CBSE Class 12 Chemistry Amines Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. Why is carbon-nitrogen bond length in aromatic amines shorter than in aliphatic amines ?
Answer:
Due to electron-donating resonance effect (+R-effect) of the NH₂ group, the carbon-nitrogen bond in aromatic amines (in the resonance structures, C=N⁺H₂ with negative charge at o- and p-positions) has some double bond character. In contrast, in aliphatic amines, the carbon-nitrogen bond has only single bond character. Therefore, carbon-nitrogen bond length in aromatic amines is shorter than in aliphatic amines.

Teacher's Note:
1. Partial double bond character always shortens a bond; the same idea explains the short C–Cl bond in chlorobenzene.

Question 2. Why is an alkylamine more basic than ammonia ? (CBSE 2009, 2011, 2012)
Answer:
Due to +I-effect of the alkyl group, the electron density on the N-atom of alkylamine (RNH₂) increases as compared to that on the N atom of ammonia (NH₃). As a result, alkylamine can more easily donate a pair electrons than NH₃ and hence alkylamine is more basic than NH₃.
R→N̈H₂ (R pushes electrons towards N) versus N̈H₃

Teacher's Note:
1. Basicity of an amine is its ability to donate the lone pair on N; anything that increases electron density on N increases basicity.

Question 3. Which is more acidic (or basic), aniline or ammonia ?
Answer:
Due to delocalization of the lone pair of electrons of the N-atom of aniline over the benzene ring, aniline is less basic than ammonia.
Conversely, loss of a proton from aniline gives ion (I) while that from NH₃ gives ion (II).
C₆H₅NH₂ → C₆H₅N⁻H (I) + H⁺; the negative charge of ion (I) is delocalised over the o- and p-positions of the ring by resonance.
NH₃ → NH₂⁻ (II) + H⁺
Since ion (I) is stabilized by resonance while ion (II) is not. Therefore, aniline is more acidic than ammonia.

Teacher's Note:
1. An acid is stronger when its conjugate base is more stable; resonance stabilises the anilide ion but not the amide ion NH₂⁻.

Question 4. Arrange the following in decreasing order of basic character : C₆H₅NH₂, (CH₃)₃N, C₂H₅NH₂. (CBSE 2019)
Answer:
(i) Due to delocalization of lone pair of electrons of the N-atom over the benzene ring, all aromatic amines are less basic than alkylamines, i.e., C₆H₅NH₂ is a weaker base than (CH₃)₃N and C₂H₅NH₂.
(ii) C₂H₅NH₂ is, however, more basic than (CH₃)₃N because the stability of the ammonium cation by H-bonding with three H-atoms in case of C₂H₅N⁺H₃ predominates over the stability due to +I-effect of three CH₃ groups in case of (CH₃)₃N⁺H.
Thus, the overall stability of the three amines decreases in the order : C₂H₅NH₂ > (CH₃)₃N > C₆H₅NH₂.

Teacher's Note:
1. In water, a 3° amine is often weaker than a 1° or 2° amine because its cation has only one N–H for H-bonding with water.

Question 5. Arrange the following in order of their increasing basicity : p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline.
Answer:
Electron-donating groups, i.e., alkyl groups in the ring and on the N-atom increase the basicity while electron-withdrawing groups such as nitro group in the ring decrease the basicity. Thus, the basicity of the given amines increases in the order : p-nitroaniline < aniline < p-toluidine < N, N-dimethyl-p-toluidine

Teacher's Note:
1. Use aniline as the reference point: put electron-withdrawing substituents below it and electron-donating ones above it.

Question 6. Arrange the basicity of o-, m- and p-nitroanilines with respect to aniline.
Answer:
The NO₂ group is a strong electron-withdrawing group (−I and −R effects). It decreases the electron density on the N-atom of NH₂ and hence all the three nitroanilines are weaker bases than aniline.
At the m-position, only the −I effect operates; at the p-position both −I and −R effects operate, so p-nitroaniline is weaker than m-nitroaniline. o-Nitroaniline is the weakest due to the ortho effect (strong −I effect from close proximity, −R effect and intramolecular H-bonding between NH₂ and NO₂).
Thus, basicity decreases in the order : aniline > m-nitroaniline > p-nitroaniline > o-nitroaniline

Teacher's Note:
1. pK_b values confirm the order: aniline 9.38, m-nitroaniline 11.5, p-nitroaniline 13.0, o-nitroaniline 14.3 (higher pK_b = weaker base).

Question 7. CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.
Answer:
Due to resonance, the lone pair of electrons on the nitrogen atom in CH₃CONH₂ is delocalized over the carbonyl group, C=O.
CH₃–C(=O)–N̈H₂ ↔ CH₃–C(–O⁻)=N⁺H₂ (Acetamide) ; CH₃CH₂→NH₂
As a result, electron density on the N-atom in acetamide decreases. In contrast, in ethylamine, due to +I-effect of the ethyl group, electron density on the N-atom increases. Consequently, CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.

Teacher's Note:
1. Amides are practically neutral; the lone pair on N is tied up in resonance with the C=O group.

Question 8. How will you convert aniline into N-phenylethanamide ? (CBSE 2014)
Answer:
C₆H₅NH₂ (Aniline) + (CH₃CO)₂O \( \xrightarrow[\mathrm{\Delta}]{\mathrm{Pyridine}} \) C₆H₅–NH–CO–CH₃ (N-Phenylethanamide) + CH₃COOH (Acetic acid)

Teacher's Note:
1. N-Phenylethanamide is the IUPAC name of acetanilide; acetyl chloride can also be used, again with pyridine to remove the acid formed.

Question 9. What is the main product of mustard oil reaction? (MP Board 2013)
Answer:
The main product of mustard oil reaction is alkylisothiocyanate having characteristic smell of mustard oil.
R–NH₂ (1° Amine) + S=C=S (Carbon disulphide) \( \xrightarrow{\mathrm{Warm}} \) R–NH–C(=S)–SH (Dithioalkyl carbonic acid) \( \xrightarrow[\mathrm{-HgS,\ -2HCl}]{\mathrm{HgCl_2}} \) R–N=C=S (Alkyl isothiocyanate)

Teacher's Note:
1. Only 1° amines give the mustard oil (Hofmann mustard oil) reaction, so it can be used as a test for 1° amines.

Question 10. Give a chemical test to distinguish between a primary and a secondary amine.
Answer:
Carbylamine reaction or Libermann nitroso reaction.

Teacher's Note:
1. In the carbylamine test only the 1° amine gives the offensive smell; with HNO₂ a 2° amine gives a yellow oily N-nitrosoamine (Libermann nitroso test).

Question 11. Electrophilic substitution in case of aromatic amines takes place more readily than in benzene. Why ? (CBSE 2010 S)
Answer:
Due to resonance, the electrons on the N-atom in aniline are delocalized over the benzene ring. As a result, the electron density on the benzene ring increases as compared to benzene itself. In other words, aniline gets activated and hence electrophilic substitution occurs more readily in aniline.

Teacher's Note:
1. The increased electron density is mainly at the ortho and para positions, which is why –NH₂ is also o, p-directing.

Question 12. Acetylation of aniline reduces its activation effect. (CBSE 2017)
Answer:
Due to strong electron-donating effect of the –NH₂ group, electron density increases in the benzene ring. As a result, aniline gets activated towards electrophilic substitution reactions. However, when aniline is acetylated, the activating effect of the amino group is reduced. This is due to the reason that the acetyl group being electron-withdrawing attracts the lone pair of electrons on the N-atom towards itself. As a result, lone pair of electrons on N-atom is not exclusively available for donation to the benzene ring and hence the activating effect of the –NH₂ group is reduced.
H–N̈(R)–C(=O)–CH₃ ↔ H–N⁺(R)=C(–O⁻)–CH₃

Teacher's Note:
1. This is why acetanilide gives only the mono-bromo product while aniline gives 2,4,6-tribromoaniline with bromine water.

Question 13. Direct nitration of aniline is not carried at all. Explain why ?
Answer:
Nitric acid being a strong oxidising agent oxidises most of aniline to form tarry oxidation products.

Teacher's Note:
1. Besides oxidation, in strongly acidic medium aniline is protonated to the anilinium ion which is m-directing, so a mixture including much m-nitroaniline would form; hence the –NH₂ group is protected by acetylation first.

Short Answer Questions (2 Marks)

Question 14. Methyl or ethylamine is soluble in water but aniline is insoluble. Explain. (CBSE 2022)
Answer:
Methylamine and ethylamine form intermolecular H-bonds with water molecules and hence are soluble in water.
In aniline, the large hydrophobic C₆H₅ group retards the tendency of the –NH₂ group to form H-bonds with water. Hence, aniline is insoluble (only sparingly soluble) in water.

Teacher's Note:
1. Lower aliphatic amines are water-soluble; solubility falls as the hydrocarbon part becomes larger.

Question 15. Why are aromatic amines weaker bases than aliphatic amines ? (CBSE 2007) or pK_b of aniline is more than that of methylamine. Explain. (Assam Board 2013)
Answer:
In aromatic amines (e.g., aniline), the lone pair of electrons on the N-atom is delocalised over the benzene ring by resonance and hence is less available for donation to a proton. In aliphatic amines (e.g., methylamine), the +I-effect of the alkyl group increases the electron density on the N-atom. Therefore, aromatic amines are weaker bases than aliphatic amines.
Since a weaker base has a higher pK_b, pK_b of aniline (9.38) is more than that of methylamine (3.38).

Teacher's Note:
1. Larger pK_b means smaller K_b and a weaker base; students often reverse this.

Question 16. Explain the observed K_b order : Me₂NH > MeNH₂ > Me₃N.
Answer:
In aqueous solution, the CH₃ group being small, there is no steric hindrance to H-bonding of the ammonium cations with water. Me₂NH₂⁺ is stabilised by +I-effect of two CH₃ groups and H-bonding through two H-atoms; MeNH₃⁺ by +I-effect of one CH₃ group and H-bonding through three H-atoms; and Me₃NH⁺ by +I-effect of three CH₃ groups but H-bonding through only one H-atom.
The combined effect of +I-effect and H-bonding (solvation) makes the stability of the cations, and hence basicity, decrease in the order : Me₂NH > MeNH₂ > Me₃N.

Teacher's Note:
1. Supporting data: pK_b values are 3.27 (Me₂NH), 3.38 (MeNH₂) and 4.22 (Me₃N).

Question 17. Justify the order of basicity of EtNH₂, Et₂NH and Et₃N.
Answer:
In aqueous solution, the basicity order is : Et₂NH > Et₃N > EtNH₂.
Ethyl groups exert a strong +I-effect but, being bulky, they cause steric hindrance to H-bonding of the ammonium cations with water. Hence, the +I-effect predominates. Et₂NH, having two ethyl groups and two N–H bonds in its cation for solvation, is the most basic. Et₃N has the largest +I-effect but poor solvation and greater steric crowding, so it comes next, while EtNH₂, with only one ethyl group, is the least basic. (In the gas phase, where solvation is absent, the order is Et₃N > Et₂NH > EtNH₂.)

Teacher's Note:
1. pK_b values (3.00, 3.25, 3.29 for Et₂NH, Et₃N, EtNH₂) show how close these three are; the order is decided by a balance of effects.

Question 18. Justify the order of basicity of 1°, 2° and 3° amines.
Answer:
(i) In the gas phase, only the +I-effect of alkyl groups operates. The greater the number of alkyl groups, the higher the electron density on N. Hence the basicity order is : 3° > 2° > 1° > NH₃.
(ii) In aqueous solution, besides the +I-effect, the stabilisation of the ammonium cation by H-bonding with water (which is greatest for 1° and least for 3°) and steric hindrance also operate. As a result of these combined effects :
for methylamines : (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
for ethylamines : (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃

Teacher's Note:
1. In every case the 2° amine is the strongest base in water, and all alkylamines are stronger than NH₃.

Question 19. Why is ethylamine more basic than aniline ? Explain. (HP Board 2011)
Answer:
In ethylamine, the +I-effect of the ethyl group increases the electron density on the N-atom, so it readily donates its lone pair to a proton.
In aniline, the lone pair of electrons on N is delocalised over the benzene ring by resonance and is less available for protonation. Moreover, aniline (five resonance structures) is more stabilised by resonance than the anilinium ion (two resonance structures). Hence, ethylamine is more basic than aniline.

Teacher's Note:
1. pK_b of ethylamine is 3.29 while that of aniline is 9.38.

Question 20. Aniline is a weaker base than cyclohexylamine. Explain.
Answer:
In aniline, the NH₂ group is attached to an sp² hybridised carbon of the benzene ring and the lone pair of electrons on N is delocalised over the ring by resonance. Hence, it is less available for protonation.
In cyclohexylamine, the NH₂ group is attached to an sp³ hybridised carbon, there is no resonance and the cyclohexyl group exerts +I-effect, increasing the electron density on N. Therefore, aniline is a weaker base than cyclohexylamine.

Teacher's Note:
1. Both have six-membered rings; the difference is only resonance with the aromatic ring, which makes this a neat test of the concept.

Question 21. Explain why aromatic amines are less basic than ammonia and aliphatic amines. (Assam Board 2012)
Answer:
In aromatic amines, the lone pair of electrons on the N-atom is delocalised over the benzene ring by resonance and the sp² hybridised carbon of the ring is electron-withdrawing. Hence, the electron density on N is lower than in NH₃ and the lone pair is less available for protonation. Further, the arylamine is more stabilised by resonance than its cation. Therefore, aromatic amines are less basic than ammonia.
In aliphatic amines, the +I-effect of the alkyl groups increases the electron density on N, so they are more basic than NH₃ and hence much more basic than aromatic amines.

Teacher's Note:
1. Basicity order to remember: aliphatic amines > NH₃ > aromatic amines.

Question 22. How do you explain that aniline is a weaker base than ethylamine ?
Answer:
In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring (+R-effect) and is not readily available for donation to a proton. In ethylamine, the +I-effect of the ethyl group increases the electron density on N, making the lone pair readily available. Further, aniline is stabilised by resonance more than the anilinium ion. Hence, aniline is a weaker base than ethylamine.

Teacher's Note:
1. Whenever the N lone pair is part of a resonance system (aryl or C=O), basicity drops sharply.

Question 23. What is quaternary ammonium salt? What happens when triethylamine is heated with ethyl chloride? (Maharashtra Board 2012)
Answer:
A salt in which all the four H-atoms of the ammonium ion (NH₄⁺) are replaced by alkyl or aryl groups, R₄N⁺X⁻, is called a quaternary ammonium salt.
Triethylamine on heating with ethyl chloride gives tetraethylammonium chloride :
(C₂H₅)₃N + C₂H₅Cl \( \xrightarrow{\mathrm{\Delta}} \) (C₂H₅)₄N⁺Cl⁻ (Tetraethylammonium chloride)

Teacher's Note:
1. A quaternary salt has no H on nitrogen and no lone pair, so it is not a base itself; its hydroxide, however, is a strong base.

Question 24. Tertiary amines do not undergo acylation reaction. or Tertiary amines are not acylated. Explain. (Manipur Board 2011)
Answer:
Acylation involves the replacement of an H-atom of the –NH₂ or >NH group by an acyl group (RCO–). Since 3° amines (R₃N) do not have any H-atom on the N-atom, they do not undergo acylation.

Teacher's Note:
1. The same reason explains why 3° amines do not react with Hinsberg's reagent.

Question 25. Describe carbylamine test for primary amines. or Write equation for carbylamine reaction. (CBSE 2018 S)
Answer:
When a 1° amine (aliphatic or aromatic) is heated with chloroform and ethanolic KOH, an isocyanide (carbylamine) having a very unpleasant smell is formed. This reaction is called carbylamine reaction.
R–NH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) R–N≡C (Alkyl isocyanide) + 3KCl + 3H₂O
e.g., C₆H₅NH₂ + CHCl₃ + 3KOH \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅NC (Phenyl isocyanide) + 3KCl + 3H₂O
2° and 3° amines do not give this reaction. Hence, it is used as a test for 1° amines.

Teacher's Note:
1. Only 1° amines have the two N–H bonds needed to form R–N≡C, so the foul smell is a positive test for a 1° amine only.

Question 26. What products are formed when chloroform reacts with aniline in presence of alcohol KOH? (Uttarakhand Board 2012)
Answer:
Phenyl isocyanide (phenyl carbylamine), along with KCl and water, is formed. It has an offensive smell.
C₆H₅NH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅N≡C + 3KCl + 3H₂O

Teacher's Note:
1. Remember all three by-products: 3KCl and 3H₂O for each mole of isocyanide.

Short Answer Questions (3 Marks)

Question 27. Account for the following :
(i) Secondary amines are more basic than primary amines.
(ii) Primary amines are more basic than ammonia.
(iii) Arylamines are weaker bases than alkylamines or vice-versa.
(iv) Alkylamines are stronger bases than arylamines. (Pb. Board 2012)
(v) Ethylamine is a stronger base than ammonia.
(vi) Benzylamine is more basic than aniline.
(vii) Trimethylamine is a weaker base than dimethylamine.
(viii) Methylamine is a stronger base than ammonia. (HP Board 2011)
(ix) Amines are basic substances while amides are neutral. (CBSE 2007)
(x) (CH₃)₂NH is more basic than CH₃NH₂ and (CH₃)₃N (HP Board 2011 ; Raj. Board 2012) or (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. (CBSE 2018)
(xi) Out of CH₃NH₂, (CH₃)₂NH and (CH₃)₃N which is the most basic. (HP Board 2011)
(xii) Amino group in ethylamine (CH₃CH₂NH₂) is more basic but not in acetamide (CH₃CONH₂) or acetanilide (C₆H₅NHCOCH₃).
(xiii) Ethylamine is more basic than aniline. (Manipur Board 2012)
(xiv) CH₃NH₂ is more basic than C₆H₅NH₂. (CBSE (Foreign) 2017)
(xv) C₆H₅N⁺(CH₃)₃OH⁻ is a stronger base than NH₄OH.
Answer:
(i) Secondary amines have two electron-donating alkyl groups (+I-effect) on the N-atom while 1° amines have only one. Hence, electron density on N is higher in 2° amines and they are more basic. In aqueous solution also, the 2° ammonium ion (R₂NH₂⁺) is well stabilised by H-bonding with two H-atoms.
(ii), (v), (viii) Due to +I-effect of the alkyl group (CH₃ or C₂H₅), the electron density on the N-atom of a 1° amine is more than that on the N-atom of NH₃. Therefore, 1° amines such as methylamine and ethylamine donate their lone pair more easily and are stronger bases than ammonia.
(iii), (iv), (xiii), (xiv) In arylamines (e.g., aniline), the lone pair of electrons on N is delocalised over the benzene ring by resonance and is less available for protonation. Further, aniline is more stabilised by resonance than the anilinium ion. In alkylamines (e.g., CH₃NH₂, C₂H₅NH₂), there is no resonance and the +I-effect of the alkyl group increases electron density on N. Hence, alkylamines are stronger bases than arylamines.
(vi) In benzylamine (C₆H₅CH₂NH₂), the NH₂ group is attached to an sp³ CH₂ group and its lone pair is not delocalised over the benzene ring, while in aniline it is. Hence, benzylamine is more basic than aniline.
(vii), (x), (xi) In aqueous solution, basicity depends upon the +I-effect, the stabilisation of the ammonium cation by H-bonding with water and steric hindrance. The ion (CH₃)₂NH₂⁺ is stabilised by H-bonding through two H-atoms, CH₃NH₃⁺ through three H-atoms but with smaller +I-effect, and (CH₃)₃NH⁺ through only one H-atom. The combined effect makes (CH₃)₂NH the most basic : (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N. Thus, trimethylamine is a weaker base than dimethylamine and dimethylamine is the most basic of the three.
(ix), (xii) In amides (CH₃CONH₂) and acetanilide (C₆H₅NHCOCH₃), the lone pair of electrons on N is delocalised over the C=O group by resonance (R–C(=O)–NH₂ ↔ R–C(O⁻)=N⁺H₂) and is not available for protonation. In amines (e.g., CH₃CH₂NH₂), there is no such delocalisation and the +I-effect of the alkyl group increases the electron density on N. Hence, amines are basic while amides are neutral.
(xv) C₆H₅N⁺(CH₃)₃OH⁻ being a salt is a strong electrolyte and hence undergoes complete dissociation while NH₄OH is a weak electrolyte and hence is only weakly ionized.

Teacher's Note:
1. In water the basicity order of methylamines is 2° > 1° > 3° (pK_b 3.27, 3.38, 4.22); do not quote the gas-phase order 3° > 2° > 1° for aqueous solution.

Question 28. Explain the observed K_b order : Et₂NH > Et₃N > EtNH₂, in aqueous solution.
Answer:
In aqueous solution, the basicity of amines depends upon (i) +I-effect of the alkyl groups, (ii) stabilisation of the conjugate acid (ammonium ion) by H-bonding with water and (iii) steric hindrance of the alkyl groups.
In case of ethyl groups, which are bigger than methyl groups, there is considerable steric hindrance to H-bonding. Therefore, the +I-effect of the ethyl groups predominates over the H-bonding effect. Et₂NH has two ethyl groups and its cation is also stabilised by two H-atoms, so it is the most basic. Et₃N, though having three ethyl groups, is less basic than Et₂NH due to poor solvation and steric hindrance, but its larger +I-effect makes it more basic than EtNH₂.
Hence, K_b order : Et₂NH > Et₃N > EtNH₂.

Teacher's Note:
1. Compare with methylamines, where H-bonding dominates and the 3° amine is the weakest: the size of the alkyl group changes the balance of effects.

Question 29. Arrange the following as directed :
(i) In an increasing order of basic strength : C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂ (Hr. Board 2013)
(ii) In a decreasing order of basic strength : aniline, p-nitroaniline and p-toluidine. (Hr. Board 2013)
(iii) In increasing order of pK_b values : C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂ (CBSE 2011)
(iv) In increasing order of their pK_b values : C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃. (CBSE 2018)
Answer:
(i) C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH
(ii) p-toluidine > aniline > p-nitroaniline
(iii) (C₂H₅)₂NH < C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
(iv) C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
Aliphatic amines are stronger bases than aromatic amines; electron-donating CH₃ groups (on N or in the ring) increase while the electron-withdrawing NO₂ group decreases basicity. Since the stronger base has the lower pK_b, the pK_b order is the reverse of the basicity order.

Teacher's Note:
1. For pK_b questions, first write the basicity order and then simply reverse it.

Question 30. Account for the following :
(i) Like ammonia, amines are good nucleophiles.
(ii) Silver chloride dissolves in an aqueous solution of methylamine. (CBSE 2007)
(iii) Methylamine in water reacts with ferric chloride to give a precipitate of ferric hydroxide. (CBSE 2008)
(iv) Methylamine or ethylamine dissolves in aqueous CuCl₂ or CuSO₄ to give a deep blue solution.
Answer:
(i) Like ammonia, the N-atom of amines carries a lone pair of electrons which it can donate to an electron-deficient centre. Hence, amines are good nucleophiles, e.g., they react with alkyl halides and acid chlorides.
(ii) Methylamine forms a soluble complex with AgCl : AgCl + 2CH₃NH₂ → [Ag(CH₃NH₂)₂]⁺Cl⁻ (soluble)
(iii) Methylamine is a stronger base than water. In water it produces OH⁻ ions which precipitate Fe³⁺ ions as hydrated ferric oxide :
CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻ ; FeCl₃ + 3OH⁻ → Fe(OH)₃↓ (brown) + 3Cl⁻
(iv) Like ammonia, methylamine (or ethylamine) acts as a ligand and forms a deep blue complex with Cu²⁺ ions :
Cu²⁺ + 4CH₃NH₂ → [Cu(CH₃NH₂)₄]²⁺ (deep blue)

Teacher's Note:
1. All four observations come from the same lone pair on N: it can attack carbon (nucleophile), take a proton (base) or bind a metal ion (ligand).

Question 31. Show the mechanism of acylation of ethanamine and write the IUPAC name of the product formed.
Answer:
Acylation of ethanamine with acetyl chloride takes place by nucleophilic substitution (addition-elimination) at the carbonyl carbon :
Step 1. The lone pair of electrons on N of CH₃CH₂NH₂ attacks the carbonyl carbon of CH₃COCl to form a tetrahedral intermediate :
CH₃CH₂N̈H₂ + CH₃COCl → CH₃–C(O⁻)(Cl)–N⁺H₂CH₂CH₃
Step 2. The intermediate eliminates Cl⁻ and then loses a proton (removed by the base pyridine) to form the amide :
CH₃–C(O⁻)(Cl)–N⁺H₂CH₂CH₃ \( \xrightarrow{\mathrm{-Cl^-}} \) CH₃CO–N⁺H₂CH₂CH₃ \( \xrightarrow{\mathrm{-H^+\ (pyridine)}} \) CH₃CONHCH₂CH₃
Overall : CH₃CH₂NH₂ + CH₃COCl \( \xrightarrow{\mathrm{Pyridine}} \) CH₃CONHCH₂CH₃ + HCl
IUPAC name of the product : N-Ethylethanamide.

Teacher's Note:
1. Pyridine is added to remove HCl; otherwise the HCl would protonate the remaining amine and stop it from acting as a nucleophile.

Question 32. Account for the following :
(i) Aniline is acetylated to prepare its monobromo derivative.
(ii) Before nitration, aniline is converted into acetanilide. or Before reacting aniline with HNO₃ for nitration, it is converted into acetanilide. Why is it done and how is nitroaniline obtained subsequently ?
(iii) Aniline does not undergo Friedel-crafts reaction ? (CBSE 2019)
Answer:
(i) Due to strong activating effect of the –NH₂ group, aniline reacts with bromine water to give 2, 4, 6-tribromoaniline. To get the monobromo derivative, the activating effect is reduced by acetylation. Acetanilide on bromination gives mainly p-bromoacetanilide which on hydrolysis gives p-bromoaniline.
C₆H₅NH₂ \( \xrightarrow{\mathrm{(CH_3CO)_2O}} \) C₆H₅NHCOCH₃ \( \xrightarrow{\mathrm{Br_2/CH_3COOH}} \) p-BrC₆H₄NHCOCH₃ \( \xrightarrow{\mathrm{H^+\ or\ OH^-,\ H_2O}} \) p-BrC₆H₄NH₂
(ii) Direct nitration of aniline is not done because (a) HNO₃ oxidises aniline to tarry products and (b) in the strongly acidic medium aniline is protonated to the anilinium ion, which is m-directing, giving much m-nitroaniline. Acetylation protects the –NH₂ group and moderates its activating effect. Acetanilide on nitration gives mainly p-nitroacetanilide which on hydrolysis gives p-nitroaniline.
C₆H₅NHCOCH₃ \( \xrightarrow[\mathrm{288\ K}]{\mathrm{Conc.\ HNO_3/H_2SO_4}} \) p-O₂NC₆H₄NHCOCH₃ \( \xrightarrow{\mathrm{H^+\ or\ OH^-,\ H_2O}} \) p-O₂NC₆H₄NH₂ (major) ; some o-nitroaniline is also formed.
(iii) Aniline being a Lewis base reacts with the Lewis acid AlCl₃ (the catalyst) to form a salt, C₆H₅NH₂⁺–AlCl₃⁻. The positively charged N-atom strongly withdraws electrons and deactivates the ring. Hence, aniline does not undergo Friedel-Crafts reaction.

Teacher's Note:
1. Acetylation is a temporary protection: it is removed by hydrolysis at the end, so the final product still has a free NH₂ group.

Question 33. How will you obtain the following from aniline? Give equations only.
(i) 2, 4, 6-Tribromoaniline (Raj. Board 2012) (ii) Benzenediazonium chloride (Raj. Board 2012) (iii) Sulphanilic acid (Chhatisgarh Board 2012) (iv) p-Benzoquinone (Chhatisgarh Board 2012)
Answer:
(i) C₆H₅NH₂ + 3Br₂ (aq) \( \xrightarrow{\mathrm{Room\ temp.}} \) 2, 4, 6-Br₃C₆H₂NH₂ (white ppt.) + 3HBr
(ii) C₆H₅NH₂ + NaNO₂ + 2HCl \( \xrightarrow{\mathrm{273\text{-}278\ K}} \) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(iii) C₆H₅NH₂ + H₂SO₄ (conc.) → C₆H₅NH₃⁺HSO₄⁻ (Anilinium hydrogensulphate) \( \xrightarrow{\mathrm{453\text{-}473\ K}} \) p-H₃N⁺C₆H₄SO₃⁻ (Sulphanilic acid) + H₂O
(iv) C₆H₅NH₂ \( \xrightarrow[\mathrm{[O]}]{\mathrm{Na_2Cr_2O_7/H_2SO_4}} \) O=C₆H₄=O (p-Benzoquinone)

Teacher's Note:
1. In (i) no catalyst is needed because –NH₂ activates all three o/p positions at once.

Question 34. How will you convert ?
(i) Propene to 1-nitropropane (Hr. Board 2013) (ii) n-Propyl chloride into n-propylamine. (iii) Methylamine into ethylamine (Chhatisgarh Board 2012) (iv) Nitrobenzene into aniline. (CBSE 2007, 2014, 2022) (v) Ethanamide to methanamine. (CBSE 2022) (vi) Aniline into acetanilide or N-phenylethanamide. (CBSE 2014) (vii) Acetamide to ethylamine. (viii) Methyl bromide to ethylamine (Assam Board 2013) (ix) Aniline to phenylisocyanide (x) Aniline to o- and p-nitroaniline. (xi) Aniline into phenylisothiocyanate (xii) Chlorobenzene into aniline. (xiii) Nitroethane to ethylamine. (xiv) Aniline into p-benzoquinone. (HP Board 2013) (xv) Acetic acid to ethylamine. (xvi) Propionamide to ethylamine. (xvii) Ethylamine to ethane. (xviii) Ethylamine to ethanol. (xix) Ethylamine to diethylamine. (xx) An alkyl halide to a quaternary ammonium salt (xxi) Ethanoic acid to methanamine (CBSE 2014) (xxii) Aniline into sulphanilic acid (CBSE Sample Paper 2018-19) (xxiii) Aniline to p-bromoaniline (CBSE 2022) (xxiv) Ethanenitrile to ethanamine.
Answer:
(i) CH₃CH=CH₂ \( \xrightarrow{\mathrm{HBr/ROOR}} \) CH₃CH₂CH₂Br \( \xrightarrow{\mathrm{AgNO_2\ (alc.)}} \) CH₃CH₂CH₂NO₂
(ii) CH₃CH₂CH₂Cl + NH₃ (alc., excess) \( \xrightarrow{\mathrm{\Delta}} \) CH₃CH₂CH₂NH₂ + HCl (or by Gabriel phthalimide synthesis)
(iii) CH₃NH₂ \( \xrightarrow{\mathrm{HNO_2}} \) CH₃OH \( \xrightarrow{\mathrm{PCl_5}} \) CH₃Cl \( \xrightarrow{\mathrm{KCN\ (alc.)}} \) CH₃CN \( \xrightarrow{\mathrm{LiAlH_4}} \) CH₃CH₂NH₂
(iv) C₆H₅NO₂ + 6[H] \( \xrightarrow{\mathrm{Sn/HCl\ or\ Fe/HCl}} \) C₆H₅NH₂ + 2H₂O
(v) CH₃CONH₂ + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) CH₃NH₂ + K₂CO₃ + 2KBr + 2H₂O
(vi) C₆H₅NH₂ + (CH₃CO)₂O \( \xrightarrow{\mathrm{Pyridine}} \) C₆H₅NHCOCH₃ + CH₃COOH
(vii) CH₃CONH₂ + 4[H] \( \xrightarrow{\mathrm{LiAlH_4/ether}} \) CH₃CH₂NH₂ + H₂O
(viii) CH₃Br \( \xrightarrow{\mathrm{KCN\ (alc.)}} \) CH₃CN \( \xrightarrow{\mathrm{LiAlH_4\ or\ Na/C_2H_5OH}} \) CH₃CH₂NH₂
(ix) C₆H₅NH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅NC + 3KCl + 3H₂O
(x) C₆H₅NH₂ \( \xrightarrow{\mathrm{(CH_3CO)_2O}} \) C₆H₅NHCOCH₃ \( \xrightarrow[\mathrm{288\ K}]{\mathrm{Conc.\ HNO_3/H_2SO_4}} \) o- and p-O₂NC₆H₄NHCOCH₃ \( \xrightarrow{\mathrm{H_2O/H^+}} \) o-O₂NC₆H₄NH₂ + p-O₂NC₆H₄NH₂ (major)
(xi) C₆H₅NH₂ + CS₂ \( \xrightarrow{\mathrm{Warm}} \) C₆H₅NH–CS–SH \( \xrightarrow[\mathrm{-HgS,\ -2HCl}]{\mathrm{HgCl_2}} \) C₆H₅N=C=S
(xii) C₆H₅Cl + 2NH₃ \( \xrightarrow[\mathrm{60\ atm}]{\mathrm{Cu_2O,\ 475\ K}} \) C₆H₅NH₂ + NH₄Cl
(xiii) CH₃CH₂NO₂ + 6[H] \( \xrightarrow{\mathrm{Sn/HCl}} \) CH₃CH₂NH₂ + 2H₂O
(xiv) C₆H₅NH₂ \( \xrightarrow{\mathrm{Na_2Cr_2O_7/H_2SO_4}} \) p-Benzoquinone (O=C₆H₄=O)
(xv) CH₃COOH \( \xrightarrow{\mathrm{NH_3}} \) CH₃COONH₄ \( \xrightarrow{\mathrm{\Delta,\ -H_2O}} \) CH₃CONH₂ \( \xrightarrow{\mathrm{LiAlH_4}} \) CH₃CH₂NH₂
(xvi) CH₃CH₂CONH₂ + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) CH₃CH₂NH₂ + K₂CO₃ + 2KBr + 2H₂O
(xvii) CH₃CH₂NH₂ \( \xrightarrow{\mathrm{HNO_2}} \) CH₃CH₂OH \( \xrightarrow{\mathrm{PCl_5}} \) CH₃CH₂Cl \( \xrightarrow{\mathrm{Zn/HCl,\ [H]}} \) CH₃CH₃
(xviii) CH₃CH₂NH₂ + HNO₂ \( \xrightarrow{\mathrm{NaNO_2/HCl}} \) CH₃CH₂OH + N₂ + H₂O
(xix) CH₃CH₂NH₂ + CH₃COCl \( \xrightarrow{\mathrm{Pyridine}} \) CH₃CH₂NHCOCH₃ \( \xrightarrow{\mathrm{LiAlH_4}} \) CH₃CH₂NHCH₂CH₃ ; or CH₃CH₂NH₂ + C₂H₅Br → (C₂H₅)₂NH + HBr
(xx) NH₃ + 4RX (excess) \( \xrightarrow{\mathrm{\Delta}} \) R₄N⁺X⁻ + 3HX, e.g., (C₂H₅)₃N + C₂H₅Cl → (C₂H₅)₄N⁺Cl⁻
(xxi) CH₃COOH \( \xrightarrow{\mathrm{NH_3,\ \Delta}} \) CH₃CONH₂ \( \xrightarrow{\mathrm{Br_2/KOH}} \) CH₃NH₂
(xxii) C₆H₅NH₂ \( \xrightarrow{\mathrm{Conc.\ H_2SO_4}} \) C₆H₅NH₃⁺HSO₄⁻ \( \xrightarrow{\mathrm{453\text{-}473\ K}} \) p-H₃N⁺C₆H₄SO₃⁻
(xxiii) C₆H₅NH₂ \( \xrightarrow{\mathrm{(CH_3CO)_2O}} \) C₆H₅NHCOCH₃ \( \xrightarrow{\mathrm{Br_2/CH_3COOH}} \) p-BrC₆H₄NHCOCH₃ \( \xrightarrow{\mathrm{H_2O/H^+}} \) p-BrC₆H₄NH₂
(xxiv) CH₃C≡N + 4[H] \( \xrightarrow{\mathrm{LiAlH_4\ or\ Na/C_2H_5OH}} \) CH₃CH₂NH₂

Teacher's Note:
1. For every conversion, first compare the carbon count of start and end: +1 carbon suggests a cyanide step, −1 carbon suggests Hofmann bromamide, no change suggests reduction or substitution.

Question 35. Complete the following reactions :
(i) C₆H₅NH₂ + HCl (aq) (CBSE 2013, 2015) (ii) C₂H₅NH₂ + C₆H₅SO₂Cl → (CBSE 2009, 2011) (iii) C₂H₅NH₂ + HNO₂ → (CBSE 2009) (iv) C₆H₅NH₂ + Br₂ (aq) \( \xrightarrow{\mathrm{r.t.}} \) (CBSE 2012, 2013, 2015, 2022) (v) C₆H₅NH₂ + CHCl₃ + 3KOH → (HP Board 2011 ; Uttarakhand Board 2012) (vi) CH₃CH₂NH₂ + CHCl₃ \( \xrightarrow{\mathrm{KOH\ (alc.)}} \) (CBSE 2013 ; HP Board 2013) (vii) C₂H₅NH₂ + AgCl → (viii) C₂H₅NH₂ + CH₃I (excess) → (ix) C₆H₅NH₂ \( \xrightarrow{\mathrm{Na_2Cr_2O_7/H_2SO_4}} \) (x) CH₃CONH₂ + Br₂ + NaOH → (xi) C₆H₅NH₂ + NaNO₂ + HCl \( \xrightarrow{\mathrm{273\ K}} \) (Uttarakhand Board 2012) (xii) C₆H₅N(CH₃)₂ + HNO₂ → (xiii) C₆H₅NH₂ + (CH₃CO)₂O \( \xrightarrow{\mathrm{Pyridine}} \) (CBSE 2015, 2018) (xiv) RNH₂ + CHCl₃ + KOH (alc.) (CBSE 2010 ; Assam Board 2012) (xv) C₆H₅NH₂ + C₆H₅COCl \( \xrightarrow{\mathrm{NaOH}} \) (West Bengal Board 2012) (xvi) C₆H₅NO₂ \( \xrightarrow[\mathrm{363\ K}]{\mathrm{Fuming\ HNO_3/conc.\ H_2SO_4}} \) (West Bengal Board 2012) (xvii) RCONH₂ + Br₂ + NaOH → (Assam Board 2012) (xviii) C₆H₅SO₂Cl + (CH₃)₂NH → (CBSE 2018)
Answer:
(i) C₆H₅NH₂ + HCl (aq) → C₆H₅NH₃⁺Cl⁻ (Anilinium chloride)
(ii) C₂H₅NH₂ + C₆H₅SO₂Cl → C₆H₅SO₂NHC₂H₅ (N-Ethylbenzenesulphonamide) + HCl
(iii) C₂H₅NH₂ + HNO₂ → C₂H₅OH + N₂↑ + H₂O
(iv) C₆H₅NH₂ + 3Br₂ (aq) → 2, 4, 6-Br₃C₆H₂NH₂ (2, 4, 6-Tribromoaniline, white ppt.) + 3HBr
(v) C₆H₅NH₂ + CHCl₃ + 3KOH \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅NC (Phenyl isocyanide) + 3KCl + 3H₂O
(vi) CH₃CH₂NH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) CH₃CH₂NC (Ethyl isocyanide) + 3KCl + 3H₂O
(vii) AgCl + 2C₂H₅NH₂ → [Ag(C₂H₅NH₂)₂]⁺Cl⁻ (soluble complex)
(viii) C₂H₅NH₂ + 3CH₃I (excess) → C₂H₅N⁺(CH₃)₃I⁻ (Ethyltrimethylammonium iodide) + 2HI
(ix) C₆H₅NH₂ \( \xrightarrow{\mathrm{Na_2Cr_2O_7/H_2SO_4}} \) O=C₆H₄=O (p-Benzoquinone)
(x) CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(xi) C₆H₅NH₂ + NaNO₂ + 2HCl \( \xrightarrow{\mathrm{273\ K}} \) C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride) + NaCl + 2H₂O
(xii) C₆H₅N(CH₃)₂ + HNO₂ → p-ON–C₆H₄–N(CH₃)₂ (p-Nitroso-N, N-dimethylaniline, green) + H₂O
(xiii) C₆H₅NH₂ + (CH₃CO)₂O \( \xrightarrow{\mathrm{Pyridine}} \) C₆H₅NHCOCH₃ (Acetanilide) + CH₃COOH
(xiv) RNH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) RNC (Alkyl isocyanide) + 3KCl + 3H₂O
(xv) C₆H₅NH₂ + C₆H₅COCl \( \xrightarrow{\mathrm{NaOH}} \) C₆H₅NHCOC₆H₅ (Benzanilide) + HCl (Schotten-Baumann reaction)
(xvi) C₆H₅NO₂ + HNO₃ \( \xrightarrow[\mathrm{363\ K}]{\mathrm{Conc.\ H_2SO_4}} \) m-C₆H₄(NO₂)₂ (m-Dinitrobenzene) + H₂O
(xvii) RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(xviii) C₆H₅SO₂Cl + (CH₃)₂NH → C₆H₅SO₂N(CH₃)₂ (N, N-Dimethylbenzenesulphonamide) + HCl

Teacher's Note:
1. In (xii), a 3° aromatic amine undergoes ring substitution (nitrosation at the para position), unlike 1° and 2° amines which react at nitrogen.

Question 36. Identify the compounds A, B and C in the following reactions.
(i) C₆H₅NH₂ \( \xrightarrow{\mathrm{conc.\ H_2SO_4}} \) A \( \xrightarrow{\mathrm{heat,\ 453\text{-}473\ K}} \) B (CBSE 2019)
(ii) C₆H₅CONH₂ \( \xrightarrow{\mathrm{Br_2/NaOH}} \) A \( \xrightarrow{\mathrm{(CH_3CO)_2O/pyridine}} \) B (CBSE 2019)
(iii) C₆H₆ \( \xrightarrow[\mathrm{<333\ K}]{\mathrm{conc.\ HNO_3,\ conc.\ H_2SO_4}} \) A \( \xrightarrow{\mathrm{H_2/Ni}} \) B (AP Board 2012)
(iv) C₆H₅NO₂ \( \xrightarrow{\mathrm{Fe/HCl}} \) A \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) B (Hr. Board 2012 ; Assam Board 2013)
(v) C₆H₅NO₂ \( \xrightarrow{\mathrm{Sn + HCl}} \) A \( \xrightarrow[\mathrm{273\ K}]{\mathrm{NaNO_2 + HCl}} \) B \( \xrightarrow[\mathrm{\Delta}]{\mathrm{H_2O/H^+}} \) C (CBSE 2013)
(vi) CH₃Cl \( \xrightarrow{\mathrm{KCN}} \) A \( \xrightarrow{\mathrm{H_2/Ni}} \) B \( \xrightarrow{\mathrm{HNO_2}} \) C (Hr. Board 2013)
(vii) CH₃Br \( \xrightarrow{\mathrm{KCN}} \) A \( \xrightarrow{\mathrm{LiAlH_4}} \) B \( \xrightarrow[\mathrm{273\ K}]{\mathrm{HNO_2}} \) C (CBSE 2014)
(viii) CH₃COOH \( \xrightarrow[\mathrm{\Delta}]{\mathrm{NH_3}} \) A \( \xrightarrow{\mathrm{Br_2 + KOH}} \) B \( \xrightarrow{\mathrm{CHCl_3 + NaOH}} \) C (CBSE 2014)
Answer:
(i) A = C₆H₅NH₃⁺HSO₄⁻ (Anilinium hydrogensulphate) ; B = p-H₃N⁺C₆H₄SO₃⁻ (Sulphanilic acid)
(ii) A = C₆H₅NH₂ (Aniline) ; B = C₆H₅NHCOCH₃ (Acetanilide)
(iii) A = C₆H₅NO₂ (Nitrobenzene) ; B = C₆H₅NH₂ (Aniline)
(iv) A = C₆H₅NH₂ (Aniline) ; B = C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride)
(v) A = C₆H₅NH₂ (Aniline) ; B = C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride) ; C = C₆H₅OH (Phenol)
(vi) A = CH₃CN (Ethanenitrile) ; B = CH₃CH₂NH₂ (Ethanamine) ; C = CH₃CH₂OH (Ethanol)
(vii) A = CH₃CN (Ethanenitrile) ; B = CH₃CH₂NH₂ (Ethanamine) ; C = CH₃CH₂OH (Ethanol)
(viii) A = CH₃CONH₂ (Ethanamide) ; B = CH₃NH₂ (Methanamine) ; C = CH₃NC (Methyl isocyanide)

Teacher's Note:
1. In (vi) and (vii) the aliphatic amine gives an alcohol with HNO₂; in (iv) and (v) the aromatic amine gives a stable diazonium salt at 273-278 K.

Question 37. Discuss briefly the separation of primary, secondary and tertiary amines.
Answer:
(i) Hinsberg's method : The mixture of amines is treated with benzenesulphonyl chloride (Hinsberg's reagent) :
RNH₂ + C₆H₅SO₂Cl → C₆H₅SO₂NHR (soluble in KOH) + HCl
R₂NH + C₆H₅SO₂Cl → C₆H₅SO₂NR₂ (insoluble in KOH) + HCl
R₃N + C₆H₅SO₂Cl → No reaction
The reaction mixture is distilled to remove the unreacted 3° amine. The residue is treated with KOH: the N, N-dialkylsulphonamide (from 2° amine) remains insoluble and is filtered off, while the N-alkylsulphonamide (from 1° amine) dissolves as its potassium salt. The filtrate is acidified to recover the N-alkylsulphonamide. Each sulphonamide on hydrolysis with acid gives back the 1° or 2° amine.
(ii) Hoffmann's method : The mixture is treated with diethyl oxalate. The 1° amine forms a solid dialkyloxamide, the 2° amine forms a liquid dialkyloxamic ester and the 3° amine does not react. The 3° amine is distilled off, the solid oxamide is filtered and the two products are separately hydrolysed with KOH to recover the 1° and 2° amines.
2RNH₂ + (COOC₂H₅)₂ → (CONHR)₂ (solid) + 2C₂H₅OH ; R₂NH + (COOC₂H₅)₂ → R₂NCOCOOC₂H₅ (liquid) + C₂H₅OH

Teacher's Note:
1. Both methods depend on the number of H-atoms on N: 1° has two, 2° has one, 3° has none.

Question 38. What is Hinsberg's reagent? How will you distinguish between primary, secondary and tertiary amines by Hinsberg's test ? (Kerala Board 2012)
Answer:
Benzenesulphonyl chloride (C₆H₅SO₂Cl) is called Hinsberg's reagent.
(i) A 1° amine reacts to form an N-alkylbenzenesulphonamide which, due to the acidic H on N, dissolves in KOH to give a clear solution.
C₆H₅SO₂Cl + H–NHR → C₆H₅SO₂NHR + HCl ; C₆H₅SO₂NHR + KOH → [C₆H₅SO₂NR]⁻K⁺ + H₂O
(ii) A 2° amine forms an N, N-dialkylbenzenesulphonamide which has no H on N and is insoluble in KOH (a precipitate remains).
C₆H₅SO₂Cl + H–NR₂ → C₆H₅SO₂NR₂ (insoluble in alkali) + HCl
(iii) A 3° amine does not react with Hinsberg's reagent since it has no H on N.

Teacher's Note:
1. Today benzenesulphonyl chloride is often replaced by p-toluenesulphonyl chloride, which works the same way.

Long Answer Questions (5 Marks)

Teacher's Note:
1. Benzylamine behaves like an aliphatic amine because its –NH₂ is not on the ring; its diazonium salt is unstable, so it gives no azo dye.

Question 40. Arrange the following:
(i) In decreasing order of pK_b values: C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
(iii) In decreasing order of basic strength: C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH, CH₃NH₂
(iv) Decreasing order of basic strength in gas phase: C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) Increasing order of boiling point: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
Answer:
(i) C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH
(ii) (a) p-nitroaniline < aniline < p-toluidine
(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂
(iii) (C₂H₅)₂NH > CH₃NH₂ > C₆H₅N(CH₃)₂ > C₆H₅NH₂
(iv) (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃
(v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

Teacher's Note:
1. Higher pK_b means weaker base – read the direction of each order carefully before writing.
2. In the gas phase there is no solvation, so basicity simply follows the +I effect: 3° > 2° > 1° > NH₃.

Question 41. How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol into ethanoic acid
(iv) Ethanamine into methanamine
Answer:
(i) CH₃COOH (ethanoic acid) + HN₃ (hydrazoic acid) Conc. H₂SO₄
⟶
heat CH₃NH₂ (methylamine) + N₂ + CO₂
(ii) CH₃(CH₂)₄CN (hexanenitrile) Conc. HCl
⟶
partial hydrolysis CH₃(CH₂)₄CONH₂ (hexanamide) Br₂/KOH
⟶ CH₃(CH₂)₄NH₂ (1-aminopentane)
(iii) CH₃OH SOCl₂
⟶ CH₃Cl KCN (alc)
⟶ CH₃CN H⁺/H₂O
⟶ CH₃COOH
(iv) CH₃CH₂NH₂ HNO₂
⟶
(NaNO₂ + HCl) CH₃CH₂OH KMnO₄/H⁺
⟶ CH₃COOH NH₄OH
⟶ CH₃COONH₄ heat
⟶ CH₃CONH₂ Br₂/KOH
⟶ CH₃NH₂

Teacher's Note:
1. (i) can also be done the NCERT way: CH₃COOH → CH₃COONH₄ → CH₃CONH₂ → CH₃NH₂ (Hoffmann bromamide degradation).

Question 42. Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotization
(iii) Hoffmann’s bromide reaction
(iv) Coupling reaction
(v) Ammonolysis
Answer:
(i) Carbylamine reaction: When a primary amine (aromatic or aliphatic) is warmed with chloroform and alc. KOH, isocyanides are formed, which can be identified by their offensive smell. This test is used to identify the presence of a primary amine or chloroform.
RCH₂NH₂ + CHCl₃ + 3KOH (alc.) Δ
⟶ RCH₂NC + 3KCl + 3H₂O
C₆H₅NH₂ (aniline) + CHCl₃ + 3KOH (alc) Δ
⟶ C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O
(ii) Diazotization: When a primary aromatic amine is treated with NaNO₂ and HCl at 273–278 K, a diazonium salt is obtained. This reaction is known as diazotization.
C₆H₅NH₂ (aniline) + NaNO₂ + 2HCl 273–278 K
⟶ C₆H₅N⁺≡NCl⁻ (benzenediazonium chloride) + NaCl + 2H₂O
Benzenediazonium chloride is a very important synthetic compound, which can be changed into haloarenes, phenol, cyanobenzene, benzene etc.
(iii) Hoffmann’s bromide reaction: When any primary amide (aliphatic or aromatic) is treated with bromine and alkali, it gives the amine with one carbon atom less.
R–CO–NH₂ + Br₂ + 4KOH → RNH₂ + K₂CO₃ + 2KBr + 2H₂O
This reaction is used to reduce one carbon atom from a compound.
(iv) Coupling reaction: When benzenediazonium chloride is treated with phenols or aromatic amines, azo dyes are produced in which the diazo (–N=N–) group is retained. Coupling reactions generally take place at the p-position of phenol or aromatic amines.
C₆H₅N⁺≡NCl⁻ (benzenediazonium chloride) + C₆H₅OH (phenol) NaOH
⟶ C₆H₅–N=N–C₆H₄–OH (p-hydroxyazobenzene, azo dye) + HCl
(v) Ammonolysis: Reaction of alkyl halides with ammonia is known as ammonolysis. Ammonolysis generally gives a mixture of 1°, 2°, 3° amines and quaternary ammonium salt.
RCH₂Cl + NH₃  
⟶
−HCl RCH₂NH₂ RCH₂Cl
⟶
−HCl (RCH₂)₂NH RCH₂Cl
⟶
−HCl (RCH₂)₃N RCH₂Cl
⟶ (RCH₂)₄N⁺Cl⁻

Teacher's Note:
1. Diazotization is done at 273–278 K because the diazonium salt decomposes at higher temperature.
2. Phenol couples in mildly alkaline medium (as phenoxide ion), aniline in mildly acidic medium.