For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne's test) is acidifie

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Q. For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne's test) is acidified with dilute HNO3. What will happen if a student acidifies the extract with dilute H2SO4 in place of dilute HNO3?

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Concept understanding — Lassaigne Test Principle

The Lassaigne Test: Why We Fuse Sodium with the Sample

Imagine you have an organic compound — say, a drug, a pesticide, or a dye — and you need to find out if it contains nitrogen, sulfur, or a halogen (chlorine, bromine, iodine). You cannot simply test the original compound directly because these elements are covalently bonded inside the molecule. They are "hidden" — not present as free ions.

The Lassaigne test solves this by breaking the covalent bonds and converting these elements into water-soluble ionic salts. That is the entire point: turn invisible atoms into detectable ions.

The Intuition

Think of the organic molecule as a locked safe. Inside the safe are the atoms you want to detect (N, S, X). The safe is the covalent framework. To get the atoms out, you need to smash the safe open. Sodium metal is the sledgehammer.

When you heat the organic compound with a piece of sodium metal, the sodium reacts violently. It rips apart the carbon‑carbon and carbon‑heteroatom bonds. In the process, the sodium itself gets oxidised (loses an electron) and the atoms from the compound get reduced (gain electrons). The result is a set of simple, stable, ionic compounds:

Nitrogen (from the compound) combines with sodium and carbon to form sodium cyanide ( $\text{NaCN}$ ).
Sulfur forms sodium sulfide ( $\text{Na}_2\text{S}$ ).
Halogens form sodium halides ( $\text{NaX}$ , where X = Cl, Br, I).

These are all water‑soluble. Once you dissolve the fused mass in water, you have a solution containing $\text{CN}^-$ , $\text{S}^{2-}$ , and $\text{X}^-$ ions — ready for standard qualitative tests.

Note

If both nitrogen and sulfur are present in the same compound, they can also form sodium thiocyanate ( $\text{NaSCN}$ ) during fusion. This is important because it can interfere with the test for sulfur — more on that later.

The Precise Statement

Lassaigne’s test (or sodium fusion test): A small piece of dry sodium metal is heated with a sample of the organic compound in a fusion tube until the tube is red hot. The fused mass is then extracted with distilled water, boiled, cooled, and filtered. The filtrate (called the Lassaigne’s extract) is used to test for the presence of nitrogen, sulfur, and halogens.

The chemical principle is:

\text{Organic compound (containing N, S, X)} + \text{Na (excess)} \xrightarrow{\Delta} \text{NaCN} + \text{Na}_2\text{S} + \text{NaX} + \text{residue}

Why This Works (and What Can Go Wrong)

For nitrogen: The cyanide ion ( $\text{CN}^-$ ) is detected by adding ferrous sulfate and ferric chloride — a deep blue precipitate of Prussian blue ( $\text{Fe}_4[\text{Fe(CN)}_6]_3$ ) confirms nitrogen.

For sulfur: The sulfide ion ( $\text{S}^{2-}$ ) gives a purple/violet colour with sodium nitroprusside ( $\text{Na}_2[\text{Fe(CN)}_5\text{NO}]$ ). Alternatively, it forms a black precipitate of lead sulfide ( $\text{PbS}$ ) with lead acetate.

For halogens: The halide ion ( $\text{X}^-$ ) is detected by acidifying the extract with nitric acid and adding silver nitrate — a curdy white precipitate (AgCl), pale yellow (AgBr), or yellow (AgI) indicates the halogen. …

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Acidifying the sodium extract with dilute $\mathrm{H_2SO_4}$ instead of dilute $\mathrm{HNO_3}$ will produce a white precipitate of $\mathrm{Ag_2SO_4}$ that interferes with the halide test, making it impossible to confirm the presence of halogens reliably.

Why We Acidify in Lassaigne's Test

When an organic compound containing halogens is fused with sodium metal, the halogen converts to sodium halide ( $\mathrm{NaX}$ , where $\mathrm{X} = \mathrm{Cl, Br, I}$ ) in the aqueous extract. To detect these halides, we add $\mathrm{AgNO_3}$ , which precipitates them as silver halides:

\mathrm{NaX + AgNO_3 \to AgX \downarrow + NaNO_3}

The precipitate colour tells us which halogen is present: $\mathrm{AgCl}$ is white, $\mathrm{AgBr}$ is pale yellow, and $\mathrm{AgI}$ is yellow.

But the sodium extract is alkaline (from excess sodium and the formation of $\mathrm{NaOH}$ , $\mathrm{Na_2CO_3}$ , $\mathrm{NaCN}$ , etc.). If we add $\mathrm{AgNO_3}$ directly to this alkaline solution, silver hydroxide and silver oxide precipitate:

\mathrm{AgNO_3 + NaOH \to AgOH \downarrow + NaNO_3}

\mathrm{2AgOH \to Ag_2O \downarrow + H_2O}

These brown/black precipitates mask the halide test. So we acidify first to neutralise the alkalinity.

Why Nitric Acid, Not Sulphuric Acid?

The choice of acid matters because we need one that does not itself interfere with the silver ion.

With dilute $\mathrm{HNO_3}$ : Nitric acid neutralises the alkaline impurities ( $\mathrm{NaOH}$ , $\mathrm{Na_2CO_3}$ , $\mathrm{NaCN}$ , $\mathrm{Na_2S}$ ) without forming any precipitate with $\mathrm{Ag^+}$ . The nitrate ion ( $\mathrm{NO_3^-}$ ) does not precipitate silver, so the only precipitate that forms is from the halide ions—exactly what we want to observe.
With dilute $\mathrm{H_2SO_4}$ : Sulphuric acid also neutralises the alkaline species, but it introduces sulphate ions ( $\mathrm{SO_4^{2-}}$ ) into the solution. When $\mathrm{AgNO_3}$ is added, these sulphate ions react:

$\mathrm{2AgNO_3 + H_2SO_4 \to Ag_2SO_4 \downarrow + 2HNO_3}$ …

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Q.For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne's test) is acidified with dilute HNO3. What will happen if a student acidifies the extract with dilute H2SO4 in place of dilute HNO3?

Why We Acidify in Lassaigne's Test

Why Nitric Acid, Not Sulphuric Acid?

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