CBSE Class 12 Chemistry Amines Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Write the IUPAC name of the following compounds :
(i) (CH₃CH₂)₂NCH₃ (CBSE 2022) (ii) CH₃NHCH(CH₃)₂ (iii) (CH₃)₂N–CH₂–CH₃ (CBSE 2017)
Answer:
(i) (CH₃CH₂)₂N–CH₃ : N-Ethyl-N-methylethanamine
(ii) CH₃–NH–CH(CH₃)₂ : N-Methyl-2-propanamine (N-Methylpropan-2-amine)
(iii) (CH₃)₂N–CH₂–CH₃ : N, N-Dimethylethanamine

Teacher's Note:
1. Choose the longest carbon chain on N as the parent alkanamine; the other groups on N are prefixed with N-, listed alphabetically.

Question 2. Write the IUPAC name for the following compounds :
(i) (CH₃)₂N–C(CH₃)(C₂H₅)–CH₂CH₃ (ii) H₂NCH₂CH₂CH=CH₂ (CBSE 2011)
Answer:
(i) 3, N, N-Trimethylpentan-3-amine (Arabic numerals are given preference over alphabets)
(ii) H₂N–C¹H₂C²H₂C³H=C⁴H₂ : But-3-en-1-amine

Teacher's Note:
1. In (i) the carbon carrying N bears one methyl and two ethyl groups, so the parent chain is pentane with N and a methyl group both on C-3.

Question 3. How will you convert benzene into aniline ?
Answer:
C₆H₆ (Benzene) \( \xrightarrow[\mathrm{333\ K}]{\mathrm{Conc.\ HNO_3 + Conc.\ H_2SO_4}} \) C₆H₅NO₂ (Nitrobenzene) \( \xrightarrow[\mathrm{(Reduction)}]{\mathrm{Sn/HCl}} \) C₆H₅NH₂ (Aniline).

Teacher's Note:
1. Nitration followed by reduction is the standard two-step route; NH₂ cannot be put on benzene directly.

Question 4. What is ammonolysis ?
Answer:
The reaction of alkyl halides with ammonia (or amines) to form a mixture of 1°, 2°, 3° amines and quaternary salts is called ammonolysis.
RX \( \xrightarrow{\mathrm{NH_3}} \) RNH₂ \( \xrightarrow{\mathrm{RX}} \) R₂NH \( \xrightarrow{\mathrm{RX}} \) R₃N \( \xrightarrow{\mathrm{RX}} \) R₄N⁺X⁻

Teacher's Note:
1. A large excess of NH₃ favours the 1° amine, but a mixture is still obtained, so this is not a clean method for pure amines.

Question 5. How will you convert an amide into an amine with the same number of carbon atoms ?
Answer:
LiAlH₄ reduction of 1°, 2° and 3° amides gives 1°, 2° and 3° amines containing the same number of carbon atoms. For example,
C₆H₅CONH₂ (Benzamide, 1° Amide) \( \xrightarrow[\mathrm{(ii)\ H_2O}]{\mathrm{(i)\ LiAlH_4/ether}} \) C₆H₅CH₂NH₂ (Benzylamine, 1° Amine)
Similarly, RCONHR' \( \xrightarrow{\mathrm{LiAlH_4}} \) RCH₂NHR' (2° amine) and RCONR'₂ \( \xrightarrow{\mathrm{LiAlH_4}} \) RCH₂NR'₂ (3° amine).

Teacher's Note:
1. Contrast with Hofmann bromamide reaction: LiAlH₄ keeps all the carbons (C=O becomes CH₂), Hofmann removes one carbon.

Question 6. How can you convert an amide into an amine having one carbon atom less than the starting compound ?
Answer:
By Hofmann-bromamide reaction, i.e., by heating a 1° amide with Br₂ and KOH.
R–CONH₂ (1° Amide) + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) R–NH₂ + K₂CO₃ + 2KBr + 2H₂O
(1° Amine having one carbon atom less than the starting amide)

Teacher's Note:
1. The lost carbon leaves as carbonate (CO₃²⁻); only 1° amides (with –CONH₂) give this reaction.

Question 7. Give the IUPAC name and structure of the compound formed when 3-chlorobutanamide is treated with Br₂ and KOH.
Answer:
CH₃–CH(Cl)–CH₂–CO–NH₂ (3-Chlorobutanamide) + Br₂ + 5KOH \( \xrightarrow{\mathrm{\Delta}} \) CH₃–CH(OH)–CH₂–NH₂ (1-Amino-2-propanol) + K₂CO₃ + 2KBr + KCl + 2H₂O
Under the basic conditions of the reaction, C–Cl also gets hydrolysed to give C–OH.

Teacher's Note:
1. Count carbons: the 4-carbon amide gives a 3-carbon amine; the extra KOH is used up in replacing Cl by OH.

Question 8. Give structures of A, B and C in the following reactions :
CH₃Br \( \xrightarrow{\mathrm{KCN}} \) A \( \xrightarrow{\mathrm{LiAlH_4}} \) B \( \xrightarrow[\mathrm{273\ K}]{\mathrm{HNO_2}} \) C (CBSE 2014)
Answer:
CH₃Br (Bromomethane) \( \xrightarrow[\mathrm{-KBr}]{\mathrm{KCN}} \) CH₃CN (Ethanenitrile, A) \( \xrightarrow{\mathrm{LiAlH_4}} \) CH₃CH₂NH₂ (Ethanamine, B) \( \xrightarrow[\mathrm{273\ K}]{\mathrm{HNO_2}} \) CH₃CH₂OH (Ethanol, C)

Teacher's Note:
1. Going through the nitrile adds one carbon; aliphatic 1° amines with HNO₂ give alcohols (with evolution of N₂), not stable diazonium salts.

Question 9. How is aminoethane (ethylamine) obtained from ethanal (acetaldehyde) ?
Answer:
CH₃CHO (Ethanal) + NH₃ \( \xrightarrow[\mathrm{(Reductive\ amination)}]{\mathrm{H_2/Ni\ or\ NaBH_3CN}} \) [CH₃CH=NH] → CH₃CH₂NH₂ (Ethylamine).

Teacher's Note:
1. Reductive amination first forms an imine (C=N) which is then reduced to the amine with the same number of carbon atoms.

Question 10. How can a carboxylic acid be converted to an amine in one step having one carbon atom less ? (Raj. Board 2011)
or What happens when a monocarboxylic acid is heated with hydrazoic acid ?
Answer:
Carboxylic acids react with hydrazoic acid (N₃H) in presence of conc. H₂SO₄ to form 1° amines containing one carbon atom less than the carboxylic acid. For example,
CH₃COOH (Acetic acid) + N₃H (Hydrazoic acid) \( \xrightarrow[\mathrm{(Schmidt\ reaction)}]{\mathrm{Conc.\ H_2SO_4,\ \Delta}} \) CH₃NH₂ (Methylamine) + CO₂ + N₂

Teacher's Note:
1. Schmidt reaction and Hofmann bromamide reaction both shorten the chain by one carbon; the lost carbon leaves as CO₂.

Question 11. Arrange the following in increasing order of their boiling point :
C₄H₉–NH₂, (C₂H₅)₂NH, C₂H₅N(CH₃)₂. (Hr. Board 2011)
Answer:
Boiling points depend upon the extent of H-bonding which, in turn, depends upon the number of H-atoms present on the N-atom. Since C₄H₉NH₂ has two, (C₂H₅)₂NH has one and C₂H₅N(CH₃)₂ has no hydrogen linked to nitrogen, therefore, boiling points decrease as the extent of H-bonding decreases, i.e., boiling points decrease in the order : C₄H₉NH₂ > (C₂H₅)₂NH > C₂H₅N(CH₃)₂.
Hence, increasing order : C₂H₅N(CH₃)₂ < (C₂H₅)₂NH < C₄H₉NH₂

Teacher's Note:
1. All three are isomers (C₄H₁₁N), so the difference comes only from the number of N–H bonds; read the question carefully for "increasing" or "decreasing".

Question 12. Give reasons for the following :
(i) (CH₃)₂NH is more basic than (CH₃)₃N
(ii) Primary amines have higher boiling point than tertiary amines. (CBSE 2016)
Answer:
(i) The basicity of (CH₃)₂NH and (CH₃)₃N mainly depends upon the stability of the ammonium cation formed when each one of them accepts a proton. Since the ammonium cation derived from (CH₃)₂NH is stabilized by H-bonding with two H-atoms while the ammonium cation derived from (CH₃)₃N is stabilized by H-bonding with only one H-atom, therefore, (CH₃)₂NH is more basic than (CH₃)₃N.
(CH₃)₂NH + H⁺ → (CH₃)₂N⁺H₂, H-bonded to two H₂O molecules (more stable);
(CH₃)₃N + H⁺ → (CH₃)₃N⁺–H, H-bonded to one H₂O molecule (less stable).
(ii) The boiling points of amines depends upon the degree of association due to H-bonding which, in turn, depends upon the number of H-atoms on the N-atom. Since 1° amines have two H-atoms linked to N-atom while 3° amines have no H-atom linked to N-atom, therefore, 1° amines exist as associated molecules (through intermolecular N–H···N H-bonds) while 3° amines exist as individual molecules. To break these H-bonds, large amount of energy is needed. Therefore, 1° amines have higher boiling point than 3° amines.

Teacher's Note:
1. In aqueous solution basicity is decided by solvation (H-bonding) of the cation as well as the +I effect; that is why the 3° methylamine is not the strongest base.

Question 13. Methylamine is soluble in water but not aniline. Explain. (Manipur Board 2011)
Answer:
Methylamine forms H-bonds with water and hence it is soluble in water (CH₃–N–H···OH₂ and CH₃–N···H–OH).
In contrast, the large hydrocarbon part of aniline retards the tendency of the –NH₂ group to form H-bonds with water and hence it is only sparingly soluble in water.

Teacher's Note:
1. Solubility of amines in water falls as the size of the hydrophobic carbon part increases.

Question 14. What amine salts are used for determining their molecular masses ?
Answer:
Chloroplatinates. For example, (RN⁺H₃)₂PtCl₆²⁻.

Teacher's Note:
1. On ignition the chloroplatinate leaves metallic platinum; from the mass of Pt the molecular mass of the amine is calculated.

Short Answer Questions (2 Marks)

Question 15. The presence of a base is needed in the ammonolysis of alkyl halides. Explain.
Answer:
During ammonolysis of alkyl halides, the acid liberated during the reaction combines with the amine formed to form the amine salt. To liberate free amine from the amine salt, a base is needed. The base used is excess of alcoholic NH₃ which serves both as a nucleophile as well as a base. For example
NH₃ (Nucleophile) + R–X (Alkyl halide) 373 K
⟶
  R–NH₃⁺X⁻ (Amine salt)
R–NH₃⁺X⁻ (Amine salt) + NH₃ (Base) ⇄ R–NH₂ (1° Amine) + NH₄⁺X⁻

Teacher's Note:
1. Using a large excess of NH₃ also reduces further alkylation to 2°, 3° amines and quaternary salts.

Question 16. Can we prepare aniline by Gabriel phthalimide reaction ? Comment.
Or
Gabriel phthalimide synthesis is not preferred for preparing aromatic primary amines. Explain. (CBSE 2019)
Answer:
Aromatic primary amines, i.e., aniline cannot be prepared by Gabriel phthalimide reaction. The reason being that preparation of aniline by Gabriel phthalimide reaction requires the treatment of pot. phthalimide with C₆H₅Cl or C₆H₅Br. Since aryl halides do not undergo nucleophilic substitution under ordinary laboratory conditions, therefore, C₆H₅Cl or C₆H₅Br does not react with pot. phthalimide to give N-phenylphthalimide and hence aniline cannot be prepared by this method.

Teacher's Note:
1. The key phrase examiners look for: aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

Question 17. Why ethanolic KOH and not aqueous KOH is used in the isocyanide test of aniline? (West Bengal Board 2013)
Answer:
The isocyanide test of aniline requires the intermediate formation of dichlorocarbene (:CCl₂).
C₆H₅NH₂ + :CCl₂ → C₆H₅–NH₂⁺–C⁻Cl₂ −HCl
⟶
  C₆H₅–NH–CHCl₂ KOH(alc.)
⟶
−2HCl C₆H₅–N⁺≡C:⁻ or C₆H₅–N≡C (Phenyl isocyanide)
To generate : CCl₂ from CHCl₃, a strong base like alc. KOH is needed. With aq. KOH only hydrolysis of CHCl₃ to HCOOK will occur.
CHCl₃ (Chloroform) + 3 KOH (aq) −3KCl
⟶
  HC(OH)₃ −H₂O
⟶
  HCOOH KOH
⟶
−H₂O HCOOK (Pot. formate)

Teacher's Note:
1. Alcoholic KOH contains the stronger base ethoxide ion, which removes H⁺ from CHCl₃ to form :CCl₂ instead of hydrolysing it.

Question 18. Why is an amide more acidic than an amine ?
Answer:
R–C(=O)–NH₂ ↔ R–C(–O⁻)=N⁺H₂
Firstly, due to delocalization of lone pair of electrons of N atom over the C = O group, amino group gets a positive charge which makes N–H bond weak. Secondly, the anion left after removal of a proton is stabilized by resonance.
R–C(=O)–NH⁻ ↔ R–C(–O⁻)=NH
In amines no such stabilization is possible.

Teacher's Note:
1. Acidity questions are best answered through the stability of the conjugate base (the anion left after losing H⁺).

Question 19. Which one is more acidic ? Explain.
C₆H₅–NH₃⁺ (anilinium ion) or p-F–C₆H₄–NH₃⁺ (p-fluoroanilinium ion)
Answer:
Due to powerful –I-effect of the F-atom, it withdraws electrons from the NH₃⁺ group. As a result, electron density in the N–H bond of p-fluoroanilinium ion decreases and hence release of a proton from p-fluoroanilinium ion is much more easier than from anilinium ion. Therefore, p- fluoroanilinium ion is more acidic than anilinium ion.

Teacher's Note:
1. Electron-withdrawing substituents make the anilinium ion more acidic and the corresponding aniline less basic.

Question 20. How will you distinguish between (CH₃)₂NH and (CH₃)₃N ? (CBSE 2015)
Answer:
These can be distinguished by Liebermann nitrosoamine reaction. (CH₃)₂NH (dimethylamine) on treatment with HNO₂ (generated in situ by the action of dil. HCl on NaNO₂) gives yellow coloured oily N-nitrosodimethylamine.
(CH₃)₂NH (Dimethylamine) + HO–N=O → (CH₃)₂N–N=O (N-Nitrosodimethylamine) + H₂O
N-Nitrosodimethylamine on warming with a crystal of phenol and conc. H₂SO₄ gives a green solution which when made alkaline with aqueous NaOH turns deep blue and then red on dilution.
(CH₃)₃N (trimethylamine), on the other hand, being a 3° amine does not give this test.

Teacher's Note:
1. Hinsberg’s test also works: dimethylamine gives an alkali-insoluble sulphonamide, trimethylamine does not react.

Question 21. Why is it necessary to add excess of mineral acid during diazotisation of amines ? (AIPMT Mains 2008)
Answer:
During diazotisation of amines, excess of mineral acid is used. In general, one mole of amine is treated with approx. three moles of acid : (i) one mole to dissolve the amine, (ii) one mole to liberate HNO₂ from NaNO₂ and (iii) one mole to maintain the proper acidity of the reaction mixture to prevent the coupling of diazonium salt thus formed with the undiazotised (free) amine to give the azo dye.

Teacher's Note:
1. Without excess acid, the diazonium salt couples with unreacted aniline to give diazoaminobenzene (a yellow solid), lowering the yield.

Question 22. (a) Give the IUPAC names of the following compounds :
(i) CH₃O–C₆H₄–NO₂ (structure shown below) (ii) C₆H₅CH=CHNO₂
(b) Write the full name of T.N.T. and its structural formula. (UP Board 2012)

Answer:
(a) (i) 1-Methoxy-4-nitrobenzene (p-nitroanisole)
(ii) 1-Nitro-2-phenylethene [(2-nitroethenyl)benzene, β-nitrostyrene]
(b) T.N.T. is 2, 4, 6-trinitrotoluene (IUPAC : 2-methyl-1, 3, 5-trinitrobenzene). Its structure is :

Teacher's Note:
1. In benzene derivatives, substituent names are listed alphabetically and the ring is numbered to give the lowest locants.

Question 23. What are primary, secondary and tertiary amines ? Give one example of each.
Answer:
Amines are classified as primary (1°), secondary (2°) and tertiary (3°) depending upon the number of H-atoms of ammonia replaced by alkyl or aryl groups.
(i) Primary amines (RNH₂) : one H-atom is replaced, e.g., CH₃NH₂ (methylamine).
(ii) Secondary amines (R₂NH) : two H-atoms are replaced, e.g., (CH₃)₂NH (dimethylamine).
(iii) Tertiary amines (R₃N) : all three H-atoms are replaced, e.g., (CH₃)₃N (trimethylamine).

Teacher's Note:
1. Count the carbon groups directly on nitrogen: one, two or three gives 1°, 2° or 3°.

Question 24. How will you prepare ethylamine by using Hofmann bromamide reaction ? or Write chemical equations to prepare ethanamine from propanamide.
Answer:
Propanamide on heating with Br₂ and aqueous KOH (or NaOH) gives ethanamine :
CH₃CH₂CONH₂ (Propanamide) + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) CH₃CH₂NH₂ (Ethanamine) + K₂CO₃ + 2KBr + 2H₂O

Teacher's Note:
1. The amine formed has one carbon less than the amide; to get ethylamine you must start with the three-carbon amide.

Question 25. How is ethylamine prepared from acetonitrile ? Compare its basic character with that of ammonia ?
Answer:
Acetonitrile on reduction with Na/C₂H₅OH (Mendius reduction) or LiAlH₄ or H₂/Ni gives ethylamine.
CH₃C≡N + 4[H] \( \xrightarrow{\mathrm{Na/C_2H_5OH}} \) CH₃CH₂NH₂
Ethylamine is more basic than ammonia. Due to +I-effect of the ethyl group, the electron density on the N-atom of ethylamine is higher than that on N-atom of NH₃ and hence it can donate its lone pair more easily.

Teacher's Note:
1. pK_b of ethylamine (3.29) is lower than that of ammonia (4.75), confirming it is the stronger base.

Question 26. Write balanced chemical equation for the preparation of methylamine from acetamide.
Answer:
CH₃CONH₂ (Acetamide) + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) CH₃NH₂ (Methylamine) + K₂CO₃ + 2KBr + 2H₂O

Teacher's Note:
1. For balancing, remember 4 mol of KOH: two form K₂CO₃ and two form KBr.

Question 27. Account for the following :
It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Or Ammonolysis of alkyl halides does not give the corresponding amine in pure state.
Answer:
The 1° amine formed by ammonolysis is a stronger nucleophile than NH₃. Therefore, it reacts further with the alkyl halide to form the 2° amine, which in turn gives the 3° amine and finally the quaternary ammonium salt.
RX \( \xrightarrow{\mathrm{NH_3}} \) RNH₂ \( \xrightarrow{\mathrm{RX}} \) R₂NH \( \xrightarrow{\mathrm{RX}} \) R₃N \( \xrightarrow{\mathrm{RX}} \) R₄N⁺X⁻
Thus, a mixture of amines is obtained and it is difficult to get a pure amine.

Teacher's Note:
1. Using a large excess of ammonia increases the yield of the 1° amine; Gabriel phthalimide synthesis gives a pure 1° amine.

Short Answer Questions (3 Marks)

Question 28. Arrange the following in the increasing order of their basic strength :
CH₃NH₂, (CH₃)₂NH, C₆H₅NH₂, C₆H₅CH₂NH₂ (CBSE (Foreign) 2017)
Answer:
(i) Due to +I-effect of two CH₃ groups as against one CH₃ group in CH₃NH₂, (CH₃)₂NH is more basic than CH₃NH₂
(ii) In C₆H₅NH₂, N is directly linked to the benzene ring and hence the lone pair of electrons on the N-atom is delocalized over the benzene ring. In contrast, N in C₆H₅CH₂NH₂ is not directly linked to the benzene ring and hence its lone pair is not delocalized over the benzene ring. In other words, the lone pair of electrons on the N-atom in C₆H₅CH₂NH₂ is more easily available for protonation than that on the N-atom in C₆H₅NH₂. Thus, C₆H₅CH₂NH₂ is more basic than C₆H₅NH₂. Further, due to –I-effect of the C₆H₅ group, electron density on the N-atom in C₆H₅CH₂NH₂ is lower than that on the N-atom in CH₃NH₂, therefore, CH₃NH₂ is more basic than C₆H₅CH₂NH₂. In view of all the facts discussed above, the overall basic strength increases in the order :
C₆H₅NH₂ < C₆H₅CH₂NH₂ < CH₃NH₂ < (CH₃)₂ NH.

Teacher's Note:
1. Benzylamine is an aliphatic amine, but the electron-withdrawing phenyl group makes it slightly weaker than methylamine.

Question 29. Write the structures of A, B and C in the following :
(a) CH₃–Cl KCN
⟶
  A LiAlH₄
⟶
  B CHCl₃ + alc.KOH
⟶
  C (CBSE 2016)
(b) CH₃ – COOH NH₃/Δ
⟶
  A Br₂/KOH(aq)
⟶
  B CHCl₃ + alc. KOH
⟶
  C (CBSE 2013, 2014)
Answer:
(a) CH₃–Cl (Chloromethane) KCN
⟶
Nucleophilic substitution (−KCl) CH₃–CN (Ethanenitrile) (A) LiAlH₄
⟶
(Reduction) CH₃–CH₂NH₂ (Ethanamine) (B) CHCl₃ + alc.KOH, Δ
⟶
(Carbylamine reaction) CH₃CH₂–N≡C (Ethyl carbylamine) (C)
(b) CH₃COOH (Acetic acid) NH₃
⟶
  CH₃COONH₄ (Amm. acetate) Δ
⟶
−H₂O CH₃–CO–NH₂ (Acetamide) (A) Br₂/KOH (aq.)
⟶
(Hofmann bromamide reaction) CH₃ – NH₂ (Methanamine) (B) CHCl₃ + alc. KOH
⟶
(Carbylamine reaction) CH₃ – N≡C (Methyl isocyanide or Methyl carbylamine) (C)

Teacher's Note:
1. KCN followed by reduction adds one carbon (ascent), while Hofmann bromamide reaction removes one carbon (descent); keep track of carbon count at each step.

Question 30. Suggest chemical reactions for the following conversions :
(i) Cyclohexanol → Cyclohexylamine
(ii) 1-Hexanenitrile → 1-Aminopentane
Answer:
(i) Cyclohexanol K₂Cr₂O₇/H₂SO₄
⟶
(Oxidation) Cyclohexanone NH₃/H₂–Ni
⟶
(Reductive amination) Cyclohexylamine
(ii) CH₃CH₂CH₂CH₂CH₂C≡N (1-Hexanenitrile) H⁺/H₂O
⟶
(Hydrolysis) CH₃(CH₂)₄COOH (Hexanoic acid) NH₃
⟶
  CH₃(CH₂)₄COONH₄
Δ
⟶
−H₂O CH₃(CH₂)₄CONH₂ (Hexanamide) NaOH/Br₂
⟶
(Hofmann bromamide reaction) CH₃CH₂CH₂CH₂CH₂NH₂ (1-Aminopentane)

Teacher's Note:
1. Hexanenitrile has 6 C and 1-aminopentane has 5 C, so a one-carbon descent (Hofmann bromamide reaction) must be part of the route.

Question 31. 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 (Foreign) 2017)
Answer:
C₆H₅NO₂ (Nitrobenzene) Sn/HCl
⟶
(Reduction) C₆H₅NH₂ (Aniline) (A) (CH₃CO)₂O
⟶
Pyridine (Acetylation) C₆H₅NHCOCH₃ (Acetanilide) (B)
HNO₃ + H₂SO₄
⟶
288 K (Nitration) p-O₂N–C₆H₄–NHCOCH₃ (p-Nitroacetanilide) (C) (major product) OH⁻ or H⁺
⟶
(Hydrolysis) p-O₂N–C₆H₄–NH₂ (p-Nitroaniline) (D)
C₆H₅NH₂ (A) H₂SO₄
⟶
  C₆H₅NH₃⁺HSO₄⁻ (Aniline hydrogen sulphate) (E)

Teacher's Note:
1. Acetylation protects –NH₂ from oxidation and protonation during nitration, so the p-nitro product is obtained in good yield.

Question 32. Identify A–D :
C₆H₅CH₂Cl A
⟶
  C₆H₅CH₂NO₂ Sn/HCl
⟶
  B
C NaOH (aq)/Br₂
⟶
  B
B D
⟶
  C₆H₅CH₂NC (CBSE 2018S)
Answer:
C₆H₅CH₂Cl AgNO₂ (A)
⟶
(Nucleophilic substitution) C₆H₅CH₂NO₂ Sn/HCl
⟶
  C₆H₅CH₂NH₂ (Benzylamine) (B)
C₆H₅CH₂CONH₂ (Phenylethanamide) (C) NaOH (aq)/Br₂
⟶
  C₆H₅CH₂NH₂ (B)
C₆H₅CH₂NH₂ (B) CHCl₃/KOH (alc.), Δ (D)
⟶
(Carbylamine reaction) C₆H₅CH₂NC (Benzyl isocyanide)
Thus, A = AgNO₂, B = C₆H₅CH₂NH₂, C = C₆H₅CH₂CONH₂ and D = CHCl₃/KOH (alc.)

Teacher's Note:
1. AgNO₂ gives nitroalkanes (attack through N), whereas KNO₂/NaNO₂ gives mainly alkyl nitrites (attack through O).

Question 33. Write the structure of A, B and C in the following :
(a) C₆H₅–CONH₂ Br₂/aq.KOH
⟶
  A NaNO₂ + HCl
⟶
0–5°C B KI
⟶
  C (CBSE 2016)
(b) C₆H₅N₂⁺BF₄⁻ NaNO₂/Cu
⟶
Δ A Fe/HCl
⟶
  B CH₃COCl/pyridine
⟶
  C (CBSE (OD) 2017)
Answer:
(a) C₆H₅–CONH₂ (Benzamide) Br₂/aq.KOH
⟶
(Hofmann bromamide reaction) C₆H₅NH₂ (Aniline) (A) NaNO₂+HCl, 0–5°C
⟶
(Diazotisation) C₆H₅–N₂⁺Cl⁻ (Benzenediazonium chloride) (B) KI
⟶
−N₂, −KCl C₆H₅–I (Iodobenzene) (C)
(b) C₆H₅N₂⁺BF₄⁻ (Benzenediazonium tetrafluoroborate) NaNO₂/Cu
⟶
−NaBF₄, −N₂ C₆H₅NO₂ (Nitrobenzene) (A) Fe/HCl
⟶
(Reduction) C₆H₅NH₂ (Aniline) (B) CH₃COCl/pyridine
⟶
(Acetylation) C₆H₅NH – COCH₃ (Acetanilide) (C)

Teacher's Note:
1. Iodine is introduced simply by warming the diazonium salt with KI; no copper catalyst is needed, unlike Cl and Br.

Question 34. Give IUPAC names of the following :
(i) (CH₃)₃N (J & K Board 2012; CBSE Sample Paper 2016) (ii) (CH₃CH₂)₂N–CH₃ (CBSE 2022)
(iii) CH₃–N(CH₃)–CH₂CH₃ (J & K Board 2012) (iv) (CH₃)₃C–NH₂
(v) H₂NCH₂CH=CH₂ (J & K Board 2012) (vi) C₆H₅CH₂NH₂
(vii) CH₃CH₂–CH(NH₂)–CH₂–CH(CH₃)–CH₃
(viii) (ix) (x) CH₃CH₂CH₂–N(CH₃)₂
(xi) (CBSE 2016) (xii) C₆H₅CH₂NH₂ (CBSE Sample Paper 2016)
Answer:
(i) N, N-Dimethylmethanamine
(ii) N-Ethyl-N-methylethanamine
(iii) N, N-Dimethylethanamine
(iv) 2-Methylpropan-2-amine
(v) Prop-2-en-1-amine
(vi) Phenylmethanamine
(vii) 5-Methylhexan-3-amine
(viii) N, N-Dimethyl-4-nitrosoaniline (N, N-Dimethyl-4-nitrosobenzenamine)
(ix) 2, 4-Dinitroaniline (2, 4-Dinitrobenzenamine)
(x) N, N-Dimethylpropan-1-amine
(xi) 2, 4, 6-Tribromoaniline (2, 4, 6-Tribromobenzenamine)
(xii) Phenylmethanamine

Teacher's Note:
1. In (vii) number the chain from the end that gives the NH₂ group the lower locant (3, not 4), then place the methyl group.

Question 35. Give the IUPAC names of the following :
(i) m-Toluidine (ii) Ethylmethyl n-propylamine (iii) n-Propylamine (iv) t-Butylamine (v) Dimethyl-sec-butylamine (vi) 2-(N-Methylamino)heptane (vii) Ethylenediamine (viii) Tri-n-propylamine (ix) Benzylamine (x) Isobutylamine (Maharashtra Board 2012) (xi) 1-Amino-2-phenylpropane (xii) ethylmethylamine (Maharashtra Board 2012)
Answer:
(i) m-CH₃C₆H₄NH₂ : 3-Methylaniline (3-Methylbenzenamine)
(ii) CH₃CH₂CH₂N(CH₃)CH₂CH₃ : N-Ethyl-N-methylpropan-1-amine
(iii) CH₃CH₂CH₂NH₂ : Propan-1-amine
(iv) (CH₃)₃CNH₂ : 2-Methylpropan-2-amine
(v) CH₃CH₂CH(CH₃)N(CH₃)₂ : N, N-Dimethylbutan-2-amine
(vi) CH₃(CH₂)₄CH(CH₃)NHCH₃ : N-Methylheptan-2-amine
(vii) H₂NCH₂CH₂NH₂ : Ethane-1, 2-diamine
(viii) (CH₃CH₂CH₂)₃N : N, N-Dipropylpropan-1-amine
(ix) C₆H₅CH₂NH₂ : Phenylmethanamine
(x) (CH₃)₂CHCH₂NH₂ : 2-Methylpropan-1-amine
(xi) H₂NCH₂CH(C₆H₅)CH₃ : 2-Phenylpropan-1-amine
(xii) CH₃NHCH₂CH₃ : N-Methylethanamine

Teacher's Note:
1. Write the structure from the common name first; only then choose the longest chain on N and name the other groups with N- prefixes.

Question 36. How will you prepare ethylamine from each of the following ?
(i) Ethyl bromide (ii) Methyl cyanide (iii) Propionamide (iv) Acetamide
Answer:
(i) CH₃CH₂Br + NH₃ (alc., excess) \( \xrightarrow[\mathrm{sealed\ tube}]{\mathrm{373\ K}} \) CH₃CH₂NH₂ + HBr
(ii) CH₃C≡N + 4[H] \( \xrightarrow{\mathrm{Na/C_2H_5OH\ or\ LiAlH_4}} \) CH₃CH₂NH₂ (Mendius reduction)
(iii) CH₃CH₂CONH₂ + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) CH₃CH₂NH₂ + K₂CO₃ + 2KBr + 2H₂O (Hofmann bromamide reaction)
(iv) CH₃CONH₂ + 4[H] \( \xrightarrow{\mathrm{LiAlH_4/ether}} \) CH₃CH₂NH₂ + H₂O

Teacher's Note:
1. Check the carbon count each time: (ii) and (iv) keep two carbons, while in (iii) the three-carbon amide loses one carbon.

Question 37. How are amines prepared from (a) amides (b) oximes and (c) aldehydes.
Answer:
(a) From amides : (i) Reduction with LiAlH₄ gives amines with the same number of carbon atoms : RCONH₂ + 4[H] \( \xrightarrow{\mathrm{LiAlH_4}} \) RCH₂NH₂ + H₂O
(ii) Hofmann bromamide reaction gives 1° amines with one carbon atom less : RCONH₂ + Br₂ + 4KOH → RNH₂ + K₂CO₃ + 2KBr + 2H₂O
(b) From oximes : Reduction of oximes with Na/C₂H₅OH or LiAlH₄ gives 1° amines.
CH₃CH=NOH (Acetaldoxime) + 4[H] \( \xrightarrow{\mathrm{Na/C_2H_5OH}} \) CH₃CH₂NH₂ + H₂O
(c) From aldehydes (reductive amination) : Aldehydes react with NH₃ and H₂ in presence of Ni to give 1° amines.
CH₃CHO + NH₃ \( \xrightarrow{\mathrm{H_2/Ni}} \) CH₃CH₂NH₂ + H₂O

Teacher's Note:
1. Both oxime reduction and reductive amination convert a C=O group into a CH–NH₂ group without changing the carbon skeleton.

Question 38. Write the IUPAC name of diethylamine. How will you obtain it from nitroethane ?
Answer:
IUPAC name of diethylamine, (CH₃CH₂)₂NH : N-Ethylethanamine.
(i) Reduction of nitroethane gives ethylamine : CH₃CH₂NO₂ + 6[H] \( \xrightarrow{\mathrm{Sn/HCl}} \) CH₃CH₂NH₂ + 2H₂O
(ii) Hydrolysis of nitroethane with dil. HCl gives ethanoic acid which is converted into acetyl chloride :
CH₃CH₂NO₂ \( \xrightarrow{\mathrm{HCl/H_2O,\ \Delta}} \) CH₃COOH \( \xrightarrow{\mathrm{SOCl_2}} \) CH₃COCl
(iii) Acetylation of ethylamine followed by reduction with LiAlH₄ gives diethylamine :
CH₃CH₂NH₂ + CH₃COCl \( \xrightarrow{\mathrm{Pyridine}} \) CH₃CH₂NHCOCH₃ \( \xrightarrow{\mathrm{LiAlH_4}} \) CH₃CH₂NHCH₂CH₃ (Diethylamine)

Teacher's Note:
1. Reducing an N-alkyl amide with LiAlH₄ gives a pure 2° amine, avoiding the mixture that alkylation with C₂H₅Br would give.

Question 39. Explain the following : (i) The boiling point of CH₃NH₂ is less than that of methanoic acid.
(ii) The boiling point of ethyl alcohol is more than that of ethylamine.
(iii) Amines have lower boiling points than alcohols of comparable molecular masses.
(iv) Amines have higher boiling points than hydrocarbons of comparable molecular masses.
(v) Primary amines have higher boiling points than corresponding tertiary amines.
(vi) The boiling points of amines having comparable molecular masses decrease in the order : primary > secondary > tertiary.
Answer:
(i) Methanoic acid (HCOOH) molecules are held together by strong O–H···O hydrogen bonds (they even exist as dimers) whereas CH₃NH₂ forms weaker N–H···N hydrogen bonds since N is less electronegative than O. Further, molecular mass of HCOOH (46) is higher than that of CH₃NH₂ (31). Hence, b.p. of CH₃NH₂ is less than that of HCOOH.
(ii) Since O is more electronegative than N, O–H···O hydrogen bonds in ethyl alcohol are stronger than N–H···N hydrogen bonds in ethylamine. Therefore, b.p. of ethyl alcohol is higher.
(iii) For the same reason as in (ii), the H-bonds in alcohols are stronger than in amines. Therefore, amines have lower b.ps than alcohols of comparable molecular masses.
(iv) Amines are polar and 1° and 2° amines are associated by intermolecular H-bonding whereas hydrocarbons are non-polar and their molecules are held by weak van der Waals forces only. Therefore, amines have higher boiling points than hydrocarbons.
(v) 1° amines have two H-atoms on N and hence are associated by intermolecular H-bonding while 3° amines have no H-atom on N and cannot form H-bonds. Therefore, 1° amines have higher b.ps.
(vi) The extent of H-bonding depends upon the number of H-atoms on the N-atom: 1° amines have two, 2° amines have one and 3° amines have none. Hence, b.ps decrease in the order : 1° > 2° > 3°.

Teacher's Note:
1. Two factors decide such comparisons: the strength of H-bonds (O beats N) and the number of H atoms available for H-bonding.

Long Answer Questions (5 Marks)

Question 40. A hydrocarbon ‘A’, (C₄H₈) on reaction with HCl gives a compound ‘B’, (C₄H₉Cl), which on reaction with 1 mol of NH₃ gives compound ‘C’, (C₄H₁₁N). On reacting with NaNO₂ and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 moles of acetaldehyde. Identify the compounds ‘A’ to ‘D’. Explain the reactions involved.
Answer:
(A) →[Ozonolysis] 2CH₃CHO
C₄H₈ (A) →[HCl] C₄H₉Cl (B): Addition of HCl has occurred on ‘A’. This implies ‘A’ is an alkene.
C₄H₉Cl (B) →[NH₃] C₄H₁₁N (C): Cl in compound ‘B’ is substituted by NH₂ to give ‘C’.
(C) →[NaNO₂/HCl, H₂O] (D): ‘C’ gives a diazonium salt with NaNO₂/HCl that liberates N₂ to give optically active alcohol. This means that ‘C’ is a primary amine. Number of carbon atoms in amine is same as compound ‘A’.
Since products of ozonolysis of compound ‘A’ are CH₃—CH=O and O=CH—CH₃. Therefore, the compound ‘A’ is CH₃—CH=CH—CH₃.
On the basis of structure of ‘A’, the reactions can be explained as follows:
CH₃—CH=CH—CH₃ (A) →[HCl] CH₃CH₂—CH(Cl)—CH₃ (B)
CH₃CH₂—CH(Cl)—CH₃ (B) →[NH₃] CH₃—CH₂—CH(NH₂)—CH₃ (C)
CH₃—CH₂—CH(NH₂)—CH₃ (C) →[NaNO₂/HCl, H₂O] CH₃—CH(OH)—CH₂CH₃ (D, Optically active)
Hence A = but-2-ene, B = 2-chlorobutane, C = butan-2-amine, D = butan-2-ol.

Teacher's Note:
1. Butan-2-ol is optically active because C-2 carries four different groups: H, OH, CH₃ and C₂H₅.

Question 41. Predict the reagents or the products in the following reaction sequence:
p-Nitrotoluene →[1] p-Toluidine (p-CH₃C₆H₄NH₂) →[(CH₃CO)₂O, Pyridine] p-CH₃C₆H₄NHCOCH₃ →[HNO₃/H₂SO₄] 2 →[3] 4-Methyl-2-nitroaniline →[NaNO₂/HCl] 4 →[5] 3-Nitrotoluene (m-nitrotoluene)
Answer:
1. Sn–HCl (conc.)
2. N-(4-Methyl-2-nitrophenyl)ethanamide (2-nitro-4-methylacetanilide): the NO₂ group enters ortho to –NHCOCH₃.
3. H₂O/H⁺
4. 4-Methyl-2-nitrobenzenediazonium chloride (the –NH₂ group is converted to –N₂⁺Cl⁻)
5. H₃PO₂/H₂O

Teacher's Note:
1. –NHCOCH₃ is a stronger o/p-director than –CH₃; with the para position blocked, nitration goes ortho to it.
2. This sequence is a classic way to get a meta product: the amino group directs and is then removed.

Question 42. Write the structure 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
OR
(i) Write the structures of the main products when benzene diazonium chloride reacts with the following reagents:
(a) CuCN
(b) CH₃CH₂OH
(c) Cu/HCl
(ii) Arrange the following in increasing order of their basic strength: CH₃NH₂, (CH₃)₂NH, C₆H₅NH₂, C₆H₅CH₂NH₂
(iii) Write one chemical test to distinguish between Aniline and Ethyl amine. [CBSE Foreign Set-1, 2, 3 2017]
Answer:
A – C₆H₅NH₂ (Aniline)
B – C₆H₅NHCOCH₃ (Acetanilide)
C – p-O₂N–C₆H₄–NHCOCH₃ (p-Nitroacetanilide)
D – p-O₂N–C₆H₄–NH₂ (p-Nitroaniline)
E – C₆H₅NH₃⁺HSO₄⁻ (Anilinium hydrogensulphate)
OR
(i) (a) C₆H₅CN   (b) C₆H₆   (c) C₆H₅Cl
(ii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < CH₃NH₂ < (CH₃)₂NH
(iii) Add NaNO₂ + HCl to both the compounds at 273 K followed by addition of phenol. Aniline gives orange dye. (Or any other correct test)

Teacher's Note:
1. Acetylation protects –NH₂ so that nitration gives mainly the para product; hydrolysis afterwards gives back the free amino group.

Question 43. (a) Write the reactions involved in the following:
(i) Hoffmann bromamide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Give reasons:
(i) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.
OR
(a) Write the structures of the main products of the following reactions:
(i) C₆H₅NH₂ → [(CH₃CO)₂O, Pyridine] →
(ii) C₆H₅SO₂Cl → [(CH₃)₂NH] →
(iii) C₆H₅N₂⁺Cl⁻ → [CH₃CH₂OH] →
(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 [CBSE Delhi/OD 2018]
Answer:
(a) (i) Ar/R–CONH₂ + Br₂ + 4NaOH → Ar/R–NH₂ + 2NaBr + Na₂CO₃ + 2H₂O
(ii) C₆H₅NH₂ + NaNO₂ + 2HCl → [273–278 K] → C₆H₅N₂Cl + NaCl + 2H₂O (or any other correct equation)
(iii) C₆H₄(CO)₂N–H (Phthalimide) → [KOH] → C₆H₄(CO)₂NK → [R–X] → C₆H₄(CO)₂N–R;   C₆H₄(CO)₂N–R → [NaOH(aq)] → C₆H₄(COO⁻Na⁺)₂ + R–NH₂
(b) (i) Because of the combined factor of inductive effect and solvation or hydration effect.
(ii) Due to resonance stabilisation (the positive charge of the –N₂⁺ group is delocalised over the benzene ring).
OR
(a) (i) C₆H₅NHCOCH₃   (ii) C₆H₅SO₂N(CH₃)₂   (iii) C₆H₆
(b) Add chloroform in the presence of KOH and heat, Aniline gives a offensive smell while N,N-dimethylaniline does not. (or any other correct test)
(c) C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂

Teacher's Note:
1. A lower pK_b means a stronger base, so the pK_b order is the reverse of the basic strength order.