Pharmaceutical Organic Chemistry 2024-2025 Yield Calculation PDF

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IdolizedDravite694

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Damanhour University

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yield calculation organic chemistry atom economy chemical reactions

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This document provides notes on yield calculation, atom economy, and reaction efficiency for pharmaceutical organic chemistry. The document covers reactions and how to calculate the yield percentages. Calculations show that efficiency also depends on the limiting factor.

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PHARMACEUTICAL ORGANIC CHEMISTRY 2024 - 2025 CALCULATION OF YIELD OF REACTION Reactant A + Reactant B Product Yield : is the amount of product obtained in a chemical reaction. Actual yield: is the weighed of product obtained from reaction Theoretical yield: the q...

PHARMACEUTICAL ORGANIC CHEMISTRY 2024 - 2025 CALCULATION OF YIELD OF REACTION Reactant A + Reactant B Product Yield : is the amount of product obtained in a chemical reaction. Actual yield: is the weighed of product obtained from reaction Theoretical yield: the quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction. The ideal or theoretical yield of a chemical reaction would be 100% ***N.B limiting reactant: the reactant which has no. of moles less than the other CALCULATION OF YIELD OF REACTION Yield Percentage, relative yield: which serve to measure the effectiveness of a synthetic procedure. Theoretical yield = no. of moles of limiting factor x molecular wt of product. (known( (Unknown) Yields% around 100% are called quantitative Yields % above about 90% are called excellent Yields % above about 80% very good Yields % above about 70-60% are called good Yields % below about 50% are called fair Yields % below about 40% are called poor ATOM ECONOMY What is atom economy in chemistry? Atom economy is a measure of the efficiency of a chemical reaction. It is calculated by dividing the total number of atoms in the desired product by the total number of atoms in all the reactants. The higher the atom economy, the more efficient the reaction is considered to be WHY IS ATOM ECONOMY IMPORTANT IN CHEMISTRY? Atom economy is important in chemistry because it provides a way to compare different reactions and determine which one is the most efficient. By choosing reactions with high atom economy, chemists can reduce the amount of waste produced and increase desired product formed. If two chemical reactions each create the desired product, but the first one has a higher atom economy than the second one, it means there is a higher percentage of the desired product formed from the first reaction than there was from the second reaction. IN CASE OF ASPIRIN SYNTHESIS, ACETYLATING AGENT WILL DETERMINE ACCORDING TO %ATOM ECONOMY. 1- Acetyl chloride: 78.49 138.12 g 189.16 g g 189.16 138.12+78.49 Atom economy = x 100 = 87.32 % 2-ACETIC ACID ANHYDRIDE 102.09 138.12 g 189.16 g g 189.16 Atom economy = 138.12+102.09 x 100 = 78.74% 3-GLACIAL ACETIC ACID 138.12 g 60 g 189.16 g 189.16 138.12+60 Atom economy = x 100 = 95.47% %atom economy: 1. Acetyl chloride = 87.32 % 2. Acetic acid anhydride = 78.74% 3. acetic acid = 95.47% Acetic acid has the highest %atom economy, so it is the best acylating agent with salicylic acid for aspirin synthesis Reaction Efficiency: The concept of reaction efficiency was developed as a measure of the mass of reactant atoms actually contained in the final product. Reaction efficiency = % yield X atom economy. One goal of green chemistry is to design synthetic pathways that improve both the atom economy of a reaction and the percentage yield in order to minimize the waste produced by chemical reactions. Example 1 Calculate the yield % and reaction efficiency obtained from reacting 120 g of acetic acid with 230 g ethanol to produce 132 g of ethyl acetate. 1st step: calculate molecular weight of reactants and product. CH3COOH = C2H4O2 = 2*12 + 4*1 + 2*16 = 60 g/mol C2H5OH = C2H6O = 2*12 + 6*1 + 1*16 = 46 g/mol CH3COOC2H5 = C4H8O2 =4*12 + 8*1 + 2*16 = 88 g/mol 2nd step: Calculation of no. of moles of reactants: No. of moles = weight in grams / molecular weight No. of moles of acetic acid = 120 g/60 = 2.00 mol No. of moles of ethanol = 230 g/46 = 5.00 mol Determine the limiting factor (the reactant which has no. of moles less than the other). acetic acid is the limiting factor. Therefore, no of moles of products= no of moles of limiting factor 3rd step: Calculate the theoretical yield : Theoretical yield = no. of moles of limiting factor x molecular wt of product. Theoretical yield = 2x88 = 176 grams 4th step: Calculate yield % % yield= Actual yield / theoretical yield *100 % yield = 132 x 100 = 75% 176 5th step: calculation of atom economy and reaction efficiency % : Atom Economy = Mass of product × 100 % Mass of reactant 88 ×100 % 46+60 = 83.01% Reaction efficiency = % yield X atom economy. 75 % × 83.01 =62.25% For example, consider the synthesis of 1-ethoxybutane from 1-bromobutane and sodium ethoxide. ▪ The procedure specifies 4.50 mL of 1-bromobutane, 3.70 g of sodium ethoxide, and 20 mL of anhydrous ethanol to produce 2.7 g of 1-ethoxybutane ▪ To calculate the theoretical yield, it is necessary to ascertain whether 1-bromobutane or sodium ethoxide is the limiting reagent by calculating the moles of each reagent present in the reaction mixture: 1st step: calculate molecular weight of reactants and product. weight CH3(CH2)3-Br = C4H9Br = 4*12+ 9*1+80 =137 g/mol Density = volume CH3CH2-O-Na+ = C2H5ONa = 2*12+ 5*1+ 16+ 23 = 68 g/mol CH3(CH2)3-O- CH2CH3 = C6H14O = 6*12+ 14*1+ 16 = 102 g/mol 2nd step: Calculation of no. of moles of reactants: No. of moles = weight in grams / molecular weight The procedure specifies 4.50 mL of 1-bromobutane, with density = 1.27 g.ml -1 and 3.70 g of sodium ethoxide, Therefore, 1-bromobutane is the limiting reagent 3rd step: Calculate the theoretical yield : Theoretical yield = no. of moles of limiting factor x molecular wt of. product Theoretical yield= 0.0417* 102=4.25 g of 1-ethoxybutane 4th step: Calculate yield % % yield = 2.7 x 100 = 63.52 = 64% 4.25 5th step: calculation of atom economy and reaction efficiency % : Atom Economy = Mass of product × 100 % Mass of reactant 102 ×100 % 68+137 = 49.75% Reaction efficiency = % yield X atom economy. 64% ×49.75 =31.84% PRACTICAL WORK Liberation of acid from Salt Example: sodium benzoate Sodium benzoate is an organic sodium salt resulting from the replacement of the proton from the carboxy group of benzoic acid by a sodium ion. The sodium salt of BENZOIC ACID. It is used as an antifungal preservative in pharmaceutical preparations and foods. It may also be used as a test for liver function. PROCEDURES 1- Dissolve 1g of sodium benzoate in 5 ml water in small beaker. 2- Add 2.5 ml of 3N HCl drop wise until you notice a precipitate. 3- filtration. 4- Recrystallization of the solid obtained by dissolving it in hot water. TOOLS 2 small beakers Conical flask Watch glass Funnel 2 filter paper Dropper Lab coat Eye goggles Towel Holder

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