification of alcohols, on the other hand, is made on the basis of the nature of carbon atom to which the –OH group is attached.
–NH₂ (1° Amine), >NH (2° Amine), >N– (3° Amine); –CH₂OH (1° Alcohol), >CHOH (2° Alcohol), >C–OH (3° Alcohol)

Teacher's Note:
1. Example: (CH₃)₃C–NH₂ is a 1° amine although its carbon is tertiary, while (CH₃)₃C–OH is a 3° alcohol.

Short Answer Questions (3 Marks)

Question 28. Identify A, B and C in the following reactions:
(i) CH₃CH₂Cl → [KCN] → A → [H₂/Ni] → B → [CH₃COCl/Base] → C
(ii) C₆H₅N₂⁺Cl⁻ → [HBF₄] → A → [NaNO₂/Cu, Δ] → B → [Sn/HCl] → C [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) A: CH₃–CH₂CN; B: CH₃–CH₂–CH₂–NH₂; C: CH₃–CH₂–CH₂–NH–COCH₃
(ii) A: Ar–N₂⁺BF₄⁻ (C₆H₅N₂⁺BF₄⁻); B: C₆H₅NO₂ (nitrobenzene); C: C₆H₅NH₂ (aniline)

Teacher's Note:
1. Heating the diazonium fluoroborate with NaNO₂/Cu replaces the diazo group by –NO₂, a route to nitro compounds that avoids direct nitration.

Question 29. Write the structures of A, B, C in the following reactions:
(i) C₆H₅–COO⁻NH₄⁺ → [Δ] → A → [Br₂/KOH] → B → [CH₃COCl/pyridine] → C
(ii) C₆H₅N₂⁺BF₄⁻ → [NaNO₂/Cu, Δ] → A → [Sn/HCl] → B → [CHCl₃ + alc. KOH] → C [CBSE OD Set-2 2016]
Answer:
(i) A ⇒ C₆H₅–CO–NH₂; B ⇒ C₆H₅–NH₂; C ⇒ C₆H₅–NH–CO–CH₃
(ii) A ⇒ C₆H₅–NO₂; B ⇒ C₆H₅–NH₂; C ⇒ C₆H₅–N≡C

Teacher's Note:
1. Heating an ammonium salt of a carboxylic acid removes water and gives the amide – the usual starting point for Hoffmann degradation.

Question 30. An organic aromatic compound 'A' with the molecular formula C₆H₇N is sparingly soluble in water. 'A' on treatment with dil HCl gives a water soluble compound 'B'. 'A' also reacts with chloroform in presence of alcoholic KOH to form an obnoxious smelling compound 'C'. 'A' reacts with benzene sulphonyl chloride to form and alkali soluble compound 'D'. 'A' reacts with NaNO₂ and HCl to form a compound 'E' which on reaction with phenol forms an orange dye 'F'. Elucidate the structures of the organic compounds from 'A' to 'F'. [CBSE SQP 2017]
Answer:
A – C₆H₅NH₂ (Aniline)
B – C₆H₅NH₃⁺Cl⁻ (Anilinium chloride)
C – C₆H₅NC (Phenyl isocyanide)
D – C₆H₅–NH–SO₂–C₆H₅ (N-Phenylbenzenesulphonamide)
E – C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride)
F – HO–C₆H₄–N=N–C₆H₅ (p-Hydroxyazobenzene)

Teacher's Note:
1. The sulphonamide from a 1° amine still has an acidic N–H, which is why D dissolves in alkali; a 2° amine's sulphonamide does not.

Question 31. Do as directed:
(i) Arrange the following compounds in the increasing order of their basic strength in aqueous solution: CH₃NH₂, (CH₃)₃N, (CH₃)₂NH.
(ii) Identify 'A' and 'B':
C₆H₅NH₂ → [NaNO₂/HCl, 273 K] → A → [H₂O/H⁺] → B
(iii) Write equation of carbylamine reaction. [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
(ii) A: C₆H₅N₂⁺Cl⁻; B: C₆H₅OH
(iii) RNH₂ + CHCl₃ + 3KOH → [Heat] → R–NC + 3KCl + 3H₂O

Teacher's Note:
1. Remember the aqueous order for methylamines, 2° > 1° > 3°; for ethylamines it is 2° > 3° > 1°.

Question 32. How would you prepare:
(i) C₆H₅NH₂ from C₆H₅NO₂
(ii) CH₃NH₂ from C₂H₅NH₂
(iii) C₂H₅NH₂ from CH₃NH₂
Answer:
(i) C₆H₅NO₂ Sn/HCl
⟶
6[H] C₆H₅NH₂ + 2H₂O
(ii) C₂H₅NH₂ HNO₂
⟶ C₂H₅OH (O)
⟶
K₂Cr₂O₇/H₂SO₄ CH₃COOH NH₃
⟶ CH₃COONH₄ Δ
⟶
−H₂O CH₃CONH₂ Br₂/KOH
⟶ CH₃NH₂
(iii) CH₃NH₂ HNO₂
⟶ CH₃OH SOCl₂
⟶ CH₃Cl KCN
⟶ CH₃CN [H]
⟶ CH₃CH₂NH₂

Teacher's Note:
1. Step-down (one carbon less) uses Hoffmann degradation; step-up (one carbon more) goes through the nitrile.

Question 33. Write the structures of A, B and C in the following:
(i) C₆H₅CONH₂ Br₂/NaOH
⟶ A NaNO₂ + HCl
⟶
0–5°C B KI
⟶ C
(ii) CH₃Cl KCN
⟶ A LiAlH₄
⟶ B CHCl₃ + alc. KOH
⟶ C
Answer:
(i) A = C₆H₅NH₂, B = C₆H₅N₂⁺Cl⁻, C = C₆H₅I
(ii) A = CH₃CN, B = CH₃CH₂NH₂, C = CH₃CH₂NC

Teacher's Note:
1. Iodobenzene is made from the diazonium salt simply with KI – no copper salt is needed, unlike Cl and Br.

Question 34. Write the structures of the reagents/organic compounds ‘A’ to ‘F’:
A NaOH/Br₂
⟶
H₂O B NaNO₂/HCl
⟶
0°C C D
⟶ E Cl₂/Fe
⟶ F
E H₂/Pt
⟶ C₆H₅CH₂NH₂
Answer:
A = C₆H₅CONH₂ (benzamide)
B = C₆H₅NH₂ (aniline)
C = C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride)
D = KCN/Cu
E = C₆H₅CN (benzonitrile), which on reduction with H₂/Pt gives C₆H₅CH₂NH₂ (benzylamine)
F = m-ClC₆H₄CN (m-chlorobenzonitrile)

Teacher's Note:
1. Work backwards from benzylamine: its reduction precursor is benzonitrile, which in turn comes from the diazonium salt.

Question 35. (i) tert-Butylamine cannot be prepared by the action of NH₃ on tert-butyl bromide. Explain why ?
(ii) Suggest a convenient method for the preparation of tert-butylamine.
Answer:
(i) tert-Butyl bromide being a 3° alkyl halide on treatment with a base (i.e. NH₃) prefers to undergo elimination rather than substitution. Therefore, the product is isobutylene rather than tert-butylamine.
(CH₃)₃C–Br NH₃, −HBr
⟶
(Substitution)× (CH₃)₃C–NH₂ (does not occur)
(CH₃)₃C–Br :NH₃
⟶
−NH₄Br (Elimination) (CH₃)₂C=CH₂ (Isobutylene)
(ii) 1° amines containing tert-alkyl groups can be prepared by action of suitable Grignard reagents on O-methylhydroxylamine. For example,
(CH₃)₃C–MgBr (tert-Butylmagnesium bromide) + NH₂–OCH₃ (O-Methylhydroxylamine) Dry ether
⟶
  (CH₃)₃C–NH₂ (tert-Butylamine) + Mg(OCH₃)Br

Teacher's Note:
1. 3° alkyl halides almost always eliminate with bases such as NH₃; so ammonolysis and Gabriel synthesis both fail for tert-alkylamines.

Question 36. Why does bromination of aniline, even under very mild conditions, gives 2, 4, 6-tribromoaniline instantaneously ?
Answer:
Due to strong electron-donating effect of the –NH₂ group, the electron density increases at the o-, p-positions. Further, when aniline is treated with Br₂, the Br⁺ attacks the benzene ring at o- and p-positions to form carbocation intermediates which are stabilized not only by the usual resonance of the benzene ring but also by the –NH₂ group as shown below :
o-Bromination : the carbocation formed by attack of Br⁺ at the o-position is a resonance hybrid of structures I, III and IV, in which the positive charge is on ring carbons ortho and para to the carbon carrying Br, and structure II, in which the lone pair of N forms a C=N bond (C=N⁺H₂) and N carries the positive charge.
p-Bromination : similarly, the carbocation formed by attack at the p-position is a hybrid of structures V, VI and VIII (positive charge on ring carbons) and structure VII (C=N⁺H₂, positive charge on N).
In case of o-bromination, the carbocation intermediate is stabilized not only by usual resonating structure (I, III and IV) but is also stabilized by resonance structure (II) in which the lone pair of electrons on the N-atom interacts with the positively charged carbon of the ring. Similarly, in case of p-bromination, the carbocation is stabilized not only by the usual resonating structures (V, VI and VIII) but is also stabilized by the resonance structure (VII) in which the lone pair of electrons on the N-atom interacts with the positively charged carbon of the ring. These additional resonating structures (II and VII) increase the stability of the carbocation to such an extent that bromination occurs instantaneously at the p- and two o- positions giving 2, 4, 6-tribromoaniline.

Teacher's Note:
1. In structures II and VII every atom has a complete octet, which is why they are especially stable; to get only the mono-bromo product, first acetylate the –NH₂ group.

Question 37. Amines are more basic than comparable alcohols.
Answer:
Amines are more basic than comparable alcohols because of the following two reasons :
(i) N being less electronegative is more willing to donate its lone pair of electrons to a proton than the more electronegative O atom. Therefore, amines are more basic than alcohols.
(ii) When an amine accepts a proton, ammonium salt is formed and when an alcohol accepts a proton oxonium salt is formed.
R–NH₂ (Amine) + H⁺ → R–NH₃⁺ (Ammonium salt, more stable) ; R–OH (Alcohol) + H⁺ → R–OH₂⁺ (Oxonium salt, less stable)
Since N being less electronegative can accommodate the positive charge better than the more electronegative O atom, therefore, ammonium salts are more stable than oxonium salts. As a result, amines are more basic than alcohols.

Teacher's Note:
1. Give both reasons: availability of the lone pair and stability of the conjugate acid.

Question 38. Arrange the following compounds in increasing order of their basic strength in aqueous solution CH₃NH₂, (CH₃)₃N, (CH₃)₂NH. (CBSE 2018S)
Answer:
The basicity of CH₃NH₂, (CH₃)₂NH and (CH₃)₃N depends upon two factors : (i) + I-effect of CH₃ groups and (ii) Stability of the ammonium salts formed when an amine accepts a proton.
(CH₃)₂NH is, however, the strongest base since the combined effect of the stability of ammonium cation by +I-effect of the two CH₃ groups and stability due to H-bonding with two H atoms is the maximum.
CH₃NH₂ is, however, more basic than (CH₃)₃N because the stability of the ammonium cation by H-bonding with three H-atom in case of CH₃NH₂ predominates over the stability due to + I-effect of the three CH₃ groups in case of (CH₃)₃N.
CH₃–N⁺H₃ H-bonded to three H₂O molecules (stability due to H-bonding) predominates over (CH₃)₃N⁺–H (stability due to + I-effect of three CH₃ groups).
Thus, the overall basicity of these three amines follows the order : (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N.
Increasing order: (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Teacher's Note:
1. The question asks for increasing order – write the weakest base first; many students lose the mark by writing the decreasing order.

Question 39. Predict, giving reasons, the order of basicity of the following compounds in gaseous phase.
(i) (CH₃)₃N, (CH₃)₂NH, CH₃NH₂, NH₃ (CBSE 2012)
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, CH₃NH₂ (CBSE 2008)
Answer:
Due to +I-effect of the alkyl groups, the electron density on the N-atom of 1°, 2° and 3° amines is higher than that on the N-atom in NH₃. Therefore, all amines are more basic than NH₃.
(i) In gaseous phase, solvation effects are absent and hence the relative basicity of amines depends only on the +I-effect of the alkyl groups. Now since the +I-effect increases in going from 1° to 2° to 3° amine, therefore, the relative basicity of amines decreases in the order : 3° amine > 2° amine > 1° amine. In other words, in the gaseous phase, the basicity of methylamines relative to NH₃ decreases in the order :
(CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
(ii) In the gas phase, solvent effects are absent and hence the relative basicity of amines depends only on the +I- effect of the alkyl groups. Now the +I-effect of the C₂H₅ group is more than that of CH₃ group and the +I-effect increases as the number of alkyl groups increases. Therefore, overall basicity of the four amines increases in the order : CH₃NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH < (C₂H₅)₃N

Teacher's Note:
1. Always check the phase: gas phase follows the +I effect alone (3° > 2° > 1°), aqueous phase does not.

Long Answer Questions (5 Marks)

Question 40. An aromatic compound ‘A’ of molecular formula C₇H₇ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions:
C₇H₇ON (A) →[Br₂ + KOH] C₆H₅NH₂ →[NaNO₂ + HCl, 273 K] B →[CH₃CH₂OH] C
C₆H₅NH₂ →[CHCl₃ + NaOH] D;   B →[KI] E [CBSE Delhi 2015]
Answer:
A = C₆H₅—CO—NH₂ (Benzamide)
B = C₆H₅—N⁺≡NCl⁻ (Benzene diazonium chloride)
C = C₆H₆ (Benzene)
D = C₆H₅—NC (Phenyl isocyanide)
E = C₆H₅—I (Iodobenzene)

Teacher's Note:
1. Start from the known compound (aniline) and work both ways: Hoffmann degradation lost one carbon, so A has seven carbons and is benzamide.

Question 41. Write the structures of A, B, C, D and E in the following reactions:
C₆H₅NO₂ →[Sn/HCl] A →[(CH₃CO)₂O, pyridine] B →[HNO₃ + H₂SO₄, 288 K] C →[OH⁻ or H⁺] D
A →[H₂SO₄] E [CBSE (F) 2017]
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) ⇄ p-H₃N⁺—C₆H₄—SO₃⁻

Teacher's Note:
1. Hydrolysis with acid or base removes only the acetyl protecting group; the nitro group stays at the para position.

Question 42. (i) How will you bring about the following conversions?
(a) Ethanamine into methanamine
(b) Aniline into 1,3,5-tribromobenzene
(c) Aniline into 4-bromoaniline
(ii) Account for the following:
(a) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(b) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. [CBSE 2019 (56/5/2)]
Answer:
(i) (a) CH₃CH₂NH₂ (Ethanamine) →[HNO₂] CH₃CH₂OH →[CrO₃/H₂SO₄, [O]] CH₃COOH →[NH₃, Δ] CH₃CONH₂ →[Br₂/KOH] CH₃NH₂ (Methanamine)
(b) C₆H₅NH₂ (Aniline) →[Br₂/H₂O, Bromination] 2,4,6-Tribromoaniline →[NaNO₂/HCl, 273–278 K] 2,4,6-Tribromobenzenediazonium chloride →[H₃PO₂/H₂O] 1,3,5-Tribromobenzene
(c) C₆H₅NH₂ (Aniline) →[(CH₃CO)₂O, Pyridine] C₆H₅NHCOCH₃ →[Br₂, CH₃COOH] p-Br—C₆H₄—NHCOCH₃ (Major) →[H⁺] p-Br—C₆H₄—NH₂ (4-Bromoaniline)
(ii) (a) Methylamine is more basic than water, so in water it accepts a proton and produces OH⁻ ions:
CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻
These OH⁻ ions combine with Fe³⁺ ions of ferric chloride to precipitate hydrated ferric oxide:
2FeCl₃ + 6OH⁻ → 2Fe(OH)₃ (or Fe₂O₃.3H₂O)↓ + 6Cl⁻
(b) In arenediazonium ions, the positive charge on the –N₂⁺ group is dispersed over the benzene ring by resonance, which stabilises the ion. Aliphatic diazonium ions have no such resonance, so they decompose readily (even at low temperature) to give carbocations and N₂.

Teacher's Note:
1. To remove the –NH₂ group after it has directed the substitution, diazotise and then reduce with H₃PO₂ – this is how the 1,3,5-pattern is obtained.