Amines • Topic 1 of 3

Classification & Preparation

Amines are organic derivatives of ammonia (NH3) in which one, two or all three hydrogen atoms are replaced by alkyl or aryl groups. They are everywhere in living systems: the neurotransmitters adrenaline and dopamine, the amino acids that build proteins, and alkaloids like quinine and nicotine are all amines.

Classification

Amines are classified by the number of hydrogen atoms of ammonia replaced by carbon-bearing groups:

  • Primary (1°): one H replaced — R–NH2 (e.g. CH3NH2, aniline C6H5NH2).
  • Secondary (2°): two H replaced — R2NH (e.g. (CH3)2NH).
  • Tertiary (3°): all three H replaced — R3N (e.g. (CH3)3N).

Note this differs from alcohols/halides, where the classification depends on the carbon. Amines are also classed as aliphatic (groups are alkyl, e.g. ethylamine) or aromatic (at least one group is aryl, e.g. aniline, N-methylaniline).

Structure

The nitrogen of an amine is sp3 hybridised and pyramidal, with a lone pair occupying one tetrahedral position. The C–N–C angle is about 108°, close to tetrahedral. This lone pair is the seat of both the basicity and the nucleophilicity of amines.

Nomenclature

Common names: name the alkyl groups attached to N alphabetically and add the suffix -amine (e.g. ethylamine, N,N-dimethylethanamine). IUPAC names: primary amines are named as alkanamines — the –NH2 is the principal characteristic group, so the e of the parent alkane is replaced by -amine (CH3CH2NH2 → ethanamine). For 2° and 3° amines the largest group is the parent and the others are prefixed with locant N (e.g. CH3NHCH2CH3 → N-methylethanamine). C6H5NH2 is aniline (IUPAC: benzenamine).

Methods of Preparation

1. Reduction of nitro compounds: R–NO2 is reduced to R–NH2 by H2/Ni (or Sn + HCl, Fe + HCl). Nitrobenzene → aniline. In acidic medium, selective reduction with metal/acid is used.

2. Reduction of nitriles: R–C≡N + 4[H] → R–CH2–NH2 (LiAlH4 or H2/Ni). This ascends the series by one carbon — a primary amine with one extra C (Mendius reaction).

3. Reduction of amides: R–CONH2 + LiAlH4 → R–CH2–NH2 (a 1° amine with the same carbon count).

4. Ammonolysis of alkyl halides: R–X + NH3 → a mixture of 1°, 2°, 3° amines and quaternary salt, because the product amine is itself a nucleophile. Excess ammonia favours the 1° amine. This is non-selective.

5. Gabriel phthalimide synthesis: gives pure 1° aliphatic amines with no 2°/3° contamination. Potassium phthalimide is alkylated by R–X, then the N-alkylphthalimide is hydrolysed (or treated with hydrazine) to release R–NH2. Aromatic 1° amines cannot be made this way because aryl halides do not undergo the required nucleophilic substitution.

6. Hofmann bromamide degradation: an amide reacts with Br2 and aqueous KOH/NaOH to give a 1° amine with one fewer carbon: R–CONH2 → R–NH2. The alkyl group migrates to electron-deficient N (intramolecular).

Physical Properties

Lower amines are gases or volatile liquids with a fishy smell. 1° and 2° amines form intermolecular N–H···N hydrogen bonds, so they boil higher than comparable alkanes but lower than alcohols (N is less electronegative than O). 3° amines have no N–H and the lowest boiling points. Lower amines dissolve in water (H-bonding to water); solubility falls as the hydrocarbon part grows.

Classification of amines into primary, secondary and tertiary by number of H atoms of ammonia replacedAmmonia and its three amine familiesNH3 (ammonia)Primary (1 deg)R-NH21 H replacede.g. CH3NH2, anilineSecondary (2 deg)R2NH2 H replacede.g. (CH3)2NHTertiary (3 deg)R3N3 H replacede.g. (CH3)3N
Solved Examples
1
Worked Example
Classify the following as 1°, 2° or 3° amines: (i) (CH3)2CHNH2, (ii) (C2H5)2NH, (iii) (CH3)3N, (iv) C6H5NHCH3.
Solution
  1. Classification of amines depends on how many H atoms of NH3 are replaced, i.e. how many carbon groups are bonded to N.
  2. (i) (CH3)2CHNH2: one group on N → 1° (the branching is on carbon, not nitrogen).
  3. (ii) (C2H5)2NH: two groups on N → 2°.
  4. (iii) (CH3)3N: three groups on N → 3°.
  5. (iv) C6H5NHCH3: two groups (phenyl + methyl) on N → 2° (aromatic).

Answer: (i) 1°, (ii) 2°, (iii) 3°, (iv) 2° aromatic.

2
Worked Example
Give the IUPAC name of CH3CH2CH(CH3)NH2 and of CH3NHCH2CH2CH3.
Solution
  1. For 1° amines, –NH2 is the principal group; replace the final ‘e’ of the parent alkane with ‘-amine’ and give it the lowest locant.
  2. CH3CH2CH(CH3)NH2: longest chain is butane, NH2 on C-2 → butan-2-amine.
  3. For 2° amines, the larger group is the parent chain; the smaller group is named as an N-substituent.
  4. CH3NHCH2CH2CH3: parent is propane (propan-1-amine), with a methyl on N → N-methylpropan-1-amine.

Answer: butan-2-amine and N-methylpropan-1-amine.

3
Worked Example
How would you convert ethanenitrile (CH3CN) into ethanamine and into propan-1-amine? Comment on the carbon count.
Solution
  1. Reduction of a nitrile R–CN adds two H to N and two to C: R–CN + 4[H] → R–CH2NH2, increasing carbon count by one.
  2. CH3CN + 4[H] ⟶{LiAlH4} CH3CH2NH2 (ethanamine, 2 carbons).
  3. Wait — here CH3CN has 2 carbons and gives ethanamine (2 carbons): reduction keeps the C of CN. To reach propan-1-amine (3 C), first make propanenitrile.
  4. So: CH3CN cannot directly give propan-1-amine. Reduce CH3CH2CN (propanenitrile) instead → CH3CH2CH2NH2.

Answer: CH3CN ⟶{LiAlH4} ethanamine; propan-1-amine comes from reducing propanenitrile (CH3CH2CN).

4
Worked Example
Why is the Gabriel phthalimide synthesis used to prepare pure primary amines, and why does it fail for aromatic amines like aniline?
Solution
  1. Potassium phthalimide is alkylated by R–X to give N-alkylphthalimide, which on hydrolysis (or hydrazinolysis) gives R–NH2.
  2. The nitrogen can carry only one alkyl group (it is locked in the ring), so no 2° or 3° amine forms — the product is a pure 1° amine.
  3. The alkylation step is an SN2 nucleophilic substitution on R–X by the phthalimide anion.
  4. Aryl halides (e.g. C6H5Br) do not undergo this nucleophilic substitution (C–X partial double-bond character; no SN2/SN1), so aniline cannot be made by this route.

Answer: It gives uncontaminated 1° amines; it fails for aniline because aryl halides resist nucleophilic substitution by phthalimide.

5
Worked Example
Identify the amine formed when CH3CH2CONH2 (propanamide) undergoes the Hofmann bromamide degradation. Why does the carbon count decrease?
Solution
  1. In Hofmann bromamide degradation, an amide is treated with Br2 and aqueous KOH: R–CONH2 → R–NH2.
  2. The carbonyl carbon is lost as carbonate (CO32–); the R group migrates to the nitrogen.
  3. Here R = CH3CH2– (ethyl), so the product is CH3CH2NH2.
  4. Propanamide (3 C) gives ethanamine (2 C): the amine has one fewer carbon than the amide.

Answer: Ethanamine (CH3CH2NH2); the carbonyl carbon is removed as carbonate.

6
Worked Example
Arrange the following in order of increasing boiling point: n-butylamine, diethylamine, n-pentane. Explain.
Solution
  1. n-Pentane (C5H12) has only weak London forces and no H-bonding.
  2. Both amines have N and similar molar mass; 1° amine (n-butylamine, RNH2) has two N–H bonds for hydrogen bonding, 2° amine (diethylamine, R2NH) has only one.
  3. More N–H bonds give stronger intermolecular H-bonding and higher boiling point, so n-butylamine boils higher than diethylamine.
  4. Order of increasing b.p.: n-pentane < diethylamine < n-butylamine.

Answer: n-pentane < diethylamine < n-butylamine.

Key Points

  • Amines = NH3 derivatives; classified 1° (RNH2), 2° (R2NH), 3° (R3N) by the number of H of ammonia replaced, and as aliphatic vs aromatic.
  • N is sp3, pyramidal with a lone pair; IUPAC names 1° amines as alkanamines (ethanamine), with N-locants for groups on nitrogen (N-methylethanamine); C6H5NH2 = aniline.
  • Reduction routes: nitro→amine (H2/Ni or Sn/HCl), nitrile→1° amine (+1 C), amide→1° amine (LiAlH4, same C).
  • Gabriel synthesis gives pure 1° aliphatic amines (fails for aryl amines); Hofmann bromamide degradation gives a 1° amine with one fewer carbon (loses CO32–); ammonolysis of R–X is non-selective.
  • 1°/2° amines form N–H···N H-bonds: b.p. alcohol > amine > alkane; 3° amines (no N–H) have the lowest b.p.; lower amines are water-soluble.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Which of the following is a tertiary amine?
Explanation: (CH3)3N has all three H of ammonia replaced by carbon groups, so it is tertiary.
Q2.The Gabriel phthalimide synthesis cannot be used to prepare:
Explanation: Aniline needs an aryl halide, which does not undergo the nucleophilic substitution with phthalimide; Gabriel makes only 1° amines from alkyl/benzyl halides.
Q3.Reduction of CH3CH2CN with LiAlH4 gives:
Explanation: R–CN → R–CH2NH2; CH3CH2CN gives CH3CH2CH2NH2 (propan-1-amine), keeping the nitrile carbon.
Q4.Hofmann bromamide degradation of CH3CONH2 gives an amine with:
Explanation: Acetamide (2 C) gives methanamine (1 C); the carbonyl carbon is lost as carbonate, so the amine has one fewer carbon.
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