Purification and Characterisation of Organic Compounds

Organic Weightage: 1-2 Questions (4-8 Marks) JEE Unit 13

“Welcome to Organic Chemistry! Before any organic compound can be studied, it has to be separated from its impurities and then identified — that is exactly what this unit is about. It stays entirely qualitative and numerical: choose the right purification technique for a given pair of properties, read a Lassaigne's-test result correctly (including its three classic traps), and convert a combustion/Carius/Dumas/Kjeldahl reading into a percentage composition and finally an empirical formula. There is no reaction mechanism or 3-D structure here — every question reduces to picking the correct formula and plugging in the given numbers carefully, so precision with formulas and units is what separates a top score from a near-miss.”

— SCORECHEM ACADEMIC TEAM

1. Purification Techniques

Sublimation: a solid passes directly to vapour (and back to solid) without melting. Works for compounds that sublime cleanly, leaving behind a non-subliming impurity — camphor, naphthalene, anthracene, benzoic acid, NH₄Cl.
Crystallisation: the most common technique for solids, based on the DIFFERENCE in solubility of the compound and its impurity at different temperatures. The impure solid is dissolved in a minimum volume of hot solvent, filtered hot (removes insoluble impurity), then cooled slowly — the pure compound crystallises out first, while soluble impurities stay behind in the mother liquor.
Simple distillation: separates a volatile liquid from a non-volatile (or much less volatile) substance, or two miscible liquids with a large difference in boiling point (Δ B.P. > ~25 K). Example: chloroform (334 K) from aniline (457 K).
Fractional distillation: used when two miscible liquids have a SMALL difference in boiling point (Δ B.P. < ~25 K) — a fractionating column provides repeated vaporisation-condensation cycles for a sharper separation. Example: separating acetone and methanol; refining crude oil into its fractions.
Distillation under reduced pressure (vacuum distillation): used for a liquid that decomposes at or near its normal boiling point — lowering the pressure lowers the boiling point below the decomposition temperature. Example: glycerol is purified from spent-lye by distilling it under reduced pressure.
Steam distillation: used for a substance that is steam-volatile and virtually immiscible with water. The mixture boils when the sum of the two separate vapour pressures equals atmospheric pressure (p_total = p_organic + p_H2O), which happens BELOW the boiling point of either pure component — so a high-boiling, heat-sensitive liquid distils over at a much lower temperature, without decomposing. Example: aniline-water mixture; extraction of essential oils.
Differential extraction: used to pull a compound OUT of a solution (typically aqueous) by shaking it with an immiscible organic solvent in a separating funnel in which the compound is more soluble than in the original solvent. The compound distributes itself between the two immiscible layers according to its distribution (partition) coefficient, K_D = (concentration in organic layer) / (concentration in aqueous layer), a constant for a given compound-solvent pair at a fixed temperature. The organic layer, now carrying most of the compound, is run off and separated from the aqueous layer. Repeated extraction with several small portions of solvent removes the compound more completely than a single extraction with the same total volume of solvent — each fresh portion re-establishes the same favourable partition ratio, so successive small extractions leave progressively less compound behind in the aqueous layer than one large extraction ever could. This is the basis of using a separating funnel to isolate a product from an aqueous reaction mixture before it is purified further (e.g. by distillation or crystallisation).

Fig. 1: Choice of technique depends on BOTH the compound's own nature (volatility, thermal stability) AND how its impurity behaves — never on one alone.

⚠️ JEE Trap: A statement question testing distillation methods almost always swaps which technique goes with which property. Match the technique to what's actually different between the compound and its impurity — volatility gap (simple/fractional distillation), thermal stability (vacuum distillation), or steam-volatility plus water-immiscibility (steam distillation) — rather than recalling technique names from memory.

⚠️ JEE Trap: "Several small extractions beat one large extraction" is a genuinely numerical fact, not just a rule of thumb. Because the distribution coefficient is fixed, splitting the same total solvent volume into multiple portions and extracting sequentially always leaves a smaller fraction of compound in the aqueous layer than a single extraction with the whole volume at once — this is testable with actual numbers, not just recalled as a slogan.

2. Chromatography

Principle: chromatography separates a mixture based on the differing rates at which its components move through a stationary phase under the influence of a mobile phase.
Adsorption chromatography: the stationary phase (silica gel or alumina, packed in a column) selectively ADSORBS different components onto its surface to different extents; the mobile phase (a suitable solvent, or "eluant") carries the less-strongly-adsorbed components through faster. Column chromatography and TLC (thin-layer chromatography) both work on this principle.
Partition chromatography: components distribute (partition) themselves continuously between two liquid phases — a stationary liquid film held on an inert solid support, and a moving liquid mobile phase — according to their relative solubility in each. Paper chromatography works on this principle.

Fig. 2: Adsorption chromatography separates by surface affinity; partition chromatography separates by relative solubility in two liquid phases. Rf never exceeds 1.

Retardation factor (R_f): R_f = (distance moved by the substance) / (distance moved by the solvent front), both measured from the same baseline (point of application). Since the substance can never outrun the solvent front, 0 < R_f < 1 always. A more polar substance (in a fixed adsorption system) is held back more strongly and has a SMALLER R_f; a less polar substance travels farther and has a LARGER R_f.
Detecting spots that aren't naturally visible: viewing under UV light (fluorescence), exposing to iodine vapour in a closed chamber (iodine forms a coloured complex with most organic compounds, especially unsaturated/aromatic ones), or spraying with a specific developing reagent such as ninhydrin (turns amino acids violet/purple).

⚠️ JEE Trap: "Higher R_f = more polar" is backwards. On a polar stationary phase (silica/alumina), a MORE polar compound is adsorbed more strongly and moves LESS — it has the SMALLER R_f. The less polar (more mobile) compound has the larger R_f.

3. Qualitative Analysis: Detection of Elements

Lassaigne's test (sodium fusion extract): the organic compound is fused with a small piece of SODIUM metal, strongly heated, then plunged into distilled water and boiled/filtered. This step is essential because it converts covalently-bound N, S, halogens and P present in the compound into water-soluble, IONIC sodium salts (NaCN, Na₂S, NaX, Na₃PO₄) — every subsequent qualitative test is really a test on these simple inorganic ions, not on the organic compound directly.

Fig. 3: Every qualitative test works by converting a covalently-bonded element into a water-soluble IONIC sodium salt first (Lassaigne's fusion), then testing that ion.

Test for nitrogen: the sodium fusion extract is boiled with freshly prepared FeSO₄, cooled, acidified with concentrated H₂SO₄. A Prussian blue colouration/precipitate (Fe₄[Fe(CN)₆]₃) confirms nitrogen, formed via Na₄[Fe(CN)₆] and then Fe³⁺ (from partial oxidation of Fe²⁺ by air/acid).
Test for sulphur: (i) extract + freshly prepared sodium nitroprusside gives a violet/purple colouration; or (ii) extract acidified with acetic acid + lead acetate gives a black precipitate of PbS.
Test for nitrogen AND sulphur together: if BOTH are present, excess sodium in the fusion can convert them into a single compound, sodium thiocyanate (NaSCN), which gives a blood-red colouration with FeCl₃ — this blood-red test, when positive, confirms both N and S are present together (and the individual Prussian-blue test for N may appear to fail, see Section on exceptions).
Test for halogens: extract + dilute HNO₃ (to decompose any NaCN/Na₂S first, see below) + AgNO₃ solution. A precipitate forms: AgCl is white (soluble in NH₄OH), AgBr is pale yellow (partially soluble in NH₄OH), AgI is bright/dark yellow (insoluble in NH₄OH).
Test for phosphorus: the compound is fused with sodium peroxide (Na₂O₂) to oxidise phosphorus to phosphate; the solution is boiled with concentrated HNO₃, then treated with ammonium molybdate — a canary-yellow precipitate of ammonium phosphomolybdate confirms phosphorus.

Fig. 4: All three exceptions trace back to how the sodium fusion extract is actually formed — check what's really in the ignition tube before reading a negative result as absence of the element.

Why these three exceptions matter: they are the single most heavily tested conceptual trap in this section (see figure above) — a compound containing nitrogen with NO carbon in it at all (hydrazine, hydroxylamine) will genuinely give a NEGATIVE Prussian-blue test, since NaCN simply cannot form without carbon.

⚠️ JEE Trap: A negative Lassaigne's test does not always mean the element is absent. Always check whether the compound could be one of the three exceptions (no carbon present; N/S not boiled off with HNO₃ before the halogen test; excess sodium used) before concluding an element is missing.

4. Quantitative Analysis: Carbon and Hydrogen

Liebig's combustion method: a known mass of the organic compound is burnt in a current of dry oxygen (over CuO, which ensures complete oxidation). Carbon converts entirely to CO₂, and hydrogen entirely to H₂O; these are absorbed separately and weighed — CO₂ in KOH solution (or soda-lime), H₂O in anhydrous CaCl₂ (absorbed FIRST, since KOH would absorb both).
Formulas:
- %C = (12 × mass of CO₂ produced × 100) / (44 × mass of compound taken)
- %H = (2 × mass of H₂O produced × 100) / (18 × mass of compound taken)
The factors 12/44 and 2/18 are simply the fraction of the molar mass of CO₂/H₂O that is contributed by C/H respectively.

5. Quantitative Analysis: Nitrogen

Fig. 5: When a compound has nitro, azo, or heterocyclic ring nitrogen, Dumas is the ONLY reliable method — Kjeldahl systematically under-reports or fails entirely.

Dumas' method (gasometric): the compound is heated with excess copper(II) oxide in an atmosphere of pure CO₂; nitrogen present is converted to free N₂ gas (any oxides of nitrogen formed are reduced back to N₂ by passing over hot copper gauze). The N₂ is collected over concentrated KOH solution, which absorbs the accompanying CO₂, leaving pure N₂ whose volume is measured at STP.
- %N = (28 × V_STP(N₂ in mL) × 100) / (22400 × mass of compound in g)
- Dumas' method works for EVERY nitrogen-containing compound without exception, including nitro-, azo- and heterocyclic-ring-nitrogen compounds, because it measures total elemental nitrogen gas directly rather than relying on a chemical conversion to ammonia.
Kjeldahl's method (titrimetric): the compound is heated with concentrated H₂SO₄ (with K₂SO₄ to raise the boiling point, and CuSO₄/Se as catalyst) — this converts the nitrogen to ammonium sulfate, (NH₄)₂SO₄. The digested mixture is then heated with excess NaOH, and the liberated NH₃ is absorbed in a known volume/normality of standard acid; the UNREACTED acid is back-titrated (or the ammonia is directly titrated) to find how much acid the ammonia neutralised.
- %N = (1.4 × Normality of acid × volume of acid used in mL) / (mass of compound in g)
- Kjeldahl's method FAILS for: nitro (–NO₂) and azo (–N=N–) compounds (the nitrogen in these groups is not reduced to NH₄⁺ under the digestion conditions), and for compounds where nitrogen is part of a ring system (pyridine, quinoline) — ring nitrogen similarly does not convert to ammonium sulfate.

⚠️ JEE Trap: Given a nitro-, azo- or pyridine-type nitrogen compound, Kjeldahl's method is the WRONG tool. Whenever a question's compound contains one of these nitrogen environments, the only usable method is Dumas' — recognise the functional group first, then pick the method.

6. Quantitative Analysis: Halogens, Sulphur and Phosphorus

Carius method (halogens and sulphur): a known mass of the compound is heated with fuming HNO₃ in the presence of AgNO₃ inside a sealed hard-glass (Carius) tube. All halogen converts to the corresponding silver halide precipitate (AgX); sulphur converts to sulfate, which is precipitated as BaSO₄ by adding BaCl₂. The precipitate is filtered, washed, dried and weighed.
- %X (halogen) = (atomic mass of X × mass of AgX formed × 100) / (molar mass of AgX × mass of compound taken)
- %S = (32 × mass of BaSO₄ formed × 100) / (233 × mass of compound taken) — since S contributes 32 out of BaSO₄'s molar mass of 233.
Estimation of phosphorus: the compound is oxidised (e.g. with fuming HNO₃), converting phosphorus to phosphoric acid; this is precipitated either as ammonium phosphomolybdate, (NH₄)₃PO₄.12MoO₃, or as Mg₂P₂O₇ (magnesium pyrophosphate), and weighed to back-calculate %P from the known stoichiometry of the precipitate.

7. Oxygen, and Empirical/Molecular Formula

Estimation of oxygen: oxygen is almost never estimated directly.
- Indirect method: %O = 100 − (sum of percentages of all other elements estimated). This is the default approach whenever every other element's percentage has already been found.
- Unterzaucher's method (direct): the compound is decomposed by pyrolysis over carbon at high temperature, converting all its oxygen quantitatively to CO, which is then oxidised to CO₂ and weighed; %O is calculated from the mass of CO₂ obtained (using oxygen's stoichiometric fraction, 32/88, of that CO₂).

Fig. 6: Every quantitative method follows the same shape — convert the target element into one measurable product, then scale by its molar-mass fraction.

From percentage composition to empirical formula — the standard sequence:
1. Divide each element's percentage by its atomic mass → gives the relative number of MOLES of each atom.
2. Divide every value obtained in step 1 by the SMALLEST value among them → gives the simplest whole-number mole ratio.
3. If any ratio is still not a whole number, multiply ALL ratios by the smallest integer that clears every fraction (e.g. multiply by 2 if one ratio comes out as x.5).
4. Write the empirical formula using these whole-number subscripts.
Empirical formula → molecular formula: find the empirical formula mass (sum of atomic masses in the empirical formula), then n = (molecular mass) / (empirical formula mass); the molecular formula is the empirical formula multiplied throughout by n. The molecular mass itself is usually supplied separately (e.g. from vapour density, VD × 2 for a gas/vapour) or given directly in the question.

⚠️ JEE Trap: Forgetting to actually multiply by n after finding it. Many students correctly compute the empirical formula and n but then report the empirical formula itself as the final answer — always apply n to every subscript to get the molecular formula.

8. How JEE Frames These Questions

Statement/assertion-reason pairs on purification technique choice: re-derive which property (volatility gap, thermal stability, steam-volatility) actually distinguishes the compound from its impurity, rather than recalling "distillation type X goes with situation Y" from memory.
Lassaigne's-test exception questions: always check the compound's structure first for the three specific traps (no carbon; N/S not destroyed with HNO₃ before the halide test; excess sodium used) before reading a negative or blood-red result at face value.
Combustion/Carius/Dumas/Kjeldahl numericals: identify exactly which product was weighed or which volume/titre was measured, plug directly into that method's formula, and watch units carefully (mL vs L for gas volumes, g vs mg for masses) — almost every arithmetic slip in this section is a units mismatch, not a wrong formula.
"Which method should be used" questions: if the compound's structure includes a nitro, azo, or ring-nitrogen group, Dumas' method is the only correct choice for nitrogen; for all other nitrogen-containing compounds, either method could in principle be used, but Kjeldahl's is the one actually described as titrimetric/volumetric.
Empirical-to-molecular-formula problems: always complete all the way through to the molecular formula (via n) when a molecular mass or vapour density is given — don't stop at the empirical formula.

⚠️ JEE Trap: A numerical that supplies BOTH a Carius halogen result and a combustion C/H result in the same question wants you to combine them into one empirical formula — compute each element's percentage independently from its own method's formula, then run the full empirical-formula sequence using all the percentages together.

9. Quick Sheet and Checklist

Element	Method	Formula
C, H	Liebig combustion	%C = 12·m(CO₂)·100/44m; %H = 2·m(H₂O)·100/18m
N	Dumas	%N = 28·V_STP·100/(22400·m)
N	Kjeldahl (fails: nitro, azo, ring-N)	%N = 1.4·Normality·V_acid/m
Halogen (X)	Carius	%X = At.mass(X)·m(AgX)·100/(M(AgX)·m)
Sulphur (S)	Carius (as BaSO₄)	%S = 32·m(BaSO₄)·100/(233·m)
Oxygen (O)	Indirect, or Unterzaucher	%O = 100 − Σ(others); or 32·m(CO₂)·100/88m
R_f	Chromatography	dist. moved by substance / dist. moved by solvent front (always ≤ 1)

Before the exam, check you can:

Match a given pair of physical-property differences (volatility, thermal stability, steam-volatility) to the correct purification technique.
State how differential extraction works (partition of a compound between two immiscible solvents by its distribution coefficient) and explain why several small extractions remove more compound than one large extraction of the same total volume.
Distinguish adsorption from partition chromatography, and state the R_f formula together with the rule that it never exceeds 1.
State the sodium-fusion basis of Lassaigne's test and correctly apply all three of its exceptions (no carbon; boil off N/S before the halogen test; excess sodium merges the N and S tests).
Recall the positive-observation colour/precipitate for each qualitative test (N, S, both N & S, halogens, P).
Apply the correct quantitative formula (Liebig, Dumas, Kjeldahl, Carius, indirect-O/Unterzaucher) given the measured mass/volume/titre, and know exactly which functional groups make Kjeldahl's method fail.
Carry a percentage-composition problem all the way through to a final molecular formula, not stopping at the empirical formula.