Chapter 13 of 18

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JEE Main Chemistry★ ~1 question (4 marks) in JEE Main, usually a technique choice or a Lassaigne's test detail, and the fastest four marks in the organic section

Purification and Characterisation of Organic Compounds

Every purification technique exploits one physical property in which the compound and its impurity differ. Every detection test does the opposite job: it converts an element locked inside a covalent molecule into a simple ion that ordinary inorganic tests can find.

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Timed practiceMCQs from this chapter with instant scoring

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JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Choose a purification technique by naming the single physical property in which compound and impurity differ
2Explain steam distillation from the independent vapour pressures of immiscible liquids, and use the distillate mass ratio to find a molar mass
3Explain why several small extractions outperform one large one, and compute the fraction remaining from a partition coefficient
4Compute and interpret $R_{f}$ values, relating retention to the polarity of compound and stationary phase
5Explain why Lassaigne's fusion uses reducing sodium, write the salt formed for each element, and give the confirmatory test for each
6Identify interferences in the halogen test and the reason phosphorus and the Dumas method run in the opposite direction

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Why this chapter matters in JEE Main

This is the shortest organic chapter and the one most often skipped, which is a mistake in a paper where every question carries four marks and this one is usually answerable in under a minute. The organising principle has two halves. For purification, every technique exploits a single physical property in which the wanted compound and its impurity differ, so asking which property differs turns technique selection from memory into reasoning. For characterisation, every detection test converts an element locked inside a covalent bond into a simple ion, so that ordinary inorganic tests can find it: once you see that Lassaigne's fusion exists purely to turn covalent nitrogen into cyanide, the whole detection scheme becomes one idea applied four times.

Before you start — revise these

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Vapour pressure, Raoult's law and boiling point from Solutions

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Freezing point depression as a purity criterion, also from Solutions

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Precipitation reactions, solubility product and complex formation from inorganic analysis

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Basic titration arithmetic including back-titration

Purification and Characterisation of Organic Compounds

Aniline boils at 457 K. Water boils at 373 K. Mix them and heat.

At what temperature does the mixture boil?

Everything learned two chapters ago says a mixture boils higher than the pure solvent, because a solute lowers the vapour pressure and elevates the boiling point. So somewhere above 373 K, and probably well above.

The mixture boils at 371 K. Below water. Below aniline. Below both of its own components.

Raoult's law never applied. It describes a solution, where the components share the same liquid and each is diluted by the other's mole fraction. Aniline and water are immiscible: they form two separate layers, each with its own free surface, each evaporating exactly as if the other were not there.

$p_{t o t a l} = p_{water}^{\circ} + p_{aniline}^{\circ}$

Not $x_{1} p_{1}^{\circ} + x_{2} p_{2}^{\circ}$ . Two full pressures added, not two fractions. A sum reaches one atmosphere sooner than either part alone, so the mixture boils below both.

That is steam distillation, and it isolates heat-sensitive natural products at a temperature they can survive.

It is also the shape of the whole chapter. Every technique here works by finding a property in which the wanted compound and its impurity differ, and exploiting only that.

Half of the chapter	The one idea
Purification	Find the single physical property in which compound and impurity differ
Characterisation	Convert an element locked in a covalent bond into a simple ion that ordinary tests can find

Ask "which property differs?" and the choice of technique stops being a memory exercise. See that Lassaigne's test exists purely to turn covalent nitrogen into cyanide ion, and the whole detection scheme becomes one idea applied four times.

1. Why Purification Comes First

An organic compound from a natural source or a synthesis is almost never pure, and its melting point, boiling point, spectra and reactions are all affected by whatever came with it.

A sharp melting point is the classic purity criterion. An impure solid melts over a range and at a lower temperature, because the impurity depresses the freezing point, which is the colligative property from the Solutions chapter turning up in a practical role.

The technique chosen depends on the physical state of the compound and on which property distinguishes it from the impurity.

Property that differs	Technique
Solubility in one solvent, with temperature	Crystallisation
Tendency to sublime	Sublimation
Volatility	The distillation family
Solubility between two immiscible solvents	Differential extraction
Strength of adsorption on a stationary phase	Chromatography

2. Crystallisation

Property exploited: difference in solubility.

Dissolve the impure solid in the minimum volume of a hot solvent in which it is sparingly soluble cold and freely soluble hot. Filter hot to remove insoluble impurities, then cool slowly.

The compound crystallises while soluble impurities stay in the mother liquor. Slow cooling gives larger, purer crystals, because rapid cooling traps impurity inside the growing lattice.

Choosing the solvent is the whole skill: the compound needs a steep solubility curve with temperature, and the impurity must be either freely soluble at all temperatures or insoluble at all of them.

Where two compounds have similar solubilities, fractional crystallisation repeats the process so the less soluble one separates first. Coloured impurities are removed by boiling with a little activated charcoal, which adsorbs them, before filtering.

Illustration 1

Benzoic acid dissolves to 6.8 g per 100 mL in water at 95 °C and to 0.34 g per 100 mL at 25 °C. A 5.0 g sample carrying 0.5 g of a freely soluble salt is crystallised from the minimum volume of hot water. Find the volume needed, the mass recovered and the yield.

The minimum volume is whatever just dissolves the benzoic acid at 95 °C:

$V = \frac{5.0}{6.8} \times 100 = 74 mL$

On cooling to 25 °C that same 74 mL can still hold

$0.34 \times \frac{74}{100} = 0.25 g$

so the crystals recovered weigh $5.0 - 0.25 = 4.75$ g, a yield of 95 per cent. The salt stays in the mother liquor throughout, being freely soluble at both temperatures.

Now see why "minimum volume" is an instruction rather than a stylistic preference. Use 150 mL instead of 74 and the cold liquor retains 0.51 g, dropping the yield to 90 per cent. Every extra millilitre of solvent is product left behind, and the loss grows in strict proportion to the excess.

3. Sublimation

Property exploited: passing directly from solid to vapour.

Only a few organic solids sublime, and that is exactly what makes the technique useful, because the impurities almost never do.

Camphor, naphthalene, anthracene and benzoic acid all sublime on gentle heating. The vapour condenses on a cool surface, leaving non-volatile impurities behind.

4. The Distillation Family

Property exploited: difference in volatility. Four variants for four situations.

Technique	Used when	Example
Simple distillation	Boiling points differ by more than about 25 K	Chloroform from aniline
Fractional distillation	Boiling points are close	Petroleum fractions, acetone from methanol
Reduced pressure	The compound decomposes before boiling	Glycerol, which chars at its normal boiling point
Steam distillation	Volatile in steam and immiscible with water	Aniline, nitrobenzene, essential oils

Fractional distillation gives the vapour many chances to condense and re-evaporate up a packed column, each cycle enriching it further in the more volatile component. That is the Raoult's law enrichment of the Solutions chapter applied repeatedly.

Reduced pressure works because boiling occurs when vapour pressure equals external pressure. Glycerol boils at 563 K at atmospheric pressure, where it also chars, and near 453 K under reduced pressure, where it does not.

Illustration 2

Choose a purification technique for each of these, and name the property being exploited.

(a) A solid that chars at its melting point but passes straight to vapour on gentle warming.
(b) A liquid boiling at 351 K contaminated with one boiling at 373 K.
(c) Glycerol, which decomposes at its normal boiling point.
(d) Aniline, immiscible with water and steam-volatile, in an aqueous mixture.
(e) An otherwise pure solid carrying a coloured impurity.

(a) Sublimation. The compound sublimes and the impurities do not, so no other property has to differ at all.

(b) Fractional distillation. The gap is 22 K, below the roughly 25 K that simple distillation needs, so a packed column is required to repeat the enrichment.

(c) Distillation under reduced pressure, which lowers the boiling point beneath the decomposition temperature — near 453 K reduced against 563 K at atmospheric.

(d) Steam distillation, which brings it over below 373 K.

(e) Crystallisation, after boiling briefly with activated charcoal to adsorb the colour.

Only (b) and (c) both turn on volatility, and even they fail differently: in (b) two things boil too close together, in (c) one thing boils too high to survive the journey. Naming the property before naming the technique makes these nearly automatic.

Illustration 3

Steam distillation of a compound gives a distillate containing 4.0 g of the organic compound for every 1.0 g of water. At the distillation temperature the vapour pressure of water is 733 mmHg and that of the compound is 27 mmHg. Find the compound's molar mass.

In the vapour, the two gases are in the same container at the same temperature, so their partial pressures are proportional to their mole numbers.

$\frac{n _{org}}{n _{water}} = \frac{p _{org}}{p _{water}} = \frac{27}{733}$

Converting moles to masses introduces the molar masses.

$\frac{w _{org} / M _{org}}{w _{water} /18} = \frac{27}{733} \Rightarrow \frac{4.0/ M}{1.0/18} = \frac{27}{733}$

$\frac{72}{M} = 0.0368 \Rightarrow M = 1955 g mo l^{- 1}$

That is not a misprint, and it is the point of the technique. The compound carries over at a rate 4 times water by mass while exerting only $27/733$ of the pressure, which is possible only because each of its molecules is very heavy.

Steam distillation therefore moves large, fragile, barely volatile molecules at 371 K, and the mass ratio in the receiver is a measurement, not just a yield.

5. Differential Extraction

Property exploited: difference in solubility between two immiscible solvents.

An organic compound dissolved in water is shaken in a separating funnel with an immiscible solvent such as ether in which it is more soluble. It transfers to the ether layer, which is run off and evaporated.

Illustration 4

A compound has a partition coefficient of 4 between ether and water, meaning it is four times as concentrated in ether at equilibrium. Starting with 1.00 g in 90 mL of water, compare extracting once with 90 mL of ether against three times with 30 mL each.

One extraction with 90 mL. Equal volumes, so the amounts split in the ratio 4:1 and $4/5 = 80$ per cent transfers. Remaining in water: 0.200 g.

Three extractions with 30 mL. With ether at one third the volume of water, the fraction remaining each time is

$fraction left = \frac{V _{w}}{V _{w} + K V _{e}} = \frac{90}{90 + 4 ( 30 )} = \frac{90}{210} = 0.4286$

After three rounds, $0.428 6^{3} = 0.0787$ , so 0.0787 g remains and 92.1 per cent is recovered.

Same solvent, same total volume, and the split method leaves less than half as much behind.

The reason is in the exponent. Each extraction removes a fixed fraction of whatever is left, so repeating the operation compounds the removal, while pouring all the solvent in at once buys only one round of it. Three small portions beat one large portion, and this is a general result rather than a laboratory superstition.

6. Chromatography

Property exploited: difference in how strongly components are held by a stationary phase while a mobile phase carries them along.

The name comes from the Greek for colour, since the technique was first used on plant pigments, but it now applies to anything.

Two mechanisms

Adsorption chromatography uses a solid stationary phase such as silica gel or alumina, and separation depends on how strongly each component sticks to the surface. Column and thin layer chromatography both work this way.

Partition chromatography uses a liquid stationary phase held on an inert support, and separation depends on how each component distributes between two liquids. Paper chromatography works this way, the stationary liquid being the water held in the cellulose fibres.

Running a column

In column chromatography the adsorbent is packed into a vertical tube, the mixture applied at the top, and solvent run through continuously. Components move down at different speeds according to how strongly they are adsorbed, and the least strongly adsorbed component leaves first.

Thin layer chromatography is the same principle in miniature, with the adsorbent as a thin film and the solvent rising by capillary action. It checks purity and follows a reaction's progress, since a single spot means a single component.

Reading a thin layer plate

$R_{f} = \frac{distance travelled by the component}{distance travelled by the solvent front}$

$R_{f}$ lies between 0 and 1 and is characteristic of a compound in a given solvent and stationary phase. A more strongly adsorbed component moves less and has a smaller $R_{f}$ ; one that does not move at all has $R_{f} = 0$ and is too strongly held for that solvent.

Illustration 5

Two compounds are run on a silica plate in a non-polar solvent. Compound A gives $R_{f} = 0.15$ and compound B gives $R_{f} = 0.80$ . One is benzoic acid and the other is naphthalene. Which is which, and what would happen in a more polar solvent?

Silica is a polar stationary phase, so it holds polar compounds tightly and lets non-polar ones travel.

Benzoic acid has a carboxyl group that hydrogen bonds strongly to silica's surface hydroxyls, so it is heavily retained and barely moves: $R_{f} = 0.15$ is benzoic acid. Naphthalene is a plain aromatic hydrocarbon with nothing to grip the surface, so the solvent carries it freely: $R_{f} = 0.80$ is naphthalene.

Increase the solvent's polarity and both $R_{f}$ values rise, because the mobile phase now competes with silica for the polar sites and can pull compounds off the surface. Benzoic acid gains most, since it had the most to gain.

Illustration 6

A silica column is loaded with a mixture of naphthalene, nitrobenzene and benzoic acid. Give the order in which they leave the column, and explain why the eluting solvent is usually made progressively more polar.

Silica retains by polarity, so the least polar compound is held most weakly and travels fastest:

$naphthalene \to nitrobenzene \to benzoic acid$

Naphthalene is a plain hydrocarbon and elutes first. Nitrobenzene is polar but cannot hydrogen bond to the surface hydroxyls. Benzoic acid can, so it is held hardest and appears last.

Note the inversion against a plate. On TLC the strongly held compound has the lowest $R_{f}$ ; on a column it is the last to emerge. Both statements say the same thing — strongly adsorbed means slow — but the reported quantity flips, and running them together reverses the answer.

Raising the solvent's polarity as the run proceeds, called gradient elution, releases each band in turn rather than waiting for the most retained one to crawl off in the original solvent. It sharpens the bands and shortens the run.

Trap. A high $R_{f}$ is not a good result and a low one is not a failure. Both are only measurements. A separation is good when the two spots are far apart, which is why choosing the solvent means tuning the gap rather than maximising travel.

7. Qualitative Analysis: The Central Problem

Detecting carbon and hydrogen

A known mass is heated with dry copper(II) oxide, which oxidises carbon to carbon dioxide and hydrogen to water. The gases pass through anhydrous calcium chloride, absorbing the water, then through lime water, which turns milky if carbon dioxide is present.

This is straightforward precisely because carbon and hydrogen become familiar inorganic products in one step. No such route exists for the rest, which is the problem the remainder of this section solves.

The difficulty: nitrogen, sulphur, phosphorus and the halogens are held by covalent bonds. They are not ions, so no ionic test can find them. Adding silver nitrate to chlorobenzene produces nothing whatsoever.

Lassaigne's test solves this by converting the covalent element into an ionic one. Sodium is fused with the compound and the red-hot mixture plunged into distilled water. Sodium, violently reducing at that temperature, breaks the covalent bonds and forms simple sodium salts.

Element present	Product formed
Nitrogen	NaCN
Sulphur	$N a_{2} S$
Nitrogen and sulphur together	NaSCN
Halogen	NaX
Phosphorus	$N a_{3} P O_{4}$

Boiling and filtering gives the sodium fusion extract, alkaline and containing all of these as free ions. Every subsequent test is ordinary inorganic analysis performed on it.

Illustration 7

Why sodium in particular? Why not fuse with potassium, or with magnesium, or simply burn the compound in oxygen and analyse the products?

Three demands must be met at once, and sodium is the only convenient thing that meets all three.

Requirement	Why it matters
Powerfully reducing	Must convert covalent N to $N^{3 -}$ -derived cyanide and covalent halogen to halide, which oxidation could never do
Products must be water-soluble	The extract has to carry every element as a free ion into aqueous solution
Must be safe enough to handle and cheap	Potassium works chemically but ignites too readily to be a teaching reagent

Burning in oxygen fails on the first count and fails badly. Oxidation sends nitrogen to $N_{2}$ gas, which escapes and reacts with nothing, and sends sulphur to $S O_{2}$ , which also leaves. Reduction is the only direction that gives ions you can keep in a beaker.

Magnesium fails on the second: many magnesium salts are sparingly soluble, so the elements would never reach solution.

So the choice of sodium is not tradition. Reduce, do not oxidise, and make everything soluble is the entire design of the test, and every step downstream depends on it.

8. The Individual Tests

Nitrogen

Add iron(II) sulphate to the extract and boil, then acidify with sulphuric acid.

Cyanide first forms hexacyanidoferrate(II), and iron(II) partly oxidises to iron(III) in air. The two combine to give Prussian blue, $F e_{4} [Fe (CN)_{6}]_{3}$ , and the blue or green colouration confirms nitrogen.

Sulphur

Two independent tests, either accepted. Sodium nitroprusside gives a deep violet colouration. Or acidify with acetic acid and add lead acetate for a black precipitate of lead sulphide.

Nitrogen and sulphur together

If both are present in the same molecule, sodium fusion may give sodium thiocyanate rather than separate cyanide and sulphide.

Adding iron(III) then gives a blood-red colouration from the thiocyanate complex, not Prussian blue. This is a useful diagnostic rather than a nuisance, since the red colour reports both elements at once.

Using excess sodium decomposes the thiocyanate into cyanide and sulphide, and the two separate tests then behave normally.

Halogens

Boil the extract with dilute nitric acid, then add silver nitrate.

Halide	Precipitate	Behaviour with ammonia
Chloride	White	Freely soluble
Bromide	Pale yellow	Sparingly soluble
Iodide	Yellow	Insoluble

Illustration 8

A student adds silver nitrate directly to the fusion extract of a compound containing both chlorine and nitrogen, and reports a white precipitate. What went wrong, and how would you know?

The nitric acid boiling was skipped, and it is not optional.

The extract contains cyanide as well as chloride. Silver cyanide, AgCN, is also a white precipitate, so the observation is real but the interpretation is worthless: nothing distinguishes it from silver chloride by eye.

Had sulphur been present too, silver sulphide would have appeared as a black precipitate and masked everything.

Boiling with dilute nitric acid first expels the interferents as gases.

$NaCN + HN O_{3} \to NaN O_{3} + HCN ↑, N a_{2} S + 2HN O_{3} \to 2NaN O_{3} + H_{2} S ↑$

Only halide survives that treatment, so only halide can precipitate afterwards.

The way to catch the error is the ammonia test. Genuine silver chloride dissolves freely in ammonia; a mixture contaminated with silver cyanide behaves inconsistently. But the real lesson is that the interference was predictable from the compound's own composition, and a question naming both elements is usually testing exactly this step.

Phosphorus

Heat with sodium peroxide, which oxidises phosphorus to phosphate. Adding ammonium molybdate in nitric acid gives a yellow precipitate of ammonium phosphomolybdate.

Note the reversal. Phosphorus is the one element detected by oxidation rather than reduction, because phosphate is a stable, soluble, easily tested anion while a phosphide would be neither.

9. A Note on Quantitative Estimation

Once the elements are known, their proportions give the empirical formula.

Carbon and hydrogen are estimated together by burning a known mass in oxygen and weighing the carbon dioxide and water absorbed. Nitrogen goes by the Dumas method, measuring the volume of nitrogen gas released, or the Kjeldahl method, converting it to ammonium sulphate and titrating the ammonia liberated. Halogens, sulphur and phosphorus go by the Carius method, heating with fuming nitric acid and silver nitrate in a sealed tube.

$% C = \frac{12}{44} \times \frac{mass of CO _{2}}{mass of compound} \times 100$

$% H = \frac{2}{18} \times \frac{mass of H _{2} O}{mass of compound} \times 100$

Illustration 9

0.246 g of a compound containing only carbon, hydrogen and oxygen gave 0.361 g of carbon dioxide and 0.148 g of water on complete combustion. Find its empirical formula.

Carbon and hydrogen come straight from the two formulas above:

$% C = \frac{12}{44} \times \frac{0.361}{0.246} \times 100 = 40.0%, % H = \frac{2}{18} \times \frac{0.148}{0.246} \times 100 = 6.7%$

Oxygen is never measured directly. It is whatever is left over:

$% O = 100 - 40.0 - 6.7 = 53.3%$

Divide each percentage by the atomic mass, then by the smallest of the three:

$C : \frac{40.0}{12} = 3.33, H : \frac{6.7}{1} = 6.7, O : \frac{53.3}{16} = 3.33$

$3.33 : 6.7 : 3.33 \Rightarrow 1 : 2 : 1 \Rightarrow C H_{2} O$

And now the limitation, which is the reason this is only an empirical formula. Formaldehyde, acetic acid and glucose all return $C H_{2} O$ , because combustion measures ratios and nothing else. Fixing the molecular formula needs a molar mass from somewhere outside the combustion — which is precisely what the steam distillation in section 4 was quietly providing.

Illustration 10

0.75 g of an organic compound was digested by the Kjeldahl method, and the ammonia liberated was absorbed in 50 mL of 0.5 M sulphuric acid. The excess acid then required 80 mL of 0.5 M sodium hydroxide. Find the percentage of nitrogen.

Work in milliequivalents, since sulphuric acid is dibasic and the arithmetic is otherwise easy to botch:

$acid taken = 50 \times 0.5 \times 2 = 50 meq, excess left = 80 \times 0.5 \times 1 = 40 meq$

$acid neutralised by ammonia = 50 - 40 = 10 meq \Rightarrow 10 mmol NH_{3}$

Each mole of ammonia carries one mole of nitrogen:

$% N = \frac{( 10 \times 1 0 ^{- 3} ) ( 14 )}{0.75} \times 100 = 18.7%$

The trap is that factor of two. Fifty millilitres of 0.5 M sulphuric acid is 25 mmol of acid but 50 meq of neutralising power, and carrying 25 forward halves the answer. Working in equivalents throughout keeps the dibasic acid and the monobasic alkali on one footing.

Illustration 11

Why is the Kjeldahl method useless for nitro compounds and azo compounds, when it works perfectly for proteins and amines?

Kjeldahl digests the compound with concentrated sulphuric acid, which converts nitrogen to ammonium sulphate. Alkali then liberates ammonia, which is trapped in standard acid and back-titrated.

Every step assumes the nitrogen can be reduced to ammonia. That holds when nitrogen starts in a negative oxidation state, as in an amine or an amide, where it is already close to ammonia and only needs the carbon skeleton stripped away.

In a nitro group the nitrogen is at $+ 3$ , and in an azo group it is at $- 1$ but locked in an $N = N$ bond. Sulphuric acid is not a reducing agent, so it cannot bring either down to ammonium, and part of the nitrogen escapes as gas instead. The result is a systematic underestimate.

Those compounds are done by the Dumas method, which oxidises everything to $N_{2}$ and measures the volume of gas, so the nitrogen's starting oxidation state does not matter.

The pattern is the same one this chapter keeps repeating. Match the direction of the conversion to where the element starts, whether choosing sodium fusion over combustion, oxidation for phosphorus, or Dumas over Kjeldahl.

The JEE Main unit names purification techniques and the qualitative detection of nitrogen, sulphur, phosphorus and halogens. Quantitative estimation is worth understanding in principle, since it is where empirical formulas come from, but the detection tests are what questions are set on.

Summary

Every purification technique exploits one physical property in which compound and impurity differ: solubility for crystallisation, sublimation tendency, volatility for the distillation family, partition between immiscible solvents for extraction, and strength of adsorption for chromatography.

Steam distillation works because immiscible liquids each exert their full vapour pressure independently, so the total exceeds either one and the mixture boils below both. Raoult's law does not apply, because there is no solution. The mass ratio in the distillate measures the molar mass.

Three small extractions beat one large one, because each removes a fixed fraction and repetition compounds it: a partition coefficient of 4 gives 80 per cent in one 90 mL portion and 92 per cent in three 30 mL portions.

$R_{f}$ is the ratio of distances travelled by component and solvent front. Silica is polar, so polar compounds are retained and travel less, and a good separation means spots far apart rather than spots far up the plate.

Covalently bound nitrogen, sulphur, phosphorus and halogen give no ionic test, so Lassaigne's fusion with sodium reduces them into soluble salts. Sodium is chosen because it is powerfully reducing, gives soluble products and is safe enough to handle; oxidation would send nitrogen and sulphur away as gases.

Nitrogen gives Prussian blue with iron(II) then acid, sulphur gives violet with nitroprusside or black with lead acetate, and both together give blood-red thiocyanate unless excess sodium was used.

Halogens need the extract boiled with dilute nitric acid first, to expel cyanide and sulphide that would otherwise give white silver cyanide and black silver sulphide. Phosphorus is the exception detected by oxidation, to phosphate, tested with ammonium molybdate.

Kjeldahl works only where nitrogen can be reduced to ammonia, so nitro and azo compounds must go by Dumas instead. The recurring principle is to match the direction of conversion to where the element starts.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The organising principle

purification exploits one differing physical property; characterisation converts a covalent element into an ion

Ask which property differs and technique choice stops being memory. See that Lassaigne's fusion exists to make cyanide, and the whole detection scheme is one idea applied four times.

Purity criterion

a pure solid has a **sharp** melting point; an impurity lowers and broadens it

This is freezing point depression from the Solutions chapter appearing in a practical role, and it is why a melting range rather than a melting point is itself the evidence of impurity.

Steam distillation

$p_{total} = p^\circ_{\text{water}} + p^\circ_{\text{organic}}$, not $x_1p_1^\circ + x_2p_2^\circ$

Immiscible liquids form separate layers, each evaporating as if the other were absent, so two **full** pressures add. The sum reaches 1 atm below either boiling point: aniline distils at 371 K rather than 457 K.

Molar mass from a steam distillate

$\dfrac{w_{\text{org}}/M_{\text{org}}}{w_{\text{water}}/18} = \dfrac{p_{\text{org}}}{p_{\text{water}}}$

Partial pressures in the shared vapour are proportional to mole numbers. A compound carrying over at 4 times water by mass on only $27/733$ of the pressure must have a molar mass near 2000.

Differential extraction

fraction left each time $= \dfrac{V_w}{V_w + KV_e}$

Each extraction removes a fixed fraction, so repetition compounds it. With $K = 4$, one 90 mL portion recovers 80 per cent while three 30 mL portions recover 92 per cent, from the same total solvent.

Choosing a distillation

simple, fractional, reduced pressure, steam

Boiling points more than about 25 K apart, close together, decomposition before boiling, and volatile-in-steam-and-immiscible respectively. Glycerol boils at 563 K and chars, but near 453 K under reduced pressure it does not.

Retardation factor

$R_f = \dfrac{\text{distance travelled by component}}{\text{distance travelled by solvent front}}$

Always between 0 and 1. Silica is polar, so polar compounds are held and travel less. A good separation means spots far **apart**, not spots far up the plate.

Why sodium fusion

powerfully **reducing**, water-soluble products, safe enough to handle

Oxidation would send nitrogen away as $\mathrm{N_2}$ and sulphur as $\mathrm{SO_2}$, so nothing stays in the beaker. Magnesium fails on solubility and potassium on safety.

Fusion products

N gives NaCN, S gives $\mathrm{Na_2S}$, both together give NaSCN, halogen gives NaX, P gives $\mathrm{Na_3PO_4}$

The extract is alkaline and holds all of these as free ions. Every subsequent test is ordinary inorganic analysis performed on that one solution.

Confirmatory tests

N: Prussian blue; S: violet with nitroprusside or black with lead acetate; N and S: **blood-red** thiocyanate

Blood-red rather than blue reports both elements at once. Repeating the fusion with excess sodium breaks the thiocyanate into cyanide and sulphide so the separate tests behave normally.

Halogen test and its interference

**boil with dilute $\mathrm{HNO_3}$ first**, then add $\mathrm{AgNO_3}$

Silver cyanide is also white and silver sulphide is black, so both would ruin the reading. Boiling expels them as HCN and $\mathrm{H_2S}$. AgCl is white and freely ammonia-soluble, AgBr pale yellow and sparingly, AgI yellow and insoluble.

Estimation formulas and their limits

$\%\mathrm{C} = \dfrac{12}{44}\dfrac{m_{\mathrm{CO_2}}}{m}\times100, \qquad \%\mathrm{H} = \dfrac{2}{18}\dfrac{m_{\mathrm{H_2O}}}{m}\times100$

Oxygen is always found by difference, never measured. Kjeldahl needs nitrogen reducible to ammonia, so nitro and azo compounds must go by Dumas, which oxidises everything to $\mathrm{N_2}$ instead.

Crystallisation recovery

w_{r eco v er e d} = w_{t ak e n} - s_{co l d} \times \frac{V}{100}, V_{min} = \frac{w _{t ak e n}}{s _{h o t}} \times 100

The mother liquor keeps whatever the cold solvent can still hold, so loss is directly proportional to the volume used. Dissolving 5.0 g of benzoic acid in the minimum 74 mL gives a 95 per cent yield; using 150 mL drops it to 90. That is why "minimum volume of hot solvent" is an instruction, not a stylistic preference.

JEE Main test series

Ready to be tested on this?

Purification and Characterisation of Organic Compounds shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~1Est. Qs

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Chemistry-only sectionals & chapter tests
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⚠️

Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Expecting a mixture to boil above its components

✓ That rule assumes a solution, where each component is diluted by the other's mole fraction. Immiscible liquids form separate layers and each exerts its full pure vapour pressure, so the two add rather than average and the total reaches 1 atm below either boiling point. Aniline and water distil at 371 K, below both 457 K and 373 K, which is the entire basis of steam distillation.

Why it happens: Boiling point elevation is drilled hard in the Solutions chapter and looks like a general rule.

WATCH OUT

✗ Adding silver nitrate to the fusion extract without boiling with nitric acid

✓ If nitrogen is present the extract also contains cyanide, and silver cyanide is white, so a white precipitate proves nothing. If sulphur is present, black silver sulphide masks everything. Boiling with dilute nitric acid expels both as HCN and

H_{2} S

gas, after which only halide can precipitate. A question naming nitrogen or sulphur alongside a halogen is usually testing exactly this step.

Why it happens: The halogen test looks like an ordinary precipitation and the extra step feels like a formality.

WATCH OUT

✗ Treating a high

R_{f}

as a good result

✓

R_{f}

is a measurement, not a score. A separation succeeds when the spots are far apart, which usually means somewhere in the middle of the plate for both. Two compounds both at

R_{f} = 0.9

have not been separated at all. Choosing the solvent means tuning the gap between components, and raising the solvent's polarity raises every

R_{f}

together without necessarily improving anything.

Why it happens: A spot near the solvent front looks like a successful, well-developed plate.

WATCH OUT

✗ Assuming a single large extraction is as good as several small ones

✓ Each extraction removes a fixed fraction of whatever remains, and repeating compounds that removal. With a partition coefficient of 4 and 90 mL of aqueous solution, one 90 mL ether portion leaves 0.200 g of 1.00 g behind while three 30 mL portions leave only 0.079 g. The fraction remaining is

[V_{w} / (V_{w} + K V_{e})]^{n}

, and the exponent is what does the work.

Why it happens: The same total volume of solvent is used, so it seems the same amount of extracting is done.

WATCH OUT

✗ Using Kjeldahl for any nitrogen-containing compound

✓ Kjeldahl assumes the nitrogen can be reduced to ammonium by digestion with sulphuric acid, which holds when nitrogen starts in a negative oxidation state. In a nitro group nitrogen is at

+ 3

and in an azo group it is locked in

N = N

, and sulphuric acid is not a reducing agent, so part of the nitrogen escapes as gas and the result is a systematic underestimate. Those compounds go by Dumas, which oxidises everything to

N_{2}

and measures the volume.

Why it happens: It is the method taught in most detail and works beautifully for proteins and amines.

WATCH OUT

✗ Expecting phosphorus to be found by sodium fusion like the rest

✓ Phosphorus is detected by oxidation with sodium peroxide to phosphate, then ammonium molybdate for a yellow precipitate. The reason is that phosphate is a stable, soluble, easily tested anion while a phosphide would be neither. The recurring principle is to match the direction of conversion to where the element starts, which is the same reason sodium fusion reduces and the Dumas method oxidises.

Why it happens: Nitrogen, sulphur and the halogens all go through Lassaigne's extract, so phosphorus looks like it should too.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Purification and Characterisation of Organic Compounds?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~4 marks in JEE Main exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Purification exploits one differing property; characterisation converts a covalent element into an ion
•A sharp melting point is the purity criterion, because impurity depresses and broadens it
•Steam distillation adds two full vapour pressures, so the mixture boils below both components; Raoult's law needs a solution
•Reduced pressure for compounds that decompose; fractional for close boiling points; simple for a gap above 25 K
•Fraction left per extraction is $V_{w} / (V_{w} + K V_{e})$ , so three small portions beat one large one
• $R_{f}$ is between 0 and 1; silica is polar so polar compounds travel less; good separation means spots far apart
•Sodium fusion is chosen for being reducing, giving soluble products and being safe; oxidation would lose N and S as gases
•N gives NaCN, S gives $N a_{2} S$ , both give NaSCN, halogen gives NaX, P gives phosphate
•Prussian blue for N, violet or black for S, blood-red for both together unless excess sodium was used
•Boil with dilute $HN O_{3}$ before silver nitrate: AgCN is white and $A g_{2} S$ is black and both ruin the test
•AgCl white and ammonia-soluble, AgBr pale yellow and sparingly, AgI yellow and insoluble
•Phosphorus is detected by oxidation; Kjeldahl needs reducible nitrogen so nitro and azo compounds go by Dumas

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: ~1 question (4 marks) of the 100-mark Chemistry section

Question style	Marks each	Typical count	What it tests
Lassaigne's test and element detection	1	1	Why sodium fusion is needed at all, the fusion products for each element, confirmatory colours and precipitates, and the cyanide and sulphide interference that makes boiling with nitric acid compulsory before the halogen test
Quantitative estimation	1	1	Percentage carbon and hydrogen from combustion masses with oxygen by difference, empirical formula from those percentages, Kjeldahl back-titration in equivalents, and why Kjeldahl fails for nitro and azo compounds
Purification techniques	1	1	Matching the technique to the property that differs, the 25 K rule for simple against fractional distillation, reduced pressure for compounds that decompose, crystallisation yield and why minimum volume matters, and molar mass from a steam distillate
Chromatography	1	1	Computing $R_f$ and ranking spots by polarity on silica, the inversion between plate position and column elution order, and why gradient elution sharpens bands

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

For a technique-choice question, name the differing property first and the technique second. The property is the mark, and it also protects against picking a plausible but wrong method.
Whenever a compound contains nitrogen or sulphur alongside a halogen, expect the question to be about boiling with dilute nitric acid before adding silver nitrate.
A blood-red colour with iron(III) is thiocyanate and reports nitrogen and sulphur together. Do not read it as a failed Prussian blue test.
Remember phosphorus runs the other way, by oxidation with sodium peroxide, and that Dumas replaces Kjeldahl whenever the nitrogen is not reducible to ammonia.
In Kjeldahl arithmetic, track the dibasic nature of sulphuric acid twice: once in converting the back-titration to moles of unreacted acid, and again in converting acid consumed to moles of ammonia.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Essential oils

Essential oils, from rose and eucalyptus to clove, are extracted commercially by steam distillation, because the fragrant compounds would decompose long before reaching their own boiling points and the immiscible-liquid trick brings them over at 371 K instead

Thin layer chromatography is how a synthetic chemist know…

Thin layer chromatography is how a synthetic chemist knows a reaction has finished: one spot at the starting material's $R_{f}$ means no progress, one spot at a new $R_{f}$ means completion, and the check takes about three minutes

The Kjeldahl method determines protein content in food by…

The Kjeldahl method determines protein content in food by measuring nitrogen and multiplying by a conversion factor, which is exactly why melamine adulteration of milk powder went undetected for so long: it is nitrogen-rich and the test cannot tell where the nitrogen came from

Where else this topic is tested

Prepare once, score in every exam that asks it.

JEE Main

JEE Advanced

NEET UG

BITSAT

CBSE Class 11 Chemistry

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because boiling point elevation describes a solution, and steam distillation deliberately avoids one. In a solution the components share a single liquid phase, so each is diluted by the other's mole fraction and contributes only $x_{i} p_{i}^{\circ}$ to the vapour, which is why the total lies between the two pure values. Aniline and water are immiscible: they form separate layers, each with its own free surface, and each evaporates exactly as if the other were not present. The total vapour pressure is therefore $p_{water}^{\circ} + p_{aniline}^{\circ}$ , a sum of two full pressures rather than a weighted average, and a sum reaches one atmosphere at a lower temperature than either term alone. The two statements are not in conflict; they apply to different physical situations.

Because the direction of the conversion matters more than the fact of it. Burning oxidises, and oxidation sends nitrogen to $N_{2}$ gas, which escapes and reacts with nothing, and sulphur to $S O_{2}$ , which also leaves. Nothing stays in the beaker to be tested. Sodium at red heat is powerfully reducing, so it drives nitrogen down to cyanide, sulphur to sulphide and halogen to halide, all of which are stable water-soluble ions that ordinary inorganic tests handle routinely. Potassium would work chemically but ignites too readily for teaching use, and magnesium fails because many of its salts are sparingly soluble and would never reach solution. Phosphorus is the deliberate exception, done by oxidation, because phosphate is the stable testable form.

No, and this is worth being clear about. $R_{f}$ measures how far a spot travelled relative to the solvent, and it tells you about retention, not about quality. Two compounds both running at 0.9 have not been separated at all, and neither have two at 0.05. What a separation needs is a large gap between the spots, which in practice usually means getting both somewhere near the middle of the plate. Raising the solvent's polarity lifts every $R_{f}$ together, because the mobile phase competes better with silica for polar sites, so it can rescue a plate where everything is stuck at the baseline but will not help one where everything is already at the front. Choosing a solvent is tuning the gap.

Because there is no clean way to convert organic oxygen into a single weighable product. Carbon becomes carbon dioxide and hydrogen becomes water, and both are trapped and weighed directly. But combustion is carried out in oxygen, so any oxygen liberated from the compound is indistinguishable from the oxygen supplied, and the products already contain far more oxygen than the sample ever held. Halogens, nitrogen, sulphur and phosphorus each have their own dedicated method. Oxygen has none in routine use, so the standard practice is to determine every other element and subtract the total from 100 per cent. The consequence is that any error in the other determinations lands entirely on the oxygen figure.

Less than the detection tests, and the JEE Main unit reflects that: it names purification techniques and the qualitative detection of nitrogen, sulphur, phosphorus and halogens explicitly. Numerical questions do appear, most often a Kjeldahl back-titration or a carbon and hydrogen combustion, and both are worth being fluent in because they are mechanical marks. What is more often tested is the reasoning around the methods: why nitric acid boiling precedes the halogen test, why phosphorus is oxidised rather than reduced, and why a nitro compound cannot go by Kjeldahl. Those questions carry the same four marks and take less time than the arithmetic.