Question

Question 1 9 pts Below is a partial data table with the hypothetical amounts of reagents you used for the oxidation of borneol, plus the hypothetic final yield of product you obtained. reagent MW conc(M) d/g/mL) vol(mL) mass (mg) borneol 154.25 118.3 NaCl 1 0.15 Oxone 307.37 232.1 EtOAC 0.897 0.4 camphor 152.24 yield camphor (mg) 99.2 Using these amounts, calculate the following values. Answers should be reported to the correct level of significant figures. Note: equivalents are usually based on the reactant with the smallest number of moles being "1 equivalent" exactly. A catalyst is not a reactant, and can be < 1 equiv. mmol borneol: mmol NaCl: mmol Oxone: equivalents borneol: equivalents NaCl: equivalents Oxone: Which reagent is the limiting reactant? What is the theoretical yield of camphor (in milligrams)? What is the percent yield of camphor?

Answer 1

1) moles of Borneol = mass / molar mass = 118.3*10^-3 g/ 154.25g/mol = 0.767 *10^-3 mol mmol of borneol = 0.767 mmol Moles of Naci = (1mol/1000mL) * 0.15 mL = 0.150 *10^-3 mol mmol of NaCl = 0.150 mmol Moles of oxone = 232.1*10^-3 g/307.37 g/mol = 0.755 *10^-3 mol mmol of oxone = 0.755 mmol Equivalents of borneol = 0.767/0.767 = 1.00 NaCl is catalyst. Equivalents of NaCl = 0.150/0.767 = 0.196 1 molecule of oxone molecule contains 2 molecules of active oxidizing agent KHSO5 (potassium peroxymonosulfate). Hence while calculating equivalents we should consider half of the moles. Equivalents of Oxone = (0.755*2)/0.767 = 1.97 Theoretical moles of camphor = 0.767 mmol Theoretical yield of camphor = 0.767*10^-3 * 152.24 = 117 mg % yield = (practical yield / theoretical yield ) *100 = (99.2/117)*100 = 84.8%

2) CI- (chloride ion) is oxidized from an oxidation number of -1 to an oxidation number of +1. S05 (peroxymonosulfate) is reduced from an oxidation number of -1 to an oxidation number of -2. 3) S is oxidized from an oxidation number of +4 to an oxidation number of +6. I is reduced from an oxidation number of 0 to an oxidation number of -1. 4) The oxidizing agent oxone is soluble in aqueous layer and it is not present in the upper organic layer. Hence to test the unreacted oxidizing agent we need to test it from lower aqueous layer.