Analysis of Amines

ChemistryAminesFor JEE aspirants

The analysis of amines answers two questions in the laboratory: is the nitrogen there at all, and is the amine primary, secondary or tertiary? Nitrous acid sorts the classes by what you see, the Hinsberg test sorts them by solubility, and the carbylamine, Liebermann and mustard oil tests confirm a single class. This page covers every test and separation used in the analysis of amines, with the reactions and the mechanism behind each one.

Key Formulas - Quick Reference

Hinsberg: 1° gives $C_{6} H_{5} S O_{2} NHR$ (soluble in KOH), 2° gives $C_{6} H_{5} S O_{2} N R_{2}$ (insoluble), 3° gives no reaction
Carbylamine (1° only): $RN H_{2} + CHC l_{3} + 3KOH alc., heat RNC + 3KCl + 3 H_{2} O$
Nitrous acid: 1° aliphatic gives $N_{2}$ ; 1° aromatic gives $Ar N_{2}^{+} C l^{-}$ ; 2° gives $R_{2} N - N = O$ ; 3° gives no visible change
Azo dye test (1° aromatic only): $Ar N_{2}^{+} C l^{-}$ with alkaline $β$ -naphthol gives an orange-red dye
Liebermann (2° only): nitrosamine + phenol + conc. $H_{2} S O_{4}$ gives a blue-green colour, red on dilution
Mustard oil test (1° only): $RN H_{2} + C S_{2} RNHCSSH$ , then $RNHCSSH + HgC l_{2} RNCS + HgS + 2HCl$
Hofmann separation: 1° gives a solid oxamide, 2° a liquid oxamic ester, 3° no reaction
Sulphonamide hydrolysis: boiling conc. HCl frees the amine again
All amines dissolve in dilute HCl as their salts; this shows basic nitrogen
Lassaigne's test: Na fusion gives $NaCN$ , which gives Prussian blue with $FeS O_{4}$ and $F e^{3 +}$

1. First: Is There Nitrogen?

Before any class test, nitrogen itself is detected by Lassaigne's test. The compound is fused with sodium, which turns the nitrogen into sodium cyanide; the extract then gives Prussian blue with iron(II) sulphate and an iron(III) salt.

Na + C + N fuse NaCN

An amine also dissolves in dilute HCl, because it is a base and forms a soluble salt. A compound that dissolves in acid but not in water, and gives no reaction with $NaHC O_{3}$ , is very likely an amine.

RN H_{2} + HCl RN H_{3}^{+} C l^{-}

Figure 1: Nitrogen first becomes

C N^{-}

, and the cyanide is then caught as Prussian blue,

F e_{4} [Fe (CN)_{6}]_{3}

2. Sorting the Classes with Nitrous Acid

Nitrous acid, made in the flask from $NaN O_{2}$ and HCl at 273-278 K, is the first class test, because each class gives a different observation (Figure 2). The chemistry behind it is set out in Properties of Amines.

Class	What you see	Product
1° aliphatic	brisk effervescence of a colourless gas	alcohol + $N_{2}$
1° aromatic	clear solution; couples with $β$ -naphthol to a dye	$Ar N_{2}^{+} C l^{-}$
2° (alkyl or aryl)	a yellow oily layer separates	$R_{2} N - N = O$
3° aliphatic	no gas, no oil; a clear solution	an ammonium nitrite salt
3° aromatic	a green solid	p-nitroso compound

Figure 2: Nitrous acid is the first test to run; the carbylamine, Hinsberg and Liebermann tests then confirm which class an amine belongs to.

The azo dye test tells an aromatic primary amine from an aliphatic one. Its cold diazonium solution poured into alkaline $β$ -naphthol gives a brilliant orange-red dye; an aliphatic amine has already lost its nitrogen as gas and gives nothing.

C_{6} H_{5} N_{2}^{+} C l^{-} + C_{10} H_{7} OH O H^{-}, 273-278 K 1-phenylazo-2-naphthol, orange-red C_{10} H_{6} (OH) N = N C_{6} H_{5} + HCl

3. The Hinsberg Test

Hinsberg's reagent is benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} Cl$ , used with aqueous KOH (or NaOH). It replaces a hydrogen on nitrogen, so the result depends on how many N-H bonds the amine has (Figure 3).

3.1 Primary Amine

One N-H is left on nitrogen. Because two S=O groups pull electron density away, that hydrogen is acidic, so the sulphonamide dissolves in KOH as its potassium salt. Acidifying the clear solution brings the sulphonamide back as a precipitate.

RN H_{2} + C_{6} H_{5} S O_{2} Cl C_{6} H_{5} S O_{2} NHR + HCl

C_{6} H_{5} S O_{2} NHR + KOH soluble potassium salt C_{6} H_{5} S O_{2} N (R) K + H_{2} O

C_{6} H_{5} S O_{2} N (R) K + HCl white precipitate again C_{6} H_{5} S O_{2} NHR + KCl

3.2 Secondary Amine

Both hydrogens on nitrogen are now replaced by alkyl groups, so no acidic N-H remains. The sulphonamide is an insoluble solid that does not dissolve in KOH.

R_{2} NH + C_{6} H_{5} S O_{2} Cl C_{6} H_{5} S O_{2} N R_{2} + HCl

3.3 Tertiary Amine

There is no N-H at all, so no sulphonamide forms. The amine stays as an insoluble layer or solid and dissolves only when the mixture is acidified, because then it forms its ammonium salt.

Figure 3: Hinsberg's reagent counts the N-H bonds: one left over (soluble salt) means

1^{\circ}

, none left (insoluble solid) means

2^{\circ}

, and no reaction means

3^{\circ}

Amine	With $C_{6} H_{5} S O_{2} Cl$ + KOH	On acidifying
1°	dissolves: clear solution	white precipitate of $C_{6} H_{5} S O_{2} NHR$
2°	insoluble solid separates	no change
3°	no reaction; amine insoluble	dissolves as the ammonium salt

Exam Trick

Hinsberg counts N-H bonds. One N-H left in the product means the amine was 1° (soluble in alkali); no N-H left means it was 2° (insoluble); no product at all means 3°. Hinsberg's reagent is benzenesulphonyl chloride, not benzoyl chloride: that is the commonest wrong option.

JEE Advanced

The test has limits. Sulphonamides of 1° amines with long chains dissolve slowly, and some N,N-dialkyl sulphonamides dissolve in hot alkali, which can make a 2° amine look like a 1° one. Tertiary amines can also react slowly with the reagent to give unstable quaternary salts. Modern laboratories often use p-toluenesulphonyl chloride (tosyl chloride) instead, which behaves the same way but gives better crystalline derivatives.

4. The Carbylamine (Isocyanide) Test

Warming an amine with chloroform and alcoholic KOH gives an isocyanide with an extremely unpleasant smell. Only primary amines, aliphatic or aromatic, respond, so this is the quickest test for a 1° amine.

RN H_{2} + CHC l_{3} + 3KOH alc., heat RNC + 3KCl + 3 H_{2} O

C_{6} H_{5} N H_{2} + CHC l_{3} + 3KOH alc., heat C_{6} H_{5} NC + 3KCl + 3 H_{2} O

The reactive intermediate is dichlorocarbene, formed from chloroform by $α$ -elimination. The amine nitrogen attacks it, and two molecules of HCl are then lost, which is why two N-H bonds are needed (Figure 4).

CHC l_{3} + KOH dichlorocarbene CC l_{2} + KCl + H_{2} O

Figure 4: The carbylamine test works because two N-H bonds are used up; that is why only

1^{\circ}

amines, aliphatic or aromatic, give the isocyanide.

The test is destructive and the product is poisonous and foul smelling, so it is carried out on a very small scale in a fume cupboard. The isocyanide can be destroyed by warming with dilute acid, which hydrolyses it back to the amine and formic acid.

5. Colour Tests

5.1 Liebermann Nitroso Reaction: Secondary Amines

A secondary amine first gives its yellow oily N-nitrosamine with nitrous acid. Warming the nitrosamine with phenol and conc. $H_{2} S O_{4}$ gives a brown to red colour that turns blue and then green; diluting with water turns it red again, and adding alkali gives a greenish blue or violet colour. Primary and tertiary amines do not give this sequence.

R_{2} NH + HN O_{2} a yellow oil R_{2} N - N = O + H_{2} O

Figure 5: The nitrosamine is the key: its colour sequence (red, blue, green, red, violet) confirms a

2^{\circ}

amine.

5.2 Mustard Oil Test: Primary Amines

A primary amine and carbon disulphide give a dithiocarbamic acid, which mercuric chloride converts to an isothiocyanate with the pungent smell of mustard oil. Secondary and tertiary amines give no such smell.

RN H_{2} + C S_{2} dithiocarbamic acid RNHCSSH

RNHCSSH + HgC l_{2} mustard oil smell RNCS + HgS + 2HCl

Figure 6: Like the carbylamine test, the mustard oil test needs two N-H bonds, so it picks out

1^{\circ}

amines.

5.3 Azo Dye Test: Aromatic Primary Amines

Covered in Section 2 and in Diazonium Salt: the cold diazonium solution coupled with alkaline $β$ -naphthol gives an orange-red dye, which no other class gives.

6. All the Tests at a Glance

Test	Reagent	1°	2°	3°
Nitrous acid	$NaN O_{2}$ /HCl, 273-278 K	$N_{2}$ gas (aliphatic) or diazonium salt (aromatic)	yellow oily nitrosamine	no visible change (green solid for 3° aromatic)
Carbylamine	$CHC l_{3}$ + alc. KOH	foul-smelling isocyanide	no reaction	no reaction
Hinsberg	$C_{6} H_{5} S O_{2} Cl$ + KOH	dissolves; precipitate on acidifying	insoluble sulphonamide	no reaction
Liebermann	$HN O_{2}$ , then phenol and conc. $H_{2} S O_{4}$	no colour	blue-green, red on dilution	no colour
Mustard oil	$C S_{2}$ , then $HgC l_{2}$	pungent mustard smell	no smell	no smell
Azo dye	$NaN O_{2}$ /HCl, then $β$ -naphthol/ $O H^{-}$	orange-red dye (aromatic only)	no dye	no dye

Exam Trick

Three tests, three classes. Carbylamine catches only 1°, Liebermann catches only 2°, and a Hinsberg test with no reaction at all points to 3°. If a question asks you to tell all three apart in one experiment, use Hinsberg; if it asks about a single amine, use the test that matches its class.

Figure 7: Carbylamine splits off the

1^{\circ}

amines, the azo dye test separates aromatic from aliphatic, and Hinsberg sorts the rest.

7. Separating a Mixture of Amines

Fractional distillation works when the boiling points are far enough apart, but the usual laboratory methods turn each class into a different kind of derivative first (Figure 8).

7.1 Hinsberg Method

The mixture is shaken with benzenesulphonyl chloride and aqueous KOH. The 1° amine dissolves as the potassium salt of its sulphonamide, the 2° amine gives an insoluble sulphonamide that is filtered off, and the 3° amine remains as an unreacted layer. Boiling each sulphonamide with conc. HCl gives the pure amine back.

7.2 Hofmann Method with Diethyl Oxalate

The mixture is warmed with diethyl oxalate. A primary amine gives a solid dialkyl oxamide, a secondary amine gives a liquid oxamic ester, and a tertiary amine does not react. The solid is filtered off and the liquid distilled; hydrolysis with KOH then returns each amine.

(COO C_{2} H_{5})_{2} + 2RN H_{2} solid dialkyl oxamide (CONHR)_{2} + 2 C_{2} H_{5} OH

(COO C_{2} H_{5})_{2} + R_{2} NH liquid oxamic ester C_{2} H_{5} OOCCON R_{2} + C_{2} H_{5} OH

Figure 8: Both separations use the same idea: turn each class into a product with a different physical property (solid, liquid, unchanged), then hydrolyse the amine back.

Figure 9: Every class test counts the N-H bonds, either by what nitrous acid does or by what a reagent can replace.

8. Solved Examples

Solved Example 1

Hinsberg's reagent is
(A) phenyl isocyanide
(B) benzenesulphonyl chloride
(C) p-toluenesulphonic acid
(D) o-dichlorobenzene

Solution:

Answer: (B). Benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} Cl$ , used with aqueous KOH. It is easy to confuse with benzoyl chloride $C_{6} H_{5} COCl$ , which acylates amines but does not separate the classes, and with tosyl chloride, the methyl-substituted version that behaves in the same way.

Solved Example 2

A positive carbylamine test is given by (i) N,N-dimethylaniline (ii) 2,4-dimethylaniline (iii) N-methyl-o-methylaniline (iv) p-methylbenzylamine.
(A) (ii) and (iv)
(B) (ii) and (iii)
(C) (i), (ii) and (iv)
(D) (ii), (iii) and (iv)

Solution:

Answer: (A). Only primary amines respond, and the class is decided by the nitrogen, not by the ring.

(i) N,N-dimethylaniline: nitrogen carries two methyl groups and the ring, so it is 3°: no reaction.
(ii) 2,4-dimethylaniline: the two methyls are on the ring, so the amine is 1° aromatic: positive.
(iii) N-methyl-o-methylaniline: one methyl is on nitrogen, so it is 2°: no reaction.
(iv) p-methylbenzylamine: $- C H_{2} N H_{2}$ on the ring, a 1° aralkyl amine: positive.

Solved Example 3

Ethylenediamine on reaction with diethyl oxalate forms
(A) an open-chain half-amide ester
(B) piperazine-2,3-dione (a cyclic diamide)
(C) an open-chain hydroxy amide
(D) an amine salt

Solution:

Answer: (B). Both $N H_{2}$ groups attack the two ester carbonyls of the same oxalate molecule, and two molecules of ethanol are lost. Closing a six-membered ring is favourable, so the product is the cyclic diamide piperazine-2,3-dione.

(COO C_{2} H_{5})_{2} + H_{2} NC H_{2} C H_{2} N H_{2} heat C_{4} H_{6} N_{2} O_{2} + 2 C_{2} H_{5} OH

Solved Example 4

How are primary, secondary and tertiary amines separated by Hofmann's method using diethyl oxalate?

Solution:

The mixture is warmed with diethyl oxalate, and each class gives a product with a different physical state (Figure 8).

1° amine: a solid dialkyl oxamide, $(CONHR)_{2}$ , which is filtered off.
2° amine: a liquid oxamic ester, $C_{2} H_{5} OOCCON R_{2}$ , which is distilled.
3° amine: no reaction; it is recovered unchanged.

Hydrolysis of the oxamide and of the oxamic ester with KOH then gives the pure 1° and 2° amines.

(COO C_{2} H_{5})_{2} + 2RN H_{2} (CONHR)_{2} + 2 C_{2} H_{5} OH

(COO C_{2} H_{5})_{2} + R_{2} NH C_{2} H_{5} OOCCON R_{2} + C_{2} H_{5} OH

Solved Example 5

The compound that gives an oily nitrosamine with aqueous nitrous acid at low temperature is
(A) methylamine
(B) ethylamine
(C) diethylamine
(D) triethylamine

Solution:

Answer: (C). A yellow oily N-nitrosamine needs exactly one N-H on nitrogen, which only a secondary amine has. (A) and (B) are 1° and give $N_{2}$ gas; (D) is 3° and gives only a soluble salt.

(C_{2} H_{5})_{2} NH + HN O_{2} (C_{2} H_{5})_{2} N - N = O + H_{2} O

Solved Example 6

The carbylamine test is performed by heating, in alcoholic KOH, a mixture of
(A) chloroform and silver powder
(B) a trihalogenated methane and a primary amine
(C) an alkyl halide and a primary amine
(D) an alkyl cyanide and a primary amine

Solution:

Answer: (B). Chloroform (a trihalogenated methane) with alcoholic KOH gives dichlorocarbene, which a primary amine converts into the isocyanide. An alkyl halide would simply alkylate the amine.

Solved Example 7

An aqueous solution of a compound (A) containing nitrogen and chlorine is acidic to litmus. (A) with aqueous NaOH gives (B), which contains nitrogen but no chlorine. (B) with

C_{6} H_{5} S O_{2} Cl

in the presence of NaOH gives an insoluble product (C),

C_{13} H_{13} N O_{2} S

. Give the structures of (A), (B) and (C).

Solution:

The Hinsberg product (C) is insoluble in alkali, so (B) is a secondary amine. Subtracting the $C_{6} H_{5} S O_{2}$ group from $C_{13} H_{13} N O_{2} S$ leaves $C_{7} H_{8} N$ , which is N-methylaniline. An aqueous solution that is acidic to litmus means (A) is its hydrochloride.

(A) = N-methylanilinium chloride, $C_{6} H_{5} N H_{2} (C H_{3})^{+} C l^{-}$
(B) = N-methylaniline, $C_{6} H_{5} NHC H_{3}$
(C) = N-methyl-N-phenylbenzenesulphonamide, $C_{6} H_{5} S O_{2} N (C H_{3}) C_{6} H_{5}$

Solved Example 8

The molecular formula of (A) is

C_{3} H_{7} NO

. On hydrolysis it gives an amine (B) and an acid (C). (B) with benzenesulphonyl chloride gives a product that is insoluble in aqueous NaOH, and (C) gives a silver mirror with Tollens' reagent. Identify (A), (B) and (C).

Solution:

(B) gives a base-insoluble Hinsberg product, so it is a 2° amine; (C) reduces Tollens' reagent, so it is formic acid, the only acid that does. Joining $HCOOH$ to a 2° amine within $C_{3} H_{7} NO$ gives N,N-dimethylformamide.

(A) = N,N-dimethylformamide, $HCON (C H_{3})_{2}$
(B) = dimethylamine, $(C H_{3})_{2} NH$
(C) = formic acid, $HCOOH$

HCON (C H_{3})_{2} + H_{2} O H^{+}, heat HCOOH + (C H_{3})_{2} NH

Solved Example 9

Compound (A),

C_{5} H_{11} NO

, is insoluble in cold acid and cold alkali. Refluxing (A) with NaOH evolves a gas (B) and leaves a salt (C). Acetyl chloride and (B) give (D),

C_{4} H_{9} NO

. (B) with

HN O_{2}

gives a yellow oil (E). Identify (A) to (E).

Solution:

Insolubility in both acid and alkali points to an amide, and refluxing with alkali splits it into an amine and the sodium salt of the acid. The yellow oil (E) shows that the amine (B) is secondary, and (D) gives its carbon count as two.

(A) = N,N-dimethylpropanamide, $C H_{3} C H_{2} CON (C H_{3})_{2}$
(B) = dimethylamine, $(C H_{3})_{2} NH$
(C) = sodium propanoate, $C H_{3} C H_{2} COONa$
(D) = N,N-dimethylacetamide, $C H_{3} CON (C H_{3})_{2}$
(E) = N-nitrosodimethylamine, $(C H_{3})_{2} N - N = O$

Solved Example 10

How would you distinguish aniline from benzylamine?

Solution:

Both are primary amines, so both give a positive carbylamine test. Use the azo dye test: diazotise each with $NaN O_{2}$ /HCl at 273-278 K and pour the solution into alkaline $β$ -naphthol.

Aniline gives a stable diazonium salt, which couples to an orange-red dye.
Benzylamine is aliphatic: its diazonium ion decomposes at once with brisk $N_{2}$ , and no dye forms.

Solved Example 11

An amine (A),

C_{7} H_{9} N

, gives brisk effervescence with

HN O_{2}

at room temperature and no dye with alkaline

β

-naphthol, but a positive carbylamine test. Identify (A).

Solution:

The carbylamine test makes (A) primary; the effervescence and the absence of a dye make it aliphatic rather than aromatic. The only primary amine of formula $C_{7} H_{9} N$ whose nitrogen is not on the ring is benzylamine, $C_{6} H_{5} C H_{2} N H_{2}$ .

Solved Example 12

Three bottles contain ethylamine, diethylamine and triethylamine. Describe one experiment that identifies all three.

Solution:

Shake each with benzenesulphonyl chloride and aqueous KOH (the Hinsberg test).

The one that dissolves and gives a precipitate on acidifying is ethylamine (1°).
The one that gives an insoluble solid at once is diethylamine (2°).
The one that does not react, and dissolves only when the mixture is acidified, is triethylamine (3°).

Confirm if needed: ethylamine alone gives the carbylamine smell, and diethylamine alone gives the Liebermann blue-green colour.

Practice Questions

Which reagent shows the acidic nature of the $- N H_{2}$ group: (a) Na, (b) $C S_{2}$ , (c) $B r_{2}$ + NaOH, (d) water?Answer: (a) sodium; an amine gives $RNHNa$ and hydrogen, which shows the N-H is weakly acidic
How would you distinguish (a) methylamine from dimethylamine and (b) aniline from N-methylaniline?Answer: (a) carbylamine test: only methylamine gives the foul-smelling isocyanide; (b) $HN O_{2}$ : aniline gives a diazonium salt that couples to a dye, N-methylaniline gives a yellow oily nitrosamine
Why does a tertiary amine give no reaction in the carbylamine test?Answer: the reaction needs two N-H bonds, which are lost as two molecules of HCl; a 3° amine has none
Name Hinsberg's reagent and give the product with a 1°, a 2° and a 3° amine.Answer: benzenesulphonyl chloride; $C_{6} H_{5} S O_{2} NHR$ (soluble in KOH), $C_{6} H_{5} S O_{2} N R_{2}$ (insoluble) and no product
Write the reactions of ethylamine with $C S_{2}$ followed by $HgC l_{2}$ .Answer: $C_{2} H_{5} N H_{2} + C S_{2}$ gives $C_{2} H_{5} NHCSSH$ , and with $HgC l_{2}$ this gives $C_{2} H_{5} NCS + HgS + 2HCl$ , the mustard oil smell
A mixture of aniline and N,N-dimethylaniline is shaken with $C_{6} H_{5} S O_{2} Cl$ and aqueous KOH. What happens to each?Answer: aniline dissolves as the potassium salt of its sulphonamide and is recovered by acidifying and then boiling with conc. HCl; N,N-dimethylaniline is 3° and stays as an unreacted layer
Describe the Liebermann nitroso reaction and say which class gives it.Answer: a 2° amine gives a nitrosamine with $HN O_{2}$ ; warming this with phenol and conc. $H_{2} S O_{4}$ gives a brown-red colour that turns blue then green, red on dilution and greenish blue with alkali
An unknown amine gives a base-insoluble solid in the Hinsberg test. What class is it, and what will it give with $HN O_{2}$ ?Answer: a secondary amine; with $HN O_{2}$ it gives a yellow oily N-nitrosamine
Complete: aniline with $NaN O_{2}$ /HCl at 273-278 K, then alkaline $β$ -naphthol.Answer: benzenediazonium chloride couples at C-1 of the naphthol to give the orange-red dye 1-phenylazo-2-naphthol
How do you tell an aliphatic primary amine from an aromatic primary amine?Answer: treat with $HN O_{2}$ in the cold: the aliphatic amine gives brisk $N_{2}$ and an alcohol, while the aromatic amine gives a stable diazonium salt that couples with $β$ -naphthol to a dye
Why is a Lassaigne's test done before any class test?Answer: it confirms that nitrogen is present in the compound at all; sodium fusion gives $NaCN$ , which gives Prussian blue with $FeS O_{4}$ and an iron(III) salt

Common Mistakes to Avoid

Watch out

Naming benzoyl chloride as Hinsberg's reagent. The reagent is benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} Cl$ .
Expecting a 2° or 3° amine to give the carbylamine test. Two N-H bonds are needed, so only 1° amines respond.
Saying the carbylamine test works only for aliphatic amines. Aromatic primary amines such as aniline also give it.
Reading a yellow oil in the nitrous acid test as a 1° amine. The oily nitrosamine means a 2° amine.
Reporting a 3° amine as 'no reaction' in the Hinsberg test without adding that it dissolves when the mixture is acidified.
Getting the Hinsberg solubilities the wrong way round. The 1° sulphonamide keeps an acidic N-H and dissolves in KOH; the 2° one has none and stays solid.
Expecting diethyl oxalate to react with a 3° amine during a Hofmann separation. It does not, which is how the 3° amine is recovered.
Using the azo dye test on an aliphatic primary amine. Its diazonium salt decomposes at once, so no dye can form.
Forgetting that the class of an amine is set by the groups on nitrogen. 2,4-Dimethylaniline is a primary amine, because both methyl groups are on the ring.

Frequently Asked Questions

What is the Hinsberg test?

The Hinsberg test uses benzenesulphonyl chloride with aqueous potassium hydroxide. A primary amine gives a sulphonamide that dissolves in the alkali, a secondary amine gives an insoluble sulphonamide, and a tertiary amine does not react and dissolves only when the mixture is acidified.

Why is the sulphonamide of a primary amine soluble in alkali?

The product still has one hydrogen on nitrogen. The two sulphonyl oxygen atoms withdraw electron density, so this hydrogen is acidic and the alkali removes it, giving a water-soluble potassium salt. Acidifying the solution returns the sulphonamide as a precipitate.

What is the carbylamine test and which amines give it?

An amine warmed with chloroform and alcoholic potassium hydroxide gives an isocyanide with a very unpleasant smell. Only primary amines give it, aliphatic and aromatic alike, because the reaction uses up two N-H bonds. Secondary and tertiary amines give no reaction.

Which amines give an azo dye test?

Only aromatic primary amines. Their diazonium salts are stable between 273 and 278 K, so they survive long enough to couple with alkaline beta-naphthol and give an orange-red dye. Aliphatic primary amines lose nitrogen at once, so they give no dye.

How can primary, secondary and tertiary amines be told apart in one experiment?

Use the Hinsberg test. The primary amine dissolves in the alkaline mixture and precipitates on acidifying, the secondary amine gives an insoluble solid straight away, and the tertiary amine does not react but dissolves when acid is added. Nitrous acid gives the same three answers by observation.

How is a mixture of amines separated by Hofmann's method?

The mixture is warmed with diethyl oxalate. The primary amine gives a solid oxamide, the secondary amine gives a liquid oxamic ester and the tertiary amine does not react. The solid is filtered, the liquid distilled and the tertiary amine recovered; hydrolysis with alkali frees the other two.

What is the Liebermann nitroso reaction used for?

It confirms a secondary amine. The nitrosamine formed with nitrous acid is warmed with phenol and concentrated sulphuric acid, giving a colour that runs from brown and red to blue and green, turns red on dilution with water and greenish blue or violet with alkali.

Which tests on amines are asked in NEET?

NEET follows NCERT closely: the carbylamine test for primary amines, the Hinsberg test for telling the three classes apart, the reactions of the classes with nitrous acid, and the azo dye test that identifies an aromatic primary amine. Questions usually give an observation and ask for the class.

What does JEE Main ask about the analysis of amines?

JEE Main sets identification problems: a molecular formula with clues such as a base-insoluble Hinsberg product, a yellow oil with nitrous acid or a silver mirror with Tollens' reagent, and asks for the structures. The mechanism of the carbylamine reaction and the separation methods also appear.

Previous year questions on Analysis of Amines

10 questions from past papers, each with a step-by-step solution.

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