IIT-JAM Qualitative Organic Analysis — Element Detection and Functional Group Tests

By Aniket Bhardwaj · 25 September 2026 · IIT-JAM Chemistry

Qualitative organic analysis is the systematic way of going from an unknown bottle to a structure: find out which elements are present, narrow the class by solubility, then confirm the functional group with a specific test. In a written examination it becomes a logic puzzle — you are given four or five observations and asked what the compound is, or given two compounds and asked for one reagent that tells them apart. Both need the same thing: knowing exactly what each test detects and what it misses. This guide covers Lassaigne's test with balanced equations, the solubility scheme, the functional-group tests in a single table, and the estimation arithmetic worked out and cross-checked.

Why sodium fusion is needed at all

Nitrogen, sulphur and halogen in an organic compound are held in covalent bonds, and covalent nitrogen does not respond to any ionic test. Fusing the compound with molten sodium converts them into ionic sodium salts, which can then be detected in aqueous solution:

Na + C + N → NaCN
2Na + S → Na₂S
Na + C + N + S → NaSCN (when both N and S are present)
Na + X → NaX (X = Cl, Br, I)

The aqueous filtrate from the fusion is the Lassaigne's extract, and it must be alkaline before use — any free sodium left over would react violently, and an acidic extract would drive off HCN and H₂S and lose the very ions you are testing for.

The four detection tests, with equations

Element	Reagent	Observation
Nitrogen	FeSO₄, boil, then acidify with dilute H₂SO₄	Prussian blue precipitate or green solution
Sulphur	Sodium nitroprusside	Deep violet colour
Sulphur (alternative)	Lead acetate in acetic acid	Black PbS precipitate
N and S together	FeCl₃	Blood-red colour from thiocyanate
Halogen	Boil with dilute HNO₃, then AgNO₃	AgCl white · AgBr pale yellow · AgI yellow

Nitrogen:
FeSO₄ + 6NaCN → Na₄[Fe(CN)₆] + Na₂SO₄
3Na₄[Fe(CN)₆] + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃ + 12Na⁺ (Prussian blue)

Sulphur:
Na₂S + Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS] (violet)
Na₂S + (CH₃COO)₂Pb → PbS↓ + 2CH₃COONa

Check the charges in the Prussian blue equation and it explains itself: four Fe³⁺ give +12 and three [Fe(CN)₆]⁴⁻ give −12, so the 3:4 ratio is forced.

Three details in this section are examined far more often than the colours themselves.

Why boil with dilute HNO₃ before the silver nitrate test. If nitrogen or sulphur is also present, the extract contains CN⁻ and S²⁻, which would give white AgCN and black Ag₂S and be mistaken for halide. Boiling with acid expels them as HCN and H₂S first.
Telling the three halides apart. Colour alone is unreliable between AgBr and AgI. Use ammonia: AgCl dissolves readily, AgBr only sparingly, and AgI not at all. Note that fluorine is not detected this way, because AgF is soluble.
When N and S are both present, thiocyanate forms and gives the blood-red FeSCN colour rather than Prussian blue. With a large excess of sodium in the fusion, the thiocyanate is broken down to CN⁻ and S²⁻ and both separate tests then work normally — which is why "use excess sodium" is part of the standard procedure.

One honest limitation to remember: Lassaigne's test fails for diazonium salts, because they lose their nitrogen as N₂ gas on heating, before the sodium can ever capture it.

Solubility classification — narrowing the field before any reagent

Behaviour	What it indicates
Soluble in water and in ether	Low molar mass, polar — a small alcohol, acid, amine, aldehyde or ketone
Insoluble in water, soluble in 5 % NaOH and in 5 % NaHCO₃	Carboxylic acid — strong enough to displace CO₂ from bicarbonate
Soluble in 5 % NaOH but not in NaHCO₃	Phenol (or another weak acid such as a nitroalkane)
Insoluble in NaOH, soluble in 5 % HCl	Amine — basic
Insoluble in acid and alkali, soluble in cold concentrated H₂SO₄	Neutral compound containing O or N, or an alkene — alcohol, ether, ester, ketone
Insoluble in everything above	Alkane, aryl halide, or another inert hydrocarbon-like compound

The NaOH-versus-NaHCO₃ line is the single most useful discrimination in the whole scheme, because it separates carboxylic acids from phenols using nothing but relative acid strength. Carbonic acid sits between them: an acid stronger than carbonic acid liberates CO₂ from bicarbonate; a phenol does not.

Functional group tests — what each one really detects

Group	Test	Positive result	Important limitation
Unsaturation	Br₂ in CCl₄; Baeyer's reagent (cold dilute alkaline KMnO₄)	Bromine colour discharged; purple → brown MnO₂	Phenols and aldehydes also decolourise KMnO₄ by being oxidised
Alcohol	Lucas reagent (conc. HCl + ZnCl₂)	Turbidity: 3° at once, 2° in a few minutes, 1° not in the cold	Only useful for alcohols soluble in the reagent, so roughly up to six carbons
Phenol	Neutral FeCl₃; coupling with a diazonium salt	Violet, green or blue colour; orange-red azo dye	Not every phenol gives the FeCl₃ colour — a negative result is not proof
Carboxylic acid	NaHCO₃ solution	Brisk effervescence of CO₂	The cleanest way to separate an acid from a phenol
Aldehyde	Tollens' reagent; Fehling's solution; Schiff's reagent	Silver mirror; red Cu₂O; magenta colour	Aromatic aldehydes do not reduce Fehling's, though they do give a silver mirror
Aldehyde or ketone	2,4-dinitrophenylhydrazine	Yellow to red precipitate	Detects the carbonyl but does not distinguish aldehyde from ketone
Methyl ketone	Iodoform test (I₂/NaOH)	Yellow CHI₃, characteristic smell	Also positive for CH₃CH(OH)– compounds such as ethanol and propan-2-ol
1° amine	Carbylamine test (CHCl₃ + alcoholic KOH)	Extremely unpleasant isocyanide smell	Only primary amines respond — 2° and 3° do not
Amine class	Hinsberg test (benzenesulphonyl chloride + KOH)	1° gives an alkali-soluble product, 2° an insoluble solid, 3° no reaction	Needs care with the alkali step or 1° and 2° look alike
1° aromatic amine	Diazotisation at 0–5 °C then coupling with 2-naphthol	Bright orange-red azo dye	Above about 5 °C the diazonium salt decomposes and the test fails
Ester	Hydroxamic acid test (NH₂OH/KOH, then FeCl₃)	Violet colour	Acid chlorides and anhydrides also respond
Amide or nitrile	Boil with NaOH	NH₃ evolved, turns moist red litmus blue	Ammonium salts do the same — check the elements first

Nitrous acid separates the three amine classes as well: an aliphatic primary amine gives brisk N₂ gas, an aromatic primary amine gives a diazonium salt that survives only in the cold, a secondary amine gives a yellow oily nitrosamine, and a tertiary aliphatic amine simply forms a salt.

Worked example 1 — a deduction problem

An organic liquid gives a Prussian blue colour in Lassaigne's test but no violet with nitroprusside and no precipitate with silver nitrate. It is insoluble in water and in NaOH but dissolves in dilute HCl. It gives an offensive smell with chloroform and alcoholic KOH, and on treatment with NaNO₂/HCl at 0–5 °C followed by alkaline 2-naphthol it gives a bright orange dye. Identify the class, and name the simplest member.

Elements. Prussian blue → nitrogen present. No violet → no sulphur. No silver halide → no halogen. So the compound contains C, H and N only, possibly with O.

Solubility. Insoluble in NaOH rules out acids and phenols. Soluble in dilute HCl → basic → an amine.

Carbylamine positive. Only a primary amine gives the isocyanide smell, so it is 1°, not 2° or 3°.

Diazotisation and coupling. A stable diazonium salt at 0–5 °C that couples to an azo dye means the –NH₂ is attached directly to an aromatic ring; an aliphatic primary amine would have released N₂ immediately instead.

Conclusion: a primary aromatic amine. The simplest is aniline, C₆H₅NH₂. Note how each observation removed one branch of the tree — that is the structure the examiner is testing, and writing the reasoning out in that order earns the marks even if the final name is not the one expected.

Worked example 2 — one reagent to tell two compounds apart

Give a single test that distinguishes each pair.

Benzoic acid and phenol. Add sodium bicarbonate solution. Benzoic acid effervesces (CO₂); phenol does not. Both dissolve in NaOH, so NaOH would tell you nothing.
Benzaldehyde and acetophenone. Tollens' reagent gives a silver mirror with benzaldehyde only. Alternatively the iodoform test is positive for acetophenone (it has a CH₃CO– group) and negative for benzaldehyde — the two tests point opposite ways, which makes the pair a favourite.
Ethanol and methanol. Iodoform test: ethanol is CH₃CH(OH)– and gives yellow CHI₃; methanol has no such group and gives nothing.
Propan-1-ol and propan-2-ol. Lucas reagent shows turbidity within minutes for the secondary alcohol and not in the cold for the primary. The iodoform test also separates them, positive for propan-2-ol.
Aniline and N-methylaniline. Carbylamine test — positive for the primary amine only.
An aliphatic and an aromatic aldehyde. Fehling's solution: the aliphatic one gives the red precipitate, the aromatic one does not.

The pattern worth internalising: a good distinguishing test is one that is positive for exactly one member of the pair. A test both compounds pass, however dramatic, is worth no marks.

Worked example 3 — nitrogen by Kjeldahl's method

0.50 g of an organic compound was digested by Kjeldahl's method and the ammonia liberated was absorbed in 50.0 mL of 0.10 M H₂SO₄. The excess acid needed 30.0 mL of 0.10 M NaOH. Find the percentage of nitrogen.

Step 1 — acid taken.
n(H₂SO₄) = 0.0500 L × 0.10 mol L⁻¹ = 5.00 × 10⁻³ mol
H₂SO₄ is diprotic, so available H⁺ = 1.00 × 10⁻² mol

Step 2 — acid left over.
n(NaOH) = 0.0300 L × 0.10 = 3.00 × 10⁻³ mol, neutralising 3.00 × 10⁻³ mol H⁺

Step 3 — acid used by ammonia.
1.00 × 10⁻² − 3.00 × 10⁻³ = 7.00 × 10⁻³ mol H⁺
NH₃ + H⁺ → NH₄⁺ is 1 : 1, so n(NH₃) = n(N) = 7.00 × 10⁻³ mol

Step 4 — mass and percentage. A_r(N) = 14.007
mass N = 7.00 × 10⁻³ × 14.007 = 0.09805 g
% N = (0.09805 ÷ 0.50) × 100 = 19.61 %

Cross-check by the standard formula % N = 1.4 × N_acid × V ÷ w, where V is the volume in mL of acid actually consumed and N_acid is its normality. The acid is 0.10 M and diprotic, so 0.20 N; the consumed volume is 7.00 × 10⁻³ eq ÷ 0.20 eq L⁻¹ = 0.0350 L = 35.0 mL.
% N = 1.4 × 0.20 × 35.0 ÷ 0.50 = 9.8 ÷ 0.50 = 19.6 % ✓

The constant 1.4 is simply A_r(N)/10 ≈ 14/10, packed into the formula along with the conversion from mL to L. Knowing where it comes from means you can rebuild the formula if you forget it. Kjeldahl's method fails for nitrogen in a ring, in a nitro group or in an azo group, because the digestion does not convert those to ammonium — a limitation that is itself a common one-mark question.

Worked example 4 — halogen by Carius' method

0.30 g of an organic compound gave 0.45 g of silver chloride in a Carius estimation. Find the percentage of chlorine. Take A_r(Ag) = 107.87, A_r(Cl) = 35.45.

M(AgCl) = 107.87 + 35.45 = 143.32 g mol⁻¹

Route 1 — through moles.
n(AgCl) = 0.45 ÷ 143.32 = 3.140 × 10⁻³ mol
Each AgCl carries one Cl, so mass Cl = 3.140 × 10⁻³ × 35.45 = 0.1113 g
% Cl = (0.1113 ÷ 0.30) × 100 = 37.10 %

Route 2 — through the mass fraction, without computing moles.
Fraction of AgCl that is chlorine = 35.45 ÷ 143.32 = 0.24735
% Cl = 0.24735 × (0.45 ÷ 0.30) × 100 = 0.24735 × 1.5 × 100 = 37.10 % ✓

The two routes agree, which is the check worth doing under exam pressure. The same logic handles sulphur, where the sulphur ends up as BaSO₄ and the fraction becomes A_r(S)/M(BaSO₄), and bromine and iodine, where AgBr and AgI replace AgCl.

Common mistakes that cost marks

Testing for halogen without first boiling off cyanide and sulphide. AgCN is white and Ag₂S is black; both are easily misread as halide.
Calling a negative test proof of absence. Some phenols give no colour with FeCl₃, and Lassaigne's test misses the nitrogen of a diazonium salt entirely.
Using NaOH to separate an acid from a phenol. Both dissolve; only NaHCO₃ tells them apart.
Treating the iodoform test as a ketone test. It detects CH₃CO– or CH₃CH(OH)–, so ethanol and propan-2-ol are positive.
Expecting Fehling's solution to work on benzaldehyde. It does not; Tollens' does.
Diazotising above 5 °C. The diazonium salt decomposes and the coupling test then fails for a reason that has nothing to do with the compound.
Applying Kjeldahl's method to nitro, azo or ring nitrogen. The result will be low, and the method is simply not valid there.
Forgetting that H₂SO₄ is diprotic in a back titration. This single slip halves or doubles the nitrogen percentage.

How to prepare this unit

Theme	What you must be able to do without hesitation
Sodium fusion	Explain why it is needed and write all four conversion equations
Detection tests	Write the Prussian blue and nitroprusside equations, and the ammonia rule for the three silver halides
Solubility scheme	Place any compound into its class from three solubility observations
Functional group tests	State the reagent, the positive observation and the limitation for each
Distinguishing pairs	Pick a test that is positive for exactly one of two given compounds
Kjeldahl	Handle a back titration with a diprotic acid, and state where the method fails
Carius	Convert AgX or BaSO₄ mass into a percentage by two independent routes

Treat that as a revision checklist, not as a prediction of the paper. For the syllabus and the current pattern, read the official IIT-JAM notification for your year.

Check the arithmetic behind the estimations. There is no single tool for qualitative analysis — the reasoning is the work — but examples 3 and 4 are ordinary mole and mass-fraction calculations, and those are exactly what the calculator suite is for. Open the suite and use the tool that fits the step you are on, rather than a bare number from memory.

Open the ABC Chemistry Calculator Suite →

Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches — classroom at the Gurugram coaching centre and live online classes for students anywhere in India. Details at abcchemistry.in.