Purification and Characterisation of Organic Compounds

Lassaigne's test on an unknown gives a blood-red colour rather than the Prussian blue expected for nitrogen. Does the compound contain nitrogen?

It almost certainly does — and sulphur as well.

Fusing an organic compound with sodium converts its nitrogen to cyanide and its sulphur to sulphide. But if both are present, the sodium reacts with them together to give thiocyanate instead:

$\text{Na}+\text{C}+\text{N}+\text{S}\to \text{NaSCN}$

Thiocyanate with iron(III) gives the blood-red ${\left[\text{Fe}\left(\text{SCN}\right)\right]}^{2+}$, not the Prussian blue of ferrocyanide. So the red colour is positive evidence for nitrogen and sulphur together, and reading it as "no nitrogen" is exactly the wrong conclusion.

Every stage of this chapter has a trap of that kind, and Advanced tests the traps rather than the procedures.

1. Choosing a purification method

Steam distillation deserves particular attention. Two immiscible liquids exert their vapour pressures independently, so the mixture boils when the sum reaches atmospheric pressure — which happens below the boiling point of either. Aniline, boiling at $184^{\circ }$C alone, distils with steam at about $98^{\circ }$C and so escapes decomposition.

Illustration 1

Explain why $o$-nitrophenol can be separated from $p$-nitrophenol by steam distillation but the two cannot be separated by ordinary distillation.

The ortho isomer forms an intramolecular hydrogen bond, so it cannot bond to water or to its neighbours and is comparatively volatile.

The para isomer bonds intermolecularly, giving it a much higher boiling point and appreciable water solubility.

Steam carries the ortho isomer over at below $100^{\circ }$C while the para isomer remains behind.

Ordinary distillation would require heating to above $200^{\circ }$C, at which both isomers begin to decompose, so the separation would fail on both counts.

2. Chromatography

All chromatography separates by differential partitioning between a stationary and a mobile phase.

Adsorption chromatography uses a solid stationary phase such as silica or alumina; column and thin-layer methods are both of this type. Partition chromatography uses a liquid held on a solid support; paper chromatography, where the stationary liquid is water held by cellulose, is the everyday example.

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

$R_f$ is always between $0$ and $1$, and a more strongly adsorbed component has a lower value. Because it depends on solvent, temperature and adsorbent, it identifies a compound only under stated conditions.

Illustration 2

In a thin-layer separation the solvent front moves $8.0$ cm while two spots move $2.4$ cm and $6.0$ cm. Find both $R_f$ values and state which component is more polar on silica.

$R_f$ values are $\frac{2.4}{8.0}=0.30$ and $\frac{6.0}{8.0}=0.75$.

Silica is a polar adsorbent, so the more polar component is held more strongly and moves less.

The spot with $R_f=0.30$ is therefore the more polar of the two.

Reversing the polarity of the stationary phase reverses the order, which is the basis of reverse-phase chromatography and a useful check that a separation is behaving as expected.

3. Lassaigne's test in detail

Sodium fusion converts covalently bound elements into ionic ones that can be tested in water:

$\text{N}\to {\text{CN}}^{-},\text{S}\to {\text{S}}^{2-},\text{X}\to {\text{X}}^{-},\text{N}+\text{S}\to {\text{SCN}}^{-}$

The nitric acid step before the halogen test is essential: it destroys any cyanide or sulphide, which would otherwise precipitate as silver cyanide or silver sulphide and give a false positive.

Illustration 3

A compound gives a black precipitate with lead acetate and a white precipitate with silver nitrate after acidification. Deduce which elements are present, and state one compound it could be.

The black lead sulphide precipitate confirms sulphur.

The white silver precipitate, obtained after boiling with nitric acid, confirms chlorine. Bromide would give pale yellow and iodide a deeper yellow.

Nitrogen is absent, since neither Prussian blue nor a red colour was reported.

Sulphanilyl chloride or a chlorinated thiol would fit. The test identifies elements present, never the structure, which is why it is only the first stage of an analysis.

4. Quantitative estimation

Each element has a standard method and a standard formula, and all of them share the same structure: weigh a derivative, take the fraction of it that is the element sought.

Oxygen is never determined directly in the ordinary scheme, which is why an accumulated error in every other element lands entirely on the oxygen figure.

Illustration 4

Combustion of $0.246$ g of an organic compound gives $0.198$ g of carbon dioxide and $0.1014$ g of water. Find the percentages of carbon, hydrogen and oxygen.

$\%\text{C}=\frac{12}{44}\times \frac{0.198}{0.246}\times 100=21.95\%$

$\%\text{H}=\frac{2}{18}\times \frac{0.1014}{0.246}\times 100=4.58\%$

$\%\text{O}=100-21.95-4.58=73.47\%$

That oxygen figure carries the error of both measurements. If either the carbon dioxide or the water were weighed carelessly, the oxygen result would absorb the whole discrepancy without any warning sign.

Illustration 5

$0.35$ g of an organic compound gave $0.466$ g of barium sulphate by the Carius method. Find the percentage of sulphur, and state what a low result would indicate.

$\%\text{S}=\frac{32}{233}\times \frac{0.466}{0.35}\times 100$

$=0.1373\times 1.331\times 100=18.3\%$

A result below the true value would mean incomplete oxidation: some sulphur remained at an intermediate oxidation state and was never precipitated as sulphate.

Nothing in the appearance of the precipitate reveals this, which is why the sealed tube and fuming nitric acid are specified rather than merely suggested.

5. Kjeldahl against Dumas

Two methods estimate nitrogen, and choosing between them is a standard question.

Dumas oxidises the compound over copper oxide, converting all its nitrogen to nitrogen gas whose volume is measured:

$\%\text{N}=\frac{28}{22400}\times \frac{V_{STP}}{w}\times 100$

Kjeldahl digests the compound with concentrated sulphuric acid, converting nitrogen to ammonium sulphate, then liberates ammonia with alkali and titrates it:

$\%\text{N}=\frac{1.4\times N_{acid}\times V_{acid}}{w}$

Kjeldahl is faster and needs no gas measurement, but it fails for nitrogen in a nitro group, an azo linkage or a ring, because digestion does not convert those to ammonium. Dumas works for everything, which is why it remains the reference method.

Illustration 6

$0.5$ g of an organic compound gives $60$ mL of nitrogen at $300$ K and $715$ mm Hg over water, the aqueous tension being $15$ mm Hg. Find the percentage of nitrogen.

Dry pressure $=715-15=700$ mm Hg.

Correcting to STP: $V_{STP}=\frac{700\times 60\times 273}{760\times 300}=50.3$ mL

$\%\text{N}=\frac{28}{22400}\times \frac{50.3}{0.5}\times 100=12.6\%$

Subtracting the aqueous tension is compulsory, since the gas was collected over water and is therefore saturated with vapour. Omitting it inflates the answer by about two per cent here.

Illustration 7

Explain why Kjeldahl's method cannot be used for nitrobenzene or for pyridine.

Digestion with sulphuric acid must reduce every nitrogen atom to the ammonium ion for the method to be quantitative.

Nitrogen in a nitro group is already at a high oxidation state and is not reduced to ammonium under those conditions.

Nitrogen in an aromatic ring such as pyridine is held in a stable delocalised system and is not liberated either.

Both would give a low result with no indication that anything was wrong, which is why Dumas is used whenever the structure is unknown.

6. From percentages to a molecular formula

Divide each percentage by the atomic mass, divide through by the smallest quotient, and clear any fractions. That gives the empirical formula. The molecular formula then needs an independent molar mass, from vapour density or a colligative measurement:

$n=\frac{\text{molar mass}}{\text{empirical formula mass}}$

Illustration 8

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen and the rest oxygen, with a vapour density of $30$. Find its molecular formula.

$\%\text{O}=100-46.7=53.3$

Ratios: $\frac{40.0}{12}=3.33$, $\frac{6.7}{1}=6.7$, $\frac{53.3}{16}=3.33$

Dividing by $3.33$: $1:2:1$, giving ${\text{CH}}_2\text{O}$ of mass $30$.

Molar mass $=2\times 30=60$, so $n=2$ and the formula is ${\text{C}}_2{\text{H}}_4{\text{O}}_2$.

Vapour density is half the molar mass, a definition that costs marks every year and is worth writing out explicitly before using.

Illustration 9

$0.20$ g of an organic compound gave $0.287$ g of silver chloride by the Carius method. Find the percentage of chlorine.

$\%\text{Cl}=\frac{35.5}{143.5}\times \frac{0.287}{0.20}\times 100$

$=0.2474\times 1.435\times 100=35.5\%$

The fraction $35.5/143.5$ is the mass of chlorine in silver chloride. Every Carius calculation has this same structure: the mass of the weighed precipitate scaled by the fraction of it that is the element sought.

Illustration 10

Explain why a Carius estimation of sulphur is carried out with fuming nitric acid rather than with dilute acid.

The sulphur must be oxidised completely to sulphate before it can be precipitated as barium sulphate.

Fuming nitric acid at high temperature in a sealed tube achieves that for every organic sulphur compound, including thiols and sulphides that resist milder oxidants.

Dilute acid would leave some sulphur at an intermediate oxidation state, which would not precipitate.

Incomplete oxidation always gives a low result, and there is no visible sign of it, which is why the forcing conditions are specified rather than merely recommended.

7. Criteria of purity

A separation is worthless without a way of knowing when it is complete, and each physical constant supplies one.

A pure solid has a sharp melting point, usually within a degree. An impurity both lowers the melting point and broadens the range, because the impurity depresses the freezing point of the melt and the composition changes as melting proceeds.

The mixed melting point exploits this. Mix the unknown with an authentic sample of the suspected compound: if they are identical the melting point is unchanged, and if they are not it falls. This identifies a compound rather than merely confirming purity.

A pure liquid has a sharp boiling point at a stated pressure, and the same broadening argument applies. Chromatography gives the most sensitive test of all: a single spot at one $R_f$ in more than one solvent system is strong evidence of a single component.

Illustration 11

An unknown solid melts at $132$ to $134^{\circ }$C. Mixing it with authentic benzoic acid (melting point $122^{\circ }$C) gives a mixture melting at $110$ to $118^{\circ }$C. What can be concluded?

The narrow original range suggests a reasonably pure substance.

Mixing with benzoic acid depressed and broadened the melting point, so the two are different compounds and the unknown is not benzoic acid.

Had they been identical, the mixture would have melted at the same temperature and just as sharply.

A mixed melting point can disprove an identification decisively, whereas a matching melting point alone only fails to disprove it.

Illustration 12

A sample gives a single spot by thin-layer chromatography in one solvent but two spots in another. Is it pure?

No. Two spots in any solvent system establishes two components.

The first solvent simply failed to resolve them, giving both the same $R_f$ by coincidence.

This is why purity by chromatography is always checked in more than one solvent. A single spot is evidence only against the systems actually tried.

Summary

• A blood-red Lassaigne result means nitrogen and sulphur, since together they give thiocyanate rather than cyanide.
• Boil the extract with nitric acid before the halogen test, or cyanide and sulphide give false silver precipitates.
• Steam distillation works because immiscible liquids exert vapour pressures independently, so the mixture boils below $100^{\circ }$C.
• Distillation under reduced pressure is for liquids that decompose before boiling, such as glycerol.
• $R_f=\frac{\text{distance by component}}{\text{distance by solvent front}}$, always between $0$ and $1$; a more strongly adsorbed component moves less.
• $\%\text{C}=\frac{12}{44}\times \frac{m_{{\text{CO}}_2}}{w}\times 100$ and $\%\text{H}=\frac{2}{18}\times \frac{m_{{\text{H}}_2\text{O}}}{w}\times 100$.
• Halogen by Carius as $\text{AgX}$; sulphur as ${\text{BaSO}}_4$ with the factor $32/233$; phosphorus as ${\text{Mg}}_2{\text{P}}_2{\text{O}}_7$ with $62/222$.
• Oxygen is always found by difference, so it absorbs the accumulated error of every other determination.
• Dumas: $\%\text{N}=\frac{28}{22400}\times \frac{V_{STP}}{w}\times 100$, and subtract the aqueous tension when collecting over water.
• Kjeldahl: $\%\text{N}=\frac{1.4N_{acid}{V}_{acid}}{w}$, faster but fails for nitro, azo and ring nitrogen.
• Empirical formula from percentage ratios; molecular formula needs an independent molar mass.
• A pure solid melts sharply; an impurity lowers and broadens the range, which is what the mixed melting point test exploits.
• Chromatographic purity must be checked in more than one solvent system; two spots anywhere means two components.
• Vapour density is half the molar mass, which is the most commonly dropped step in the whole chapter.