"Purification and Characterisation of Organic Compounds" — Practice

Question 1

State with reasons the purification technique you would use for: a solid acid contaminated with sand; glycerol, which decomposes at its boiling point; and aniline in a tarry reaction mixture.

Accepted Answer

For the acid and sand, crystallisation, because the acid dissolves in a suitable hot solvent while sand does not, so hot filtration removes the sand and slow cooling recovers pure crystals. For glycerol, distillation under reduced pressure, because lowering the external pressure lowers the boiling point from 563 K to about 453 K, below the temperature at which it chars. For aniline, steam distillation, because aniline is volatile in steam and immiscible with water, so it comes over at 371 K rather than at 457 K, leaving the involatile tar behind.

Question 2

Explain why a mixture of aniline and water distils at 371 K when aniline alone boils at 457 K and water at 373 K.

Accepted Answer

Aniline and water are immiscible, so neither dissolves in the other and each exerts its own vapour pressure independently, as though the other were absent. The mixture boils when the **sum** of the two vapour pressures reaches atmospheric pressure, and because both contribute in full, that sum is reached below the boiling point of either liquid alone. Raoult's law does not apply, since there is no solution and no mole fractions to scale the pressures down. This is why steam distillation suits heat-sensitive natural products: the compound is never heated near its own boiling point.

Question 3

Why does a fractionating column separate liquids of close boiling point when simple distillation cannot?

Accepted Answer

Because each condensation and re-evaporation cycle enriches the vapour only slightly in the more volatile component, and one cycle is not enough when the boiling points are close. A fractionating column is packed to provide a large surface, so vapour rising through it condenses and re-evaporates many times, each cycle repeating the Raoult's law enrichment. By the top of the column the vapour is almost entirely the more volatile component. The number of effective cycles is what determines whether a given separation is achievable at all.

Question 4

On a thin layer plate, the solvent front travels 8.0 cm while two spots travel 6.0 cm and 2.4 cm. Calculate both retardation factors and identify which component is more polar, given a silica stationary phase.

Accepted Answer

The retardation factors are $6.0/8.0 = 0.75$ and $2.4/8.0 = 0.30$. The component with the smaller value travelled less and is therefore held more strongly by the stationary phase. Silica gel is polar, so it adsorbs polar compounds most strongly, meaning the component with $R_f = 0.30$ is the more polar of the two. A value near 1 would indicate a compound barely retained at all, and a value of 0 one too strongly held for that solvent.

Question 5

Distinguish adsorption chromatography from partition chromatography, giving one example of each.

Accepted Answer

In adsorption chromatography the stationary phase is a solid such as silica gel or alumina, and separation depends on how strongly each component sticks to its surface; column chromatography and thin layer chromatography both work this way. In partition chromatography the stationary phase is a liquid held on an inert support, and separation depends on how each component distributes itself between two liquids. Paper chromatography is the standard example, the stationary liquid being the water held within the cellulose fibres rather than the paper itself.

Question 6

Why can chlorine in chlorobenzene not be detected by simply adding silver nitrate solution?

Accepted Answer

Because the chlorine is covalently bonded to the ring and is not present as chloride ion at all. Silver nitrate precipitates silver chloride only from free chloride ions, so adding it to chlorobenzene produces nothing whatsoever. The element must first be converted into an ionic form, which is exactly what Lassaigne's sodium fusion does: sodium at red heat is powerfully reducing, breaks the covalent bond and forms sodium chloride, after which the ordinary silver nitrate test works normally on the aqueous extract.

Question 7

A student adds silver nitrate directly to the sodium fusion extract of a compound containing both nitrogen and chlorine and obtains a white precipitate. Why is this result unreliable, and what step was omitted?

Accepted Answer

Because the extract contains cyanide ions as well as chloride, and silver cyanide is also a white precipitate, so the student cannot tell which has formed and would report chlorine even if none were present. The omitted step is boiling the extract with dilute nitric acid, which decomposes the cyanide and expels it as hydrogen cyanide gas; the same treatment removes sulphide, which would otherwise give a black silver sulphide precipitate. Only after that boiling is a white precipitate unambiguous evidence of chloride.

Question 8

Lassaigne's test on a compound gives a blood-red colouration with iron(III) rather than Prussian blue. What does this indicate and how would you confirm each element separately?

Accepted Answer

A blood-red colour with iron(III) indicates thiocyanate, which forms when nitrogen and sulphur are present together in the same molecule and the fusion gives sodium thiocyanate instead of separate cyanide and sulphide. The compound therefore contains both elements, and the red colour reports both at once. To confirm them individually, repeat the fusion using a large excess of sodium, which decomposes the thiocyanate into cyanide and sulphide. The iron(II) test then gives Prussian blue for nitrogen and sodium nitroprusside gives a violet colour for sulphur.

Question 9

How would you distinguish between silver chloride, silver bromide and silver iodide precipitates obtained in a halogen test?

Accepted Answer

By colour and by their behaviour with aqueous ammonia. Silver chloride is white and dissolves freely in ammonia, forming the soluble diamminesilver(I) complex. Silver bromide is pale yellow and only sparingly soluble in ammonia. Silver iodide is distinctly yellow and insoluble in ammonia altogether. The trend follows the solubility products, which fall steeply from chloride to iodide, so ammonia can supply enough complexing power to dissolve the most soluble but not the least.

Question 10

Describe how phosphorus is detected in an organic compound.

Accepted Answer

The compound is heated with sodium peroxide, a strong oxidising agent, which converts the phosphorus into sodium phosphate. The resulting solution is acidified with nitric acid and treated with ammonium molybdate, which gives a yellow precipitate of ammonium phosphomolybdate. Note that this route uses oxidation rather than the reduction that sodium fusion performs, because phosphorus needs raising to $+5$ as phosphate rather than reducing, phosphate being a stable soluble easily tested anion while a phosphide would be none of those.

Question 11

1.40 g of an organic compound gave ammonia by the Kjeldahl method, absorbed in 60.0 mL of 0.100 M $\mathrm{H_2SO_4}$, and the excess acid required 20.0 mL of 0.100 M NaOH. Find the percentage of nitrogen.

Accepted Answer

Total acid taken is $0.0600 	imes 0.100 = 6.00 	imes 10^{-3}$ mol. The back titration used $2.00 	imes 10^{-3}$ mol of NaOH, and since $\mathrm{H_2SO_4}$ is dibasic this corresponds to $1.00 	imes 10^{-3}$ mol of unreacted acid. Acid consumed by ammonia is therefore $5.00 	imes 10^{-3}$ mol, and each mole of the dibasic acid neutralises two moles of ammonia, giving $1.00 	imes 10^{-2}$ mol of nitrogen. That is 0.140 g, so the percentage is $0.140/1.40 	imes 100 = 10.0$.

Question 12

Combustion of 0.246 g of an organic compound gave 0.352 g of carbon dioxide and 0.144 g of water. Calculate the percentages of carbon and hydrogen.

Accepted Answer

For carbon, the mass is $	frac{12}{44}$ of the carbon dioxide mass, that is 0.0960 g, so the percentage is $0.0960/0.246 	imes 100 = 39.0$. For hydrogen, the mass is $	frac{2}{18}$ of the water mass, that is 0.0160 g, so the percentage is $0.0160/0.246 	imes 100 = 6.5$. Carbon and hydrogen together account for 45.5 per cent, so the remaining 54.5 per cent is presumably oxygen, which is always found by difference rather than measured directly.

Practice — "Purification and Characterisation of Organic Compounds"

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