Chemistry
Purification and Characterisation of Organic Compounds
Purification & Characterisation of Organic Compounds Formula Sheet
All key formulas, tests, and reactions for organic purification and characterisation - Rf, Raoult, Carius, Kjeldahl, Lassaigne - JEE Main & Advanced quick revision.
Last-minute revision sheet for purification and characterisation of organic compounds. This chapter is mostly technique- and reaction-based, so it mixes a small set of true formulas (percentage composition, Rf, Raoult’s law, Clausius-Clapeyron) with high-yield qualitative tests, key reactions, and selection rules. Everything below is drawn only from the chapter pages.
Purification Methods - Selection Rules
Pick the right technique first
Match the method to the physical nature of the sample before anything else. This single decision wins most “which technique” MCQs.
| Method | Use when | Key requirement |
|---|---|---|
| Crystallization | Solid + temperature-dependent solubility difference | High solubility hot, low solubility cold |
| Simple distillation | Liquid + solid impurity, or two liquids | B.P. difference > 30 °C |
| Fractional distillation | Miscible liquids, close boiling points | B.P. difference < 30 °C (even 5-10 °C) |
| Steam distillation | Heat-sensitive, water-immiscible compound | Appreciable vapour pressure at ~100 °C |
| Vacuum distillation | High-B.P. compound that decomposes | Lower pressure -> lower B.P. |
| Chromatography | Closely related compounds / mixtures / trace amounts | Differential adsorption or partition |
Crystallization
Principle: purification by difference in solubility of compound in hot vs cold solvent; pure crystals separate on cooling, impurities stay in mother liquor.
| Quantity | Relation | Notes |
|---|---|---|
| Amount crystallized | $m_{\text{crystals}} = m_{\text{dissolved}} - (S_{\text{cold}} \times V)$ | $S$ = solubility per unit volume at the low temperature |
| % recovery | $\dfrac{m_{\text{crystals}}}{m_{\text{pure}}} \times 100$ | mother liquor retains $S_{\text{cold}} \times V$ |
Condition for complete separation of A and B in one step:
$$\boxed{\dfrac{S_A(T_H)}{S_A(T_L)} \gg \dfrac{S_B(T_H)}{S_B(T_L)}}$$A must have a much steeper solubility-temperature curve than B.
High-yield crystallization rules
- Use minimum hot solvent -> maximizes yield.
- Slow cooling -> large, pure crystals; rapid cooling traps impurities.
- Hot filtration removes insoluble impurities; activated charcoal removes coloured impurities.
- Wash crystals with cold solvent only (hot solvent dissolves product).
- Heat organic/volatile solvents on a water bath, never a direct flame.
- Sharp melting point = pure compound; a wide range signals impurities.
Distillation
Raoult’s Law (ideal solutions - fractional distillation)
$$\boxed{P_A = X_A \, P_A^{0}} \qquad P_{\text{total}} = P_A + P_B$$Vapour-phase mole fraction (more volatile component is enriched):
$$Y_A = \dfrac{P_A}{P_{\text{total}}}$$| Quantity | Formula | Notes |
|---|---|---|
| Partial pressure | $P_A = X_A P_A^{0}$ | $X_A$ = liquid mole fraction |
| Total pressure | $P_{\text{total}} = P_A + P_B$ | two-component ideal mix |
| Vapour composition | $Y_A = P_A / P_{\text{total}}$ | always richer in volatile component |
| Theoretical plate | one complete vaporization-condensation cycle | good column has 10-20 plates |
| Multi-plate enrichment | $\left(\dfrac{A}{B}\right)_n = \left(\dfrac{A}{B}\right)_0 \, \alpha^{\,n}$ | $\alpha$ = enrichment factor/plate, $n$ = plates |
Steam Distillation
Boils when the sum of vapour pressures equals atmospheric pressure (below the B.P. of either pure liquid):
$$\boxed{P_{\text{total}} = P_{\text{water}} + P_{\text{organic}} = P_{\text{atm}}}$$| Quantity | Formula | Notes |
|---|---|---|
| Mole ratio in vapour | $\dfrac{n_{\text{org}}}{n_{\text{water}}} = \dfrac{P_{\text{org}}}{P_{\text{water}}}$ | independent of amount present |
| Mass ratio in distillate | $\dfrac{m_{\text{org}}}{m_{\text{water}}} = \dfrac{P_{\text{org}} \, M_{\text{org}}}{P_{\text{water}} \, M_{\text{water}}}$ | $M$ = molar mass |
Vacuum Distillation - Clausius-Clapeyron
$$\boxed{\ln\!\left(\dfrac{P_2}{P_1}\right) = \dfrac{\Delta H_{vap}}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)}$$Used to find the lowered boiling point at reduced pressure. Rule of thumb: roughly every halving of pressure lowers B.P. by 10-15 °C.
Apparatus rules that get tested
- Thermometer bulb sits at the side-arm level (measures vapour temperature, not liquid).
- Condenser water enters from the bottom, exits at the top (countercurrent cooling).
- Boiling chips added before heating (prevent superheating/bumping); never to a hot liquid.
Azeotropes (constant-boiling mixtures - cannot be separated by simple fractional distillation)
| Type | Example | Composition / B.P. |
|---|---|---|
| Minimum-boiling | Ethanol-water | 95.6% ethanol, 4.4% water; B.P. 78.2 °C |
| Maximum-boiling | HCl-water | 20.2% HCl, 79.8% water; B.P. 108.6 °C |
Break the ethanol-water azeotrope chemically: $\ce{CaO + H2O -> Ca(OH)2}$ removes water, giving absolute ethanol; or use 3 Å molecular sieves.
Detection of Elements (Qualitative)
Lassaigne’s (Sodium Fusion) Reactions
Sodium fusion converts covalent N, S, X into water-soluble ionic salts:
$$\ce{C, N + Na ->[\Delta] NaCN} \qquad \ce{C, S + Na ->[\Delta] Na2S} \qquad \ce{C, X + Na ->[\Delta] NaX}$$When both N and S are present:
$$\ce{NaCN + S ->[\Delta] NaSCN}$$Element Tests Summary
| Element | Reagent / test | Positive result | Key reaction |
|---|---|---|---|
| Nitrogen | FeSO₄ then FeCl₃ (Prussian blue test) | Intense blue ppt | $\ce{3Na4[Fe(CN)6] + 4FeCl3 -> Fe4[Fe(CN)6]3 + 12NaCl}$ |
| Sulfur | Sodium nitroprusside | Purple/violet colour | $\ce{Na2S + Na2[Fe(CN)5NO] -> Na4[Fe(CN)5NOS]}$ |
| Sulfur | Lead acetate | Black ppt (PbS) | $\ce{Na2S + (CH3COO)2Pb -> PbS + 2CH3COONa}$ |
| N + S together | FeCl₃ | Blood-red colour | $\ce{NaSCN + FeCl3 -> [Fe(SCN)]^{2+}}$ |
| Halogen | dil. HNO₃ then AgNO₃ | Coloured ppt (see below) | $\ce{NaX + AgNO3 -> AgX + NaNO3}$ |
| Halogen (quick) | Beilstein (Cu wire) | Green flame | volatile CuX₂ |
| Carbon | Heat with CuO, lime water | Lime water turns milky | $\ce{C + 2CuO -> CO2 + 2Cu}$; $\ce{CO2 + Ca(OH)2 -> CaCO3 + H2O}$ |
| Hydrogen | Heat with CuO, anhyd. CuSO₄ | White CuSO₄ turns blue | $\ce{2H + CuO -> H2O + Cu}$ |
| Phosphorus | conc. HNO₃, then ammonium molybdate | Canary-yellow ppt | $\ce{Organic-P ->[HNO3] H3PO4}$ |
Silver Halide Identification
| Halogen | AgX colour | Solubility in NH₃ |
|---|---|---|
| Cl | White (AgCl) | Soluble in dilute NH₃ |
| Br | Pale yellow (AgBr) | Soluble only in conc. NH₃ |
| I | Yellow (AgI) | Insoluble even in conc. NH₃ |
Dissolution of AgCl: $\ce{AgCl + 2NH3 -> [Ag(NH3)2]+ + Cl-}$
Interference removal - examiners love this
- Remove S²⁻ before the N test (add lead acetate, filter off PbS) - otherwise blood-red SCN⁻ masks the blue.
- Remove CN⁻ and S²⁻ before the halogen test by boiling with dilute HNO₃: $\ce{NaCN + HNO3 -> HCN ^ + NaNO3}$ and $\ce{Na2S + 2HNO3 -> H2S ^ + 2NaNO3}$ (fume hood - HCN is toxic).
- Use freshly prepared FeSO₄ (Fe²⁺); oxidised Fe³⁺ gives a false negative for nitrogen.
- Use HNO₃, not HCl/H₂SO₄, before AgNO₃ (HCl adds Cl⁻; H₂SO₄ gives Ag₂SO₄).
Quantitative Estimation
Carbon & Hydrogen (Liebig combustion)
$$\boxed{\%\,C = \dfrac{12}{44} \times \dfrac{\text{mass of CO}_2}{\text{mass of compound}} \times 100}$$$$\boxed{\%\,H = \dfrac{2}{18} \times \dfrac{\text{mass of H}_2\text{O}}{\text{mass of compound}} \times 100}$$CO₂ absorbed in KOH; H₂O absorbed in anhydrous CaCl₂ (or Mg(ClO₄)₂).
Nitrogen
Dumas method (works for all N; collects N₂ gas):
$$\text{moles of N}_2 = \dfrac{PV}{RT}, \qquad \%\,N = \dfrac{2 \times (\text{moles N}_2) \times 14}{\text{mass of compound}} \times 100$$Kjeldahl method (amino-N only - amines, amides, proteins):
$$\ce{Organic-N ->[H2SO4] (NH4)2SO4} \;\xrightarrow{\text{NaOH}}\; \ce{NH3}$$$$\%\,N = \dfrac{1.4 \times \text{(normality} \times \text{volume of acid used by NH}_3)}{\text{mass of compound}}$$
Kjeldahl fails for
Nitro (–NO₂), nitroso (–NO), azo (–N=N–), and ring/heterocyclic nitrogen (pyridine). These N atoms are not reduced to NH₃ by H₂SO₄ - use the Dumas method instead.
Sulfur & Halogens (Carius method - oxidise in sealed tube with fuming HNO₃)
Sulfur, weighed as BaSO₄ (M = 233):
$$\boxed{\%\,S = \dfrac{32}{233} \times \dfrac{\text{mass of BaSO}_4}{\text{mass of compound}} \times 100}$$Halogen, weighed as AgX:
$$\boxed{\%\,X = \dfrac{\text{at. mass of }X}{M_{AgX}} \times \dfrac{\text{mass of AgX}}{\text{mass of compound}} \times 100}$$| Halide | $M_{AgX}$ | At. mass of X |
|---|---|---|
| AgCl | 143.5 | Cl = 35.5 |
| AgBr | 188 | Br = 80 |
| AgI | 235 | I = 127 |
Empirical & Molecular Formula
- Find percentage composition of each element.
- Empirical formula: divide each % by atomic mass, then by the smallest ratio.
- Molecular formula:
Chromatography
Principle: differential migration of components between a stationary and a mobile phase (adsorption for TLC/column; partition for paper).
Retention factor
$$\boxed{R_f = \dfrac{\text{distance travelled by compound}}{\text{distance travelled by solvent front}}}$$| Property | Value / meaning |
|---|---|
| Range | $0 < R_f < 1$ (ideal 0.3-0.7) |
| $R_f = 0$ | compound not moved (too strongly adsorbed) |
| $R_f = 1$ | moved with solvent front (not adsorbed) |
| Polar compound | low $R_f$ (strong adsorption on silica) |
| Non-polar compound | high $R_f$ |
| More polar mobile phase | higher $R_f$ |
Partition coefficient (paper chromatography)
$$\boxed{R_f = \dfrac{K}{K + \beta}} \qquad K = \dfrac{R_f \, \beta}{1 - R_f}$$where $K = C_{\text{mobile}}/C_{\text{stationary}}$ and $\beta = V_s/V_m$ is the phase ratio.
Resolution & column efficiency
| Quantity | Formula | Notes |
|---|---|---|
| Resolution | $R_s = \dfrac{2(d_2 - d_1)}{w_1 + w_2}$ | $R_s > 1.5$ = baseline separation |
| Theoretical plates | $N = 16\left(\dfrac{d}{w}\right)^2$ | higher $N$ = better efficiency |
| Plate height (HETP) | $HETP = \dfrac{L}{N}$ | lower = more efficient |
| Van Deemter | $H = A + \dfrac{B}{u} + Cu$ | optimal flow $u$ gives minimum $H$ |
Eluotropic series (increasing polarity)
$$\text{Pet. ether} < \text{hexane} < \text{CCl}_4 < \text{benzene} < \text{CHCl}_3 < \text{Et}_2\text{O} < \text{EtOAc} < \text{acetone} < \text{MeOH} < \text{water}$$
Chromatography exam pointers
- TLC = analytical (mg scale, fast); column = preparative (g scale); paper = partition, best for polar compounds (amino acids, sugars).
- Spot must sit above the solvent level; mark only with pencil (ink dissolves and runs).
- Saturate the chamber with vapour for a straight solvent front and reproducible $R_f$.
- Never let a column run dry (cracking, channelling, sample loss); flow ~1-2 drops/sec; adsorbent:sample ratio ≈ 30:1.
- Visualisation: UV (254/366 nm), iodine vapour, ninhydrin (amino acids).