Amines

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Amines — Key Facts for JEE Advanced

Classification:

• 1° amine (Primary): R–NH₂ (one C attached to N)
• 2° amine (Secondary): R₂N–H (two C attached to N)
• 3° amine (Tertiary): R₃N (three C attached to N)
• Quaternary ammonium salt: R₄N⁺ X⁻ (four C attached to N, permanent charge)
• Aromatic amine: Aniline = C₆H₅–NH₂ (–NH₂ attached directly to benzene ring)

⚠️ JEE Warning: Don’t confuse organic chemistry nomenclature systems:

• Alcohols: 1° = carbon bearing –OH attached to one R
• Amines: 1° = nitrogen attached to one R (and two H)
• A 1° amine has TWO hydrogens on N!

Nomenclature:

• Aliphatic: alkyl + amine (methylamine, ethylamine, dimethylamine)
• Aromatic: aniline, toluidine (methyl aniline), anisidine (methoxy aniline)
• IUPAC: aminoalkane (CH₃CH₂CH₂NH₂ = propan-1-amine)
• For substituted amines: priority order: –NH₂ > –OH > –SH when numbering

Key Physical Property:

• Lower members (C₁–C₃) are gases or low-boiling liquids, water-soluble
• All amines can act as H-bond acceptors (lone pair on N)
• 1° and 2° amines can H-bond with themselves; 3° amines cannot (no N–H bond) → lower BP than isomeric 1° or 2°
• Boiling point order for isomeric C₃H₉N: CH₃CH₂CH₂NH₂ (propan-1-amine, BP 48°C) > CH₃CH₂NHCH₃ (ethylmethylamine, BP 37°C) > (CH₃)₃N (trimethylamine, BP 3°C)

⚡ Exam Tip: Basicity order in water: Aliphatic amines > NH₃ > Aromatic amines (aniline). Aniline is ~10⁴ times weaker base than NH₃ because the lone pair on N is delocalized into the benzene ring (resonance), making it less available for protonation.

⚡ Exam Tip: In the gas phase (without solvation), the order can reverse. This is because solvation (H-bonding with water) stabilizes the conjugate acid. Aromatic amines have less solvation because the positive charge on the conjugate acid is delocalized into the ring. In water: aliphatic > aromatic. In gas phase: aromatic may be comparable.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Amines — Chemistry Study Guide

1. Structure & Bonding:

Hybridization of Nitrogen:

• In ammonia and amines, N is sp³ hybridized
• Three sp³ orbitals form σ bonds; fourth sp³ orbital holds the lone pair
• Ideal tetrahedral angle: 109.5°; actual C–N–C angle in trimethylamine: 108.7° (slightly compressed by lone pair repulsion)
• In aniline: N is sp² hybridized (lone pair in p orbital, conjugated with benzene ring)
• C–N bond in aniline (~140 pm) is shorter than C–N in alkylamines (~147 pm) due to partial double bond character

Basicity — pKb Values:

Amine	pKb	pKa of conjugate acid	Relative Basicity
Methylamine	3.36	10.64	Strong base
Dimethylamine	3.27	10.73	Stronger
Trimethylamine	4.20	9.80	Weaker
Aniline	9.37	4.60	Weak base
Ammonia	4.75	9.25	Moderate

⚡ Key factors affecting amine basicity:

1. Inductive effect: Electron-donating groups (+I) increase basicity (alkyl groups donate → amine is more basic). Order: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N (in water).
2. Steric hindrance: Bulky groups hinder solvation of the conjugate acid (decrease basicity). Explains why (CH₃)₃N is weaker than (CH₃)₂NH even though +I of three methyls should donate more.
3. Resonance: Aniline’s lone pair is delocalized into benzene ring → less available → much weaker base (pKb ~9.4).
4. Solvent effects: In water, H-bonding stabilization of conjugate acid matters greatly.

⚡ Substituted Anilines:

• p-Chloroaniline: Cl has –I (withdraws) but also +M (donates via resonance). Net effect: Cl withdraws inductively but donates through resonance. The electron-withdrawing effect makes it slightly less basic than aniline, but the –I effect is partially offset by +M.
• p-Nitroaniline: NO₂ is strongly –M and –I. Very weak base (pKb ~13). Almost non-basic.
• p-Methoxyaniline: OCH₃ has +M (donates) and –I. Overall donating → more basic than aniline.

Order of Basicity in Aqueous Solution:

Aliphatic 2° > Aliphatic 1° > Aliphatic 3° > Aromatic
(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > PhNH₂

⚡ Note: In the gas phase, the order is (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃ (pure inductive ordering). Solvation reverses this for 3° amines.

2. Preparation Methods:

1. Ammonolysis of Alkyl Halides:

R–X + NH₃ (excess) → R–NH₂ + (R)₂NH + (R)₃N (mixture)
Problem: Polyalkylation always occurs
Mechanism: SN2; NH₃ is nucleophile; successive alkylation of the amine product

⚡ Use excess NH₃ to favor primary amine. The primary amine product is also nucleophilic → can react further. This is why a mixture forms.

2. Gabriel Synthesis:

Phthalimide + KOH → K⁺ ⁻N(Phthaloyl)
⁻N(Phthaloyl) + R–X → R–N(Phthaloyl)
→ H₂N–NH₂ (hydrazine) or acid hydrolysis → R–NH₂ (pure 1° amine)

⚡ Gabriel synthesis gives pure primary amines (no polyalkylation). Used for making amino acids. The key is that the intermediate secondary amine is not nucleophilic enough to react again because the nitrogen is part of the imide ring (adjacent carbonyls make it very acidic).

3. From Nitro Compounds:

R–NO₂ + Sn/HCl → R–NH₂ (reduction of nitro group)
Aromatic nitro + Sn/HCl → aromatic amine
Industrial: Fe/HCl is cheaper

⚡ The nitro group must be reduced to –NH₂. For aromatic nitro compounds, this is the main method to make aniline.

4. From Amides (Hofmann Bromamide Reaction):

R–CONH₂ + Br₂ + 4NaOH → R–NH₂ + CO₃²⁻ + 2NaBr + H₂O
酰胺 → 1° amine with loss of one carbon
Mechanism: Bromination → dehydrohalogenation → isocyanate → hydrolysis → amine

⚡ This is THE go-to method for making pure primary amines from carboxylic acids (via amide). The product has one fewer carbon than the starting amide. Important in amino acid chemistry.

5. Reductive Amination (Ammonia + Carbonyl):

R–CHO + NH₃ → R–CH=NH (imine) → NaBH₃CN/H⁺ → R–CH₂–NH₂
R₂C=O + R'NH₂ → imine → reduce → 2° amine

⚡ In JEE problems: Reacting a ketone with NH₃ under reductive conditions gives an amine. The carbonyl carbon becomes the carbon attached to N in the amine.

6. Schmidt Reaction:

R–COOH + N₃H → R–NH₂ + CO₂ + N₂ (carboxylic acid to amine)

⚡ One carbon is lost as CO₂. The nitrogen of the amine comes from hydrazoic acid.

7. Curtius Reaction:

R–COCl + NaN₃ → R–CON₃ (acyl azide)
→ Δ → R–N=C=O (isocyanate) → R–NH₂ (hydrolysis)
Also loses one carbon

3. Important Reactions:

1. Acylation (1° and 2° Amines):

R–NH₂ + (CH₃CO)₂O → R–NHCOCH₃ + CH₃COOH
R₂NH + (CH₃CO)₂O → R₂NCOCH₃ + CH₃COOH
3° Amines: no reaction (no N–H to acylate)

⚡ Acylation distinguishes 1°/2° from 3° amines. 3° amines have no N–H bond to react with acetyl chloride or acetic anhydride. ⚡ Acylated amines are no longer basic (lone pair is delocalized into the carbonyl). This is used to “protect” the amine in synthesis.

2. Carbylamine Reaction (Isocyanide Test):

R–NH₂ + CHCl₃ + 3KOH (alc) → R–NC (isocyanide) + 3KCl + 3H₂O
1° amine → foul-smelling isocyanide (red-orange color with picryl chloride)
2° amine: no reaction
3° amine: no reaction

⚡ This is THE distinguishing test for 1° amines. The product is an isocyanide (carbylamine) with an extremely foul odor — considered a confirmatory test for primary amines. Only aliphatic 1° amines give this (aromatic 1° amines like aniline also give it).

3. Hofmann Elimination:

R–CH₂–CH₂–NH₂ + CH₃I (excess) → R–CH₂–CH₂–N(CH₃)₃ I⁻ (quaternary ammonium)
→ Ag₂O/heat → R–CH₂–CH₂–N⁺(CH₃)₃ OH⁻ → Δ → R–CH=CH₂ + N(CH₃)₃
Hofmann elimination: β-hydrogen elimination from quaternary ammonium
Less substituted alkene (Hofmann product) is formed, NOT Zaitsev

⚡ This is a SYN-ELIMINATION. The β-hydrogen and the leaving group (N⁺R₃) must be anti-periplanar for E2 elimination. This is JEE’s way of making alkenes with Hofmann regioselectivity (less substituted alkene predominates).

4. Diazotization:

Ar–NH₂ + NaNO₂ + HCl (0-5°C) → Ar–N₂⁺Cl⁻ (diazonium salt)

⚡ Temperature is critical: above 5°C, diazonium salt decomposes to phenol. This reaction is the basis of ALL transformations from aniline to other aromatic compounds.

Reactions of Diazonium Salts:

a) Sandmeyer Reaction:
   Ar–N₂⁺ + CuCl → Ar–Cl (aryl chloride)
   Ar–N₂⁺ + CuBr → Ar–Br (aryl bromide)
   Ar–N₂⁺ + CuCN → Ar–CN (aryl cyanide)
   Ar–N₂⁺ + KI → Ar–I (aryl iodide)
   Catalyst: Cu (cuprous) salts

b) Gattermann Reaction:
   Ar–N₂⁺ + HCl/Cu → Ar–Cl (similar to Sandmeyer but uses H⁺ not Cu²⁺)
   Ar–N₂⁺ + H₂O (heat) → Ar–OH (phenol)
   Ar–N₂⁺ + H₃PO₂ → Ar–H (reductive deamination)
   Ar–N₂⁺ + NH₄₂S → Ar–SH (aryl thiol)

5. Electrophilic Aromatic Substitution of Aniline:

• Aniline is strongly activating (–NH₂ is o/p director)
• Nitration: direct nitration of aniline gives oxidation products (nitroaniline is hard to make directly because aniline is too reactive)
• To nitrate aniline: protect –NH₂ as acetanilide (acetylation) → nitration → hydrolyze → p-nitroaniline
• Halogenation: Br₂/H₂O gives 2,4,6-tribromoaniline (white precipitate) — too activated
• Friedel-Crafts: aniline + AlCl₃ gives complex (N coordinates to Al) → doesn’t work cleanly

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Amines — Comprehensive Chemistry Notes

1. Detailed Mechanisms:

Carbylamine Formation (Isocyanide Synthesis):

Step 1: CHCl₃ + OH⁻ ⇌ :CCl₃⁻ (dichlorocarbene)
Step 2: :CCl₃⁻ → :CCl₂ + Cl⁻
Step 3: R–NH₂ attacks :CCl₂ → R–N=CHCl₂ (intermediate)
Step 4: Repeated dehydrohalogenation → R–N≡C

⚡ The intermediate is an imine dichloride. Three moles of base are required per mole of amine. The reaction is irreversible.

Hofmann Rearrangement (Bromamide Reaction):

Step 1: R–CONH₂ + Br₂ → R–CONHBr + HBr (bromination at N)
Step 2: Base removes H from N → R–CONHBr⁻ (anion)
Step 3: Rearrangement: R–N=C=O + Br⁻ (isocyanate forms, this is the key step)
        The R group migrates from C to N while Br leaves — this is a rearrangement
Step 4: R–N=C=O + H₂O → R–NH₂ + CO₂ (hydrolysis of isocyanate)

⚡ This is a true rearrangement: the R group migrates WITH its bonding electrons. The migrating group must be able to stabilize a positive charge in the transition state. This is important for understanding why some groups migrate better than others.

⚡ Stereochemistry note: In the Hofmann rearrangement, if the migrating carbon is chiral, retention of configuration is observed (suprafacial shift).

Diazo Coupling Reaction:

Ar–N₂⁺ + Ar'–H (activated ring) → Ar–N=N–Ar' (azo compound)
Aniline diazonium + phenol (pH > 7) → para-hydroxyazobenzene (orange dye)
Aniline diazonium + aniline (pH ~5-7) → aminoazobenzene (para coupling)

⚡ This is a classic electrophilic aromatic substitution where the diazonium ion (–N₂⁺) is the electrophile. It attacks electron-rich aromatic rings (phenol, aniline derivatives). The azo compounds are intensely colored — used as dyes.

Mechanism of Azo Dye Formation:

Step 1: Ar–N₂⁺ + H₂O ⇌ Ar–N=N–OH (diazotic acid) — pH dependent
Step 2: Ar–N=N–OH ⇌ Ar–N=N–O⁻ (diaotate) — alkaline conditions
Step 3: The diazonium/diazo species attacks the activated aromatic ring
Step 4: Loss of H⁺ → azo compound

2. Stereochemistry of Amines:

Chiral Amines:

• Tertiary amines with three different substituents and a lone pair are technically chiral (pyramidal nitrogen)
• However, nitrogen inverts rapidly (umbrella inversion) at room temperature (barrier ~25 kJ/mol)
• Unlike carbon (tetrahedral, stable configuration), nitrogen pyramidal flips rapidly: R-NH₂ enantiomers interconvert
• Quaternary ammonium salts CAN be chiral if all four substituents are different (no lone pair inversion)
• Example: [CH₃EtPrN]⁺I⁻ is chiral — used in resolving chiral acids

Nucleophilic Attack on Chiral Center:

• If amine attacks a chiral carbonyl compound, diastereomeric products form
• The approach is less hindered from one face → diastereomeric excess

3. Comparative Study — Amine Basicity in Different Media:

Aqueous Solution (pKa of conjugate acid):

(n-Bu)₂NH > (n-Bu)NH₂ > (n-Bu)₃N > aniline > ammonia

Reason for 3° being weaker than 1° in water: Solvation of the conjugate acid is hindered by three alkyl groups (steric).

Gas Phase (Absolute Basicity — proton affinity):

Me₃N > Me₂NH > MeNH₂ > NH₃

Pure inductive effect: more alkyl groups = more electron donation = stronger base.

Solvent Effects on Basicity:

In water: Aliphatic 2° > 1° > 3° > aromatic
In protic solvents: solvation stabilizes ions; smaller conjugate acid = better H-bonding
In aprotic solvents: inductive dominates

Heteroaromatic Amines:

• Pyridine (C₅H₅N): N is sp², lone pair in sp² orbital (not in ring π-system) → weakly basic (pKb ~8.8)
• Pyrrole (C₄H₄NH): N is sp², lone pair in p orbital (part of aromatic sextet) → NOT basic (pKb ~13, almost non-basic)
• Imidazole: has two N atoms, one basic (pKa ~7) — part of histidine and biological systems

⚡ JEE常常考 Basicity Order Problems: Arrange: aniline, p-nitroaniline, p-methylaniline, p-chloroaniline Answer: p-methylaniline > aniline > p-chloroaniline > p-nitroaniline

• p-NO₂ (strongly –M, –I): dramatically decreases basicity → weakest base
• p-CH₃ (+I, hyperconjugation): increases basicity → strongest base
• p-Cl (–I dominates over +M): decreases basicity slightly
• Aniline: baseline aromatic amine basicity

4. Synthetic Utility — Advanced Synthesis:

Distinguishing 1°, 2°, 3° Amines — Hinsberg’s Test:

1° Amine + Benzenesulfonyl chloride → R–NHSO₂Ph (soluble in alkali, acidified → precipitate)
2° Amine + Benzenesulfonyl chloride → R₂N–SO₂Ph (insoluble in everything)
3° Amine + Benzenesulfonyl chloride → No reaction (no N–H to attack)

⚡ This is JEE’s method for distinguishing primary, secondary, and tertiary amines. The benzenesulfonyl chloride attacks the N–H in 1° and 2° amines.

Tertiary Amine Reactions:

R₃N + R'–X → R₃N⁺–R' X⁻ (quaternary ammonium salt)
R₃N + R–COCl → R₃N–CO–R' (amide, unstable, reversible)
R₃N + metal ions → coordinate complexes

Mannich Reaction:

R–CH₂–CO–R' + HCHO + R''₂NH → R–CH₂–CO–CH(R'')–N(R'')₂
Ketone + formaldehyde + 2° amine → β-amino ketone
This is an iminium ion intermediate: R₂N⁺=CH₂ is the electrophile

5. Important Named Reactions Summary:

Reaction	Starting Material	Product	Key Feature
Gabriel	Alkyl halide	Pure 1° amine	No over-alkylation
Hofmann Bromamide	Amide	1° amine (–1C)	Rearrangement
Curtius	Acyl azide	1° amine (–1C)	Similar to Hofmann
Schmidt	Carboxylic acid	1° amine (–1C)	Uses HN₃
Sandmeyer	Diazonium	Ar–Cl, Br, CN	Cuprous catalyst
Carbylamine	1° amine	Isocyanide (R–NC)	Foul odor test
Hofmann Elimination	Quaternary ammonium	Alkene	Less substituted alkene

6. Biological Significance:

• Methylamine (CH₃NH₂): Produced in decaying fish; contributes to fish odor
• Ethanolamine: Found in phospholipids (cell membranes); precursor to choline
• Choline (HOCH₂CH₂N⁺(CH₃)₃): Essential nutrient; neurotransmitter precursor
• Epinephrine (adrenaline): Contains secondary amine; hormone/neurotransmitter
• Histamine: Imidazole ring + ethylamine; causes allergic responses, gastric acid secretion
• Serotonin: Tryptophan derivative; neurotransmitter
• Dopamine, norepinephrine: Catecholamine neurotransmitters (aromatic amine + catechol)
• Aniline: Industrial solvent and chemical intermediate; toxic (methemoglobinemia)
• Putrescine (NH₂(CH₂)₄NH₂): Polyamine from decay; diamine
• Cadaverine (NH₂(CH₂)₅NH₂): Pentane-1,5-diamine; odor of decaying flesh
• Amino acids: All have –NH₂ and –COOH; amphoteric behavior

⚡ Drug chemistry note: Many CNS drugs are amines because the basic nitrogen can be protonated to form water-soluble salts (e.g., morphine, codeine, amphetamine). The amine group is crucial for receptor binding via hydrogen bonding.

7. Multi-Step Synthesis Problems:

Synthesize p-nitroaniline from benzene:

Step 1: Benzene + HNO₃/H₂SO₄ → Nitrobenzene (nitration)
Step 2: Nitrobenzene + Sn/HCl → Aniline (reduction)
Step 3: Aniline + (CH₃CO)₂O → Acetanilide (protection of –NH₂)
Step 4: Acetanilide + HNO₃/H₂SO₄ → p-Nitroacetanilide (nitration at para, ortho is minor)
Step 5: p-Nitroacetanilide + H₃O⁺/Δ → p-Nitroaniline (deprotection of amide)

⚡ Direct nitration of aniline gives oxidation products. That’s why we protect the amine as acetanilide first.

Synthesize methylamine from acetic acid:

Acetic acid → Acetamide (NH₃, heat)
Acetamide → Methylamine + CO₂ via Hofmann rearrangement (Br₂/NaOH)
CH₃–CONH₂ → Br₂/4NaOH → CH₃–NH₂

⚡ This is a classic sequence: carboxylic acid → amide → primary amine (one carbon lost). This pathway is how we make pure 1° amines.

Why can’t we just reduce acetamide with LiAlH₄? Reduction of amide with LiAlH₄ gives amine with ALL carbons intact: CH₃–CONH₂ + LiAlH₄ → CH₃–CH₂–NH₂ (ethylamine, two carbons). To get one-carbon shorter amine, use Hofmann rearrangement.

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