Class 12 Chemistry | Chapter 9
Amines
Classification • Basicity • Name Reactions • Diazonium Salts
1. Introduction, Structure and Classification
Amines are organic derivatives of ammonia (NH3), obtained by replacing one, two, or all three hydrogen atoms by alkyl/aryl groups.
1.1 Classification
- Primary (1°) Amines: One H replaced (R-NH2). Example: Methylamine, Aniline.
- Secondary (2°) Amines: Two H replaced (R-NH-R'). Example: Dimethylamine, N-Methylaniline.
- Tertiary (3°) Amines: Three H replaced (R3N). Example: Trimethylamine, N,N-Dimethylaniline.
1.2 Structure
The nitrogen atom in amines is sp³ hybridized. The geometry is pyramidal due to the presence of an unshared pair of electrons (lone pair), causing the bond angle to be slightly less than 109.5° (around 108° in trimethylamine).
1.3 Boiling Points
Primary and secondary amines can form intermolecular hydrogen bonds because they have N-H bonds. Tertiary
amines cannot.
Boiling point order (isomeric amines): Primary > Secondary > Tertiary.
Since oxygen is more electronegative than nitrogen, alcohols have much stronger H-bonds and higher B.P.
than amines of comparable mass.
2. Preparation of Amines
2.1 Reduction of Nitro Compounds
Nitro compounds are reduced to primary amines by passing hydrogen gas in presence of finely divided Ni/Pd/Pt or by reduction with metals in acidic medium.
1. R-NO2 + H2/Pd (Ethanol) → R-NH2
2. R-NO2 + Sn/HCl (or Fe/HCl) → R-NH2 (Fe/HCl is preferred as FeCl2 gets hydrolyzed to release HCl, making it cost-effective).
2. R-NO2 + Sn/HCl (or Fe/HCl) → R-NH2 (Fe/HCl is preferred as FeCl2 gets hydrolyzed to release HCl, making it cost-effective).
2.2 Ammonolysis of Alkyl Halides
Nucleophilic substitution of alkyl halide by ammonia. Forms a mixture of 1°, 2°, 3° amines and quaternary ammonium salt. (Not good for preparing pure primary amine unless huge excess of NH3 is used).
NH3 + R-X → R-NH2 → R2NH →
R3N → R4N+X−
2.3 Reduction of Nitriles and Amides
Nitriles (R-C≡N) on reduction with LiAlH4 or catalytic hydrogenation produce 1°
amines.
Amides (R-CONH2) on reduction with LiAlH4 yield 1° amines.
💡 Crucial Name Reactions for Preparation:
1. Gabriel Phthalimide Synthesis: Used to prepare PURE Aliphatic Primary Amines ONLY. Aromatic amines (aniline) cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution easily.
2. Hoffmann Bromamide Degradation: Amide (R-CONH2) is treated with Br2 and aqueous/ethanolic NaOH. Yields a primary amine with one carbon atom LESS than the parent amide.
R-CONH2 + Br2 + 4NaOH → R-NH2 + Na2CO3 + 2NaBr + 2H2O
1. Gabriel Phthalimide Synthesis: Used to prepare PURE Aliphatic Primary Amines ONLY. Aromatic amines (aniline) cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution easily.
2. Hoffmann Bromamide Degradation: Amide (R-CONH2) is treated with Br2 and aqueous/ethanolic NaOH. Yields a primary amine with one carbon atom LESS than the parent amide.
R-CONH2 + Br2 + 4NaOH → R-NH2 + Na2CO3 + 2NaBr + 2H2O
3. Chemical Reactions & Basicity
3.1 Basicity of Amines
Amines act as Lewis bases due to the lone pair of electrons on the nitrogen atom. They react with acids to
form salts.
Larger Kb (or smaller pKb) means a stronger base.
Aliphatic Amines vs Ammonia
All aliphatic amines are stronger bases than ammonia. Alkyl groups are electron-donating (+I effect). They push electron density towards nitrogen, making the lone pair more available for donation and stabilizing the conjugate acid (alkylammonium ion) formed.
Order of Basicity in Aqueous Solution
In aqueous phase, basicity relies on: (1) Inductive Effect (+I), (2) Solvation Effect (hydrogen bonding with water), and (3) Steric Hindrance. The order gets anomalous depending on the size of the alkyl group:
- Methyl substituted (-CH3): 2° > 1° > 3° > NH3 (Secondary > Primary > Tertiary).
- Ethyl substituted (-C2H5): 2° > 3° > 1° > NH3 (Secondary > Tertiary > Primary).
Note: In gaseous phase, where solvation effect is absent, the order strictly follows Inductive effect: 3° > 2° > 1° > NH3.
Aromatic Amines (Aniline) vs Ammonia
Aniline is a weaker base than ammonia. The lone pair of electrons on the nitrogen atom in aniline is delocalized over the benzene ring through resonance, making it less available for protonation.
Substituent Effects on Aniline:
EWG (-NO2, -X, -CN) withdraw electrons, decreasing basicity.
ERG (-CH3, -OCH3) donate electrons, increasing basicity.
3.2 Alkylation and Acylation
Amines react with acid chlorides, anhydrides, and esters to form amides. This reaction is carried out in the presence of a base stronger than the amine (like pyridine) to remove HCl formed.
C2H5NH2 + CH3COCl → (Pyridine)
→ C2H5-NH-CO-CH3 (N-Ethylethanamide) + HCl
3.3 Carbylamine Reaction (Isocyanide Test)
Exclusively for Primary Amines (aliphatic and aromatic): When heated with chloroform (CHCl3) and alcoholic KOH, 1° amines form isocyanides (carbylamines) which have a highly foul/offensive smell.
R-NH2 + CHCl3 + 3KOH(alc) → R-NC (Foul smell) + 3KCl +
3H2O
Secondary and tertiary amines DO NOT give this test.
3.4 Reaction with Nitrous Acid (HNO2)
Prepared in situ from NaNO2 + HCl :
- Primary Aliphatic Amines: Form highly unstable aliphatic diazonium salts which immediately decompose to release Nitrogen gas (N2) bubbles and form an alcohol. Useful for qualitative estimation of amino acids.
- Primary Aromatic Amines (Aniline): Form arene diazonium salts which are stable at low temperatures (0-5°C). R-N2+Cl−.
3.5 Reaction with Hinsberg's Reagent
Hinsberg's Reagent: Benzenesulphonyl chloride (C6H5SO2Cl).
- 1° Amines: Form N-alkylbenzenesulphonamide. The nitrogen has an acidic hydrogen attached, making the product soluble in aqueous KOH.
- 2° Amines: Form N,N-dialkylbenzenesulphonamide. No acidic hydrogen is present, making the product insoluble in aqueous KOH.
- 3° Amines: Do not react with Hinsberg's reagent.
3.6 Electrophilic Substitution of Aniline
The -NH2 group is highly activating and ortho/para directing. The reactivity is so high that bromination yields 2,4,6-tribromoaniline directly.
- To get mono-substituted products: The -NH2 group must be "protected" first by acetylation (reacting with acetic anhydride). This forms an amide, which is less activating due to resonance of lone pair with the carbonyl group. After bromination/nitration, hydrolysis restores the -NH2 group.
- Nitration Anomaly: Direct nitration of aniline with conc. HNO3/H2SO4 gives ~47% meta product. Why? In strongly acidic medium, aniline is protonated to form the anilinium ion (-NH3+), which is strongly deactivating and meta directing.
4. Diazonium Salts
Formula: Ar-N2+X−. Prepared by Diazotisation (Aniline + NaNO2 + HCl at 0-5°C). They are highly versatile intermediates for creating various substituted benzenes.
4.1 Reactions Replacing Nitrogen
- Sandmeyer's Reaction: Treatment with Cu(I) salts (Cu2Cl2, Cu2Br2, CuCN) to place -Cl, -Br, -CN on the ring.
- Gattermann's Reaction: Treatment with Copper powder / HCl or HBr. (Sandmeyer uses Cuprous halide, Gattermann uses Cu powder. Sandmeyer produces higher yields).
- Replacement by Iodine: No copper needed. Simply warm the salt with aqueous KI.
- Replacement by Fluorine (Balz-Schiemann): React salt with HBF4 to get Ar-N2+BF4−, which upon heating yields Ar-F + N2 + BF3.
- Replacement by Hydrogen (Reduction): Mild reducing agents like Hypophosphorous acid (H3PO2) or Ethanol reduce the salt to Benzene. (C6H5N2Cl + H3PO2 + H2O → C6H6 + N2 + H3PO3 + HCl).
- Replacement by Hydroxyl (-OH): Warming the salt solution with water (up to 283K) yields Phenol.
4.2 Coupling Reactions (Retention of Diazo Group)
The diazonium ion acts as a weak electrophile and attacks electron-rich aromatic rings (like phenol and aniline) usually at the para position to form colorful Azo compounds (dyes).
- With Phenol (in basic medium, pH 9-10): Forms p-Hydroxyazobenzene (Orange dye).
- With Aniline (in acidic medium, pH 4-5): Forms p-Aminoazobenzene (Yellow dye).
🎓 NEET Previous Year Questions
Q1. [NEET 2022] The primary amine which does not give Gabriel phthalimide synthesis
reaction is: (a) Aniline (b) Ethylamine (c) Isopropylamine (d) Methylamine
Answer Gabriel phthalimide synthesis is used
exclusively for aliphatic primary amines. Aromatic primary amines (Aniline) cannot be
prepared because the sp² C-X bond in aryl halides does not undergo nucleophilic substitution
required in step 2. Correct is (a) Aniline.
Q2. [NEET 2021] A given nitrogen-containing aromatic compound A reacts with Sn/HCl,
followed by HNO2 to give an unstable compound B. B, on treatment with phenol, forms an orange dye. A is:
Answer A is reduced by Sn/HCl to a primary amine
→ A is Nitrobenzene. The amine (Aniline) reacts with HNO2 to form Diazonium salt
(B). B couples with phenol to form Orange dye (p-hydroxyazobenzene).
Q3. [NEET 2020] Which of the following amines will give the carbylamine test? (a)
N,N-Dimethylaniline (b) Dimethylamine (c) N-Methylaniline (d) Aniline
Answer Carbylamine test (foul-smelling isocyanide) is
given only by primary amines (both aliphatic and aromatic). Aniline is a 1°
aromatic amine. The others are secondary or tertiary. Correct is (d) Aniline.
Q4. [NEET 2019] The correct increasing order of basic strength for the following
compounds in gas phase is: Methylamine, Dimethylamine, Trimethylamine, Ammonia.
Answer In the gas phase, there is no solvation effect
to cause anomalies. Basicity depends strictly on the +I effect of alkyl groups. Three methyls push more
electrons than two, than one. Correct order: NH3 < CH3NH2 < (CH3)2NH <
(CH3)3N.
Q5. [NEET 2018] Nitration of aniline in strong acidic medium also gives m-nitroaniline
because:
Answer In strong acidic medium, the basic -NH2 group
gets protonated to form the anilinium ion (-NH3+). The positively charged -NH3+ group
is strongly electron-withdrawing and is a meta-directing group. This leads to
substantial (~47%) formation of m-nitroaniline.
💡 Rapid Revision
- Hinsberg Test: 1° Amine + Reagent → Base-Soluble product. 2° Amine + Reagent → Base-Insoluble product. 3° Amine → No reaction.
- Carbylamine Test: Exclusively given by 1° Amines (both alkyl and aryl) giving a foul-smelling isocyanide.
- Aqueous Basicity Order (Methyl): 2°(Dimethyl) > 1°(Methyl) > 3°(Trimethyl) > Ammonia. (Note: 213 pattern).
- Aqueous Basicity Order (Ethyl): 2°(Diethyl) > 3°(Triethyl) > 1°(Ethyl) > Ammonia. (Note: 231 pattern).
- Hoffmann Bromamide: Converts Amide (RCONH2) to 1° Amine (RNH2) by dropping the carbonyl carbon. Step-down reaction.
CLASS 12 CHEMISTRY | NCERT SOLUTIONS
Chapter 9 — Amines
22 Solved Questions — Basicity Order, Distinguishing Tests & Conversions
Note: Amines chapter focuses heavily on arranged basicity (pKb values) in
different phases, structural reasoning, characteristic chemical tests (Carbylamine, Hinsberg), and organic
synthesis pathways.
📝 Basicity & Physical Properties (Q1 – Q7)
3 MarksQ1. Arrange the following in decreasing order
of basic strength in gas phase: C₂H₅NH₂, (C₂H₅)₂NH,
(C₂H₅)₃N, NH₃.
✓ Solution
In the gas phase, there is no solvation effect (no water to hydrogen bond with the ions). The basicity depends purely on the +I (inductive) effect of the alkyl groups.
More alkyl groups = higher electron density on nitrogen = stronger base.
Order: (C₂H₅)₃N (3°) > (C₂H₅)₂NH (2°) > C₂H₅NH₂ (1°) > NH₃
In the gas phase, there is no solvation effect (no water to hydrogen bond with the ions). The basicity depends purely on the +I (inductive) effect of the alkyl groups.
More alkyl groups = higher electron density on nitrogen = stronger base.
Order: (C₂H₅)₃N (3°) > (C₂H₅)₂NH (2°) > C₂H₅NH₂ (1°) > NH₃
3 MarksQ2. Arrange the following in decreasing order
of basic strength in aqueous solution: CH₃NH₂, (CH₃)₂NH, (CH₃)₃N,
NH₃.
✓ Solution
In aqueous solution, basicity is governed by a combination of the +I effect, hydration (solvation) effect, and steric hindrance.
For methyl substituted amines, the hydration and steric factors make the secondary amine the strongest, followed by primary, then tertiary.
Order: (CH₃)₂NH (2°) > CH₃NH₂ (1°) > (CH₃)₃N (3°) > NH₃ (2>1>3 rule for methyl).
In aqueous solution, basicity is governed by a combination of the +I effect, hydration (solvation) effect, and steric hindrance.
For methyl substituted amines, the hydration and steric factors make the secondary amine the strongest, followed by primary, then tertiary.
Order: (CH₃)₂NH (2°) > CH₃NH₂ (1°) > (CH₃)₃N (3°) > NH₃ (2>1>3 rule for methyl).
3 MarksQ3. Arrange the following in decreasing order
of basic strength in aqueous solution: C₂H₅NH₂, (C₂H₅)₂NH,
(C₂H₅)₃N, NH₃.
✓ Solution
For ethyl substituted amines in aqueous solution, the larger ethyl groups increase steric hindrance but offer a stronger +I effect.
The delicate balance of Inductive effect > Hydration effect > Steric hindrance results in the 3° amine being stronger than the 1° amine, but 2° remains the strongest.
Order: (C₂H₅)₂NH (2°) > (C₂H₅)₃N (3°) > C₂H₅NH₂ (1°) > NH₃ (2>3>1 rule for ethyl).
For ethyl substituted amines in aqueous solution, the larger ethyl groups increase steric hindrance but offer a stronger +I effect.
The delicate balance of Inductive effect > Hydration effect > Steric hindrance results in the 3° amine being stronger than the 1° amine, but 2° remains the strongest.
Order: (C₂H₅)₂NH (2°) > (C₂H₅)₃N (3°) > C₂H₅NH₂ (1°) > NH₃ (2>3>1 rule for ethyl).
3 MarksQ4. Arrange the following in decreasing order
of their pKb values: Aniline, N-methylaniline, p-toluidine, p-nitroaniline.
✓ Solution
High basic strength = Low pKb value. To arrange in decreasing pKb means arranging from weakest base to strongest base.
1. p-Nitroaniline: -NO₂ is a strong EWG (-R, -I), withdrawing electron density and making it the weakest base (highest pKb).
2. Aniline: Base case, lone pair delocalized over ring.
3. p-Toluidine: -CH₃ is ERG (+I, hyperconjugation), increases electron density, stronger than aniline.
4. N-methylaniline: The -CH₃ group directly on nitrogen increases electron density considerably (+I effect), making it the strongest base (lowest pKb).
Order (decreasing pKb, increasing basicity): p-nitroaniline > aniline > p-toluidine > N-methylaniline.
High basic strength = Low pKb value. To arrange in decreasing pKb means arranging from weakest base to strongest base.
1. p-Nitroaniline: -NO₂ is a strong EWG (-R, -I), withdrawing electron density and making it the weakest base (highest pKb).
2. Aniline: Base case, lone pair delocalized over ring.
3. p-Toluidine: -CH₃ is ERG (+I, hyperconjugation), increases electron density, stronger than aniline.
4. N-methylaniline: The -CH₃ group directly on nitrogen increases electron density considerably (+I effect), making it the strongest base (lowest pKb).
Order (decreasing pKb, increasing basicity): p-nitroaniline > aniline > p-toluidine > N-methylaniline.
2 MarksQ5. Why is aniline less basic than ethylamine?
✓ Solution
1. In Aniline: The lone pair of electrons on the nitrogen atom is in conjugation with the π-electrons of the benzene ring (Resonance). Because of this delocalization, the lone pair is less available for donation to a proton.
2. In Ethylamine: The ethyl group exerts a +I (inductive) effect, pushing electron density towards nitrogen, making the lone pair more available for protonation. Thus, ethylamine is stronger.
1. In Aniline: The lone pair of electrons on the nitrogen atom is in conjugation with the π-electrons of the benzene ring (Resonance). Because of this delocalization, the lone pair is less available for donation to a proton.
2. In Ethylamine: The ethyl group exerts a +I (inductive) effect, pushing electron density towards nitrogen, making the lone pair more available for protonation. Thus, ethylamine is stronger.
2 MarksQ6. Why are primary amines higher boiling than
tertiary amines?
✓ Solution
Primary amines (R-NH₂) have two hydrogen atoms attached to the highly electronegative nitrogen atom, allowing them to form extensive intermolecular hydrogen bonds.
Tertiary amines (R₃N) have no hydrogen atoms attached directly to nitrogen, so they cannot form intermolecular hydrogen bonds. Therefore, they have significantly lower boiling points.
Primary amines (R-NH₂) have two hydrogen atoms attached to the highly electronegative nitrogen atom, allowing them to form extensive intermolecular hydrogen bonds.
Tertiary amines (R₃N) have no hydrogen atoms attached directly to nitrogen, so they cannot form intermolecular hydrogen bonds. Therefore, they have significantly lower boiling points.
2 MarksQ7. Arrange the following in increasing order
of boiling point: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂.
✓ Solution
1. (CH₃)₂NH (Secondary amine): Has only one N-H bond for hydrogen bonding. Weakest intermolecular forces here.
2. C₂H₅NH₂ (Primary amine): Has two N-H bonds, forms more extensive hydrogen bonding than secondary amines, higher b.p.
3. C₂H₅OH (Alcohol): O is more electronegative than N, so O-H bonds are more polar than N-H bonds. The hydrogen bonding in alcohols is significantly stronger than in primary amines.
Order: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
1. (CH₃)₂NH (Secondary amine): Has only one N-H bond for hydrogen bonding. Weakest intermolecular forces here.
2. C₂H₅NH₂ (Primary amine): Has two N-H bonds, forms more extensive hydrogen bonding than secondary amines, higher b.p.
3. C₂H₅OH (Alcohol): O is more electronegative than N, so O-H bonds are more polar than N-H bonds. The hydrogen bonding in alcohols is significantly stronger than in primary amines.
Order: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
💡 Mechanism, Reactivity & Distinguishing Tests (Q8 – Q14)
3 MarksQ8. Give a chemical test to distinguish
between Methylamine (1°) and Dimethylamine (2°).
✓ Solution
1. Carbylamine Test: Warm with Chloroform (CHCl₃) and alcoholic KOH.
Methylamine (1° amine) gives an extremely foul-smelling gas (Methyl isocyanide). Dimethylamine (2° amine) gives no reaction.
2. Hinsberg's Test: React with Benzenesulphonyl chloride.
Methylamine forms N-methylbenzenesulphonamide, which is soluble in aqueous KOH/NaOH.
Dimethylamine forms N,N-dimethylbenzenesulphonamide, which is insoluble in aqueous KOH/NaOH.
1. Carbylamine Test: Warm with Chloroform (CHCl₃) and alcoholic KOH.
Methylamine (1° amine) gives an extremely foul-smelling gas (Methyl isocyanide). Dimethylamine (2° amine) gives no reaction.
2. Hinsberg's Test: React with Benzenesulphonyl chloride.
Methylamine forms N-methylbenzenesulphonamide, which is soluble in aqueous KOH/NaOH.
Dimethylamine forms N,N-dimethylbenzenesulphonamide, which is insoluble in aqueous KOH/NaOH.
2 MarksQ9. Give a chemical test to distinguish
between Aniline and N-methylaniline.
✓ Solution
Carbylamine Test:
Aniline is a 1° aromatic amine. When heated with CHCl₃ and alc. KOH, it forms phenyl isocyanide, giving an offensive/foul smell.
N-methylaniline is a 2° aromatic amine. It does not give the carbylamine test (no foul smell).
Carbylamine Test:
Aniline is a 1° aromatic amine. When heated with CHCl₃ and alc. KOH, it forms phenyl isocyanide, giving an offensive/foul smell.
N-methylaniline is a 2° aromatic amine. It does not give the carbylamine test (no foul smell).
2 MarksQ10. Give a chemical test to distinguish
between Ethylamine and Aniline.
✓ Solution
Azo Dye Test (Diazotisation followed by coupling):
Dissolve the amine in dil. HCl. Cool to 0-5°C, add NaNO₂ solution, then add cold alkaline β-naphthol solution.
Aniline: Forms a brilliant orange/red dye (an azo dye) because it forms a stable diazonium salt.
Ethylamine: Does not form a dye (it releases brisk effervescence of N₂ gas because aliphatic diazonium salts are highly unstable).
Azo Dye Test (Diazotisation followed by coupling):
Dissolve the amine in dil. HCl. Cool to 0-5°C, add NaNO₂ solution, then add cold alkaline β-naphthol solution.
Aniline: Forms a brilliant orange/red dye (an azo dye) because it forms a stable diazonium salt.
Ethylamine: Does not form a dye (it releases brisk effervescence of N₂ gas because aliphatic diazonium salts are highly unstable).
3 MarksQ11. Account for the following: Although amino
group is o, p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a
substantial amount of m-nitroaniline.
✓ Solution
Nitration is carried out using a nitrating mixture (conc. HNO₃ + conc. H₂SO₄), which is a strongly acidic medium.
In this acidic medium, the basic -NH₂ group of aniline is protonated to form the anilinium ion (-NH₃⁺).
Unlike the -NH₂ group which is activating and o/p directing, the positively charged -NH₃⁺ group is strongly deactivating and meta-directing. This leads to the formation of a significant amount (~47%) of m-nitroaniline.
Nitration is carried out using a nitrating mixture (conc. HNO₃ + conc. H₂SO₄), which is a strongly acidic medium.
In this acidic medium, the basic -NH₂ group of aniline is protonated to form the anilinium ion (-NH₃⁺).
Unlike the -NH₂ group which is activating and o/p directing, the positively charged -NH₃⁺ group is strongly deactivating and meta-directing. This leads to the formation of a significant amount (~47%) of m-nitroaniline.
2 MarksQ12. How will you control the multiple
substitution during bromination of aniline? Explain with equations.
✓ Solution
Aniline is highly activated; direct bromination yields 2,4,6-tribromoaniline.
To get mono-bromoaniline, we protect the -NH₂ group by acetylation.
1. Aniline + Acetic anhydride/Pyridine → Acetanilide (The lone pair on N now resonates with the C=O group, decreasing its activating effect on the ring).
2. Bromination: Acetanilide + Br₂/CH₃COOH → p-Bromoacetanilide (major) + o-isomer.
3. Hydrolysis: p-Bromoacetanilide + H₂O/H⁺ → p-Bromoaniline.
Aniline is highly activated; direct bromination yields 2,4,6-tribromoaniline.
To get mono-bromoaniline, we protect the -NH₂ group by acetylation.
1. Aniline + Acetic anhydride/Pyridine → Acetanilide (The lone pair on N now resonates with the C=O group, decreasing its activating effect on the ring).
2. Bromination: Acetanilide + Br₂/CH₃COOH → p-Bromoacetanilide (major) + o-isomer.
3. Hydrolysis: p-Bromoacetanilide + H₂O/H⁺ → p-Bromoaniline.
2 MarksQ13. Write the reaction of Gabriel phthalimide
synthesis.
✓ Solution
1. Phthalimide + ethanolic KOH → Potassium phthalimide + H₂O.
2. Potassium phthalimide + R-X (Alkyl halide) → N-Alkylphthalimide + KX.
3. Hydrazinolysis (or alkaline hydrolysis): N-Alkylphthalimide + NH₂NH₂ → Phthalhydrazide + R-NH₂ (Pure primary aliphatic amine).
1. Phthalimide + ethanolic KOH → Potassium phthalimide + H₂O.
2. Potassium phthalimide + R-X (Alkyl halide) → N-Alkylphthalimide + KX.
3. Hydrazinolysis (or alkaline hydrolysis): N-Alkylphthalimide + NH₂NH₂ → Phthalhydrazide + R-NH₂ (Pure primary aliphatic amine).
2 MarksQ14. Why cannot aromatic primary amines be
prepared by Gabriel phthalimide synthesis?
✓ Solution
The second step of Gabriel synthesis requires an alkyl halide (R-X) to undergo nucleophilic substitution (SN2) with the nucleophilic potassium phthalimide.
To prepare an aromatic amine (like aniline), we would need to use an aryl halide (Ar-X).
Aryl halides generally do not undergo nucleophilic substitution easily because the C-X bond has partial double bond character due to resonance. Hence, the synthesis fails.
The second step of Gabriel synthesis requires an alkyl halide (R-X) to undergo nucleophilic substitution (SN2) with the nucleophilic potassium phthalimide.
To prepare an aromatic amine (like aniline), we would need to use an aryl halide (Ar-X).
Aryl halides generally do not undergo nucleophilic substitution easily because the C-X bond has partial double bond character due to resonance. Hence, the synthesis fails.
📈 Organic Conversions (Q15 – Q22)
3 MarksQ15. Benzene to Aniline.
✓ Solution
1. Nitration: Benzene + Conc. HNO₃ + Conc. H₂SO₄ (Heat) → Nitrobenzene (C₆H₅NO₂).
2. Reduction: Nitrobenzene + Sn/HCl (or Fe/HCl) followed by NaOH → Aniline (C₆H₅NH₂).
1. Nitration: Benzene + Conc. HNO₃ + Conc. H₂SO₄ (Heat) → Nitrobenzene (C₆H₅NO₂).
2. Reduction: Nitrobenzene + Sn/HCl (or Fe/HCl) followed by NaOH → Aniline (C₆H₅NH₂).
3 MarksQ16. Aniline to Bromobenzene.
✓ Solution
Requires diazotisation.
1. Diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl (0-5°C) → C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride).
2. Sandmeyer's Reaction: C₆H₅N₂⁺Cl⁻ + Cu₂Br₂/HBr → C₆H₅Br (Bromobenzene) + N₂.
Requires diazotisation.
1. Diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl (0-5°C) → C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride).
2. Sandmeyer's Reaction: C₆H₅N₂⁺Cl⁻ + Cu₂Br₂/HBr → C₆H₅Br (Bromobenzene) + N₂.
3 MarksQ17. Benzene to p-Nitroaniline.
✓ Solution
Direct nitration of aniline gives massive meta product. We must protect the group.
1. Benzene → Nitrobenzene (Nitration: HNO₃/H₂SO₄).
2. Nitrobenzene → Aniline (Reduction: Sn/HCl).
3. Aniline + Acetic Anhydride → Acetanilide (Protection).
4. Acetanilide + Conc. HNO₃/H₂SO₄ → p-Nitroacetanilide (major product).
5. p-Nitroacetanilide + H₂O/H⁺ → p-Nitroaniline.
Direct nitration of aniline gives massive meta product. We must protect the group.
1. Benzene → Nitrobenzene (Nitration: HNO₃/H₂SO₄).
2. Nitrobenzene → Aniline (Reduction: Sn/HCl).
3. Aniline + Acetic Anhydride → Acetanilide (Protection).
4. Acetanilide + Conc. HNO₃/H₂SO₄ → p-Nitroacetanilide (major product).
5. p-Nitroacetanilide + H₂O/H⁺ → p-Nitroaniline.
2 MarksQ18. Chlorobenzene to p-Chloroaniline.
✓ Solution
1. Nitration: Chlorobenzene + Conc. HNO₃/H₂SO₄ → p-Chloronitrobenzene (major).
2. Reduction: p-Chloronitrobenzene + Sn/HCl → p-Chloroaniline.
1. Nitration: Chlorobenzene + Conc. HNO₃/H₂SO₄ → p-Chloronitrobenzene (major).
2. Reduction: p-Chloronitrobenzene + Sn/HCl → p-Chloroaniline.
2 MarksQ19. Aniline to Sulphanilic Acid.
✓ Solution
Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate, which on heating (453-473 K) yields p-aminobenzenesulphonic acid (sulphanilic acid).
Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate, which on heating (453-473 K) yields p-aminobenzenesulphonic acid (sulphanilic acid).
C₆H₅NH₂ + H₂SO₄(conc) →
C₆H₅NH₃⁺HSO₄⁻
C₆H₅NH₃⁺HSO₄⁻ → (Heat 453-473 K, -H₂O) → H₂N-C₆H₄-SO₃H
Note: Sulphanilic acid exists largely as a dipolar Zwitterion:
H₃N⁺-C₆H₄-SO₃⁻C₆H₅NH₃⁺HSO₄⁻ → (Heat 453-473 K, -H₂O) → H₂N-C₆H₄-SO₃H
2 MarksQ20. Ethanamine to Methanamine.
✓ Solution
A step-down conversion (loss of one carbon). We must reach an amide, then use Hoffmann bromamide degradation.
1. C₂H₅NH₂ + HNO₂ → C₂H₅OH (Ethanol).
2. C₂H₅OH + Alkaline KMnO₄ (Oxidation) → CH₃COOH (Ethanoic acid).
3. CH₃COOH + NH₃ (Heat) → CH₃CONH₂ (Ethanamide + water).
4. CH₃CONH₂ + Br₂ + NaOH (Hoffmann Bromamide) → CH₃NH₂ (Methanamine).
A step-down conversion (loss of one carbon). We must reach an amide, then use Hoffmann bromamide degradation.
1. C₂H₅NH₂ + HNO₂ → C₂H₅OH (Ethanol).
2. C₂H₅OH + Alkaline KMnO₄ (Oxidation) → CH₃COOH (Ethanoic acid).
3. CH₃COOH + NH₃ (Heat) → CH₃CONH₂ (Ethanamide + water).
4. CH₃CONH₂ + Br₂ + NaOH (Hoffmann Bromamide) → CH₃NH₂ (Methanamine).
3 MarksQ21. Methanamine to Ethanamine.
✓ Solution
A step-up conversion (gain of one carbon). We use cyanide substitution.
1. CH₃NH₂ + HNO₂ → CH₃OH (Methanol).
2. CH₃OH + PCl₅ → CH₃Cl (Chloromethane).
3. CH₃Cl + KCN(alc) → CH₃CN (Ethane nitrile / Methyl cyanide).
4. CH₃CN + LiAlH₄ (or Na/C₂H₅OH, Reduction) → CH₃CH₂NH₂ (Ethanamine).
A step-up conversion (gain of one carbon). We use cyanide substitution.
1. CH₃NH₂ + HNO₂ → CH₃OH (Methanol).
2. CH₃OH + PCl₅ → CH₃Cl (Chloromethane).
3. CH₃Cl + KCN(alc) → CH₃CN (Ethane nitrile / Methyl cyanide).
4. CH₃CN + LiAlH₄ (or Na/C₂H₅OH, Reduction) → CH₃CH₂NH₂ (Ethanamine).
2 MarksQ22. Aniline to Benzene.
✓ Solution
1. Diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl (0-5°C) → C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride).
2. Reduction: C₆H₅N₂⁺Cl⁻ + H₃PO₂ (Hypophosphorous acid) + H₂O → C₆H₆ (Benzene) + H₃PO₃ + HCl + N₂.
1. Diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl (0-5°C) → C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride).
2. Reduction: C₆H₅N₂⁺Cl⁻ + H₃PO₂ (Hypophosphorous acid) + H₂O → C₆H₆ (Benzene) + H₃PO₃ + HCl + N₂.
✍ Score Guide — 22 Questions
Essential board/NEET concepts mastered: aqueous/gas phase basicity charts, specific detection tests (Azo Dye, Carbylamine, Hinsberg), Diazonium salt replacements, and Step-up/Step-down amide degradation pathways.
Essential board/NEET concepts mastered: aqueous/gas phase basicity charts, specific detection tests (Azo Dye, Carbylamine, Hinsberg), Diazonium salt replacements, and Step-up/Step-down amide degradation pathways.
High-Yield Facts & Formulas: Amines
Amine Structure
Nitrogen atom in amines is sp3 hybridized with pyramidal geometry. Bond angle is slightly less than 109.5° (due to lone pair).
Nitrogen atom in amines is sp3 hybridized with pyramidal geometry. Bond angle is slightly less than 109.5° (due to lone pair).
Basicity of Amines
Amines are Lewis bases. Basic strength: Substituted amines > Ammonia (due to +I effect of alkyl groups).
Amines are Lewis bases. Basic strength: Substituted amines > Ammonia (due to +I effect of alkyl groups).
Basicity in Gas Phase
3° > 2° > 1° > NH3. (Follows pure inductive effect).
3° > 2° > 1° > NH3. (Follows pure inductive effect).
Basicity in Water (Methyl)
2° > 1° > 3° > NH3. (Balanced by inductive and solvation effects).
2° > 1° > 3° > NH3. (Balanced by inductive and solvation effects).
Basicity in Water (Ethyl)
2° > 3° > 1° > NH3.
2° > 3° > 1° > NH3.
Aromatic Amine Basicity
Aniline is weaker than NH3 due to resonance (lone pair delocalization).
Aniline is weaker than NH3 due to resonance (lone pair delocalization).
pKb Correlation
Smaller pKb = Stronger Base.
Smaller pKb = Stronger Base.
Gabriel Phthalimide Synthesis
Phthalimide + KOH → Potassium phthalimide + RX → N-Alkylphthalimide → (NaOH) → 1° Amine.
Phthalimide + KOH → Potassium phthalimide + RX → N-Alkylphthalimide → (NaOH) → 1° Amine.
Hoffmann Bromamide Reaction
Amide → 1° Amine (with one less Carbon) using Br2 and alkali.
Amide → 1° Amine (with one less Carbon) using Br2 and alkali.
Carbylamine Test
Primary amines + CHCl3 + alc. KOH → Isocyanide (Foul smell).
Primary amines + CHCl3 + alc. KOH → Isocyanide (Foul smell).
Hinsberg Reagent
Benzenesulphonyl chloride (C6H5SO2Cl).
Benzenesulphonyl chloride (C6H5SO2Cl).
Reaction with Nitrous Acid (1° Aliphatic)
RNH2 + HNO2 → Alcohol + N2 ↑ (used to estimate nitrogen).
RNH2 + HNO2 → Alcohol + N2 ↑ (used to estimate nitrogen).
Reaction with Nitrous Acid (1° Aromatic)
ArNH2 + HNO2 → Diazonium salt (stable at 273-278 K).
ArNH2 + HNO2 → Diazonium salt (stable at 273-278 K).
Nitration of Aniline
Conc. H2SO4/HNO3 gives m-nitroaniline (47%) due to formation of anilinium ion.
Conc. H2SO4/HNO3 gives m-nitroaniline (47%) due to formation of anilinium ion.
Acetylation of Aniline
Decreases activation of the ring, allowing monobromination at para position.
Decreases activation of the ring, allowing monobromination at para position.
Sandmeyer Reaction (Cl/Br)
ArN2X + Cu2Cl2/HCl → ArCl.
ArN2X + Cu2Cl2/HCl → ArCl.
Sandmeyer Reaction (CN)
ArN2X + CuCN/KCN → ArCN.
ArN2X + CuCN/KCN → ArCN.
Balz-Schiemann Reaction
ArN2Cl + HBF4 → ArN2BF4 → (Heat) → ArF.
ArN2Cl + HBF4 → ArN2BF4 → (Heat) → ArF.
Replacement by -OH
Diazonium salt + H2O (Warm) → Phenol.
Diazonium salt + H2O (Warm) → Phenol.
Reduction of Diazonium
ArN2Cl + H3PO2/H2O → Benzene (ArH).
ArN2Cl + H3PO2/H2O → Benzene (ArH).
Dye Coupling (Phenol)
Diazonium + Phenol → p-Hydroxyazobenzene (Orange dye). pH 9-10.
Diazonium + Phenol → p-Hydroxyazobenzene (Orange dye). pH 9-10.
Dye Coupling (Aniline)
Diazonium + Aniline → p-Aminoazobenzene (Yellow dye). pH 4-5.
Diazonium + Aniline → p-Aminoazobenzene (Yellow dye). pH 4-5.
Boiling points
Primary > Secondary > Tertiary amines (due to extent of H-bonding).
Primary > Secondary > Tertiary amines (due to extent of H-bonding).
Mendius Reaction
Reduction of alkyl cyanides with sodium and ethanol to primary amines.
Reduction of alkyl cyanides with sodium and ethanol to primary amines.
Curtius Rearrangement
Acid azide → Isocyanate → Primary Amine.
Acid azide → Isocyanate → Primary Amine.
Schmidt Reaction
Acid + Hydrazoic acid (H/N3) → Primary Amine.
Acid + Hydrazoic acid (H/N3) → Primary Amine.
Reaction with COCl2
Amines react with phosgene to form alkyl isocyanates.
Amines react with phosgene to form alkyl isocyanates.
Schotten-Baumann Reaction
Acylation of amines using benzoyl chloride in presence of NaOH.
Acylation of amines using benzoyl chloride in presence of NaOH.
Sulfanilic acid
p-aminobenzenesulfonic acid. Exists as Zwitter ion.
p-aminobenzenesulfonic acid. Exists as Zwitter ion.
Gattermann reaction improvement
Higher yield than Sandmeyer in some specific cases, uses Cu powder.
Higher yield than Sandmeyer in some specific cases, uses Cu powder.
Hofmann's exhaustion methylation
Conversion of amine to quaternary ammonium salt.
Conversion of amine to quaternary ammonium salt.
Hofmann Elimination
Quaternary ammonium hydroxide → Less substituted alkene (major).
Quaternary ammonium hydroxide → Less substituted alkene (major).
Solubility
Lower aliphatic amines are soluble in water. Aniline is almost insoluble.
Lower aliphatic amines are soluble in water. Aniline is almost insoluble.
Olfactory detection
Amines have characteristic fishy odors.
Amines have characteristic fishy odors.
Diazonium stability
Resonance with aryl group makes benzene diazonium salts stable at low temp.
Resonance with aryl group makes benzene diazonium salts stable at low temp.
Aliphatic diazonium
Very unstable, decompose immediately to give alcohols and nitrogen.
Very unstable, decompose immediately to give alcohols and nitrogen.
Hofmann mustard oil reaction
1° amine + CS2 + HgCl2 → Alkyl isothiocyanate (pungent smell).
1° amine + CS2 + HgCl2 → Alkyl isothiocyanate (pungent smell).
Alkylation of ammonia
Nucleophilic substitution of RX by NH3. Leads to mixture of products.
Nucleophilic substitution of RX by NH3. Leads to mixture of products.
Identification of 1°, 2°, 3° amines
Using Hinsberg's reagent or nitrous acid.
Using Hinsberg's reagent or nitrous acid.
Electrophilic substitution of aniline
Very rapid; -NH2 group is highly activating.
Very rapid; -NH2 group is highly activating.
Friedel-Crafts on Aniline
Does not occur; AlCl3 forms complex with nitrogen lone pair.
Does not occur; AlCl3 forms complex with nitrogen lone pair.
Protection of -NH2 group
Done by acetylation (using acetic anhydride) to control reactivity.
Done by acetylation (using acetic anhydride) to control reactivity.
N-Nitrosamines
Formed by reaction of secondary amines with HNO2. Yellow oily liquids.
Formed by reaction of secondary amines with HNO2. Yellow oily liquids.
Tertiary amine + HNO2
Form soluble nitrite salts which decompose on heating.
Form soluble nitrite salts which decompose on heating.
Replacement by -H in diazonium
Heat with Ethanol or H3PO2.
Heat with Ethanol or H3PO2.
Replacement by -NO2
ArN2BF4 + NaNO2 (Cu/Heat) → ArNO2.
ArN2BF4 + NaNO2 (Cu/Heat) → ArNO2.
Industrial importance of diazonium salts
Intermediate for synthesizing variety of aromatic compounds not easily made by direct substitution.
Intermediate for synthesizing variety of aromatic compounds not easily made by direct substitution.
Quaternary ammonium salts
[R4N]+X-. Used as phase transfer catalysts and detergents.
[R4N]+X-. Used as phase transfer catalysts and detergents.
Basicity of Pyridine
Stronger base than pyrrole but weaker than aliphatic amines.
Stronger base than pyrrole but weaker than aliphatic amines.
Saponification of amides
Leads to formation of carboxylic acid salts and evolution of ammonia/amines.
Leads to formation of carboxylic acid salts and evolution of ammonia/amines.
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