CALCULATIONS FOR VANILLYL ALCOHOL SYNTHESIS

 The calculations for this exercise are particularly simple.  We measure the success of our synthesis by the PERCENT YIELD.  The percent yield is defined as the percent of the maximum weight of product that could be obtained represented by what is actually obtained.  I.e., PERCENT YIELD = 100 X Actual Weight of Product / Maximum Possible Weight of Product This calculation is dependent on only two features -- the stoichiometry of the equation that represents the synthesis, namely: 4 C8H8O3 + BH4- + 4 H2O = 4 C8H10O3 + OH- + H3BO3 and knowledge of the limiting reagent. We use the following quantities of the reactants:

 Substance Amount mmoles Available mmoles Required Vanillin 400 mg 400 mg / 152 mg/mmol = 2.63 mmol 2.63 mmol Sodium Borohydride 80 mg 80 mg / 38 mg/mmol = 2.1 mmol 1/4 X 2.63 = 0.66 mmol Water 2.5 mL of 1 M NaOH 2.5 mL X 1000 mg/mL / 18 mg/mmol = 139 mmol 2.63 mmol

 In order to make comparisons of the "chemical amounts" of each substance, we must convert all of the available amounts to moles (or millimoles), which we do in the third column of the above table.  The fourth column shows how much of each reagent is needed to form the product.  It should be apparent that the limiting reagent is vanillin.   All of the other reagents are present in large excess. Since the stoichiometry of the reaction shows that 1 mole of sodium borohydride reduces 4 moles of vanillin, we require only 1/4th as many mmol. If we begin with 2.63 mmol of vanillin, the maximum amount of vanillyl alcohol we can produce s 2.63 mmol. What remains to be calculated is the corresponding weight of vanillyl alcohol. The molar mass of vanillyl alcohol is 154 (2 mass units more than vanillin). Therefore, 2.63 mmol of alum will weigh 2.63 X 154 = 405 mg. This is our theoretical maximum weight. Suppose our synthesis actually produces 348 mg.  Our PERCENT YIELD will then be % YIELD = 100 X 348 / 405 = 85.9 %
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