(A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform

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John Walter Farmer
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1 PURIFICATION METHODS Distillation Techniques : Ty p e C ondi ti ons E x a m p l e s (A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform non volatile impurities (BP = 334K) and Aniline (BP = 457K) (ii) When boiling point (ii) Mixture of Ether (BP = 308K) & difference is 80 K or more. Toluene (BP = 384K) (iii) Hexane (34K) and Toulene(384K) (B) Fractional distillation When BP difference is 10 (i) Crude oil in petroleum industry (ii) Acetone (39) and Methyl alcohol(338k) (C) Distallation under When liquid boils at higher (i) Concentration of sugar juice reduced pressure temperature and it may (ii) Recovery of glycerol from spent lye. (Vacuum distillation) decompose before BP is attained. (iii) Glycerol (D) Steam distillation When the substance is immiscible (i) Aniline is separated from water with water and steam volatile. (ii) Turpentine oil (iii) Nitro Benzene (iv) Bromo Benzene (v) Naphthalene (vi) o-nitrophenol SIMPLE DISTILL ATION TECHNIQUE

2 FR ACTIONAL DISTILL ATION DISTILL ATION UNDER REDUCED PRESSURE STE A M DISTILL ATION

3 LASSAIGNE METHOD : 1. Nitrogen (N) : Na + C + N NaCN (Sodium extract) Test : NaCN + Fe + NaOH boil & cool FeCl 3 + conc. HCl Blue or green colour [Fe 4 [Fe(CN) 6 ] 3 Prussian blue. Sulphur (S) : Na + S Na S (Sodium extract) Test (1) Na S Sodium nitro prusside A deep voilet colour Test () Na S + CH 3 COOH + (CH 3 COO) Pb Black colour (PbS) 3. Halogen (X) : Na + Cl NaCl (Sodium extract) Test : NaCl + HNO 3 + AgNO 3 (a) White ppt soluble in aq. NH 3 confirms Cl (b) Yellow ppt partially soluble in aq. NH 3 confirms Br (c) yellow ppt insoluble in aq. NH 3 confirms I 4. Nitrogen a nd Sulphur toget her : Na + C + N + S NaCNS (Sodium thiocynate) As in test of nitrogen, instead of green or blue colour, blood red coloration confirms presence of N and S both. 5. Phosphorous (P) : P Na O boils, Na 3 PO 4 THE QUANTITATIVE ANALYSIS : Solution in boiled with nitric acid and then treated with ammonium molybdate (NH 4 ) MoO 4 Formation of yellow ppt indicate presence of phosphate hence phosphorous present in organic compounds. Quantitative analysis involves the estimation of percentage composition of various element by suitable method. The molecular mass of the compound is also determined by suitable methods. The knowledge of molecular mass and percentage composition help us to determine the molecular formula of the compound. ESTIMATION OF CARBON AND HYDROGEN OR LIEBIG COMBUSTION METHOD : Organic compound + dry, black CuO (w g) Heat strongly in combustion tube mixture of oxide Pass through (i) anhydrous Cu (ii) anhydrous CaCl

4 say increase in weight = a g Pass through (i) KOH (aq) (ii) Saphnolite resin say increase in weight = b g For % of H - The increase in weight is 'a' g due to the formation of H O 18 g H O has H = g a a g H O has H g 18 Since w g organic compound has = a 18 g hydrogen. 100 g organic compound has= a w g hydrogen weight of H O formed 18 weight of subs tan ce % H 100 For % of C - The increase in wt is b g due to formation of CO 44 g CO has C = 1 g b g CO has C = 1 b 44 g Since w g organic compound has 1 b 44 g carbon 100 g organic comound has b g carbon 44 w 1 weight of CO formed 44 weight of subs tan ce % C 100 Note : This method is suitable for estimation if organic compound contains C and H only. In case if other elements e.g., N,S, halogens are also present the organic compound will also give their oxides which on being absorbed in KOH will increase the percentage of carbon and therefore following modification should be made. (a) If N is also present : The gaseous oxide mixture is first passed over heated Cu gauze which reduces all oxides of N to N which are neither absorbed by H O nor by KOH NO + Cu N + CuO (b) If halogens are present : A roll of silver gauze is placed at the exit end of combustion tube which retains all the halogens on itself. Ag + X AgX Ag + CuX AgX + Cu (c) If S is present : A layer of fused lead chromate is placed near the exit end, all the oxides of sulphur are oxidised to Pb & left in tube it self. Cr e Cr +3 S +4 S +6 + e (in SO ) (in SO 4 )

5 GENER AL CONCEPT FOR DRYING AND DEHYDR ATING AGENTS (1) Acidic oxides ( non metals oxides such as CO, NO, SO P O 5 etc.) are absorbed by alkali (KOH, NaOH, Ca(OH) etc.) (ii) Basic oxides (Na O, K O etc.) are dissolved in acids (H, HCl etc.) (iii) Acidic oxides can not be dried by basic dehydrating agent (e.g CaO) but they are dried by acid dehydrating agent (e.g H, P O 5 etc). However H S can be dried by H because it reacts with H. (iv) H S + H H O + S + SO Basic substances can not be dried by acidic dehydrating agent but are dried by basic dehydrating agent (e.g CaO). It is because NH 3 is dried by CaO but not by H. ESTIMATION OF NITROGEN (a) Duma' s method : Organic compound CuO heated strongly Mixture of oxides Pass over Cu gauze N + Other oxides (w g) (Let Vml at N.T.P.) Dry N 400 ml N weigh at N.T.P. = 8 g drying agent N + Moisture Pass through KOH (aq.) V ml N weigh at NTP = V g since w g organic compound gives V w 100 g 8 Volume of N at NTP 400 weight of the compound % N 100 Note : This method can be used to estimate nitrogen in all types of organic compounds. KJELDAHL'S METHOD : This method is based on the principle that a nitrogenous organic compound on heating with conc. H converts all its nitrogen quantitatively to ammonium sulphate ; organic compound + conc. H + K + Cu (w g) (raises the B.P. of H ) (catalyst) Heating in Kjeldahl flask add KOH (NH ) SO NH is formed 4 4 drop by drop 3 Pass this gas through V 1 ml of N 1 H Now titrate the H by another alkali say of N 1 H (NH 3 is absorbed here) ml of N,KOH are used Meq of H taken = N 1 V 1 Meq. of KOH used for H = N Meq. of H for NH 3 = N 1 V 1 N Meq. of NH 3 formed = (N 1 V 1 N ) or say NV) (the difference obtained or V ml of N H are used to absorb NH 3 )

6 W NH NV WNH NV Now 17 g NH 3 has 14 g Nitrogen 17 NV NV g NH 3 has w g organic compound has 14NV g organic compound has g Nitrogen = 14NV 1000 g Nitrogen g nitrogen 14NV 100 w 1000 = 1.4NV w g nitrogen. g nitrogen Normality of the acid volume of the acid % Nitrogen 1.4 weight of the compound Note : This method is simpler and more convenient and is mainly used for finding out the percentage of nitrogen in food stuffs, soil, fertilizers and various agricultural products. This method cannot be used for compound having nitro groups, azo gp. ( N = N ) and nitrogen in the ring (pyridine, quinole etc.). Since nitrogen in these compounds is not quantitatively converted in to ammonium sulphate. ESTIMATION OF SULPHUR (CARIUS METHOD) : Organic compound + con. HNO 3 + BaCl ppt (Ba ) is dried and weighed say "a" g (w g) Furning.HNO3 S BaCl H Ba + 33 g Ba has 3 g sulphur (a g) HCl a g Ba has 3 a 33 g sulphur % of Sulphur = 3 a w 3 weight of BaSO formed 33 weight of compound 4 % S 100 ESTIMATION OF HALOGENS ( CARIUS METHOD) : Organic compound + con HNO 3 + AgNO 3 ppt (AgX) is dried and weighed say a g % of Cl = % of Br = 35.5 weight of AgCl weight of compound 80 weight of AgBr weight of compound % of I = 17 weight of AgI weight of compound

7 ESTIMATION OF PHOSPHORUS - (CARIUS METHOD) : Organic compound + con. HNO 3 + Magnesia mixture (Mg + NH 4 OH + NH 4 Cl) (w g) MgNH 4 PO 4 Mg P O 7 + NH 3 + H O Precipitate is dried and ignited to give magnesium pyrophosphate (Mg P O 7 ) weighed say a g 6 weight of Mg P O 7 formed % of P 100 weight of compound Atomic weight of P = 31 = 6 g molecular weight of Mg P O 7 = g ESTIMATION OF OXYGEN : There is no direct formula and method for estimation of oxygen however, percentage of oxygen is calculated by % of oxgyen = 100 [Sum of the % of all other elements] Note - A direct method for estimation of oxygen has been given by Alusive in 1947 known as Alusive method.

Chemistry Notes Class 11 Chapter 12 Organic Chemistry Some Basic Principles and Techniques part 2

1 P a g e Chemistry Notes Class 11 Chapter 12 Organic Chemistry Some Basic Principles and Techniques part 2 Organic compounds extracted from a natural source or synthesized in the laboratory requires purification.