Question

Chemical Table (4 pts): 1) Fill in the reaction table below. Make sure you correctly calculate the molar amounts of your reactants. name formula mol.-eq. Mw mmol amount Meso-Stilbene Dibromide 1.00 300 mg Potassium Hydroxide 5.7 290 mg Triethylene glycol 2 mL product 2) Potassium hydroxide is hygroscopic. Will it be easy to weigh exactly 290 mg? Will the exact amount of potassium hydroxide be critical to this reaction and why?

Answer 1

88 B T*"Bo + KOH + KOH Tsiethylene, glycogene glycol = meso Stillbene diboomide Diphenylacetyione (Product) (mol-wt. = 340.05) Name / formula 1 mol. eg. mol.cot. mmol amount me so Shilbere oiboomide | 1.00 / 340.05 0.882 300 mg KOH KOH poti hydroxide 5.7 56.11 / 5.168 | 290 mg GH2-OH rolethylene glyco) (Solveni) CH2 Сн 270L CH, . Ch-ch-01 product - Paeda Ph-cac-ph | 1 o Ph: Cặc- Pb 18:29 - 889 | 154.19 ° 157.1

i mol of Stillbene diboomide = 340.05 g i mmol of Stillbene diboomide = 340.05 mg : 300 mg = ( 300 x 1) = 0.882 mmol 340.05 i mole of KOH = 56.11 g Of KOH I motos ol of KOH = 56.16 mg of KOH : 290 mg Of KOH = (290x1) = 5.168 mmol 56:11 According to equation, I emole of Shilbene dibromide gives Imole of product :. 0.882 mmol of Shillbene dibromide gives 0.882 mmol of product i l mole of eliphenylacetylene (Pooduct ) = 178.23 g 1 mmol of product = 178.23 mg 0.882 mmol of product = (0.882*178123) = 157.19 mg