📘 CodingMarble Learn

Amines: Structure, Naming, Preparation, Basicity and Reactions

An amine is ammonia with one, two or three H atoms swapped for carbon groups. The N keeps a lone pair, so amines are bases and nucleophiles. Count the C atoms on N to get 1°, 2° or 3°. Make them by reducing nitro compounds, nitriles or amides, from alkyl halides + NH₃, by Gabriel (1° only) or Hofmann bromamide (one C fewer). Alkyl groups push electrons and raise basicity; the ring in aniline pulls the pair away and lowers it. Key reactions: salts with acids, acylation, carbylamine test (1°), Hinsberg test (1°/2°/3°), nitrous acid and ring substitution in aniline.

🎬 Step-by-step story

  1. Start with ammonia, NH₃. Swap one H for a CH₃ group and you get methylamine, an amine. The purple blob is the lone pair on N.
  2. Count the carbons touching N: one C is primary (1°), two is secondary (2°), three is tertiary (3°).
  3. Naming: find the longest chain that holds N, number from the end nearest NH₂, drop the final 'e' of the alkane and add 'amine': propan-1-amine.
  4. Making an amine: reduce a nitro group. The two O atoms leave as water and H atoms come in: R–NO₂ → R–NH₂.
  5. Why amines are bases: the lone pair grabs an H⁺. Alkyl groups push electrons in, so they grab more strongly; the ring in aniline pulls them away.
  6. Free play: pick an amine and a reagent. Watch which part reacts: the N–H hydrogens or the lone pair.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why is the shape pyramidal and not flat?

N has three bonds and one lone pair. The four electron groups spread out like a tetrahedron; you only 'see' the three atoms, so it looks like a pyramid.

Why is (CH₃)₃CNH₂ primary when its carbon is tertiary?

For amines only the N counts. Only one C touches N, so it is primary. In the 3D, count the C atoms touching the yellow N.

When do I use the prefix N- in names?

When a group sits on the N instead of the main chain, e.g. N-methylethanamine.

Why use Fe/HCl in industry instead of Sn/HCl?

Iron is cheap, and the FeCl₂ formed hydrolyses to give back HCl, so only a small amount of acid is needed.

Why are alkylamines stronger bases than ammonia?

Alkyl groups push electron density onto N, so the lone pair grabs H⁺ more easily and the cation formed is more stable.

Why is aniline a weaker base than ammonia?

Its lone pair is shared with the benzene ring by resonance, so less of it is free to catch H⁺.

Why does the 1° Hinsberg product dissolve in KOH but the 2° one does not?

The 1° product still has an H on N that is acidic (next to the SO₂ group), so KOH removes it and forms a soluble salt. The 2° product has no H on N.

Structure of amines

An amine is a derivative of ammonia (NH₃). Replace one or more H atoms by alkyl or aryl groups and you get an amine, e.g. CH₃NH₂.

The N atom uses three bonds and keeps one lone pair (two electrons not in a bond). So the shape round N is pyramidal, like ammonia. The C–N–C or C–N–H angle is about 108°, a little less than a perfect tetrahedron (109.5°) because the lone pair pushes the bonds closer.

In aniline (C₆H₅NH₂) the N is joined to a benzene ring. Its lone pair partly spreads into the ring. Keep this in mind: it explains why aniline is a weaker base.

Classification: primary, secondary, tertiary

Count the carbon groups joined to the N:

Simple amines have all groups the same; mixed amines have different ones. If N is joined straight to a benzene ring it is an aryl (aromatic) amine (aniline). Benzylamine C₆H₅CH₂NH₂ is still an alkyl amine, because N sits on a CH₂, not on the ring.

Careful: for alcohols we count C on the carbon that holds OH. For amines we count C on the N itself. So (CH₃)₃C–NH₂ is a primary amine.

Nomenclature (naming amines)

Common names: name the alkyl groups in alphabetical order and add 'amine' in one word: methylamine, ethylmethylamine, trimethylamine.

IUPAC names:

  1. Take the longest chain holding the N. Drop the final 'e' of the alkane and add 'amine': ethane → ethanamine.
  2. Number from the end nearest the NH₂: CH₃CH₂CH₂NH₂ is propan-1-amine; CH₃CH(NH₂)CH₃ is propan-2-amine.
  3. Other groups on N get the prefix N-: CH₃NHC₂H₅ is N-methylethanamine; (CH₃)₂NC₂H₅ is N,N-dimethylethanamine.
  4. Two NH₂ groups: keep the 'e': H₂N–CH₂CH₂–NH₂ is ethane-1,2-diamine.
  5. Aromatic: C₆H₅NH₂ is aniline, also the IUPAC name benzenamine. With a methyl on the ring: 2-methylaniline (o-toluidine).

Preparation of amines

  1. Reduction of nitro compounds: R–NO₂ or Ar–NO₂ with H₂/Pd, or Sn + HCl, or scrap iron + HCl (preferred in industry, because the FeCl₂ formed gets hydrolysed and releases HCl again, so only a little acid is needed). Nitrobenzene → aniline.
  2. Ammonolysis of alkyl halides: R–X + NH₃ → R–NH₂ (as salt). The amine formed is itself a nucleophile, so it reacts again → a mixture of 1°, 2°, 3° amines and the quaternary salt R₄N⁺X⁻. A big excess of NH₃ favours the 1° amine.
  3. Reduction of nitriles: R–C≡N + 4[H] (LiAlH₄ or H₂/Ni) → R–CH₂–NH₂. One C is added to the chain (ascent).
  4. Reduction of amides: R–CONH₂ + LiAlH₄, then water → R–CH₂–NH₂ (same number of C).
  5. Gabriel phthalimide synthesis: phthalimide + KOH → potassium phthalimide; + R–X → N-alkylphthalimide; alkaline hydrolysis → R–NH₂. Gives pure 1° amines only. It cannot make aniline, because aryl halides do not do the substitution step.
  6. Hofmann bromamide degradation: R–CONH₂ + Br₂ + 4NaOH → R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The amine has one C fewer than the amide (descent).

Physical properties

Small aliphatic amines are gases or liquids with a fishy smell; aniline is a liquid that turns brown in air (it gets oxidised).

Hydrogen bonding: 1° and 2° amines have N–H, so their molecules hydrogen-bond to each other. 3° amines have no N–H and cannot. So for similar mass, boiling point: 1° > 2° > 3°.

N is less electronegative than O, so N–H···N bonds are weaker than O–H···O. Amines boil lower than alcohols of similar mass but higher than alkanes.

Small amines dissolve in water (they H-bond with water). Solubility falls as the carbon part grows. Order of boiling points for similar mass: alkane < amine < alcohol < carboxylic acid.

Basicity of amines

A base accepts H⁺. The lone pair on N does this: R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻. The strength is measured by Kb or pKb (pKb = −log Kb). Smaller pKb = stronger base.

Chemical reactions of amines

  1. With acids (salt formation): C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻. Salts dissolve in water; NaOH gives the amine back. Used to separate amines from neutral compounds.
  2. Alkylation: with R–X, 1° → 2° → 3° → quaternary ammonium salt.
  3. Acylation: 1° and 2° amines (with N–H) react with acid chlorides/anhydrides to give amides; pyridine removes the HCl. C₆H₅NH₂ + CH₃COCl → C₆H₅NHCOCH₃ (acetanilide). With benzoyl chloride it is benzoylation. 3° amines do not react (no N–H).
  4. Carbylamine reaction (test for 1°): R–NH₂ + CHCl₃ + 3KOH (heat) → R–NC (isocyanide, very bad smell) + 3KCl + 3H₂O. 2° and 3° amines do not give it.
  5. Nitrous acid (NaNO₂ + HCl): 1° aliphatic amines give N₂ gas (quantitatively) and alcohols; aniline at 273–278 K gives the benzenediazonium salt (next lesson).
  6. Hinsberg test (benzenesulphonyl chloride, C₆H₅SO₂Cl): 1° gives a sulphonamide that still has an acidic N–H and dissolves in KOH; 2° gives a sulphonamide with no N–H, insoluble in alkali; 3° does not react.
  7. Ring substitution in aniline: –NH₂ strongly activates the ring at ortho and para positions. Bromine water gives a white precipitate of 2,4,6-tribromoaniline at once. To get one Br only, first acetylate (acetanilide), then brominate, then hydrolyse.
  8. Nitration: direct nitration gives lots of meta product (about 47%), because in strong acid aniline becomes anilinium (–NH₃⁺, a meta director), plus tarry oxidation products. Protect as acetanilide to get mainly p-nitroaniline.
  9. Sulphonation: aniline + conc. H₂SO₄ → anilinium hydrogensulphate → on heating at 453–473 K → sulphanilic acid (a zwitterion).
  10. Aniline does not do Friedel–Crafts reactions: AlCl₃ (a Lewis acid) grabs the lone pair and N becomes positive, which deactivates the ring.

Try it: the lemon-and-fish test

At home: smell a piece of raw fish (or the fish counter). Then squeeze lemon on it and smell again. The smell drops because the amines (bases) turn into salts with citric acid, and salts do not evaporate. That is 'amine + acid → salt' in your kitchen.

In the 3D: in free play, pick each amine with Hinsberg's reagent. Predict first: will it dissolve in KOH? Then press Show and check.

Board exam focus

The unit carries about 6 marks. Common questions: arrange amines by basicity or boiling point; give reasons (aniline less basic, Gabriel cannot make aniline, meta product in nitration, no Friedel–Crafts); distinguish 1°/2°/3° (Hinsberg, carbylamine); write name reactions (Hofmann bromamide, Gabriel, carbylamine); and conversion chains (A → B → C).

Key formulas and definitions

Worked examples

1. Classify (CH₃)₃C–NH₂ and C₆H₅NHCH₃ as 1°, 2° or 3°.

Step 1: Look at the N, not the carbon. Step 2: In (CH₃)₃C–NH₂ the N has one C attached → primary (1°) amine, even though that C is a tertiary carbon. Step 3: In C₆H₅NHCH₃ the N has two C groups (phenyl and methyl) and one H → secondary (2°) aromatic amine.

2. Give the IUPAC name of CH₃–CH(NH₂)–CH₂–CH₃ and of (C₂H₅)₂N–CH₃.

First: longest chain = 4 C (butane). Number so NH₂ gets the lowest number: C2. Drop 'e', add 'amine' → butan-2-amine. Second: the longest group on N is ethyl → parent ethanamine. The other ethyl and the methyl sit on N → N-ethyl-N-methylethanamine.

3. An amide C₂H₅CONH₂ is treated with Br₂ and NaOH. Name the amine and state how many C atoms it has.

Step 1: This is Hofmann bromamide degradation; the C=O carbon is lost as carbonate. Step 2: The amide has 3 C, so the amine has 3 − 1 = 2 C. Step 3: C₂H₅CONH₂ → C₂H₅NH₂ (ethanamine).

4. Convert benzene into aniline. Write both steps.

Step 1 (nitration): C₆H₆ + conc. HNO₃ / conc. H₂SO₄ (≈ 330 K) → C₆H₅NO₂ (nitrobenzene). Step 2 (reduction): C₆H₅NO₂ + Fe/HCl (or Sn/HCl) → C₆H₅NH₂ (aniline); add NaOH to free the amine from its salt.

5. Arrange in increasing order of basicity in water: C₆H₅NH₂, NH₃, CH₃NH₂, (CH₃)₂NH.

Step 1: Aniline is weakest: its lone pair is spread into the ring (pKb ≈ 9.4). Step 2: NH₃ (pKb ≈ 4.75) is next. Step 3: Methyl groups push electrons, so CH₃NH₂ (pKb ≈ 3.4) is stronger. Step 4: (CH₃)₂NH (pKb ≈ 3.3) is strongest in water (good +I and still good hydration). Order: C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH.

6. The pKb of an amine is 3.30. Find Kb and compare it with NH₃ (Kb = 1.8 × 10⁻⁵).

Step 1: Kb = 10^(−pKb) = 10^(−3.30). Step 2: 10^(−3.30) = 10^(0.70) × 10⁻⁴ ≈ 5.0 × 10⁻⁴. Step 3: 5.0 × 10⁻⁴ ÷ 1.8 × 10⁻⁵ ≈ 28. The amine is about 28 times stronger a base than ammonia.

7. Three bottles hold ethanamine, N-methylethanamine and N,N-dimethylethanamine. How will you tell them apart with Hinsberg's reagent?

Step 1: Shake each with C₆H₅SO₂Cl and KOH. Step 2: Ethanamine (1°) gives C₆H₅SO₂NHC₂H₅; its N–H is acidic, so it dissolves in KOH (clear solution). Step 3: N-methylethanamine (2°) gives a sulphonamide with no N–H → a solid that does not dissolve in KOH. Step 4: The 3° amine does not react at all (it dissolves only when acid is added).

8. Why does direct nitration of aniline give a large amount of m-nitroaniline, and how can we get mostly p-nitroaniline?

Step 1: The nitrating mixture is strongly acidic. Step 2: Aniline takes H⁺ and becomes anilinium, C₆H₅NH₃⁺. Step 3: –NH₃⁺ pulls electrons and directs to the meta position → about 47% meta product. Step 4: Fix: acetylate first (acetanilide, –NHCOCH₃ is a milder o/p director and is not protonated), nitrate (mainly para), then hydrolyse back → p-nitroaniline.

Common mistakes

Practice quiz

1. Which is a secondary amine?
2. The carbylamine test is given by:
3. Which is the weakest base?
4. Hofmann bromamide degradation of CH₃CONH₂ gives:
5. Aniline + bromine water gives:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What are amines in simple words?

Amines are ammonia molecules where one or more H atoms are replaced by carbon groups. They keep a lone pair on N, so they behave as bases.

Why is aniline less basic than ammonia?

In aniline the lone pair on N is spread into the benzene ring by resonance, so it is less available to take an H⁺. Ammonia has no ring, so its pair is fully free.

How do you distinguish primary, secondary and tertiary amines?

Use the Hinsberg test: the 1° amine product dissolves in KOH, the 2° product does not, and 3° amines do not react. The carbylamine (bad smell) test also picks out 1° amines.

Where this is taught

PolandLiceum ogólnokształcące, klasa IIINitrogen-containing organic compounds
PolandLiceum ogólnokształcące, klasa IVNitrogen-containing organic compounds
Ukraine10 класNitrogen-containing organic compounds
Ukraine10 класNitrogen-containing organic compounds
CBSE (India)Class 12Amines
England (GCSE, A level)Year 133.3 Organic chemistry
Russia10 классNitrogen-containing compounds

Learn first

Learn next

Related lessons

All Chemistry lessons