Amines
Classification, basicity (electronic and steric effects), synthesis (reduction of nitro compounds, NH₃ alkylation, nitrile reduction), diazonium salts (diazotization, coupling).
Complete Theory
5Amines are organic compounds derived from ammonia (NH₃) by replacement of one, two, or three hydrogen atoms with alkyl or aryl groups. They are classified into:
Aromatic amines: the main example is aniline (C₆H₅NH₂). In aniline, the nitrogen lone pair is partially conjugated with the benzene ring (resonance), making it less basic and less nucleophilic than aliphatic amines.
Real-world applications: amines are present in many drugs (antidepressants, antihistamines, local anesthetics), dyes (azo dyes), and are the building blocks of amino acids and proteins. Trimethylamine is responsible for the smell of decomposing fish.
- Primary amines (1°): R—NH₂ (e.g., methylamine CH₃NH₂)
- Secondary amines (2°): R₂NH (e.g., dimethylamine (CH₃)₂NH)
- Tertiary amines (3°): R₃N (e.g., trimethylamine (CH₃)₃N)
- Quaternary ammonium salts: R₄N⁺ X⁻ (e.g., tetramethylammonium chloride)
Aromatic amines: the main example is aniline (C₆H₅NH₂). In aniline, the nitrogen lone pair is partially conjugated with the benzene ring (resonance), making it less basic and less nucleophilic than aliphatic amines.
Real-world applications: amines are present in many drugs (antidepressants, antihistamines, local anesthetics), dyes (azo dyes), and are the building blocks of amino acids and proteins. Trimethylamine is responsible for the smell of decomposing fish.
Amines are Brønsted bases: the nitrogen lone pair accepts a proton to form a quaternary ammonium ion.
The pKₐ of the conjugate base (RNH₃⁺) measures basicity: the higher the pKₐ, the stronger the base.
Basicity order in water (aliphatic amines): (2° > 1° > 3° > NH₃)
The order may be surprising one would expect 3° > 2° > 1° from the +I inductive effect of alkyl groups. In water, basicity is determined by two factors:
Aniline and aromatic amines: they are much less basic than aliphatic amines (pKₐ of aniline = 4.6, pKₐ of CH₃NH₂ = 10.6). The nitrogen lone pair is delocalized by resonance into the aromatic ring, reducing availability for protonation.
Real-world applications: the different basicities are exploited in separating amine mixtures (acid-base extraction) and in drug design: many drugs act as weak bases that are protonated in the acidic stomach and absorbed in the intestine.
Basicity order in water (aliphatic amines): (2° > 1° > 3° > NH₃)
The order may be surprising one would expect 3° > 2° > 1° from the +I inductive effect of alkyl groups. In water, basicity is determined by two factors:
- Inductive effect (+I): alkyl groups donate electrons to nitrogen, increasing electron density on the lone pair and therefore basicity. Expected order: 3° > 2° > 1°.
- Solvation: the ammonium ion RNH₃⁺ is stabilized by hydrogen bonding with water. The more H atoms bonded to N⁺, the greater the solvation. 1° (3 H) is better solvated than 2° (2 H) and 3° (1 H).
- Steric effect: bulky groups hinder solvent approach to the ammonium ion, reducing solvation.
Aniline and aromatic amines: they are much less basic than aliphatic amines (pKₐ of aniline = 4.6, pKₐ of CH₃NH₂ = 10.6). The nitrogen lone pair is delocalized by resonance into the aromatic ring, reducing availability for protonation.
Real-world applications: the different basicities are exploited in separating amine mixtures (acid-base extraction) and in drug design: many drugs act as weak bases that are protonated in the acidic stomach and absorbed in the intestine.
The main synthetic methods for amines are:
1. Reduction of aromatic nitro compounds: Aromatic nitro compounds (obtained by SEAr nitration) are reduced to anilines with Sn/HCl, Fe/HCl, or H₂/Pd-C. This is the most common method for preparing aromatic amines.
2. Alkylation of ammonia (Hofmann reaction): SN2 reaction: ammonia (nucleophile) attacks the alkyl halide. Problem: the primary amine produced is more nucleophilic than ammonia and reacts further, giving mixtures of 1°, 2°, 3° amines and quaternary salts. To limit polyalkylation, a large excess of NH₃ is used.
3. Reduction of nitriles and amides:
Real-world applications: reduction of nitro compounds is the industrial method for producing aniline (over 4 million tonnes/year), an intermediate for dyes, polymers (polyurethanes), and drugs. The Gabriel synthesis is used to prepare amino acids in the laboratory.
1. Reduction of aromatic nitro compounds: Aromatic nitro compounds (obtained by SEAr nitration) are reduced to anilines with Sn/HCl, Fe/HCl, or H₂/Pd-C. This is the most common method for preparing aromatic amines.
2. Alkylation of ammonia (Hofmann reaction): SN2 reaction: ammonia (nucleophile) attacks the alkyl halide. Problem: the primary amine produced is more nucleophilic than ammonia and reacts further, giving mixtures of 1°, 2°, 3° amines and quaternary salts. To limit polyalkylation, a large excess of NH₃ is used.
3. Reduction of nitriles and amides:
- R—C≡N + LiAlH₄ (or H₂/Pd) → R—CH₂—NH₂ (primary amine, chain extension by 1 C)
- RCONH₂ + LiAlH₄ → R—CH₂—NH₂ (primary amine)
- RCONR'R'' + LiAlH₄ → R—CH₂—NR'R'' (2° or 3° amines)
Real-world applications: reduction of nitro compounds is the industrial method for producing aniline (over 4 million tonnes/year), an intermediate for dyes, polymers (polyurethanes), and drugs. The Gabriel synthesis is used to prepare amino acids in the laboratory.
Diazonium salts (Ar—N≡N⁺ X⁻) are extremely versatile aromatic compounds in organic synthesis. They are prepared by diazotization of primary aromatic amines:
The reaction must be carried out at 0–5°C because diazonium salts are unstable at higher temperatures (they decompose releasing N₂). The mechanism involves formation of nitrous acid (HNO₂) from NaNO₂ + HCl, which nitrosates the amine.
Substitution reactions (with N₂ loss):
Real-world applications: azo dyes constitute over 60% of world dye production (methyl red, methyl orange, butter yellow). Methyl orange is also a pH indicator (transition at pH 3.1–4.4). The Sandmeyer reaction allows introduction of halogens at specific ring positions not accessible via direct SEAr.
Substitution reactions (with N₂ loss):
- Hydrolysis: ArN₂⁺ + H₂O (Δ) → Ar—OH + N₂ + H⁺ (phenols)
- Sandmeyer substitution with halogens: CuCl → ArCl, CuBr → ArBr, CuCN → ArCN (aromatic nitriles, then hydrolyzable to acids)
- Substitution with KI: ArN₂⁺ + I⁻ → Ar—I + N₂ (aromatic iodides, difficult to obtain otherwise)
- Schiemann substitution with HBF₄: ArN₂⁺ BF₄⁻ (Δ) → Ar—F + N₂ + BF₃ (aromatic fluorides)
- Reductive deamination with H₃PO₂: ArN₂⁺ + H₃PO₂ → Ar—H + N₂ + H₃PO₃
Real-world applications: azo dyes constitute over 60% of world dye production (methyl red, methyl orange, butter yellow). Methyl orange is also a pH indicator (transition at pH 3.1–4.4). The Sandmeyer reaction allows introduction of halogens at specific ring positions not accessible via direct SEAr.
Amines are key intermediates in many syntheses, thanks to their nucleophilicity, basicity and the ways they can be transformed.
Protecting amines:
Protecting amines:
- The —NH₂ group is very reactive (nucleophilic). To "protect" it while reacting on other groups, it is converted into an amide (acetylation: R—NHCOCH₃) or a carbamate (Boc₂O → R—NH—Boc).
- Deprotection is by acid hydrolysis (Boc → HCl/CH₂Cl₂; Fmoc → mild base).
- Primary amines → isocyanates (R—N=C=O) reacting with alcohols → urethanes (polyurethanes).
- Secondary amines → nitrosamines (R₂N—NO), carcinogenic compounds formed with nitrites.
- Hofmann rearrangement: an amide treated with Br₂/NaOH → primary amine with loss of CO₂.
- Hofmann elimination: R₄N⁺OH⁻ (Δ) → the least-substituted alkene (Hofmann rule, opposite to Zaitsev).
- Amino acids contain the —NH₂ (or NH₃⁺) group.
- The DNA nitrogenous bases (adenine, guanine, cytosine) are heterocyclic amines.
- Dopamine, serotonin, adrenaline are biogenic amines.
- Alkaloids (morphine, quinine, strychnine) contain amine nitrogens.
Worked Examples
2Example 1Comparative basicity: aniline vs aliphatic amine
Given
Aniline (C₆H₅NH₂): pKₐ (conjugate base) = 4.6
Methylamine (CH₃NH₂): pKₐ (conjugate base) = 10.6
Find
Which is the stronger base?
Explain the 6-order-of-magnitude difference
Step-by-step solution
1Methylamine is a much stronger base than aniline. The pKₐ of protonated methylamine (CH₃NH₃⁺) is 10.6, while that of protonated aniline (C₆H₅NH₃⁺) is 4.6. Kₐ differs by a factor of 10⁶!
2In aniline, the nitrogen lone pair is in conjugation with the aromatic ring: it is partially delocalized into the ring (resonance structures with negative charge on the ring at ortho and para positions). This delocalization reduces the lone pair availability for protonation.
3In methylamine, the +I inductive effect of CH₃ increases electron density on nitrogen. There is no lone pair delocalization (no nearby orbitals). Consequently, the lone pair is fully available for protonation.
4Synthetic consequence: aniline is not significantly protonated at neutral pH (can be extracted with organic solvents), while methylamine exists as a salt in water.
✓ Final result: Methylamine is 10⁶ times more basic than aniline (lone pair delocalization in the aromatic ring)
Example 2Diazotization and coupling: synthesis of an azo dye
Given
Aniline (C₆H₅NH₂) + NaNO₂/HCl at 0°C
Then addition of phenol (C₆H₅OH) under basic conditions
Find
Final product of the reaction sequence
Step-by-step solution
1First step: diazotization of aniline at 0°C: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N≡N⁺ Cl⁻ (benzenediazonium chloride) + 2H₂O + NaCl. The temperature must stay below 5°C to prevent decomposition.
2Second step: coupling with phenol. The diazonium salt is a weak electrophile. Phenol, under basic conditions (pH 8–10), exists as phenoxide (C₆H₅O⁻), strongly activated toward electrophilic aromatic substitution at the para position (or ortho if para is occupied).
3Product: 4-hydroxyazobenzene (p-hydroxyazobenzene): C₆H₅—N=N—C₆H₄—OH, a yellow-orange azo dye. The azo group (—N=N—) conjugates the two aromatic rings, extending the system and producing absorption in the visible region.
✓ Final result: C₆H₅N₂⁺ + C₆H₅OH → C₆H₅—N=N—C₆H₄—OH (4-hydroxyazobenzene, azo dye)
Exercises with Solutions
3Exercise 1Amine classificationMedium
Problem to solve
Classify the following amines as primary, secondary, or tertiary: (a) (CH₃)₂CHNH₂, (b) (C₂H₅)₂NH, (c) C₆H₅N(CH₃)₂, (d) NH₂CH₂CH₂NH₂.
Given data
(a) isopropylamine(b) diethylamine(c) N,N-dimethylaniline(d) ethylenediamine
Step-by-step solution
1(a) (CH₃)₂CHNH₂: nitrogen is bonded to 1 C and 2 H → primary.
2(b) (C₂H₅)₂NH: nitrogen is bonded to 2 C and 1 H → secondary.
3(c) C₆H₅N(CH₃)₂: nitrogen is bonded to 3 C (one phenyl and two methyls) and 0 H → tertiary.
4(d) NH₂CH₂CH₂NH₂: two NH₂ groups, both primary (diamine).
✓ Final answer: (a) 1°, (b) 2°, (c) 3°, (d) 1° (diamine)
Exercise 2BasicityHard
Problem to solve
Order by increasing basicity: aniline, p-nitroaniline, p-methoxyaniline, p-toluidine.
Given data
p-NO₂-C₆H₄NH₂C₆H₅NH₂ (aniline)p-CH₃O-C₆H₄NH₂p-CH₃-C₆H₄NH₂ (p-toluidine)
Step-by-step solution
1The basicity of substituted aromatic amines depends on the electronic effect of the substituent at the para (or ortho) position.
2NO₂: strong electron-withdrawing (−M, −I) → reduces electron density on nitrogen → reduces basicity. CH₃O: resonance donor (+M) → increases basicity. CH₃: weak inductive effect (+I) donor → slightly increases basicity.
3Increasing basicity order: p-nitroaniline (weakest) < aniline < p-toluidine < p-methoxyaniline (strongest).
✓ Final answer: p-NO₂-aniline < aniline < p-CH₃-aniline < p-CH₃O-aniline
Exercise 3Gabriel synthesisVery Hard
Problem to solve
Describe how to prepare benzylamine (C₆H₅CH₂NH₂) using the Gabriel synthesis.
Given data
Target: C₆H₅CH₂NH₂ (benzylamine)
Step-by-step solution
1Step 1: preparation of the potassium salt of phthalimide: phthalimide + KOH → phthalimide-K⁺.
2Step 2: Gabriel alkylation: phthalimide-K⁺ + C₆H₅CH₂Br (benzyl bromide, 1° halide) → N-benzylphthalimide + KBr. SN2 reaction, the phthalimide anion is a good nucleophile.
3Step 3: amine liberation: N-benzylphthalimide + NH₂NH₂ (hydrazine) → phthalimide (recycled) + C₆H₅CH₂NH₂ (benzylamine). Hydrazine attacks the imide carbonyl, releasing the pure primary amine.
4Advantage: no polyalkylation (the phthalimide anion has only one nucleophilic site).
✓ Final answer: Phthalimide-K + C₆H₅CH₂Br → N-benzylphthalimide + NH₂NH₂ → benzylamine (pure 1° amine)
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