12 min readPathway sheet1-mark MCQ or assertion-reason on test results; 2 or 3-mark carbylamine reaction, acylation or Hinsberg test; 'distinguish primary, secondary and tertiary amines' is a board favourite; also a NEET favourite

Reactions of amines: alkylation, acylation, carbylamine, nitrous acid and the Hinsberg test

The N lone pair attacks alkyl halides and acyl chlorides. The carbylamine test picks out primary amines, and the Hinsberg test separates 1°, 2° and 3° amines.

Carbylamine reactionRNH₂ + CHCl₃ + 3KOH (ethanolic) → (heat) RNC + 3KCl + 3H₂O; foul smell

Hinsberg's reagentBenzenesulphonyl chloride, C₆H₅SO₂Cl

Nitrous acid1° aliphatic amine → alcohol + N₂; aniline at 273 to 278 K → benzenediazonium chloride

Acylation1° and 2° amines + RCOCl, (RCO)₂O or esters → amides; pyridine removes the HCl

Alkylation and acylation

1
Alkylation: RNH₂ + R'-X → RNH-R' + HX → (R'X) RNR'₂ → (R'X) RNR'₃⁺X⁻
Nucleophilic substitution by the N lone pair gives secondary and tertiary amines and finally a quaternary ammonium salt. The HX formed is taken up by the amine as a salt
2★
Acylation: C₂H₅NH₂ + CH₃COCl → (pyridine) C₂H₅NHCOCH₃ + HCl
N-Ethylethanamide. Acid chlorides, anhydrides and esters replace an H on N by an acyl group. Pyridine, a stronger base than the amine, removes HCl and pushes the equilibrium forward
3
(C₂H₅)₂NH + CH₃COCl → (pyridine) (C₂H₅)₂NCOCH₃ + HCl
N,N-Diethylethanamide, from a secondary amine
4
C₆H₅NH₂ + (CH₃CO)₂O → (pyridine) C₆H₅NHCOCH₃ + CH₃COOH
Acetanilide, N-phenylethanamide
5
Benzoylation: CH₃NH₂ + C₆H₅COCl → C₆H₅CONHCH₃ + HCl
N-Methylbenzamide. Tertiary amines have no H on N, so they are not acylated

★ key step

Carbylamine reaction (isocyanide test)

1★
R-NH₂ + CHCl₃ + 3KOH (ethanolic) → (heat) R-NC + 3KCl + 3H₂O
Primary amines, aliphatic and aromatic, give isocyanides (carbylamines) with a very foul smell
2
C₆H₅NH₂ + CHCl₃ + 3KOH → (heat) C₆H₅NC + 3KCl + 3H₂O
Phenyl isocyanide
3
Secondary and tertiary amines: no isocyanide
So this is a test for primary amines. Do it in a fume cupboard, because isocyanides are poisonous

★ key step

Reaction with nitrous acid

1
NaNO₂ + HCl → HNO₂ + NaCl
Nitrous acid is unstable, so it is made in the reaction mixture
2★
R-NH₂ + NaNO₂ + 2HCl → [R-N₂⁺Cl⁻] + NaCl + 2H₂O; then [R-N₂⁺Cl⁻] + H₂O → R-OH + N₂ + HCl
Primary aliphatic amines give unstable alkyldiazonium salts that at once give an alcohol and N₂ gas. The N₂ comes off quantitatively, which is used to estimate amino acids and proteins
3
C₆H₅NH₂ + NaNO₂ + 2HCl → (273 to 278 K) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Aromatic primary amines give arenediazonium salts, stable for a short time in the cold. This is diazotisation
4
Secondary and tertiary amines react with nitrous acid in a different way
No N₂ gas is given off

★ key step

Hinsberg test (benzenesulphonyl chloride)

1★
Primary: C₆H₅SO₂Cl + H-NH-C₂H₅ → C₆H₅SO₂NHC₂H₅ + HCl
N-Ethylbenzenesulphonamide. The H left on N is strongly acidic (the sulphonyl group pulls electrons), so the product dissolves in alkali
2★
Secondary: C₆H₅SO₂Cl + H-N(C₂H₅)₂ → C₆H₅SO₂N(C₂H₅)₂ + HCl
N,N-Diethylbenzenesulphonamide. No H on N, so it is not acidic and does not dissolve in alkali
3
Tertiary: (C₂H₅)₃N + C₆H₅SO₂Cl → no reaction
There is no H on N to replace
4
Use: tells 1°, 2° and 3° amines apart, and separates a mixture of them
Today p-toluenesulphonyl chloride is often used in place of benzenesulphonyl chloride

★ key step

Primary, secondary or tertiary? The tests side by side

Test	Primary (1°)	Secondary (2°)	Tertiary (3°)
Carbylamine: CHCl₃ + alcoholic KOH, heat	Foul smell of isocyanide	No reaction	No reaction
Hinsberg: C₆H₅SO₂Cl, then KOH	Sulphonamide dissolves in alkali	Sulphonamide does not dissolve in alkali	No reaction
Nitrous acid: cold NaNO₂ + HCl	Aliphatic: N₂ bubbles, alcohol forms; aromatic: diazonium salt at 273 to 278 K	No N₂	No N₂
Acylation: CH₃COCl or (CH₃CO)₂O	Amide forms	Amide forms	No reaction

Worked example: identify the amines

Problem

Three amines A, B and C all have the formula C₃H₉N. With benzenesulphonyl chloride, A gives a product that dissolves in KOH, B gives a product that does not dissolve in KOH, and C does not react. A also gives an offensive smell when heated with CHCl₃ and alcoholic KOH. Suggest structures for A, B and C.

Working

A: sulphonamide soluble in alkali, and positive carbylamine test, so A is primary: CH₃CH₂CH₂NH₂ (propan-1-amine) or (CH₃)₂CHNH₂ (propan-2-amine).
B: sulphonamide insoluble in alkali, so B is secondary. The only secondary amine C₃H₉N is CH₃NHCH₂CH₃ (N-methylethanamine).
C: no reaction, so C is tertiary. The only tertiary amine C₃H₉N is (CH₃)₃N (N,N-dimethylmethanamine).
Check the formula: each has 3 C, 9 H and 1 N.

Answer

A: propan-1-amine (or propan-2-amine); B: N-methylethanamine; C: N,N-dimethylmethanamine

In real life

Paracetamol, the common fever tablet, is an amide made by acetylating the amino group of p-aminophenol, the same reaction that turns aniline into acetanilide.

Exam angle

2 marks: “Write the carbylamine reaction of aniline. Why is it a test for primary amines only?” C₆H₅NH₂ + CHCl₃ + 3KOH → (heat) C₆H₅NC + 3KCl + 3H₂O; phenyl isocyanide smells foul. Only an NH₂ group, with two H on N, can form the isocyanide; 2° and 3° amines cannot.

3 marks: “What is Hinsberg's reagent? How does it distinguish primary, secondary and tertiary amines? Write the equations.” Benzenesulphonyl chloride. 1°: C₆H₅SO₂NHR, with an acidic N-H, dissolves in alkali. 2°: C₆H₅SO₂NR₂ has no N-H and is insoluble in alkali. 3°: no reaction.

2 marks: “Why is acylation of amines carried out in the presence of pyridine? Why do tertiary amines not undergo acylation?” Pyridine is a stronger base than the amine; it removes the HCl formed and shifts the equilibrium forward. Tertiary amines have no H on N to be replaced by an acyl group.

Remember it

Carbylamine: only 1° stinks. Hinsberg: 1° dissolves in alkali, 2° stays solid, 3° ignores it.