CHE 176 Lecture 2 PDF

DETERMINATION OF MOLECULAR COMPOSITION AND STRUCTURE OF ORGANIC COMPOUNDS This involves many steps viz; (1) Isolation and Purification of the compound (2) Qualitative Analysis of the compound which involves: (a) Test for the elements in the compound (b) Test for the functional Groups ...

DETERMINATION OF MOLECULAR COMPOSITION AND STRUCTURE OF ORGANIC COMPOUNDS This involves many steps viz; (1) Isolation and Purification of the compound (2) Qualitative Analysis of the compound which involves: (a) Test for the elements in the compound (b) Test for the functional Groups (3) Quantitative Analysis which includes: (a) Combustion experiments to determine the amount of each element present in the compound (b)Determination of Empirical Formulae (c)Determination of molecular mass/weight (d) Determination of molecular formulae (4) Structure Determination which involves: (a) Use of the information in 1 to 3 above (b) Spectroscopic Analysis – IR, UV, NMR, Mass Spectrometry e.t.c. (c) Preparation of Derivatives e.g. esters (d) Degradation reactions (e) Synthesis. ISOLATION AND PURIFICATION OF THE COMPOUND The organic compound could be isolated from different natural sources: plants, soil, microorganisms etc. Process of isolation and purification involves extraction with different solvents (solvents of different polarities) and various chromatographic techniques e.g. Thin Layer Chromatography (TLC), Column chromatography (CC), Paper Chromatography (PC), High Performance or (Pressure) Liquid Chromatography (HPLC) etc.  QUALITATIVE ANALYSIS This involves test for the elements present in the compound and the different functional groups. Test for C and H is unnecessary since most organic compounds contain C and H, (However their presence can be detected (confirmed) by combustion experiment in which the organic compound is heated with copper (II) oxide [CuO]. The C is converted to CO2 and the H to H2O. CO2 turns lime water [Ca(OH)2] milky while H2O turns anhydrous CuSO4 blue and anhydrous Cobalt(II)Chloride [CoCl2] pink. 2CHO + 3CuO Δ 2CO2 + H2O + 3Cu The presence of N, S and Halogens (X) is detected by Sodium Fusion test known as [LASSAIGNE’S] TEST. In this test, the organic compound (small portion) is fused (heated strongly) with small piece of sodium metal in a combustion tube. The products of the fusion reaction are extracted with water and analyzed (tested) accordingly. During fusion, the Nitrogen [N] is converted to sodium cyanide [NaCN], the Sulphur [S] to sodium sulphide [Na2S] and the Halogens [X] to sodium halide [NaX] It is important to use a little excess of the sodium metal to prevent formation of sodium thiocynate [NaCNS] which may hinder detection of the N and S. In the presence of the excess Na, any thiocynate formed is converted to NaCN and Na2S i.e. NaCNS + 2Na Δ NaCN + Na2S At the end of the reaction, the combustion tube and contents are added to water, mixed together and filtered. The filtrate is used to test for the elements.  TEST FOR NITROGEN This is the first test to be carried out. To a portion of the filtrate in a test tube, add freshly prepared Iron (II) sulphate [FeSO4] solution. Heat, cool and add few drops of Iron (III) Sulphate solution and dilute H2SO4. A blue or green colouration with or without precipitate indicates presence of CN i.e. (N) in the compound. 6NaCN + FeSO4 → Na2SO4 + Na4Fe(CN)6 Na4Fe(CN)6 + 2Fe2(SO4)3 → 6Na2SO4 + Fe4[Fe(CN)6]3 Prussian blue  TEST FOR SULPHUR To a portion of the filtrate in a test tube, add dilute HNO3 to acidify and few drops of lead ethanoate (acetate) solution. A black precipitate of lead sulphide confirms presence of S. Na2S + Pb(OCOCH3)2 → PbS + 2CH3COONa TEST FOR HALOGENS If N and /or S are (is) present, a small portion of the filtrate is heated with dil. HNO3 first in a beaker/ flask to about half its volume. The solution is cooled, then AgNO3 is added. The purpose of heating with dil. HNO3 is to remove the N/S present in the compound as HCN and H2S so as not interfere with the detection of the halide. NaCN + HNO3 Δ HCN + NaNO3 Na2S + 2HNO3 Δ H2S + 2NaNO3 Otherwise, the cyanide and sulphide can react with AgNO3 to give insoluble AgCN and Ag2S which will give a false positive test for the halides. CN- + S2- + 3Ag+ → AgCN + Ag2S If N/S is absent in the compound, dilute HNO3 solution and AgNO3 is added to a portion of the filtrate right away. The different precipitates produced indicate the presence of the halogens. (a) A white precipitate, soluble in excess NH4OH indicates the presence of chloride [Cl-] (b) A cream (light yellow) precipitate slightly soluble in excess NH4OH shows Bromide [Br -]. (c) A yellow precipitate, insoluble in excess NH4OH shows presence of Iodide [I-].  TEST FOR FUNCTIONAL GROUP The various functional groups [viz; alkenes and alkynes, alcohols, alkanals, alkanones, carboxylic acids, esters, amides, phenols, amines, nitriles, nitro compounds aromatic compounds etc can also be tested for (DETAILS LATER).  QUANTITATIVE ANALYSIS The goal of quantitative analysis is to know OR determine the relative amount of the different elements present in the organic compound i.e. to determine the percentage of each element present in the compound. Conventional Method of Determination of the Percentage Composition of Elements in an Organic Compound This is outlined below: (a) Determination of Quantity of C and H A known weight of the organic compound is burnt (heated) in excess oxygen to give CO2 and H2O. The CO2 is absorbed in a known weight of soda lime (NaOH) or KOH while the H2O is absorbed in a known weight of magnesium perchlorate [Mg(ClO4)2] or Calcium chloride [CaCl2]. The weight of CO2 and H2O is obtained by difference or subtraction and the weights of C and H can be calculated from these: Wt of Carbon (C) = RAM of carbon × Wt ofCO2 RMM of CO2 i.e. 12 x wt of CO2 44 Wt of Hydrogen (H) = RAM of H2× Wt of H2O RMM of H2O i.e. 2 x wt of H2O 18 RAM – Relative Atomic Mass RMM – Relative Molecular Mass Determination of Quantity of Oxygen The percentage of oxygen is always determined as the difference between 100% and the sum of the percentage values for all other elements present in the compound i.e. %O =100% -(Percentage of all other elements).  CALCULATION OF EMPIRICAL FORMULA Empirical Formula (E.F.) is the ratio of the number of atoms of each element present in the compound. It is usually calculated from the percentage composition of the elements. WORKED EXAMPLES Question 1: An organic compound X on quantitative analysis was found to contain 40.0% carbon and 6.7% hydrogen. Calculate its empirical formula. [RAM: H = 1, C = 12, O = 16] Solution: 1st, assume the remaining element is O and calculate its percentage composition by difference or subtraction: % O = 100 – (40.0 + 6.7)= 100 – 46.7= 53.3 % The E.F. can be calculated using a simple table as outlined below: Element Carbon [C] Hydrogen[H] Oxygen[O] %Composition OR Weight 40.0 6.7 53.3 No of Moles 40 6.7 53.3 12 1 16 Relative No of Moles 3.33 6.7 3.33 Divide the moles with the 3.33 = 1 6.7 = 2.01 3.33 = 1 3.33 3.33 3.33 lowest Simplest Mole Ratio 1 2 1 Empirical Formula (E.F.) of compound X = CH2O DETERMINATION OF MOLECULAR MASS AND MOLECULAR FORMULA The molecular formula (M.F.) is the actual number of each kind of element present in one molecule of the compound. M.F = (E.F)n where n = whole number. In order to determine the molecular formula, the molecular mass (MM) of the compound must be known. There are many methods of determining the MM. The modern method of choice is by High Resolution Mass Spectrometry [HRMS]. This technique gives the exact mass of the compound. Once the MF is known, then the structural formula or structure of the compound can be determined. Hence for compound X, If Molecular Mass (MM) = 120, what is the Molecular Formula? MF = (E.F.)n = MM MF of cpd. X = (CH2O)n =120 (12+2+16)n =30n = 120, n=120/30 = 4 MF = (CH2O)4 = C4H8O4 Hence Molecular Formula (MF) of compound X = C4H8O4 Structural formula can then be determined. Question 2: 6.51mg of an organic compound, B gave on combustion 8.36 mg of H2O and 20.47 mg of CO2. Quantitative analysis indicated absence of N,S and halogens. Determine the molecular formula [M.F.] of the compound if the MM is 126. SOLUTION 1st calculate the weight of each element in the compound and their percentage composition. Weight of carbon in sample = 12 × 20.47 = 5.58mg 44 % of Carbon = 5.58 × 100 = 85.7% 6.51 Weight of Hydrogen in sample = 2 × 8.36 = 0.93mg 18 % of H = 0.93 × 100 = 14.3% 6.51 % C + %H = 85.7 + 14.3 = 100%, hence B is an hydrocarbon, it has no oxygen.  Empirical Formula(E.F.) Determination Element C H C H % 85.7 14.3 5.58 0.93 composition OR mass Number of 85.7 = 7.14 14.3 = 14.3 5.58 = 0.47 0.93 = 0.93 moles 12 1 12 1 Relative No 7.14 = 1 14.3 = 2 0.47 = 1 0.93 = 2 of moles 7.14 7.14 0.47 0.47 Simplest 1 2 1 2 Mole Ratio Therefore, Empirical formula of cpd. B = CH2 MM = 126 MF = (EF)n = MM MF = (CH2)n = 126 14n = 126 n = 126 = 9, MF = (CH2)9 = C9H18 14 Hence, MF of cpd. B = C9H18 Question 3: Quantitative analysis of a liquid organic compound D, furnished 48.76% C and 8.07% H. No other element was found. Calculate it E.F.? If the MM is 74, what is the MF? SOLUTION % C + % H = 48.76 + 8.07 = 56.83 Therefore, the compound must contain Oxygen to make 100% % O = 100 – 56.83 = 43.17  E.F. Determination Element C H O % composition 48.76 8.07 43.17 No of Moles 48.76 8.07 43.17 12 1 16 Since atoms 4.06 8.07 2.70 combine in whole numbers, Relative No of 4.06 8.07 2.70 simple mole 2.70 2.70 2.70 ratio is Moles multiplied by 2. x2 Simplest Mole Ratio 1.50 2.99 1.0 3.0 5.98 2.0 3.0 6.0 2.0 Empirical Formula of D = C3H6O2 MM = 74 MF = (EF)n = MM MF = (C3H6O2)n = 74 (36+6+32)n = 74 74n = 74, n=1 MF = (C3H6O2)1= C3H6O2 Hence, MF of cpd. D = E.F. = C3H6O2 Note: If on dividing by the smallest no/or calculating simplest mole ratio, fractions are obtained as above, multiply the ratios by small numbers (e.g. 2,3) until a ratio of nearly whole numbers is obtained. This is because atoms combine in whole numbers. Practice questions1. Compound H has a composition of 78.5% C, 8.4% H and 13.1% N. Determine its Empirical formula. If Molecular Mass of H = 107, what is the Molecular Formula?