QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 8.5NCERT Class 11 · Chemistry · Chapter 8

Purification and Qualitative Analysis of Organic Compounds

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 9 min read

An extensive examination of physical purification techniques (sublimation, crystallization, multiple distillation regimes, chromatography) alongside rigorous chemical protocols for detecting carbon, hydrogen, nitrogen, sulphur, halogens, and phosphorus via Lassaigne's sodium fusion test.

1. Methods of Purification of Organic Compounds

Natural extraction or chemical synthesis yields crude organic compounds admixed with byproducts, unreacted starting materials, and solvents. Selection of purification technique depends strictly on physical state, thermal stability, solubility characteristics, and vapor pressure behaviors.

A. Sublimation

Employed to separate substances that change directly from solid to vapor state on heating without passing through the liquid phase from non-sublimable impurities. Applicable to: camphor, benzoic acid, naphthalene, anthracene, and iodine.

B. Crystallization

Based on differences in solubilities of the compound and impurities in a suitably chosen solvent. The ideal solvent dissolves a small amount of the compound at room temperature but large amounts at elevated temperatures, while impurities either remain insoluble (filtered hot) or remain completely dissolved in the mother liquor even upon cooling. Activated charcoal is added to decolorize colored resinous impurities.

C. Distillation Techniques

Technique Underlying Principle Criteria for Application Standard Examples
Simple Distillation Volatile liquid vaporizes upon heating and condenses into a separate receiver. Liquids that boil without decomposition and possess a boiling point difference > 25 K. Separation of Ether (b.p. 308 K) & Toluene (b.p. 384 K); Chloroform (b.p. 334 K) & Aniline (b.p. 457 K).
Fractional Distillation Successive vaporizations and condensations within a fractionating column packed with glass beads. Separation of miscible liquids having boiling point differences < 25 K. Crude petroleum fractions; Acetone (b.p. 329 K) and Methyl alcohol (b.p. 338 K).
Vacuum Distillation (Under Reduced Pressure) Lowering external pressure lowers the boiling point below normal decomposition temperature. Liquids that decompose at or below their normal boiling points. Purification of Glycerol (normal b.p. 563 K, decomposes; distills without decomposition at 453 K under 12 mm Hg); concentration of H2O2.
Steam Distillation Mixture boils when: Ptotal = PH2O + Porganic = 1 atm. Substances that are steam volatile, water-immiscible, with high vapor pressure near 373 K. Purification of Aniline, Nitrobenzene, o-Nitrophenol (separated from non-volatile p-nitrophenol), essential oils.
Steam Distillation Ratio Law: The mass ratio of organic compound (w1) to water (w2) condensing in the distillate is directly proportional to the product of their respective partial vapor pressures (P1, P2) and molar masses (M1, M2):
w1 / w2 = (P1 × M1) / (P2 × M2)

D. Differential Extraction

An organic compound dissolved in an aqueous solution is recovered by shaking with an immiscible organic solvent (e.g., ether, chloroform) in a separatory funnel. Based on the Nernst Distribution Law (Kd = Corganic / Caqueous). Repeated extractions with smaller solvent volumes are mathematically proven to be far more efficient than a single extraction using the entire volume.

E. Chromatography

Separation based on differential distribution of components between a stationary phase and a mobile phase:

  • Adsorption Chromatography (Column & TLC): Differential adsorption of components on a stationary adsorbent (silica gel or alumina). In Thin Layer Chromatography (TLC), separation is quantified by the Retention Factor (Rf):
    Rf = (Distance traveled by substance from baseline) / (Distance traveled by solvent front from baseline)
  • Partition Chromatography (Paper Chromatography): Stationary phase is water trapped inside cellulose fibers of the chromatography paper. Mobile phase ascends via capillary action, partitioning components based on relative solubilities.

2. Qualitative Elemental Analysis of Organic Compounds

A. Detection of Carbon and Hydrogen (Copper Oxide Test)

The organic substance is mixed thoroughly with dry cupric oxide (CuO) and heated in a hard glass tube. Carbon is oxidized to CO2, and hydrogen is oxidized to H2O:

C + 2CuO → CO2↑ + 2Cu
2H + CuO → H2O + Cu
  • Carbon Detection: CO2 gas turns lime water turbid due to CaCO3 precipitation:
    CO2 + Ca(OH)2 → CaCO3↓ (White ppt) + H2O
  • Hydrogen Detection: H2O vapor condenses on cooler walls and turns anhydrous white copper sulphate blue:
    CuSO4 (White) + 5H2O → CuSO4•5H2O (Hydrated Blue)

B. Lassaigne's Test (Sodium Fusion Extract / SFE)

Organic compounds covalently bond nitrogen, sulphur, and halogens. Fusion with molten sodium metal in an ignition tube converts these heteroatoms into ionic water-soluble sodium salts:

Na + C + N → NaCN (Sodium cyanide)
2Na + S → Na2S (Sodium sulphide)
Na + X → NaX (Sodium halide, where X = Cl, Br, I)
Na + C + N + S → NaSCN (Sodium thiocyanate, when both N & S are present)

C. Chemical Detection Tests from Sodium Fusion Extract

Element Reagent & Procedure Observation & Reaction Equation
Nitrogen (N) Boil SFE with freshly prepared FeSO4, cool, and acidify with concentrated H2SO4. Prussian Blue precipitate/coloration:
Fe2+ + 6CN− → [Fe(CN)6]4−
Fe2+ →[H2SO4] Fe3+
4Fe3+ + 3[Fe(CN)6]4− → Fe4[Fe(CN)6]3•xH2O (Prussian Blue)
Sulphur (S) (a) Add sodium nitroprusside Na2[Fe(CN)5NO] solution.
(b) Acidify SFE with acetic acid and add lead acetate (CH3COO)2Pb.		 (a) Deep Violet / Purple coloration:
S2− + [Fe(CN)5NO]2− → [Fe(CN)5NOS]4−
(b) Black precipitate of PbS:
S2− + Pb2+ → PbS↓ (Black)
Both N & S present Acidify the SFE and add FeCl3 solution. Blood Red coloration due to ferric thiocyanate:
Fe3+ + 3SCN− → [Fe(SCN)]2+ or Fe(SCN)3 (Blood Red)
Halogens (Cl, Br, I) Boil SFE with concentrated HNO3 (to decompose NaCN & Na2S), cool, and add AgNO3 solution. • Chlorine: Curdy white precipitate of AgCl, readily soluble in dilute NH4OH.
• Bromine: Pale yellow precipitate of AgBr, sparingly soluble in concentrated NH4OH.
• Iodine: Bright yellow precipitate of AgI, completely insoluble in NH4OH.
Phosphorus (P) Fuse compound with sodium peroxide (Na2O2) to form Na3PO4. Extract with water, boil with conc. HNO3, and add ammonium molybdate. Canary Yellow precipitate of ammonium phosphomolybdate:
Na3PO4 + 3HNO3 → H3PO4 + 3NaNO3
H3PO4 + 12(NH4)2MoO4 + 21HNO3 →
(NH4)3PO4•12MoO3↓ (Canary Yellow) + 21NH4NO3 + 12H2O
Critical Pitfall • Why boil SFE with Conc. HNO3 prior to Halide testing?
If nitrogen or sulphur is present, the unboiled SFE contains NaCN and Na2S. Upon adding AgNO3, these generate interfering white AgCN or black Ag2S precipitates that obscure the silver halide tests:
NaCN + AgNO3 → AgCN↓ (White precipitate)
Na2S + 2AgNO3 → Ag2S↓ (Black precipitate)
Boiling with concentrated HNO3 oxidizes and expels cyanide as HCN↑ and sulphide as H2S↑ gas, eliminating interference!

3. Interactive Practice & JEE Focus Questions

Q1. In Lassaigne's test for nitrogen, a blood-red coloration is observed upon adding neutral FeCl3. What does this signify about the elemental composition of the sample?
Significance: The sample contains BOTH nitrogen and sulphur.
Explanation: Molten sodium reacts simultaneously with carbon, nitrogen, and sulphur to form sodium thiocyanate (NaSCN). Upon reaction with ferric ions (Fe3+), it yields the intense blood-red complex [Fe(SCN)]2+ or Fe(SCN)3, rather than Prussian blue.
Q2. A mixture of ortho-nitrophenol and para-nitrophenol is to be separated. Which distillation technique should be used, and why?
Technique: Steam Distillation.
Rationale: ortho-Nitrophenol exhibits strong intramolecular hydrogen bonding (chelation), which prevents intermolecular association, resulting in high volatility (steam volatile). In contrast, para-nitrophenol forms extensive intermolecular hydrogen bonds, causing molecular association, higher boiling point, and rendering it non-steam volatile.
Q3. Why is calcium chloride (CaCl2) not suitable for drying ethyl alcohol?
Answer: Anhydrous CaCl2 reacts chemically with ethyl alcohol (ethanol) to form an insoluble crystalline addition complex (alcoholate) of formula CaCl2•4C2H5OH, thereby destroying the alcohol instead of merely drying it.
Q4. In a thin-layer chromatography (TLC) run, the solvent front traveled 8.0 cm from the baseline. Compound A traveled 4.8 cm, and Compound B traveled 6.4 cm. Calculate the Rf values and deduce which compound is more strongly adsorbed by the stationary phase.
Calculation:
Rf(A) = 4.8 cm / 8.0 cm = 0.60
Rf(B) = 6.4 cm / 8.0 cm = 0.80
Deduction: Compound A has a lower Rf value, meaning it migrated more slowly and is therefore more strongly adsorbed by the polar stationary phase.
Q5. In the layer test for bromide and iodide ions using chlorine water and chloroform (or CCl4), state the colors produced in the organic layer.
Observations:
• Bromide (Br−): Chlorine oxidizes Br− to molecular bromine (Br2), which dissolves in the non-polar CHCl3/CCl4 layer imparting an orange / reddish-brown color.
• Iodide (I−): Chlorine oxidizes I− to molecular iodine (I2), which dissolves in the organic layer imparting an intense violet / purple color. Excess chlorine bleaches this violet color by oxidizing I2 to colorless iodic acid (HIO3).
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