Purification & Characterisation of Organic Compounds Notes

Page # 1 PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS 1. PURIFICATION OF ORGANIC COMPOUNDS The methods to be employed depend on the physical state of the compound. 1.1 Crystallisation : This method is based on the differences in the solubility of the organic compound and its impurities in a solvent. (a) Preparation of the solution : Organic substance is powdered and is dissolved in a suitable solvent by heating. The amount of solvent should be just sufficient to dissolve the whole of the solid on heating. Choice of Solvent. The choice of solvent is very important in the crystallisation process. The main conditions of the solvent are: (i) The organic substance should dissolve in the solvent upon heating and it should get separated on cooling. (ii) The sol vent shoul d not di ssol ve t he impurities. (iii) The solvent should not react chemically with the substance. For example, suppose we want to purify sugar containing an impurity of common salt. This can be done by treating the mixture with ethanol around 350 K. The sugar will dissolve whereas common salt remains insoluble. (b) Filtration of the solution. The hot saturated solution is filtered preferably through a fluted filter paper placed in a glass funnel. The use of the fluted filter paper makes the filtration rapid. The jacketer of the hot water funnel is heated from outside and this keeps the solution hot in the glass funnel. This will prevent the formation of crystals during filtration. (c) Crystallisation. The hot filtration is allowed to cool slowly and undisturbed in a beaker or in a crystallising dish. After some time the crystals of the pure compound are formed. (d) Separation of the crystals. The crystals formed are separated from the mother liquor by filtration. The filtration is normally done by use of Buckner funnel and a suction pump. This enables the filtration under reduced pressure and is therefore, quite fast. (e) Drying of crystals. The crystals are dried by pressing between the folds of filter paper and then placed in a steam of air oven for some time. The crystals are finally dried over sulphuric acid or calcium chloride in a desiccator. 1.2 Sublimation : Certain organic solids directly change from solid to vapour state on heating. This process is called sublimation. The vapours on cooling change back to the solid form Heat Solid Vapours Cool The subli mation process is used for the separation of those solids which sublime on heating from non-volatile solids. The process is generally used for the purification of camphor, naphthalene, anthracene, benzoic acid, etc. containing non-volatile impurities. 1.3 Distillation : This method is used for the puri fication of li quids which boil without decompositi on and contain n on-vol at ile impurities. The simple distillation involves its boiling point so that it is converted into vapours. On cooling the vapours, pure liquid is obtained. The distillate contains pure liquid while the impurities are left behind in the distillation flask e.g.Ether from ethyl alcohol. 1.4 Fractional Distillation : This process is used to separate a mixture of two or more miscible liquids which have boiling points close to each other. The fractionating column is a long tube provided with obstructions to the passage of the vapours moving upwards and liquid moving downwards. This method may be used to separate a mixture of acetone (b.p. 330K) and methyl alcohol (b.p. 338. K). 1.5 Distillation under Reduced pressure (Vaccum Distillation) : Certain liquids have a tendency to decompose at a temperature below their boiling points. Such liquids cannot be purified by ordinary distillation. Therefore vacuum distillation is used for liquids which decompose at a temperature below their normal boiling points. Ex. glycerol boils with decomposition at 563K. 1.6 Steam distillation : The process of steam distillation is used for the separation and purification of liquid which is appreciably volatile in steam, from non-volatile components of a mixture. Thus, the process of steam distillations is used to purify the substances which (i) are volatile in steam but are not miscible with water (ii) possess sufficiently high vapour pressure at the boiling point temperature of water (1000C) (iii) contain non-volatile impurities. Page # 2 PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS The process of steam distillation can be applied for the separat ion of a mixture of o-nitrophenol and p-nitrophenol. In this process, water vapours carry along with them vapours of o-nitrophenol which is more volatile and they get cond ensed in the recei ver ; p-nitrophenol with higher b.p. remains in the distillation flask. The method can also be used for the purification of impure sample of aniline. 1.7 Chromatography : This method is based on the differences in the rates at which the components of a mixture are adsorbed on a suitable adsorbent. There are many forms of chromatography such as column chromatograhy, pa per chromatography, thin layer chromatography (TLC), gas chromatography, etc. The simplest method is column chromatography. Applications of chromatographic method. This method has been used (i) To separate ortho and para nitro-anilines. (ii) To separate blue and red dyes. (iii) To separate and purify plant pigments and other natural products. Types of chromatography : Based on the principle involved chromatography is classified into different categories. Two of these are. (a) Adsorption chromatography, and (b) Partition chromatography (a) Adsorption chromatography : Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina, When a mobile phase is allowed to move over a stationary phase (adsorbent), the components of the mixture move by varying distances over the stationary phase. Following are two main types of chromatographic techniques based on the principle of differential adsorption. (i) Column chromatography, and (ii) Thin layer chromatography. Column chromatography: Solvent Adsorbent (stationary phase) Glass wool a b+c a b c Column chromatography. Different stages of separation of components of a mixture Column chromatography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end. The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube. An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column Thin layer chromatography : jar adsorbent coated on glass plate sample dot base line solvent Thin layer chromatography chromatogram being developed solvent front spot y x base line Developed chromatogram Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin layer (about 0.2 mm thick) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate or chromaplate. The solution of the mixture to be separated is applied as a small spot about 2 cm above one end of TLC plate. The glass plate is then placed in a closed jar containing the eluant. As the eluant rises up the plate, the components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and separation takes place. The ralative adsorption of each component of the mixture is expressed in terms of its retardation factor i.e. Rf value D istance moved by the substance from base line (x) Rf  Distance moved by the solvent from base line (y) Page # 3 PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS The spots of coloured compounds are visible on TLC plate due to their original colour. The spots of colourless compounds which are invisible to the eye but fluoresce, can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine spots of compounds, which adsorb iodine will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate. For example, Amino acids may be detected by spraying the plate with ninhydrin solution (b) Partition Chromatography : card board chromatography paper jar spot base line appropriate spray reagent as discussed under thin layer chromatography. 2. QUALITATIVE ANALYSIS The qualitative analysis of an organic compound involves the detection of all the elements present in it. 2.1 Detection of Carbon and Hydrogen : A small amount of the dry and powdered substance is mixed with about double the amount of pure and dry copper oxide. The mixture is heated in a well dried hard glass tube (fig) delivery tube is packed with glass wool containing anhydrous copper sulphate (white). When the mixture is heated , the carbon present in the compound is oxidised to carbon dioxide which turns lime water milky. The hydrogen present in the organic compound is oxidised to water which turns anhydrous copper sulphate in the bulb to blue. Anhydrous copper sulphate solvent card board chromatography paper jar spot base line solvent Paper chromatography chromatography paper in two different shapes Partition chromatography is based on continuous differential partitioning of components of mixture between stationary and mobile phases. Paper chromatography is a typ e of partiti on chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents This solvent rises up the paper by capillary action and flows over the spot. The paper selectively ratains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an Organic compound + Copper oxide Lime water fig. Detection of carbon and hydrogen in an organic compound C + 2CuO from organic compound Heat   CO2 + 2Cu  CaCO3 + H2O CO2 + Ca(OH)2   lime water milky 2H + CuO Heat  H2O + Cu from organic compound CuSO4 + 5H2O    CuSO4 . 5H2O anhydrous hydrated copper sulphate copper sulphate (blue) (white) This method is known as copper oxide test. 2.2 Detection of Nitrogen : Nitrogen in an organic compound is detected by the following tests : (a) Soda lime test. A pinch of an organic compound is heated strongly with soda lime (NaOH + CaO) in a test tube. If ammonia gas evolves, it indicates nitrogen. CaO CH3CONH2 + NaOH acetamide CH3COONa + NH3 Limitation : This method has a limitation. A large number of organic compounds such as Page # 4 PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS nitro and diazo compounds do not liberated ammonia on heating with sodalime. (b) Lassaigne’s method : A small piece of a dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. Then , a small amount of organic substance is added and the tube is heated strongly till it becomes red hot. The red hot tube is then cooled and filtered. The filtered liquid is known as sodium extract or Lassaigne’s extract. The Lassaigne’s extract is usually alkaline. If not, it may be made alkaline by adding a few drops of a dilute solution of sodium hydroxide added to a part of sodium extract a small amount of a freshly prepared ferrous sulphate solution is and the contents are warmed. A few drops of ferric chloride solution are then added to the contents and the resulting solution is acidified with dilute hydrochloric acid. The appearance of a bluish green or a blue colouration confirms the presence of nitrogen in the organic compound. The following chemical reactions occur during the test : Na + C + N  NaCN from organic compound FeSO4 + 2NaCN Fe(CN)2 + Na2SO4 Fe(CN)2 + 4NaCN  Na4[Fe(CN)6 ] sodium ferrocyanide 3Na4[Fe(CN)6]+4FeCl3Fe4[Fe(CN)6]3+ 12NaCl ferric ferrocyanide (blue colour) 2.3 Detection of Sulphur : The sulphur in the compound reacts with sodium metal to form sodium sulphide. 2Na + SNa2S from organic compound The Lassaigne’s extract is divided into two parts and following tests are performed. (i) Sodium nitroprusside test. The one portion of the extract, a few drops of sodium nitroprusside are added. The appearance of violet colouration indicates sulphur. Na2S + Na2 [Fe(CN)5NO]  Na4[Fe(CN)5NO.S] sod. nitroprusside violet colouration (ii) Lead acetate test : The other part of the Lassaigne’s extract is acidified with acetic acid and then lead acetate solution is added. Formation of black precipitate confirms the presence of sulphur. Na2S + Pb(CH3COO)2  PbS + 2CH3COONa lead acetate black 2.4 Detection of Halogens : Lassaigne’s test. Sodium extract is prepared as already. During fusion , sodium will combine with the halogen (from the organic compound) to form sodium halide Na + X Fusion   NaX (X = Cl, Br, I) (i) A white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound NaCl + AgNO3  AgCl + NaNO3 white ppt. (ii) A dull yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound. NaBr + AgNO3  AgBr + NaNO3 dull yellow ppt. (iii) A bright yellow precipitate, completely insoluble in ammonium hydroxide solution, indicates the presence of iodine in the organic compound. NaI + AgNO3  AgI + NaNO3 bright yellow ppt Special test for bromine and iodine. 2NaBr + Cl 2  2NaCl + Br2 (turns CS2 layer orange) 2NaI + Cl2  2NaCl + I2 (turns CS2 layer violet) Test of phosphorus : The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipi tate indicat es the presence of phosphorus. Na 3PO 4  3HNO3  H3PO 4  3NaNO3 H3PO 4  12(NH4 )2 MoO 4  21HNO3  Ammonium molybdate (NH4 )3 PO 4 .12MoO 3  21NH4NO3  12H2O Ammonium phosphomolybdate 3. QUANTITATIVE ANALYSIS 3.1 Estimation of Carbon and Hydrogen : Principle. A known weight of the given dry organic compound is heated strongly with dry cupric oxide in an atmosphere of air or oxygen free from CO2. The carbon and hydrogen of the organic compound are oxidised to CO2 and water vapour as : C + 2CuO ––––– CO2 + 2Cu (from compound) Page # 5 PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS 2H + CuO ––––– H2O + Cu (from compound) Procedure : Let the mass of organic compound taken = mg Mass of water formed = m1 g (increase in CaCl 2 U-tube) Mass of carbon dioxide formed = m2 g increase in potash tubes) (a) Percentage of Carbon We know that 1 mole of carbondioxide (44 g) contains 1 gram atom of carbon (12g). CO2 C 44 g of CO2 contain C = 12 g 12 m2 g of CO2 contains C = × m2 g 44 12m 2 Hence Percentage of carbon = 44  m × 100 (b) Percentage of Hydrogen We know that one mole of water (18g) contains 2 gram atom of hydrogen (2g) H2O=2H 18g of H2O contain H = 2g 2  m 1 g of H2O contains H = × m1 g 18 2m 1 Percentage of hydrogen = 18  m × 100 Percentage of C Mass of CO 2 formed 12 = × x 100 Mass of compound taken 44 Percentage of H Mass of H2 O formed 2 = × x 100 Mass of compound taken 18 3.2Estimation of Nitrogen : There are two methods for the estimation of nitrogen (i) Duma’s method (ii) Kjeldahl’s method (i) Duma’s method : Principle of the method. A known mass of an organic compound is heated with dry cupric oxide in an atmosphere of carbon dioxide. The carbon and hydrogen are oxidised to carbon dioxide and water respectively while nitrogen is set free. If any oxide of nitrogen is produced during this process, it is reduced to nitrogen by passing over a heated copper gauze. The gaseous mixture is collected over an aqueous solution of KOH when all the gases except nitrogen are absorbed. The volume of nitrogen produced is measured at room temperature and atmospheric pressure. From the volume of N2 produced, percentage of nitrogen in the sample can be calculated. C + 2CuO  CO2 + 2Cu 2H + CuO  H2O + Cu 2N + CuO  N2 + oxides of nitrogen Oxides of nitrogen + Cu  CuO + N2 Let the mass of organic compound = w g The volume of nitrogen collected = Vcm3 Atm ospheric pressure (from barometer) = P mm of Hg Room temperature = t0C Aqueous tension at t0C = a mm of Hg Pressure of dry nitrogen = (P – a) mm of Hg Let us first convert the volume of nitrogen to volume at S.T.P. Experimental conditions S.T.P. conditions P1 = (P – a) mm Hg P2 = 760 mm T1 = (273 + t) K T2 = 273 V1 = V cm3 V2 = ? Applying gas equation  V2 = P2 V2 P1V1 = T2 T1 (P  a) x V x 273 P1V1T2 = = x cm3 (say) 760 (273  t) P2 T1 Now 22400 cm3 of N2 at S.T.P. weight = 28 g 28 x cm3 of N2 at S.T.P. will weigh = × xg 22400  Percentage of nitrogen = 28 x x 100 Mass of nitrogen x 100 = 22400 x w Mass of organic compound (ii) Kjeldahl’s method : This method cannot be used for (i) Organic compounds containing nitrogen in the ring such as pyridine, quinoline, etc. (ii) Organic compounds containing nitrogen (-NO2) and diazo (-N = N-) groups. Principle : A known weight of the organic compound is heated with conc. H2SO4 so that ni trogen is quantitatively converted into ammonium sulphate. The solution is then heated with excess of sodium hydroxide. The ammonia gas evolved is passed into a known but excess volume of standard acid (HCl or H2SO4). The acid left unused is estimated by titrating the solution with standard alkali. From the amount of acid left unused the amount of acid used for neutralisation of ammonia can be calculated. From this percentage of nitrogen can be calculated. . H2 SO 4 C, H, S  Conc   CO2 + H2O + SO2 (from organic compound) Page # 6 PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS Conc. H SO 2 4   (NH4)2SO4 N   (from organic ammonium compound) sulphate (NH4)2SO4 + 2NaOH Heat  Na2SO4 + 2NH3 + 2H2O 2NH3 + H2SO4  (NH4)2SO4 NH3 + HCl  NH4Cl Calculations Let the mass of organic compound = w g Volume of standard acid taken = V cm3 Normality of acid = N1 Let vol. of standard alkali used for neutralisation of unused acid = V2 cm3 Normality of standard alkali = N2 N N2 V2 1V1   =  alkali acid Let this volume of v cm3  Vol. of acid used for neutralisation of ammonia = (V – v) cm3 of N1 normality  Ammonia liberated = (V – v) cm3 of N1 solution Now, 1000 cm3 of 1N NH3 solution contains nitrogen = 14 g (V – v) cm3 of N1 NH3 solution contains nitrogen 14 ( V  v) x N1 g 1000 Percentage of nitrogen = = Mass of nitrogen x 100 Mass of organic compound = 14 (V  v) x N 1 1000 × 1. 4 ( V  v) x N1 100 = w w 3.3 Estimation of Halogens : Carius method : Principle : A known mass of the organic substance is heated with fuming HNO3 in a Carius tube. The silver halide so obtained is separated, washed, dried and weighed. From th e weight of sil ver hal i de formed, the percentage of halogen can be calculated. X + AgNO3   AgX Halogen Carbon, hydrogen or sulphur present in the compound will be oxidised to CO2, H2O and H2SO4 respectively. HNO 3  CO2 C + 2O   HNO 3  H2O 2H + O   HNO 3  H2SO4 S + H2O + 3O   Calculations : Let the mass of organic compound be w g Mass of silver halide formed = a g Now, AgX = X 108 + X parts by weight of silver halide contains X parts by weight of halogen (X is its atomic mass) (108 + X) g of silver halide give halogen = X a g of silver halide will give halogen = X × ag (108  X)  Percentage of halogen = Mass of ha log en x 100 Mass of organic compound = Xxa 100 x (108  X ) w Here X is the atomic mass of halogen, e.g. Cl = 35.5 , Br = 80 (79.9 exact), I = 127 (126.9 exact) Percentage of halogen = Atomic mass of halogen × Mass of silver halide (108 + At.mass of halogen) × Mass of organic substance × 100 3.4 Estimation of Sulphur Sulphur is estimated by Carius method. Principle : A known mass of the organic compound is heated with fuming HNO3 in a sealed tube when sulphur is quantitatively converted into sulphuric acid. It is then precipitated with barium chloride as barium sulphate. The precipitate is filtered, washed, dried and weighed. From the weight of BaSO4 formed, the percentage of sulphur can be calculated. The main reactions are : 3 S + H2O + 3O HNO    H2SO4 H2SO4 + BaCl 2  BaSO4 ppt. Calculations : Let the mass of organic compound = w g Mass of BaSO4 formed = a g BaSO4  S 137 + 32 + 64 = 32 = 233 233 g of BaSO4 contain sulphur = 32 g a g of BaSO4 will contain sulphur = Percentage of sulphur = 32 a 100 × 233 w 32 x a g 233