Chemistry · Ch 12 — Organic Chemistry – Some Basic Principles and Techniques

Detection of Other Elements

12.9.2

Detection of Other Elements

Detection of Other Elements — Lassaigne’s Test

Organic compounds are built from carbon and hydrogen, but they often carry other elements — nitrogen, sulphur, halogens (chlorine, bromine, iodine), and phosphorus. These elements are covalently bonded inside the molecule, so they cannot be detected directly by simple ionic reactions. The first job is to break those covalent bonds and convert the element into an ionic form that can be tested in solution.

That conversion is done by fusing the organic compound with a piece of sodium metal. The process is called sodium fusion, and the resulting test is known as Lassaigne’s test (after the French chemist who developed it).

The Sodium Fusion — What Happens Inside

When the organic compound is heated strongly with sodium metal, the sodium atoms donate electrons to break the covalent bonds. The elements from the compound form the following sodium salts:

Carbon and nitrogen together give sodium cyanide:

\ce{Na + C + N ->[\Delta] NaCN}

Sulphur gives sodium sulphide:

\ce{2Na + S ->[\Delta] Na2S}

A halogen (Cl, Br, or I) gives the corresponding sodium halide:

\ce{Na + X ->[\Delta] NaX} \quad (X = \ce{Cl},\ \ce{Br},\ \ce{I})

The symbols C, N, S, and X here come from the organic compound itself. The fused mass — a hard, dark residue — is then boiled with distilled water. The sodium salts dissolve, while unreacted carbon and other debris remain behind. This aqueous solution is called the sodium fusion extract, and it is the starting material for all the tests that follow.

Watch out

If the fusion is not done properly — if the sodium is not fresh or the heating is insufficient — the conversion to ionic form may be incomplete. That leads to false negative results. Always use a clean, dry sodium piece and heat the test tube until it is red hot.

(A) Test for Nitrogen

The sodium fusion extract contains cyanide ions ( $\ce{CN-}$ ) if nitrogen was present in the original compound. The test exploits the formation of a deep blue complex called Prussian blue.

Procedure:

Boil the sodium fusion extract with a freshly prepared solution of iron(II) sulphate. Then acidify the mixture with concentrated sulphuric acid and heat gently. A Prussian blue colour or precipitate confirms nitrogen.

The chemistry in two steps:

Step 1 — Formation of hexacyanidoferrate(II) ion:

Cyanide ions from the extract react with iron(II) ions to form the stable complex ion hexacyanidoferrate(II):

\ce{6CN- + Fe^{2+} -> [Fe(CN)6]^{4-}}

Step 2 — Oxidation and formation of Prussian blue:

When concentrated sulphuric acid is added and the mixture is heated, some of the iron(II) ions are oxidised to iron(III) ions by the air (or by the nitric acid that may be present from the acid). These iron(III) ions then react with the hexacyanidoferrate(II) ions to produce a deep blue insoluble compound — iron(III) hexacyanidoferrate(II), commonly called Prussian blue:

\ce{3[Fe(CN)6]^{4-} + 4Fe^{3+} ->[xH2O] Fe4[Fe(CN)6]3·xH2O}

The formula $\ce{Fe4[Fe(CN)6]3·xH2O}$ is the classic Prussian blue. The water of crystallisation ( $x$ is variable) is part of the structure.

Note

The Prussian blue colour is so intense that even a trace of nitrogen gives a visible result. This is one of the most sensitive tests in qualitative organic analysis.

(B) Test for Sulphur

Sulphur, if present, appears in the sodium fusion extract as sulphide ions ( $\ce{S^{2-}}$ ). Two independent tests confirm its presence.

Test (i) — Lead acetate test

Procedure:

Acidify the sodium fusion extract with acetic acid (not sulphuric or nitric acid, because those would oxidise the sulphide). Then add a few drops of lead acetate solution.

Reaction:

Sulphide ions react with lead(II) ions to form a black precipitate of lead(II) sulphide:

\ce{S^{2-} + Pb^{2+} -> PbS (black)}

The black colour is unmistakable. If the precipitate is only faintly grey, it may still indicate a small amount of sulphur.

Watch out

Do not use mineral acids like $\ce{HCl}$ or $\ce{H2SO4}$ to acidify before adding lead acetate. They produce hydrogen sulphide gas ( $\ce{H2S}$ ) which escapes, and the test may fail. Acetic acid is weak enough to keep the sulphide in solution until the lead salt is added.

Test (ii) — Sodium nitroprusside test

Procedure:

To a fresh portion of the sodium fusion extract, add a few drops of sodium nitroprusside solution.

Reaction:

Sulphide ions react with the nitroprusside ion to form a violet-coloured complex:

\ce{S^{2-} + [Fe(CN)5NO]^{2-} -> [Fe(CN)5NOS]^{4-} (violet)}

This test is even more sensitive than the lead acetate test. A violet colour — ranging from pinkish-violet to deep violet — confirms sulphur.

Tip

Use the sodium nitroprusside test as a quick preliminary check. If it gives a clear violet, you can skip the lead acetate test. But if the colour is doubtful, the black precipitate from the lead acetate test is more definitive.

When Nitrogen and Sulphur Are Both Present — The Thiocyanate Complication

If the organic compound contains both nitrogen and sulphur, something interesting happens during the sodium fusion. Instead of forming separate sodium cyanide and sodium sulphide, the three elements (Na, C, N, S) combine to form sodium thiocyanate:

\ce{Na + C + N + S ->[\Delta] NaSCN}

The thiocyanate ion ( $\ce{SCN-}$ ) is present in the sodium fusion extract. This ion does not give the Prussian blue test for nitrogen, because there are no free cyanide ions. Instead, when iron(III) ions are added (from the ferric ions produced during the nitrogen test), thiocyanate forms a blood-red complex:

\ce{Fe^{3+} + SCN- -> [Fe(SCN)]^{2+} (blood red)}

This blood-red colour is a classic test for thiocyanate — and it tells you that both nitrogen and sulphur were present together.

Important

If you see a blood-red colour instead of Prussian blue when doing the nitrogen test, do not conclude that nitrogen is absent. It means both N and S are present. The Prussian blue test fails because the cyanide is tied up as thiocyanate.

How to get around this:

If the sodium fusion is carried out with excess sodium, the thiocyanate decomposes:

\ce{NaSCN + 2Na ->[\Delta] NaCN + Na2S}

Now the extract contains free cyanide and free sulphide ions, and both the usual tests (Prussian blue for N, black PbS for S) work normally. So if you suspect both elements are present, use a larger piece of sodium and fuse for a longer time.

(C) Test for Halogens

The sodium fusion extract contains halide ions ( $\ce{X-}$ ) if chlorine, bromine, or iodine was present. The test uses silver nitrate to precipitate the silver halide, and the colour and solubility of the precipitate distinguish the three halogens.

Procedure:

Acidify the sodium fusion extract with dilute nitric acid. Then add a few drops of silver nitrate solution.

Reaction:

\ce{X- + Ag+ -> AgX (precipitate)}

The results are:

Halogen	Precipitate colour	Solubility in ammonium hydroxide
Chlorine ( $\ce{Cl}$ )	White ( $\ce{AgCl}$ )	Soluble
Bromine ( $\ce{Br}$ )	Pale yellow ( $\ce{AgBr}$ )	Sparingly soluble
Iodine ( $\ce{I}$ )	Yellow ( $\ce{AgI}$ )	Insoluble

The solubility in ammonium hydroxide is a key distinguishing feature. Silver chloride dissolves readily in dilute $\ce{NH4OH}$ to form the soluble complex $\ce{[Ag(NH3)2]+}$ . Silver bromide dissolves only in concentrated $\ce{NH4OH}$ , and silver iodide does not dissolve at all. …

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