8. Equivalencies and yields
CHEM 231 Lab
Technique Primer
8. Equivalencies and yields When a reaction is carried out, the components are measured out in grams or milliliters. However, these quantities reveal nothing about the underlying chemistry. Therefore, the real metrics of a reaction lie in molar quantities and their relationship to each other (that is, equivalencies). To illustrate, consider the reaction of benzyl bromide (1) with sodium cyanide in dimethylformamide (DMF, a solvent) to form benzonitrile (2), as shown in Scheme 1. Even a cursory inspection should reveal that one molecule of benzyl bromide reacts with one molecule of sodium cyanide to generate one molecule of benzonitrile. Thus, there is a 1:1:1 molar relationship among those components.
product that can be formed is 19.8 mmol. However, it is also useful to know exactly how much more (or less) of a reagent there is in a reaction mixture. This relationship is expressed as equivalency, that is, the molar ratio of one component with respect to the other. In this example, the equivalency of sodium cyanide is 1.2, as the following calculation shows: 23.7 mmol NaCN 19.8 mmol BnBr = 1.2 eq NaCN Another useful parameter to calculate is the molarity (or concentration) of the reaction. In this case, there is 19.8 mmol of benzyl bromide (the limiting reagent) and the total volume of the reaction is about 20 mL (this value is not explicitly given, but is rather a reasonable estimate extrapolated from the information provided). Thus, the molarity of the reaction is about 1 M, as illustrated in the following calculation: 19.8 mmol BnBr 20 mL total reaction volume = 0.99 mmol/mL = 0.99 mol/L = 0.99 M
Scheme 1. Preparation of benzonitrile. The following passage describes one experiment run to prepare benzonitrile: Benzyl bromide (2.35 mL) is added to DMF (18 mL) at room temperature, after which sodium cyanide (1.16 g) is added in one portion. The reaction is stirred at room temperature until TLC shows the disappearance of starting material (2.5 h). The reaction mixture is transferred into a separatory funnel and partitioned between water (20 mL) and dichloromethane (20 mL). After removal of solvent under vacuum, benzonitrile is left as an oily residue (1.97 g). Since the stoichiometry of the reactions depends ultimately upon moles, not grams, it is useful to convert these mass quantities, as follows:
Finally, it is always necessary to calculate the yield of a reaction, and this is usually expressed as percent yield (i.e., percent of theory yield). This requires two steps, namely, calculating the theory yield and then comparing the actual yield to theory. In this example, the yield is 85%, as shown in the following calculations: theory yield 19.8 mmol product predicted x 117 g/mol* = 2317 mg = 2.32 g *note the unit of g/mol is the same as mg/mmol
percent yield 1.97 g actual 2.32 g theory x 100% = 84.9% yield
benzyl bromide 2.35 mL x 1.438 g/mL = 3.379 g* 3.379 g 171 g/mol = 0.01976 mol = 19.8 mmol sodium cyanide 1.16 g 49.0 g/mol = 0.02367 mol = 23.7 mmol *one extra significant figure retained to prevent rounding error
Right away it is evident that the sodium cyanide is in excess; therefore, the benzyl bromide is the limiting reagent. In other words, the maximum amount of
photo of analytical balance from vanleestantiques.com
Technique Primer
8. Equivalencies and yields When a reaction is carried out, the components are measured out in grams or milliliters. However, these quantities reveal nothing about the underlying chemistry. Therefore, the real metrics of a reaction lie in molar quantities and their relationship to each other (that is, equivalencies). To illustrate, consider the reaction of benzyl bromide (1) with sodium cyanide in dimethylformamide (DMF, a solvent) to form benzonitrile (2), as shown in Scheme 1. Even a cursory inspection should reveal that one molecule of benzyl bromide reacts with one molecule of sodium cyanide to generate one molecule of benzonitrile. Thus, there is a 1:1:1 molar relationship among those components.
product that can be formed is 19.8 mmol. However, it is also useful to know exactly how much more (or less) of a reagent there is in a reaction mixture. This relationship is expressed as equivalency, that is, the molar ratio of one component with respect to the other. In this example, the equivalency of sodium cyanide is 1.2, as the following calculation shows: 23.7 mmol NaCN 19.8 mmol BnBr = 1.2 eq NaCN Another useful parameter to calculate is the molarity (or concentration) of the reaction. In this case, there is 19.8 mmol of benzyl bromide (the limiting reagent) and the total volume of the reaction is about 20 mL (this value is not explicitly given, but is rather a reasonable estimate extrapolated from the information provided). Thus, the molarity of the reaction is about 1 M, as illustrated in the following calculation: 19.8 mmol BnBr 20 mL total reaction volume = 0.99 mmol/mL = 0.99 mol/L = 0.99 M
Scheme 1. Preparation of benzonitrile. The following passage describes one experiment run to prepare benzonitrile: Benzyl bromide (2.35 mL) is added to DMF (18 mL) at room temperature, after which sodium cyanide (1.16 g) is added in one portion. The reaction is stirred at room temperature until TLC shows the disappearance of starting material (2.5 h). The reaction mixture is transferred into a separatory funnel and partitioned between water (20 mL) and dichloromethane (20 mL). After removal of solvent under vacuum, benzonitrile is left as an oily residue (1.97 g). Since the stoichiometry of the reactions depends ultimately upon moles, not grams, it is useful to convert these mass quantities, as follows:
Finally, it is always necessary to calculate the yield of a reaction, and this is usually expressed as percent yield (i.e., percent of theory yield). This requires two steps, namely, calculating the theory yield and then comparing the actual yield to theory. In this example, the yield is 85%, as shown in the following calculations: theory yield 19.8 mmol product predicted x 117 g/mol* = 2317 mg = 2.32 g *note the unit of g/mol is the same as mg/mmol
percent yield 1.97 g actual 2.32 g theory x 100% = 84.9% yield
benzyl bromide 2.35 mL x 1.438 g/mL = 3.379 g* 3.379 g 171 g/mol = 0.01976 mol = 19.8 mmol sodium cyanide 1.16 g 49.0 g/mol = 0.02367 mol = 23.7 mmol *one extra significant figure retained to prevent rounding error
Right away it is evident that the sodium cyanide is in excess; therefore, the benzyl bromide is the limiting reagent. In other words, the maximum amount of
photo of analytical balance from vanleestantiques.com
