**Definition:** Amines are organic compounds derived by replacing one, more, or all three hydrogen atoms of ammonia (NH₃) by alkyl (R) or aryl (Ar) groups.
**General formula:** RNH₂, R₂NH, R₃N (where R = alkyl group or Ar = aryl group)
**Structural characteristics:**
**Example:** In trimethylamine, N(CH₃)₃, the three methyl groups and lone pair occupy sp³ orbitals, with bond angle ~108°.
Amines are classified based on the number of hydrogen atoms replaced in ammonia:
**1. Primary Amines (1°):** RNH₂
**2. Secondary Amines (2°):** R₂NH or RNHR'
**3. Tertiary Amines (3°):** R₃N or RNHR'R''
**Simple vs Mixed amines:**
**For aliphatic amines:** Alkyl groups are named as prefixes followed by "amine" as one word.
**For arylamines:**
**For primary amines:** The longest carbon chain is named as "alkanamine" (replace final 'e' of alkane with 'amine').
**For secondary and tertiary amines:** Use **locant N** to indicate substituents on nitrogen.
**For diamines:** Use "damine" with locants and prefixes for multiple amino groups.
**For arylamines:** Suffix of arene is replaced with 'amine'.
**Principle:** Nitro compounds (RNO₂ or ArNO₂) are reduced to primary amines.
**Reagents and conditions:**
**General reaction:**
RNO₂ + 6[H] → RNH₂ + 2H₂O
**Specific examples:**
**Reduction with H₂/Ni:**
C₆H₅NO₂ + 3H₂ →(Ni, Δ) C₆H₅NH₂ + 2H₂O
**Reduction with Fe/HCl:**
CH₃C₆H₄NO₂ + 6[H] →(Fe/HCl) CH₃C₆H₄NH₂ + 2H₂O
**Mechanism:** Fe is oxidized to Fe²⁺ which then reduces NO₂ to NO, then to other intermediates finally to NH₂. FeCl₂ hydrolyzes to release HCl, making only initial HCl necessary.
**Advantage:** This is the preferred industrial method for preparing aromatic amines because it's economical and Fe is easily available.
**Principle:** Alkyl or benzyl halides undergo nucleophilic substitution by ammonia, replacing halogen with amino group.
**Reaction conditions:**
**General reaction:**
R-X + 2NH₃ → R-NH₂ + NH₄X
(excess ammonia used to obtain primary amine as major product)
**Specific examples:**
CH₃CH₂Cl + 2NH₃ →(373 K, sealed tube) CH₃CH₂NH₂ + NH₄Cl
C₆H₅CH₂Cl + 2NH₃ →(373 K, sealed tube) C₆H₅CH₂NH₂ + NH₄Cl
**Mechanism (SN2 mechanism):**
**Important limitation:** Mixture of primary, secondary, tertiary amines, and quaternary ammonium salt is produced because:
1. Primary amine (RNH₂) acts as nucleophile and reacts with more alkyl halide
2. Secondary amine (R₂NH) further reacts with alkyl halide to give tertiary amine (R₃N)
3. Tertiary amine reacts with alkyl halide to give quaternary ammonium salt [R₄N]⁺X⁻
**Solution:** To obtain primary amine as major product, take **large excess of ammonia** (ratio 1:100 or more). This ensures most alkyl halide molecules collide with NH₃ rather than formed RNH₂.
**Reactivity order of halides:** RI > RBr > RCl >> RF
**Separation of product:** Ammonium salt is water soluble and ionic; free amine is obtained by treating the salt with strong base (NaOH):
RNH₃⁺Cl⁻ + NaOH → RNH₂ + NaCl + H₂O
**Principle:** Nitriles (R-CN) are reduced to primary amines containing one more carbon than original alkyl group. This is used for **ascent of amine series**.
**Reagents:**
**General reaction:**
R-C≡N + 4[H] → R-CH₂-NH₂
**Specific examples:**
**Using LiAlH₄:**
CH₃CH₂C≡N + 4[H] →(LiAlH₄) CH₃CH₂CH₂NH₂
(Propanenitrile → Propan-1-amine)
C₆H₅C≡N + 4[H] →(LiAlH₄) C₆H₅CH₂NH₂
(Benzonitrile → Phenylmethanamine)
**Using catalytic hydrogenation:**
R-C≡N + 2H₂ →(Ni, Pt, Pd) R-CH₂-NH₂
**Mechanism with LiAlH₄:**
**Important exam point:** This is the preferred method when **ascent of amine series is desired** (i.e., when amine with one additional carbon atom is required).
**Principle:** Primary, secondary, and tertiary amides can be reduced to corresponding amines by LiAlH₄. The number of carbon atoms remains unchanged.
**General reaction:**
RCONH₂ + 4[H] →(LiAlH₄) RCH₂NH₂
**Specific examples:**
**Primary amides → Primary amines:**
CH₃CONH₂ + 4[H] →(LiAlH₄) CH₃CH₂NH₂
(Acetamide → Ethanamine)
**Secondary amides → Secondary amines:**
CH₃CONHC₂H₅ + 4[H] →(LiAlH₄) CH₃CH₂NHCH₂CH₃
(N-ethylacetamide → N-ethylethanamine)
**Tertiary amides → Tertiary amines:**
(CH₃)₂NCH₂COCH₃ + 4[H] →(LiAlH₄) (CH₃)₂NCH₂CH₂CH₃
**Principle:** This is the **most important method for selective synthesis of primary amines** because it avoids formation of secondary and tertiary amines.
**Steps:**
**Step 1:** Phthalimide is treated with ethanolic KOH to form potassium salt of phthalimide (nucleophile).
**Step 2:** The potassium salt is heated with alkyl halide (RX). The imide anion attacks carbon bearing halogen in SN2 mechanism, displacing halide:
**Step 3:** The N-alkylphthalimide is hydrolyzed with aqueous alcoholic KOH or dilute HCl to give primary amine:
**Overall reaction for primary amine formation:**
**Specific example:**
Starting with CH₃Cl:
**Advantages:**
**Limitation:** **Aromatic primary amines cannot be prepared** by Gabriel synthesis because C-Ar bonds are very strong and do not undergo SN2 substitution.
**Why this method is superior:** Traditional ammonolysis gives mixture of products; Gabriel synthesis is selective for primary amines only.
**Principle:** Primary amines are prepared by degrading (breaking down) amides using Br₂ and NaOH. The amine produced contains **one carbon atom less** than the starting amide (descent of amine series).
**Reaction conditions:**
**General mechanism (simplified):**
1. Bromine converts amide NH₂ to NHBr
2. NaOH deprotonates to form N=Br (nitrene)
3. **Alkyl/aryl group migration** from adjacent carbonyl carbon to nitrogen with simultaneous C-N bond formation and C-C bond breaking
4. Rearranged product (isocyanate, R-N=C=O) is hydrolyzed by water to give amine
**General reaction:**
RCONH₂ + Br₂ + NaOH → R-NH₂ + CO₂ + NaBr + H₂O
**Specific examples:**
**Primary aliphatic amines:**
CH₃CH₂CONH₂ + Br₂ + NaOH → CH₃CH₂NH₂ + CO₂ + NaBr + H₂O
(Propanamide → Ethanamine, loses one carbon)
**Aromatic amines:**
C₆H₅CONH₂ + Br₂ + NaOH → C₆H₅NH₂ + CO₂ + NaBr + H₂O
(Benzamide → Aniline, loses one carbon)
**Mechanism in detail:**
Step 1 - Bromination:
RCONH₂ + Br₂ → RCONHBr + HBr
Step 2 - Deprotonation by NaOH:
RCONHBr + NaOH → RCONBr⁻ + H₂O + Na⁺
Step 3 - Rearrangement (nitrene formation and migration):
RCONBr⁻ → R-N=C=O (isocyanate) + Br⁻
(Alkyl group migrates from C to N while C-C bond breaks)
Step 4 - Hydrolysis:
R-N=C=O + H₂O → R-NH₂ + CO₂
**Key point:** The rearrangement step is the **Hoffmann rearrangement** — the carbon skeleton is retained, but the carbon atom attached to carbonyl group is lost as CO₂.
**Advantages:**
**Comparison with other methods:**
**Water solubility:**
**Organic solvent solubility:**
**Comparative solubility:** Out of butan-1-ol and butan-1-amine:
**Intermolecular hydrogen bonding:**
**Primary amines (RNH₂):** Form **strong intermolecular hydrogen bonds**
**Secondary amines (R₂NH):** Form **moderate intermolecular hydrogen bonds**
**Tertiary amines (R₃N):** **No intermolecular hydrogen bonding**
**Order of boiling points for isomeric amines:**
**Primary > Secondary > Tertiary**
**Hydrogen bonding diagram (primary amine):**
R-NH₂···N-H-R' (linear H-bonding chain)
**Comparison table (similar molar masses):**
| Compound | Molar mass | b.p. (K) | Reason |
|----------|-----------|----------|---------|
| n-C₄H₉NH₂ | 73 | 350.8 | Primary amine, strong H-bonding |
| (C₂H₅)₂NH | 73 | 329.3 | Secondary amine, moderate H-bonding |
| C₂H₅N(CH₃)₂ | 73 | 310.5 | Tertiary amine, no H-bonding |
| C₂H₅CH(CH₃)₂ | 72 | 300.8 | Alkane, only van der Waals |
| n-C₄H₉OH | 74 | 390.3 | Alcohol, stronger H-bonding than amine |
**Note:** Alcohols have higher boiling points than amines of similar molar mass because O-H···O hydrogen bonds are stronger than N-H···N bonds (O is more electronegative).
**Fundamental principle:** Amines are **Brønsted-Lowry bases** and **Lewis bases**.
**Reaction with acids:**
R-NH₂ + HCl → [R-NH₃]⁺Cl⁻
(Primary amine hydrochloride)
R₂NH + HBr → [R₂NH₂]⁺Br⁻
(Secondary amine hydrobromide)
R₃N + HNO₃ → [R₃NH]⁺NO₃⁻
(Tertiary amine nitrate)
**General reaction:**
Amine + Acid → Ammonium salt (water soluble, ionic)
**Regeneration of free amine:**
[R-NH₃]⁺Cl⁻ + NaOH → R-NH₂ + NaCl + H₂O
**Practical application:** This acid-base reaction forms the basis for **separation of amines from non-basic organic compounds**:
1. Dissolve mixture in organic solvent (ether/ether)
2. Shake with dilute HCl → amine becomes salt, dissolves in aqueous layer
3. Non-basic compounds remain in organic layer
4. Separate aqueous layer, add NaOH → regenerate free amine
**Dissociation in water:**
R-NH₂ + H₂O ⇌ [R-NH₃]⁺ + OH⁻
**Base dissociation constant:**
**Kb = [RNH₃⁺][OH⁻]/[RNH₂]**
**pKb = -log Kb**
**Interpretation:**
**pKb values (Table 9.3):**
| Amine | pKb | Classification |
|-------|-----|----------------|
| Methanamine | 3.38 | Strong base (aliphatic) |
| N-Methylmethanamine | 3.27 | Strong base |
| N,N-Dimethylmethanamine | 4.22 | Strong base |
| Ethanamine | 3.29 | Strong base |
| N-Ethylethanamine | 3.00 | Very strong base |
| Benzenamine (aniline) | 9.38 | Weak base (aromatic) |
| N-Methylaniline | 9.30 | Weak base |
| N,N-Dimethylaniline | 8.92 | Weak base |
| NH₃ (ammonia) | 4.75 | Reference base |
#### (A) Aliphatic Amines vs Ammonia
**General trend (gaseous phase):**
**Tertiary > Secondary > Primary > NH₃**
(Based on inductive effect: more alkyl groups → more +I effect → stronger base)
**In aqueous phase** (which is exam-relevant):
**Primary > Secondary > Tertiary > NH₃**
(Based on solvation effect overriding inductive effect)
**Explanation for aliphatic amines being stronger bases than ammonia:**
1. **Protonation reaction:**
R-NH₂ + H⁺ → [R-NH₃]⁺
NH₃ + H⁺ → [NH₄]⁺
2. **Why RNH₂ is more basic:**
3. **Why the order is different in aqueous solution:**
4. **Steric hindrance factor:**
**Basicity order for methyl-substituted amines (aqueous):**
**(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃**
**Basicity order for ethyl-substituted amines (aqueous):**
**(C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃**
#### (B) Aromatic Amines (Arylamines) vs Ammonia and Aliphatic Amines
**Basicity order:**
**Aliphatic primary > NH₃ >> Aromatic primary**
**pKb values:**
**Why aromatic amines are much weaker bases:**
1. **Resonance stabilization of aniline (neutral form):**
2. **Anilinium ion (protonated form):**
3. **Comparison:**
4. **Effect of substituents on aromatic amines:**
**Electron-releasing groups increase basicity:**
**Electron-withdrawing groups decrease basicity:**
**Ortho > para > meta** (for same substituent) due to resonance effects
**Exam-critical point:** Aromatic amines are **much weaker bases than aliphatic amines** because the resonance stabilization of the neutral form (with extended conjugation) dominates over the stabilization of the protonated cation.
**Principle:** Amines react with alkyl halides through nucleophilic substitution.
**Primary amines:**
RNH₂ + R'Cl → R-NHR' + HCl
(Primary amine → Secondary amine)
**Secondary amines:**
R₂NH + R'Cl → R₂N-R' + HCl
(Secondary amine → Tertiary amine)
**Tertiary amines:**
R₃N + R'Cl → [R₃NR']⁺Cl⁻
(Quaternary ammonium salt — no further alkylation)
**Example:**
CH₃NH₂ + CH₃Cl → CH₃-NH-CH₃ (N-methylmethanamine) + HCl
**Mechanism:** SN2 nucleophilic substitution (amine as nucleophile)
**Industrial application:** **Quaternary ammonium salts** (like [R₄N]⁺X⁻) are used as surfactants in detergents and fabric softeners.
**Principle:** Primary and secondary amines react with acid derivatives (acid chlorides, anhydrides, esters) through nucleophilic acyl substitution, replacing H on N with acyl group (R-CO–).
**Reagents used:**
1. **Acid chlorides (R-COCl):** Most reactive
RNH₂ + R'-COCl → RNH-CO-R' + HCl (amide)
2. **Anhydrides (R-CO-O-CO-R):**
RNH₂ + (R'-CO)₂O → RNH-CO-R' + R'-COOH
3. **Esters (R-COOR'):**
RNH₂ + R'-COOR'' → RNH-CO-R' + R''OH
(Slower than chlorides and anhydrides)
**Reaction conditions:**
**General mechanism (with acid chloride):**
Step 1 - Nucleophilic attack:
RNH₂ + R'-COCl → RNH-CO-R' (tetrahedral intermediate forms)
Step 2 - Chloride departure:
Tetrahedral intermediate → RNH-CO-R' + Cl⁻
Step 3 - Base abstraction (if pyridine present):
Base + HCl → Base-H⁺ + Cl⁻ (removes acid, drives equilibrium)
**Product:** **Amide** (primary amide from primary amine, secondary amide from secondary amine)
**Examples:**
**Primary amine acylation:**
C₂H₅NH₂ + CH₃COCl →(pyridine) CH₃-CO-NH-C₂H₅ (N-ethylethanamide/propanamide)
**Secondary amine acylation:**
C₂H₅NH-CH₃ + C₆H₅COCl →(pyridine) C
Q1. Which of the following is NOT a correct IUPAC name?
Answer: C — (CH₃)₂CHNH₂ is propan-2-amine (isopropylamine), not propan-1-amine, because the –NH₂ group is attached to the 2nd carbon atom of the propane chain.
Q2. In ammonolysis of alkyl halides, the primary amine formed can further react with the alkyl halide. Which condition prevents excessive secondary and tertiary amine formation?
Answer: C — Excess ammonia increases the concentration of ammonia relative to the primary amine, so ammonia competes more effectively as a nucleophile, favouring the formation of primary amine as the major product.
Q3. The reduction of nitrobenzene to aniline by iron scrap and dilute HCl is preferred over other reducing agents because:
Answer: B — FeCl₂ formed during the reaction undergoes hydrolysis to regenerate HCl, ensuring that acidic conditions are maintained throughout without requiring continuous addition of more HCl.
Q4. Which statement correctly explains the basicity order: aliphatic amines > ammonia > aromatic amines?
Answer: B — Electron-donating alkyl groups in aliphatic amines stabilise the conjugate acid (RNH₃⁺), increasing basicity; in arylamines, resonance delocalisation of the lone pair into the benzene ring reduces basicity.
Q5. The Gabriel phthalimide synthesis is preferred over direct ammonolysis for preparing primary amines because:
Answer: B — Phthalimide reacts with alkyl halides once, preventing over-alkylation; hydrolysis then yields pure primary amine, unlike ammonolysis which gives a mixture of 1°, 2°, and 3° amines.
Q6. In the Carbylamine test, a primary amine reacts with CHCl₃ and KOH to produce a foul-smelling isocyanide. Which statement is NOT correct about this reaction?
Answer: D — The Carbylamine test is specific to primary amines (both aliphatic and aromatic), not aromatic amines only; both R–NH₂ and Ar–NH₂ produce the foul-smelling isocyanide (R–N≡C or Ar–N≡C).
Q7. Reduction of nitriles using LiAlH₄ is described as 'ascent of amine series.' This is because:
Answer: B — Reduction of R–C≡N by LiAlH₄ produces R–CH₂–NH₂, which has one more carbon atom than the starting material, effectively moving up one step in the homologous series of amines.
Q8. Assertion (A): In amines, the C–N–C bond angle is approximately 108°, which is less than the tetrahedral angle of 109.5°. Reason (R): The lone pair on nitrogen occupies more space than a bonding pair, causing greater repulsion and compression of bond angles.
Answer: A — The assertion is correct: the C–N–C angle is ~108°. The reason correctly explains this: the lone pair on sp³-hybridised nitrogen exerts greater repulsion than bonding pairs, compressing the bond angles below 109.5°.
Q9. Which of the following conversions is NOT possible using the reduction methods described in this unit?
Answer: D — The Carbylamine test (CHCl₃/KOH) does not work on primary aliphatic amines; it works on primary amines to produce isocyanides (R–N≡C), but the reverse conversion shown is not a reduction method and produces a different product type.
Q10. The coupling reaction of diazonium salts with phenols or anilines produces azo dyes. If 0.5 mol of aniline is coupled with 0.5 mol of benzene diazonium salt, how many moles of azo dye product are formed (assuming 100% conversion)?
Answer: B — The coupling reaction is Ar–N₂⁺ + Ar'–OH (or Ar'–NH₂) → Ar–N=N–Ar'–OH (or Ar–N=N–Ar'–NH₂); 1 mole of diazonium salt reacts with 1 mole of phenol/aniline to form 1 mole of azo dye, so 0.5 mol each gives 0.5 mol product.
What is the hybridisation state of nitrogen in amines and why is the C–N–C angle not 109.5°?
Nitrogen is sp³ hybridised but the C–N–C angle is ~108° (less than 109.5°) because the lone pair occupies more space than a bonding pair, compressing the bond angles.
How are primary, secondary and tertiary amines classified?
Primary amines (RNH₂) have one alkyl/aryl group; secondary amines (R₂NH) have two; tertiary amines (R₃N) have three alkyl/aryl groups attached to nitrogen.
Write the IUPAC name of CH₃CH₂NHCH₃.
The IUPAC name is N-methylethanamine because the longest carbon chain has 2 carbons (ethanamine) and the methyl substituent on nitrogen is prefixed with 'N-'.
Why are aliphatic amines more basic than aniline?
Aliphatic amines are more basic because the lone pair on nitrogen is freely available; in aniline, the lone pair is delocalised into the benzene ring, reducing its basicity.
In ammonolysis of alkyl halides, why is a large excess of ammonia used?
Excess ammonia favours the formation of primary amine as the major product because the primary amine formed is less likely to react further when ammonia is in large excess.
What is the Gabriel phthalimide synthesis and when is it preferred over ammonolysis?
Gabriel phthalimide synthesis converts phthalimide to phthalamide using alkyl halide, then hydrolyses to give pure primary amines; it is preferred because it avoids the mixture of 1°, 2°, and 3° amines from ammonolysis.
Describe the Carbylamine test for primary amines.
A primary amine reacts with chloroform (CHCl₃) and potassium hydroxide (KOH) to produce a foul-smelling isocyanide (R–N≡C), confirming the presence of a primary amine.
Write the equation for reduction of nitrobenzene by iron scrap and dilute HCl.
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O; iron reduces the nitro group and HCl released from hydrolysis of FeCl₂ maintains acidic conditions.
What is the order of reactivity of alkyl halides in ammonolysis?
The reactivity order is RI > RBr > RCl because the C–I bond is weakest and most easily cleaved by the nucleophilic ammonia molecule.
Why is reduction of nitriles by LiAlH₄ called 'ascent of amine series'?
Reduction of nitriles (R–C≡N) by LiAlH₄ produces primary amines (R–CH₂–NH₂) with one additional carbon atom, effectively ascending the homologous series of amines by one carbon.
Write the structure and IUPAC name of the primary amine with molecular formula C₃H₉N. [2 marks]
There are two structural isomers for C₃H₉N as primary amines; identify the straight-chain and branched structures, then apply IUPAC naming rules where the longest carbon chain containing the –NH₂ group determines the base name.
Explain why aliphatic amines are more basic than aromatic amines. Also show the conjugate acid formation for both types and explain how electron effects stabilise or destabilise the conjugate acid. [5 marks]
Compare the lone pair availability on nitrogen: in aliphatic amines, alkyl groups donate electron density (inductive effect) stabilising the conjugate acid RNH₃⁺; in aromatic amines, the lone pair delocalises into the benzene ring via resonance, reducing basicity. Write the conjugate acid structures (RNH₃⁺ and [C₆H₅NH₃]⁺) and explain the stabilisation mechanism in each case.
Starting from nitrobenzene, outline the complete preparation of aniline using iron scrap and hydrochloric acid. Explain why this method is preferred over other reducing agents like tin or zinc. Also describe a chemical test that can be used to confirm the presence of a primary amino group in the product. [6 marks]
Write the reduction equation C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O; explain that FeCl₂ hydrolyses to release HCl continuously (Fe²⁺ + 2H₂O ⇌ Fe(OH)₂ + 2H⁺), maintaining acidic conditions with minimal acid input; then describe the Carbylamine test (RNH₂ + CHCl₃ + KOH → R–N≡C with foul smell) to confirm the primary amine functional group.
Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly
Try StudyOS Free →