PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1p5qBJPB6vANn92LxAyFMgBD0ty120G6l/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Purification and Characterisation of Organic Compounds Purification - Crystallization, sublimation, distillation, differential extraction and chromatography - principles and their applications. Qualitative analysis - Detection of nitrogen, sulphur, phosphorus and halogens. Quantitative analysis (basic principles only) - Estimation of carbon, hydrogen, nitrogen, halogens, sulphur, phosphorus. Calculations of empirical formulae and molecular formulae; Numerical problems in organic quantitative analysis. For identifying any given organic compound, a systematic approach is required, which involves following steps: PURIFICATION OF ORGANIC COMPOUNDS C H A P T E R THIS CHAPTER INCLUDES In order to study the properties of the organic compound for its identification, ● Purification of the given organic compound must be obtained in the purest form. The purification of the compound can be achieved by any of the following methods: organic compounds Crystallisation : It is based on the fact that certain organic substances ● Detection of dissolve partly in a solvent and their solubility increases with increase in temperature. The excess of solid substance is heated to a higher Elements temperature with a selected solvent i.e. near its boiling point. The ● quantitative saturated solution thus obtained is filtered, while compound separates out in the form of fine crystals. The pure solid is filtered out and dried. The suitable solvent selected is one (i) which dissolve more substance at Estimation higher temperature than at room temperature (ii) in which impurities are ● Empirical and insoluble or dissolve to such an extent that they remain in solution in the mother liquor upon crystallisation (iii) which is not highly inflammable (iv) which does not react chemically with the compound to be purified. Fractional crystallisation : Fractional crystallisation is used for the separation of a mixture of two components which are soluble in the same solvent but their solubilities are different. The hot saturated solution of the mixture is allowed to cool the less soluble component crystallises out earlier than the more soluble component. The various fractions are separated from time to time. These fractions are now separately put to crystallisation. Sublimation : Certain organic substances change directly from solid to vapour state without passing through the liquid state and vice-versa. This is called sublimation. Only the substances having vapour pressure equal to the atmospheric pressure much below their respective melting points sublime and can be purified by this method. This method is useful for purification and separation of organic substances such as camphor, naphthalene, benzoic acid, anthracene, phthalic anhydride which sublime on heating from non-volatile impurities. Distillation: Distillation involves the process of heating a liquid to convert it into the vapours and then condensing the vapours to get back the liquid. The simple distillation is applied only for the purification of liquids which boil without decomposition at atmospheric pressure and contain non volatile impurities. Molecular formula A mixture of two liquids can be separated and purified if their boiling points differ by 30 – 50 K. For example a mixture of hexane (bp 342 K) and toluene (bp 384 K) (ii) a mixture of benzene (bp 353 K) and aniline (bp 457 K). Fractional Distillation: This method is used for the separation and purification of organic liquids from volatile impurities, or for separating two or more volatile liquids, from a liquid mixture whose boiling point differ by 40 K or less. When the liquids present in the mixture have their boiling points close to each other (i.e., differ by 10–15 K only), the separation into pure liquids is difficult. The fractions obtained are always contaminated. The fractions are further purified by repeated distillations. To decrease the number of distillations, the separation is done by fitting the distillation flask with a fractionating column. Fractionating column is a special type of long glass tube provided with obstructions to the passage of the rising vapour upwards and that of the liquids downwards. For example. (i) Separation of acetone (bp 325 K) from methyl alcohol (bp 338 K) (ii) Separation of petroleum into gasoline, kerosene oil, diesel oil. Vacuum Distillation: In case of organic liquids which decompose before their boiling point is reached, the distillation is carried out under reduced pressure. Glycerol is one such compound. A liquid boils, when its vapour pressure is equal to the atmospheric pressure. When the outer pressure is reduced the liquid boils at a lower temperature without undergoing decomposition. Steam Distillation: It is co-distillation with water. This method is used for the purification and separation of the organic solid or liquid compounds which are volatile in steam, immiscible with water, contain non volatile impurities and have high vapour pressure at 373 K. A liquid boils, when its vapour pressure is equal to the atmospheric pressure. Since in steam distillation, the distilling mixture consists of steam and vapour of organic substance, the sum of the vapour pressure of the organic substance ( = P1) and that of steam (= P2) becomes equal to the atmospheric pressure (= P) at the temperature of the distillation. It follows from this that, at the temperature of distillation the vapour pressure of organic substance is lower than the atmospheric pressure by an amount of vapour pressure of steam. In other words the organic substance distills below its normal boiling point without decomposition. For example aniline which normally boils at 457 K boils at 371.5 K with the help of steam distillation. Differential Extraction : Differential Extraction is the process of separation of an organic compound from its aqueous solution by shaking with a suitable solvent such as ether, chloroform and benzene. The solvent should be miscible with water and the organic compound to be separated should be highly soluble in it. With a given volume of extracting solvent available, the extraction is more efficient (i.e. more complete) if it is used in several installments, than, if it is used all in one installment. For a given volume of the solvent, greater the number of the extractions, more is the amount of the organic substance extracted. Chromatography : Chromatography (discovered by Russian Botanist Tswett) is a modern technique for the separation, isolation, purification and identification of the constituents of the mixture. Chromatography is based on the principle of selective distribution of compounds of a mixture between two phases, a stationary phase (or a fixed phase) and moving (or a mobile) phase. The stationary phase can be a solid or a liquid while the moving phase is a liquid or gas. When the stationary phase is a solid, the basis is adsorption and when it is a liquid, the basis is partition. Types of Chromatography Based on the nature of the stationary phase and the moving phase, different types of chromatographic technique have been developed. The most common ones are tabulated. Table Types of chromatography Mobile phase Stationary phase Applications 1. Adsorption or column Liquid Solid Separations on large scale chromatography 2. Thin layer chdromatography (T.L.C.) (Qualitative analysis) Liquid Solid Identification of organic compounds 3. High performance liquid chromatography (HPLC) Liquid Solid Qualitative and Quantitative analysis 4. Gas liquid chromatography (GLC) Gas Liquid Qualitative and Quantitative analysis 5. Paper or Partition chromatography Liquid Liquid Qualitative and Quantitative analysis of polar organic compounds e.g.sugars α-amino acids and inorganic compound DETECTION OF ELEMENTS (Qualitative Analysis) Element Sodium Fusion Extract (S.E) Confirmed Test Reactions Nitrogen Na + C + N Δ NaCN (S.E) S.E + FeSO4 + NaOH, boil and cool, + FeCl3 FeSO4 + 6NaCN → Na4[Fe(CN)6] + Na2SO4 + conc. HCl → Prussian 3Na4[Fe(CN)6] + 2Fe2(SO4)3 blue or green → Fe4[Fe(CN)6]3 + 6 Na2SO4 Prussian blue (Ferric Ferrocyanide) Sulphur 2Na + S Δ Na S 2 (S.E) S.E + Sodium nitroprusside → Deep violet colour S.E + (CH3COO)2Pb → Na2S + Na2[Fe(CN)5NO] → Na4 [Fe(CN)5NOS] deep violet Black ppt. Halogens Na + Cl Δ NaCl S.E + AgNO3 → AgX (X=Cl, Br, I) Na2S + (CH3COO)2Pb → (i) White ppt soluble in NH3(liq) Cl confirms PbS↓ + 2CH3COONa (black ppt) (S.E) (ii) Yellow ppt partial soluble in NH3(liq) NaX + AgNO3 → AgX ppt Br confirms (iii) Pale yellow insoluble in NH3(liq) I confirms Nitrogen and Sulphur together Na + C + N + S Δ NaCNS (S.E) As in test for nitrogen instead of green or blue colour, blood red colouration confirms presence of N and S both 3NaCNS + FeCl3 → Fe(CNS)3 + 3NaCl blood red (Ferric thiocyanate) Detection of C and H Organic comp C + 2CuO → 2Cu + CO2 2H + CuO → Cu + H2O CO2 turns lime water milky whereas H2O turns only CuSO4 to blue. Detection of Phosphorus Organic compound + Na2O2 ⎯⎯Fus⎯e → Na3PO4 Na3PO4 + 3HNO3 ⎯⎯Δ →H3PO4 + 3NaNO3 H3PO4 + 12(NH4 )2 MoO4 + 21HNO3 → (NH4 )3 PO4 .12MoO3 + 21NH4NO3 + 12H2O Canary yellow ppt QUANTITATIVE ESTIMATION OF ELEMENTS IN ORGANIC COMPOUNDS Element Technique Formula Carbon and Hydrogen Nitrogen Sulphur Halogens Oxygen Phosphorus (Method) Leibig's Method Dumas method Kjeldahl method Carius method Carius method Carius method C → CO % C = 12 × wt. of CO2 × 100 2 12g 44g 44 × weight of organic compound 2H → H O % H = 2 × wt. of H2O × 100 2 2g 18g 18 × wt. of organic compound (i) 2N → N2(g) 28 × V × 100 22.4 L at S.T.P. % N = 22.4 × wt. of organic compound where V is the volume of N2 gas in L as S.T.P. (ii) N → NH3 ≡ H2SO4 1.4 N V % N = 1 1 wt. of organic compound where, N1V1 is the m.eq. of H2SO4 used S → H2SO4 → BaSO4 32 × wt. of BaSO4 32g 233g % S =233 × wt. of org. comp × 100 . Cl —→ AgCl 35.5 × wt. of AgCl × 100 35.5g 143.5g % Cl = 143.5 × wt. of organic comp. Br —→ AgBr 80 × wt. of AgBr × 100 80g 188g % Br = 188 × wt. of organic comp. I —→ AgI 127 wt. of AgI 127g 235g % I = 235 × wt. of organic comp. × 100 100 – (sum of % of all elements) 62 wt. of Mg P O formed % P = × 2 2 7 × 100 222 wt. of organic comp. EMPIRICAL AND MOLECULAR FORMULA Empirical Formula of a compound is the simplest whole number ratio of the atoms of elements constituting its one molecule. The sum of atomic masses of the atoms representing empirical formula is called empirical formula mass. Molecular Formula of a compound shows the actual number of the atoms of the elements present in its one molecule. The sum of atomic masses of the atoms representing molecule is called molecular mass. Relationship between Empirical Formula and Molecular Formula Molecular formula = n × empirical formula where n is a simple whole number having values of 1, 2, 3... etc. Also, n = Molecular formula mass/Empirical formula mass. An organic compound contains 40% C, 6.66% H and rest Oxygen. Its vapour density is 30. Calculate empirical and molecular formula. Solution : Calculation of empirical formula Element % At. mass Relative no. of atoms Simplest ratio C 40 12 40 = 3.33 3.33 = 1 12 3.33 H 6.66 1 6.66 = 6.66 6.66 = 2 1 3.33 O 53.34 53.34 3.33 16 16 = 3.33 3.33 = 1 Empirical formula = CH2O Empirical formula mass = 12 + 2 + 16 = 30 Molecular mass = 2 × 30 = 60 n = 60 = 2 . 30 Molecular formula = 2 × empirical formula = 2 × (CH2O) = C2H4O2 ❑ ❑ ❑