Purification and Characterisation of Organic Compounds

Organic Chemistry Weightage: 1–2 Questions (4–8 Marks) NMC Unit 13

“Every organic synthesis ends with the same two questions: is it pure, and what is it? This unit answers both: separation techniques (sublimation to chromatography), Lassaigne's sodium-fusion tests, Liebig/Dumas/Kjeldahl/Carius estimations, and the arithmetic that turns percentages into empirical and molecular formulae. NEET asks it almost every year.”

— SCORECHEM ACADEMIC TEAM

1. Purification Methods

Purity is judged by a sharp, reproducible melting point (solids) or boiling point (liquids). Choose the method from the physical property that differs between the compound and its impurity.

Technique	Principle	Typical use
Sublimation	Solid changes directly to vapour on heating, without melting	Camphor, naphthalene, anthracene, benzoic acid, $\text{NH}_4\text{Cl}$ from non-sublimable impurities
Crystallisation	Solubility differs with temperature: sparingly soluble cold, highly soluble hot	Sugar, benzoic acid from hot water; mother liquor keeps the impurities
Simple distillation	Volatile liquid from non-volatile matter, or liquids with $\Delta \text{b.p.} > 25$ K	Chloroform (334 K) and aniline (457 K)
Fractional distillation	Fractionating column separates liquids with $\Delta \text{b.p.} < 25$ K	Crude oil fractions, acetone and methanol
Distillation under reduced pressure	Lower external pressure lowers the boiling point, so no decomposition	Glycerol from spent-lye
Steam distillation	Steam-volatile, water-immiscible liquid boils when $p = p_1 + p_2 =$ atmospheric pressure, so below 373 K	Aniline-water, essential oils
Differential extraction	Compound distributes between water and an immiscible organic solvent by its distribution coefficient (separating funnel)	Organic compound from an aqueous solution; low coefficient needs repeated or continuous extraction
Chromatography	Components move at different rates between a stationary and a mobile phase	Amino acids, sugars, pigments

Crystallisation extras: activated charcoal adsorbs coloured trace impurities. A solvent pair is used when the solute is too soluble in one solvent and insoluble in the other.
Steam distillation logic: a mixture boils when the total vapour pressure reaches atmospheric pressure, so a high-boiling liquid distils below 373 K and avoids thermal decomposition.

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.

Chromatography

Adsorption chromatography: differential adsorption on silica gel or alumina. Column chromatography uses a vertical packed tube and an eluant. TLC uses a $0.2$ mm adsorbent layer on a glass or aluminium plate.
Partition chromatography: continuous partitioning between a liquid stationary phase and a liquid mobile phase. Paper chromatography is partition chromatography: water held in the cellulose fibres is the stationary phase and the developing solvent is the mobile phase.

R_f = \frac{\text{distance moved by the substance from the base line } (x)}{\text{distance moved by the solvent front from the base line } (y)} \le 1

Visualising colourless spots: UV light (fluorescent compounds), iodine chamber (brown spots), ninhydrin spray (amino acids, purple).

2. Qualitative Detection of Elements

Carbon and hydrogen: heat with dry $\text{CuO}$ . $\text{C} + 2\text{CuO} \to 2\text{Cu} + \text{CO}_2$ (lime water turns milky) and $2\text{H} + \text{CuO} \to \text{Cu} + \text{H}_2\text{O}$ (white anhydrous $\text{CuSO}_4$ turns blue). This reaction is not part of Lassaigne's test.
Lassaigne's sodium fusion: fuse the compound with sodium metal, then extract with water. Covalent N, S and X become water-soluble ionic salts:

\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN} \qquad 2\text{Na} + \text{S} \xrightarrow{\Delta} \text{Na}_2\text{S} \qquad \text{Na} + \text{X} \xrightarrow{\Delta} \text{NaX}

Element	Test on the extract	Reaction	Observation
Nitrogen	Boil with $\text{FeSO}_4$ , acidify with conc. $\text{H}_2\text{SO}_4$	$6\text{CN}^- + \text{Fe}^{2+} \to [\text{Fe(CN)}_6]^{4-}$ ; $3[\text{Fe(CN)}_6]^{4-} + 4\text{Fe}^{3+} \to \text{Fe}_4[\text{Fe(CN)}_6]_3 \cdot x\text{H}_2\text{O}$	Prussian blue
Sulphur	(a) acetic acid + lead acetate; (b) sodium nitroprusside	$\text{S}^{2-} + \text{Pb}^{2+} \to \text{PbS}$ ; $\text{S}^{2-} + [\text{Fe(CN)}_5\text{NO}]^{2-} \to [\text{Fe(CN)}_5\text{NOS}]^{4-}$	(a) black ppt; (b) violet colour
N and S together	Fuse with insufficient Na, acidify with dil. HCl, add $\text{FeCl}_3$	$\text{Na} + \text{C} + \text{N} + \text{S} \to \text{NaSCN}$ ; $\text{Fe}^{3+} + \text{SCN}^- \to [\text{Fe(SCN)}]^{2+}$	Blood red, no Prussian blue
Halogens	Boil with dil. $\text{HNO}_3$ , add $\text{AgNO}_3$	$\text{Ag}^+ + \text{X}^- \to \text{AgX}$	AgCl white (soluble in $\text{NH}_4\text{OH}$ ); AgBr pale yellow (sparingly soluble); AgI yellow (insoluble)
Phosphorus	Fuse with $\text{Na}_2\text{O}_2$ , boil with conc. $\text{HNO}_3$ , add ammonium molybdate	$\text{H}_3\text{PO}_4 \to (\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3$	Canary yellow ppt

Exceptions that NEET loves

No carbon, no NaCN: hydrazine ( $\text{NH}_2\text{NH}_2$ ) and hydroxylamine contain nitrogen but give no Prussian blue.
Boil with $\text{HNO}_3$ before the halogen test whenever N or S is present, otherwise white $\text{AgCN}$ or black $\text{Ag}_2\text{S}$ gives a false result.
Excess sodium splits thiocyanate: $\text{NaSCN} + 2\text{Na} \to \text{NaCN} + \text{Na}_2\text{S}$ , so N and S are then detected separately (Prussian blue and black/violet) and no blood-red colour appears.
Phthalein dye test identifies the phenolic group (phenol heated with phthalic anhydride and conc. $\text{H}_2\text{SO}_4$ gives phenolphthalein, which turns pink with NaOH).

3. Quantitative Estimation

Let $m$ be the mass of the organic compound (g).

Carbon and hydrogen (Liebig combustion): the compound burns in excess $\text{O}_2$ over $\text{CuO}$ . $\text{CO}_2$ is absorbed in a KOH tube and $\text{H}_2\text{O}$ in an anhydrous $\text{CaCl}_2$ tube.

\%\text{C} = \frac{12 \times m_{\text{CO}_2} \times 100}{44 \times m} \qquad \%\text{H} = \frac{2 \times m_{\text{H}_2\text{O}} \times 100}{18 \times m}

Nitrogen: Dumas method. Heat with $\text{CuO}$ in a $\text{CO}_2$ atmosphere, pass over hot Cu gauze to reduce nitrogen oxides, collect $\text{N}_2$ over conc. KOH. Applies to all nitrogen compounds.

V_{\text{STP}} = \frac{p_1 V_1 \times 273}{760 \times T_1},\ p_1 = p_{\text{atm}} - \text{aqueous tension} \qquad \%\text{N} = \frac{28 \times V_{\text{STP}} \times 100}{22400 \times m}

Nitrogen: Kjeldahl method. Digest with conc. $\text{H}_2\text{SO}_4$ (with $\text{CuSO}_4$ catalyst and $\text{K}_2\text{SO}_4$ ) to $(\text{NH}_4)_2\text{SO}_4$ , add excess NaOH, trap $\text{NH}_3$ in a known excess of standard acid and back-titrate.

\%\text{N} = \frac{1.4 \times M \times 2\,(V - V_1/2)}{m}\ \text{ or }\ \frac{1.4 \times N \times V_{\text{acid used}}}{m}

Kjeldahl does not work for nitro, azo and ring nitrogen (pyridine, quinoline), because that nitrogen does not become $(\text{NH}_4)_2\text{SO}_4$ .

Halogens and sulphur: Carius method. Heat with fuming $\text{HNO}_3$ in a sealed hard-glass Carius tube. Halogen is weighed as AgX, sulphur as $\text{BaSO}_4$ (precipitated by excess $\text{BaCl}_2$ ).

\%\text{X} = \frac{\text{at. mass of X} \times m_{\text{AgX}} \times 100}{\text{molar mass of AgX} \times m} \qquad \%\text{S} = \frac{32 \times m_{\text{BaSO}_4} \times 100}{233 \times m}

(AgCl $= 143.5$ , AgBr $= 188$ , AgI $= 235$ .)

Phosphorus: oxidise to $\text{H}_3\text{PO}_4$ with fuming $\text{HNO}_3$ , then weigh as ammonium phosphomolybdate (molar mass 1877) or as $\text{Mg}_2\text{P}_2\text{O}_7$ (molar mass 222):

\%\text{P} = \frac{31 \times m_1 \times 100}{1877 \times m} \qquad \%\text{P} = \frac{62 \times m_{\text{Mg}_2\text{P}_2\text{O}_7} \times 100}{222 \times m}

Oxygen: normally by difference, $\%\text{O} = 100 - \sum(\text{all other elements})$ . Direct (Unterzaucher) method: pyrolysis over red-hot coke, then $\text{CO}$ oxidised by $\text{I}_2\text{O}_5$ to $\text{CO}_2$ , so $\%\text{O} = \dfrac{32 \times m_{\text{CO}_2} \times 100}{88 \times m}$ .

Method	Estimates	Weighed or measured as	Trap
Liebig	C, H	$\text{CO}_2$ (KOH), $\text{H}_2\text{O}$ ( $\text{CaCl}_2$ )	Absorb water first, then $\text{CO}_2$
Dumas	N	$\text{N}_2$ volume at STP	Subtract aqueous tension
Kjeldahl	N	$\text{NH}_3$ by back titration	Fails for $-\text{NO}_2$ , $-\text{N}=\text{N}-$ , ring N
Carius	X, S	AgX, $\text{BaSO}_4$	Heavy sealed tube

Modern laboratories use automated CHN elemental analysers that need only 1 to 3 mg of sample.

4. Empirical & Molecular Formula

Percentages from Section 3 become a formula in five steps:

Take the percentage of each element (use $100 - \sum$ for the one not measured, often oxygen) as grams in a 100 g sample.
Divide each by its atomic mass to get moles.
Divide all by the smallest value to get the simplest ratio.
If a ratio is not a whole number, multiply every term by 2, 3 or 4 (for example $1 : 1.5 \to 2 : 3$ ). The whole-number ratio gives the empirical formula.
Find $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$ , then molecular formula $= (\text{empirical formula})_n$ .

Worked example (NEET 2024). A compound has 32% A, 20% B and the rest C (A = 64, B = 40, C = 32 u). C is $100 - 52 = 48\%$ .

Element	%	Moles	Ratio
A	32	$32/64 = 0.5$	$1$
B	20	$20/40 = 0.5$	$1$
C	48	$48/32 = 1.5$	$3$

Empirical formula $= \text{ABC}_3$ .

Vapour density shortcut: molar mass $= 2 \times \text{V.D.}$
$n$ is always a whole number. If it is not, an arithmetic slip has crept in.
The two formulae coincide when $n = 1$ . Glucose has empirical formula $\text{CH}_2\text{O}$ and molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$ , so $n = 6$ .