JEE Chemistry Purification And Characterisation Of Organic Compounds Notes Set 01

Purification Methods and Their Uses

Extraction is more efficient when the solvent is used in several small portions rather than all at once.

Chromatography in Brief

1. A mobile phase (liquid or gas) carries the mixture over a stationary phase of large surface area (solid or liquid). Components move at different rates because they stick to the stationary phase to different extents.
2. Adsorption (column) and thin-layer chromatography use a liquid mobile phase over a solid. Gas-liquid chromatography uses a gas over a liquid. Paper chromatography is partition between two liquids, water held in the paper being the stationary phase.
3. The retention factor is Rf = distance moved by the component / distance moved by the solvent front. It is constant for a given substance under fixed conditions and so identifies components.

Some Chemical Purification Methods

1. Commercial benzene contains thiophene. Shaking with cold conc. \( \ce{H2SO4} \) sulphonates the thiophene, which dissolves in the acid layer while benzene does not.
2. Rectified spirit (an ethanol-water azeotrope) is refluxed over quicklime \( \ce{CaO} \) and distilled to give absolute alcohol.
3. Carboxylic acids dissolve in aqueous \( \ce{NaHCO3} \) with evolution of \( \ce{CO2} \), but most phenols do not, so the two can be separated.
4. Primary, secondary and tertiary amines are separated with benzenesulphonyl chloride (Hinsberg's reagent).

Criteria of Purity

A pure solid melts sharply at a fixed temperature, while an impure one melts lower and over a range. A pure liquid boils at a constant temperature. In a mixed melting point test the sample is ground with the pure compound; no lowering of the melting point shows the sample is the same pure compound.

Qualitative Analysis

1. Carbon and hydrogen: heat with dry \( \ce{CuO} \). \( \ce{CO2} \) turns lime water milky and water turns white anhydrous \( \ce{CuSO4} \) blue.
2. Lassaigne's test: fuse the compound with sodium so that N, S and halogens become \( \ce{NaCN} \), \( \ce{Na2S} \) and \( \ce{NaX} \); boil with water and filter to get the sodium extract.
3. Nitrogen: sodium extract with \( \ce{FeSO4} \) and \( \ce{NaOH} \), boil, then add \( \ce{FeCl3} \) and conc. \( \ce{HCl} \). A Prussian blue colour is due to \( \ce{Fe4[Fe(CN)6]3} \). A very small amount of nitrogen gives a green colour.
4. Sulphur: sodium nitroprusside gives a violet colour (\( \ce{Na4[Fe(CN)5NOS]} \)); lead acetate in acetic acid gives black \( \ce{PbS} \).
5. Nitrogen and sulphur together: the extract gives a blood red colour with \( \ce{FeCl3} \) because \( \ce{Fe(SCN)^{2+} \)} forms.
6. Halogens: acidify the extract with \( \ce{HNO3} \) (after boiling off \( \ce{HCN} \) and \( \ce{H2S} \)) and add \( \ce{AgNO3} \). \( \ce{AgCl} \) is white and soluble in \( \ce{NH3} \), \( \ce{AgBr} \) pale yellow and sparingly soluble, \( \ce{AgI} \) yellow and insoluble.
7. Beilstein's test: a green or blue-green flame from a copper wire coated with the compound shows halogen, but it is unreliable for halogen-free compounds that give volatile copper compounds.
8. Lassaigne's test fails for compounds with N-N bonds such as diazo compounds and for compounds with no carbon such as hydrazine.

Quantitative Analysis: Formulas

1. Kjeldahl's method does not work for compounds with nitro, azo or azoxy groups, or for nitrogen in a ring (pyridine), because the nitrogen is not fully converted to ammonium sulphate. Dumas' method works for all nitrogen compounds.
2. In Kjeldahl's method \( \ce{CuSO4} \) acts as a catalyst and \( \ce{K2SO4} \) raises the boiling point of the digestion mixture.

Molecular Mass Determination

Empirical and Molecular Formula

1. Divide the percentage of each element by its atomic mass to get relative moles.
2. Divide all by the smallest value.
3. If a value is not whole, multiply every number by the smallest integer that makes all of them whole (for example 1.5 gives a factor of 2, 1.33 gives 3).
4. Write the symbols with these numbers as subscripts: this is the empirical formula.
5. Molecular formula = (empirical formula) x n, where n = molecular mass / empirical formula mass.

Example: a compound has C = 40%, H = 6.7%, O = 53.3%. Moles: 3.33, 6.7, 3.33, so the ratio is 1 : 2 : 1 and the empirical formula is \( \ce{CH2O} \) (mass 30). If the molecular mass is 180, n = 6 and the molecular formula is \( \ce{C6H12O6} \).

Eudiometry: a gaseous hydrocarbon is exploded with excess oxygen. The contraction after cooling gives the volume of water vapour, and the further contraction on absorbing \( \ce{CO2} \) in KOH gives the volume of \( \ce{CO2} \), from which x and y of \( \ce{CxHy} \) follow. The nitrogen rule says a compound with an odd number of nitrogen atoms has an odd molecular mass.