Question

6. Please fill this table completely Molar equiv. Substance Water Pentane Acetone d (g/mL) 0: 997 0.626 0.784 TEMPO M. Wt. (g/mol) 18.02 72.15 58.08 156.25 156,19 143.45 82.10 201.06 0.96 Amount 25ML 25 mL 15 mL 40 mg 40mg 25 mg 4 dropš 2.5 mmol 11.03 approx Bipyridine CuBr NMI Starting alcohol: (4-bromophenylmethanol Product aldehyde: 4-bromobenzaldehyde 1.44 185.02 1.85

Answer 1

Water mass= density & volume = 0.9979 x 25 ml = 24.925 9 nal moles mass 24.925 -1.383 mola mass 18.02 g mol mo! Pentare mass= 0.6269 x 25m = 15.65 g m Moles = 15.65g = 0. 217 mol 72.15 gmot Acetone Mass = 0.7049 X 15 ml =11.769 ML Moles = 11.769 = 0.2025 mol 58.08 glamo TEMPO Mass = 40mg moles hong 156.25 gmot = 0.256X163 mol Bipyridine = 0.256 mmol Mass = 40mg moles = tony = 0.256 mmol 156-18 g/mol = 0.25 6 xlo mol CUBE Mass=35mg moles = 5mg = 0.244 nmol 143 · 45 g/mol = 0.244 x10 ml Starting Alcohol 2.5 mmol. Molar equwalere TEMPO = 0.256 mmol = 1.05) Bipyridine = 0.256 mmol = 105 0.244 mmol Cubr= 0.244 mohol = 0-244m 16 TEMPO = 214 mmol

Starting Alcohol = 2.5 mmol = 10.24 | o. 244 mmol 10.243€ Chemical Molar equivalence 1.05 1:05 TEMPO Bipyridine GUBY Starting Alcohol Product aldehyde 10.24 10.24

Water, acetone, pentane are solvents and hence molar equivalence would not be for them.