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ChemistryOrganic Analysis
During Lassaigne's test, the elements present in an organic compound are converted from :
Options
1
Covalent form to covalent form
2
Ionic form to ionic form
3
Covalent form to ionic form
4
Ionic form to covalent form
Correct Answer
Option 3 : Covalent → Ionic
Step-by-Step Solution
1

In organic compound — elements are covalently bonded:

C-N bond (in amines, amides) — covalent

C-S bond (in thioethers) — covalent

C-Cl, C-Br, C-I (in halocompounds) — covalent

2

Fusion with sodium metal converts to ionic form:

C≡N + Na → Na⁺CN⁻ (sodium cyanide — ionic)

S + 2Na → Na₂S (sodium sulphide — ionic)

C-X + Na → Na⁺X⁻ (sodium halide — ionic)

3

Summary: Covalent bonds in organic compound → broken by sodium → ionic salts formed. The ionic salts dissolve in water (sodium fusion extract = SFE) and are easily detected by simple chemical tests.

Theory: Lassaigne's Test — Complete Guide
1. Purpose of Lassaigne's Test

Lassaigne's test (sodium fusion test) detects the presence of nitrogen, sulphur, and halogens in organic compounds. The key challenge: these elements are covalently bonded in organic molecules and cannot be directly detected by simple ionic tests. Fusion with sodium converts them to ionic salts (NaCN, Na₂S, NaX) which dissolve in water and can be detected by standard confirmatory tests. The method was developed by J. L. Lassaigne in 1843.

2. Reactions During Sodium Fusion

📌 Nitrogen: C + N (covalent in compound) + Na → NaCN (ionic sodium cyanide)

📌 Sulphur: C-S (covalent) + 2Na → Na₂S (ionic sodium sulphide)

📌 Chlorine: C-Cl (covalent) + Na → NaCl (ionic sodium chloride)

📌 Bromine: C-Br (covalent) + Na → NaBr (ionic sodium bromide)

📌 Iodine: C-I (covalent) + Na → NaI (ionic sodium iodide)

📌 N + S together: NaCN + Na₂S → NaSCN (sodium thiocyanate) — changes N detection

3. Confirmatory Tests on Sodium Fusion Extract (SFE)

After fusion and dissolving in water (SFE = sodium fusion extract): Test for N: add FeSO₄ solution → boil → add FeCl₃ + dil. HCl → Prussian blue colour (Fe₄[Fe(CN)₆]₃) = N confirmed. Test for S: add sodium nitroprusside (Na₂[Fe(CN)₅NO]) → purple/violet colour = S confirmed. OR add lead acetate → black PbS precipitate. Test for Cl: acidify with dil. HNO₃ → add AgNO₃ → white curdy precipitate (AgCl) soluble in NH₃ = Cl confirmed. Test for Br: AgNO₃ → pale yellow precipitate (AgBr) sparingly soluble in NH₃. Test for I: AgNO₃ → yellow precipitate (AgI) insoluble in NH₃.

4. Special Case: N and S Present Together

When both N and S are present in the organic compound, during fusion they react with each other: NaCN + S → NaSCN (sodium thiocyanate). As a result, less NaCN is formed → the Prussian blue test for N may fail. Detection of N when S is also present: acidify SFE with acetic acid → add excess FeCl₃ → blood red colour = thiocyanate (SCN⁻) confirms N (since thiocyanate contains N). So even though the Prussian blue test fails, the thiocyanate test detects N indirectly. This is an important modification for compounds containing both N and S (e.g., thiourea, cysteine).

5. Why Sodium Metal is Used

Sodium is an extremely reactive metal (oxidation state goes from 0 to +1). It readily reduces halogens and reacts with C, N, S in the organic compound at high temperature. The high temperature during fusion (red-hot conditions) breaks C-N, C-S, C-X covalent bonds and allows Na to react. No other commonly available metal would work as efficiently. Sodium is chosen because: (1) Very reactive — strong reducing agent. (2) Sodium salts are water-soluble. (3) Sodium is cheap and available. (4) Na reacts cleanly without introducing interfering ions.

6. Prussian Blue Formation — Detailed Chemistry

Step 1: Na + C,N → NaCN (from fusion). Step 2: NaCN + FeSO₄ → Fe(CN)₂ + Na₂SO₄ (or complex formation). Step 3: On boiling: Fe²⁺ + 6CN⁻ → [Fe(CN)₆]⁴⁻ (ferrocyanide, hexacyanoferrate(II)). Step 4: Add FeCl₃: 4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃ (Prussian blue = ferric ferrocyanide). This intense blue colour is the positive test for nitrogen. If N is absent → no CN⁻ → no blue complex.

7. Distinguishing Halogens in Lassaigne's Test

📌 Cl⁻: AgNO₃ → white AgCl precipitate → dissolves in NH₃

📌 Br⁻: AgNO₃ → pale yellow AgBr → sparingly soluble in NH₃

📌 I⁻: AgNO₃ → yellow AgI → insoluble in NH₃

📌 All halogens: SFE + dil. HNO₃ (to remove CN⁻ and S²⁻ that would interfere) → then AgNO₃

📌 Layer test: SFE + Cl₂ water + CHCl₃ → Cl₂ displaces I⁻ → violet/purple layer in CHCl₃ (I₂)

📌 Baeyer's reagent test can differentiate Br and I in presence of each other

8. Limitations of Lassaigne's Test

The test has some limitations: (1) Cannot detect F (fluorine) — NaF doesn't give precipitate with AgNO₃ (AgF is soluble). (2) N and S present together interfere with Prussian blue test. (3) Some organohalogen compounds (like CCl₄, CHCl₃) react slowly with Na — special precautions needed. (4) The test is qualitative only — doesn't tell how many N/S/halogen atoms are present. Quantitative analysis requires combustion analysis (C, H, N by Dumas method; halogens by Carius method; S by Carius method).

Frequently Asked Questions
1. Why must elements be converted to ionic form for detection? ⌄
Covalent compounds like C-N, C-S, C-Cl don't ionise in water — they can't be detected by ionic precipitation tests. For example: C₂H₅Cl dissolved in water doesn't give Cl⁻ ions — it's a covalent compound. AgNO₃ test needs free Cl⁻ ions. After fusion with Na: C₂H₅Cl + Na → NaCl (ionic). NaCl dissolved in water → Na⁺ + Cl⁻ → AgNO₃ test gives white AgCl precipitate. Converting covalent to ionic form is the entire purpose of Lassaigne's test.
2. What is the role of FeSO₄ in the nitrogen test? ⌄
FeSO₄ provides Fe²⁺ which reacts with CN⁻ from SFE: Fe²⁺ + 6CN⁻ → [Fe(CN)₆]⁴⁻ (ferrocyanide complex). This ferrocyanide complex then reacts with Fe³⁺ (added as FeCl₃): 4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃↓ (Prussian blue). The FeSO₄ must be freshly prepared (Fe²⁺ oxidises to Fe³⁺ in air over time). If Fe³⁺ is present in the FeSO₄ solution, the test may fail or give incorrect results.
3. Why is HNO₃ added before the silver nitrate test for halogens? ⌄
The SFE may contain CN⁻ (if N was present) and S²⁻ (if S was present). Both would interfere with the AgNO₃ test: AgCN (white) and Ag₂S (black) would also precipitate and give false positives. Adding dilute HNO₃: CN⁻ + H⁺ → HCN (gas, escapes); S²⁻ + 2H⁺ → H₂S (gas, escapes). This removes interfering ions before adding AgNO₃. Only the genuine halide ions (Cl⁻, Br⁻, I⁻) remain → clean precipitation with AgNO₃.
4. Why can't Lassaigne's test detect fluorine? ⌄
During fusion: C-F + Na → NaF (ionic sodium fluoride). NaF dissolves in water → F⁻ ions. But AgF (silver fluoride) is SOLUBLE in water — so adding AgNO₃ doesn't give a precipitate with F⁻. This is because F⁻ is very small and highly hydrated → AgF lattice energy is overcome by hydration energy. Therefore, the standard Lassaigne test halide detection (using AgNO₃ precipitation) cannot detect F. Special tests for F: Ca(OH)₂ → CaF₂ precipitate; or etch glass (HF etches SiO₂ in glass — HF + SiO₂ → SiF₄).
5. What is the sodium nitroprusside test for sulphur? ⌄
Sodium nitroprusside = Na₂[Fe(CN)₅NO] = sodium pentacyanonitrosylferrate(II). Test: add a few drops of fresh sodium nitroprusside to the SFE → violet/purple colour = sulphur confirmed. Reaction: S²⁻ + [Fe(CN)₅NO]²⁻ → [Fe(CN)₅NOS]⁴⁻ (violet complex). This is more sensitive than the lead acetate test. Important: the nitroprusside solution must be freshly prepared (it decomposes on standing). The same reagent is used in the legal test for illicit drug detection (presence of secondary amines gives blue colour).
6. How is the Carius method used for quantitative halogen analysis? ⌄
Carius method (quantitative halogens): weighed organic compound + fuming HNO₃ + AgNO₃ sealed in a Carius tube → heated strongly (250–300°C). Halogens are oxidised and precipitate as AgX (AgCl, AgBr, AgI). The AgX precipitate is filtered, washed, dried, and weighed. Mass of X = mass of AgX × (atomic mass of X / molar mass of AgX). %X = (mass of X / mass of compound) × 100. This is the standard quantitative method for C-X bond determination in organic compounds.
7. What is the Dumas method for nitrogen analysis? ⌄
Dumas method (quantitative N): weighed organic compound + CuO (oxidising agent) → heated → CO₂, H₂O, N₂ produced. CO₂ and H₂O absorbed. N₂ collected over KOH solution. Volume of N₂ measured. N (%) = (volume of N₂ × 28 × 100) / (22400 × mass of compound) [at STP]. This is more accurate than Kjeldahl method (which can't detect N in rings/azo compounds). Kjeldahl converts N → NH₃ (only works for amine/amide N).
8. What is the significance of ionic vs covalent form in analytical chemistry? ⌄
Ionic compounds: dissociate in water → free ions → can be detected by precipitation, colour reactions. Covalent compounds: don't ionise → ions not free → can't be directly detected by ionic tests. This distinction is fundamental to analytical chemistry. Lassaigne's test beautifully exploits this: converts non-ionisable covalent C-N, C-S, C-X to ionisable ionic NaCN, Na₂S, NaX. Then standard ionic tests detect these. This principle — converting covalent bonds to ionic form — is the key concept tested by this question.
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