Purification of organic compounds

Purification methods are not a random list to memorise; they fall out of asking one question about a mixture: which physical property separates the compound from its impurities, and how cleanly?

• Crystallisation — solubility. The impure solid is dissolved in the minimum volume of a hot solvent, filtered hot to remove insoluble matter, then cooled slowly. Because most solids are less soluble in cold solvent than in hot, the desired compound crystallises out first while soluble impurities remain dissolved in the mother liquor. It works best when the compound's solubility changes sharply with temperature and the impurity's does not.
• Sublimation — direct solid-to-vapour transition. A few solids (naphthalene, camphor, benzoic acid, iodine) pass straight from solid to vapour on gentle heating, without an intervening liquid phase, and resolidify on a cool surface. Non-sublimable impurities are simply left behind.
• Distillation — boiling point (volatility). All four variants exploit the same property but handle different constraints:
• Simple distillation separates a volatile liquid from a non-volatile or much-higher-boiling impurity, generally when the boiling-point gap exceeds about $25\ \text{K}$.
• Fractional distillation separates two miscible liquids whose boiling points are close, using a fractionating column that behaves like many simple distillations stacked in series.
• Steam distillation suits a compound that is steam-volatile, insoluble in water, and thermally sensitive — it distils over below $373\ \text{K}$, well under its own boiling point.
• Vacuum distillation lowers the external pressure so a high-boiling liquid reaches its (now lower) boiling point at a much lower temperature, again protecting a thermally unstable compound.
• Differential extraction — relative solubility in two immiscible solvents. Shaking the mixture with a solvent such as ether in a separating funnel pulls the organic solute preferentially into the organic layer, which is drawn off from the aqueous layer.
• Chromatography — rate of adsorption or partition. A mobile phase carries the mixture over a stationary phase (alumina in a column, or paper/a thin adsorbent layer); components that adsorb more weakly or partition more into the mobile phase travel farther/faster, spreading the mixture into separated spots or bands.

The one technique in this list with a genuine quantitative handle at this level is steam distillation, because "why does it boil below $373\ \text{K}$" has a precise answer in Dalton's law of partial pressures — worked out below.

The claim that a steam-volatile liquid distils over "below $373\ \text{K}$" is not a rule to accept on faith; it follows directly from how the total vapour pressure of two immiscible, mutually insoluble liquids adds up, and working through it explains why steam distillation protects a thermally sensitive compound that simple distillation would decompose.

Step 1. Two liquids that do not dissolve in each other, such as water and most organic liquids used in steam distillation, each generate their own vapour pressure at a given temperature independently of the other — neither liquid's presence lowers or raises the other's individual vapour pressure, because they are not in solution together.

Step 2. By Dalton's law of partial pressures applied to this pair, the total vapour pressure above the mixture is simply the sum of the two individual (pure-liquid) vapour pressures at that temperature:

where: $P_{\text{total}}$ is the combined vapour pressure of the boiling mixture, $p_{\text{water}}$ is the vapour pressure pure water would have alone at that temperature, and $p_{\text{organic}}$ is the vapour pressure the pure organic liquid would have alone at that same temperature.

Step 3. A liquid (or liquid mixture) boils at the temperature where its total vapour pressure equals the surrounding atmospheric pressure, $P_{\text{total}} = P_{\text{atm}} \approx 101.3\ \text{kPa}$. Because two vapour pressures are now adding up rather than one, the mixture reaches $101.3\ \text{kPa}$ at a temperature lower than either pure liquid's own boiling point — in particular lower than $373\ \text{K}$ (water's own boiling point) and far lower than a high-boiling organic liquid's own boiling point.

Step 4. Concretely: at any temperature below $373\ \text{K}$, pure water alone has $p_{\text{water}} < 101.3\ \text{kPa}$ (it has not yet reached its own boiling point), so it cannot boil by itself. But if the organic liquid contributes even a modest extra $p_{\text{organic}}$ at that same temperature, the sum can reach $101.3\ \text{kPa}$ well before water would have boiled on its own — and long before the organic liquid, heated alone, would have reached its own (much higher) boiling point.

Step 5. This is why the technique protects thermally sensitive compounds: sustained heating near a high boiling point (or above a decomposition threshold) is never required, because co-distillation with steam lets the mixture boil at whichever lower temperature makes the two partial pressures sum to atmospheric pressure — always below $373\ \text{K}$ for any steam-volatile, water-insoluble organic liquid.

Problem. A mixture contains chlorobenzene (own boiling point $405\ \text{K}$, insoluble in water, steam-volatile) and a resinous impurity that decomposes above $420\ \text{K}$. At the temperature where the mixture actually boils, water's own vapour pressure is about $85\ \text{kPa}$. (a) Estimate chlorobenzene's contribution to the total vapour pressure at that boiling point. (b) Explain why simple distillation is unsuitable here.

Step 1. The mixture boils when the sum of the two partial pressures equals atmospheric pressure: $P_{\text{total}} = p_{\text{water}} + p_{\text{chlorobenzene}} = 101.3\ \text{kPa}$.

Step 2. Rearranging for the unknown contribution: $p_{\text{chlorobenzene}} = 101.3 - 85 = 16.3\ \text{kPa}$.

Step 3. This is the vapour pressure chlorobenzene alone would show at that boiling temperature — a modest figure compared with the $101.3\ \text{kPa}$ it would need to reach on its own at $405\ \text{K}$. Because water is already supplying $85\ \text{kPa}$, chlorobenzene needs to contribute only the remaining $16.3\ \text{kPa}$, which happens at a temperature well under $373\ \text{K}$, not at $405\ \text{K}$.

Step 4. Simple distillation, by contrast, would require heating the flask up toward chlorobenzene's own boiling point of $405\ \text{K}$ to make $p_{\text{chlorobenzene}}$ alone reach $101.3\ \text{kPa}$ — comfortably past the $420\ \text{K}$ decomposition threshold's neighbourhood in practice, and certainly with no safety margin. The resinous impurity would decompose before or during that heating.

Conclusion: steam distillation is the correct choice, since Dalton's law guarantees the mixture boils below $373\ \text{K}$ — safely under the $420\ \text{K}$ decomposition limit — while simple distillation would demand temperatures close to chlorobenzene's own $405\ \text{K}$ boiling point.

Competitive stretch

If the resinous impurity instead decomposed above $500\ \text{K}$ (well clear of $405\ \text{K}$) and chlorobenzene needed to be recovered at high purity from a miscible liquid with a boiling point of $415\ \text{K}$, the better choice would flip to fractional distillation, since the boiling-point gap ($10\ \text{K}$) is too small for simple distillation to resolve cleanly, and there is no longer a decomposition risk that steam distillation is needed to avoid.

• Treating simple distillation as good for separating any two liquids — it only resolves mixtures with a substantial boiling-point gap (roughly $> 25\ \text{K}$); closer boiling points need fractional distillation instead.
• Confusing sublimation with ordinary evaporation or boiling — sublimation is a direct solid-to-vapour transition with no liquid phase in between, and only certain solids (naphthalene, camphor, benzoic acid, iodine) show it.
• Assuming steam distillation and vacuum distillation solve the identical problem because both suit heat-sensitive compounds — steam distillation adds a second, independent vapour pressure (Dalton's law) at fixed atmospheric pressure, while vacuum distillation instead lowers the external pressure the compound's own vapour pressure must match; they are mechanistically different routes to the same goal.
• Forgetting that steam distillation requires the compound to be insoluble in water as well as steam-volatile — a water-soluble steam-volatile liquid would just dissolve into the aqueous phase instead of distilling over as a separate layer.

Ask the tutor: if you would like to work through a fresh mixture — solubility data, boiling points, or a thermal-stability limit — and decide which purification technique fits, just ask the AI tutor to reason through it with you.