JEE Chemistry Purification and Characterisation of Organic Compounds Notes Set 02

Methods of Purification

Crystallisation purifies solids by differing solubility; sublimation separates a sublimable solid from non-volatile impurities; chromatography separates by differing adsorption.
Simple distillation: non-volatile impurities, or liquids whose boiling points differ by about 40 K or more (ether and toluene).
Fractional distillation: liquids with close boiling points (acetone and methyl alcohol, crude oil).
Distillation under reduced pressure: liquids that decompose near their boiling point (glycerol, sugar juice).
Steam distillation: substances that are immiscible with water and steam volatile (aniline, nitrobenzene, turpentine oil).

Qualitative Analysis (Lassaigne's Test)

The compound is fused with sodium to convert N, S and halogens into NaCN, Na₂S and NaX, which are tested in the water extract.
Nitrogen: FeSO₄, NaOH, boil, then FeCl₃ and conc. HCl gives Prussian blue (Fe₄[Fe(CN)₆]₃).
Sulphur: deep violet with sodium nitroprusside, or black PbS with lead acetate and acetic acid. Nitrogen and sulphur together give NaSCN and a blood red colour with Fe³⁺.
Halogens: boil with HNO₃, then AgNO₃: white (soluble in NH₃) for Cl, pale yellow (partly soluble) for Br, yellow (insoluble) for I.
Phosphorus: oxidise to phosphate and warm with HNO₃ and ammonium molybdate for a yellow precipitate.

Quantitative Analysis

Carbon and hydrogen (Liebig): % C = \( \frac{12}{44}\times\frac{\text{mass of CO}_2}{\text{mass of compound}}\times100 \) and % H = \( \frac{2}{18}\times\frac{\text{mass of H}_2O}{\text{mass of compound}}\times100 \). CaCl₂ absorbs water and KOH absorbs CO₂.
Nitrogen by Dumas: % N = \( \frac{28}{22400}\times\frac{V_{STP}}{w}\times100 \). Nitrogen by Kjeldahl: % N = \( \frac{1.4\,NV}{w} \), where NV is the milliequivalents of acid neutralised by the ammonia; it fails for nitro, azo and ring nitrogen.
Carius method: % S = \( \frac{32}{233}\times\frac{\text{mass of BaSO}_4}{w}\times100 \), % Cl = \( \frac{35.5}{143.5}\times\frac{\text{mass of AgCl}}{w}\times100 \), % Br = \( \frac{80}{188}\times\frac{\text{mass of AgBr}}{w}\times100 \), % P = \( \frac{62}{222}\times\frac{\text{mass of Mg}_2P_2O_7}{w}\times100 \).
Oxygen is obtained by difference: % O = 100 − (sum of the percentages of all other elements).

Illustration 1. 0.30 g of a compound containing C, H and O gave 0.44 g of CO₂ and 0.18 g of H₂O. Find its empirical formula.
Answer:
C = 12/44 × 0.44 = 0.12 g (40 %); H = 2/18 × 0.18 = 0.02 g (6.67 %); O = 0.16 g (53.33 %).
Moles C : H : O = 3.33 : 6.67 : 3.33 = 1 : 2 : 1, so the empirical formula is CH₂O.

Illustration 2. 0.50 g of an organic compound was Kjeldahl digested and the ammonia absorbed in 50 mL of 0.5 N H₂SO₄. The excess acid needed 60 mL of 0.25 N NaOH. Find the percentage of nitrogen.
Answer:
Milliequivalents of acid used by NH₃ = 25 − 15 = 10.
% N = 1.4 × 10/0.5 = 28 %.