Question

Tab Caps Loc Postlab Exercises Record all data during the experiment in your the lab notebook. When the lab is complete Summarize your results on this sheet and tum it in, along with the copy sheets from your notcbook Shitt 1 Ctri Synthesis Results (show all calculations in lab notebook) 0-D459 moles 0,0534 moles 0.01 (3 moles 1-Pentanol mass 5.50q 2.019 Sodium bromide mass 1-Bromopentane massi Theoretical yield mass 49.l631 % Yield Postlab Questions Give your answers and calculations on a separate piece of paper, or clearly and legibly in your lab note book! 1) What is the theoretical yield, the actual yield and the percentage yield of your reaction? calculations! Show 2) What is the function of the following compounds in the formation of I-bromopentane reaction from 1-pentanol? a) Concentrated sulfuric acid b) Sodium hydroxide c) Magnesium sulfate 3) Is the reaction an SNl or SN2reaction? Explain. 4) How much starting material do you need to use if the percentage yield of the reaction is 70 % and you want to obtain 10 gram of final product? Show calculations. /0/D./14.37 5) Calculate the limiting reactant and the theoretical yield when you use 3.0 g of NaBr, 5.0 mL of 1-butanol and 5.0 mL of concentrated sulfuric acid (18.0M) to synthesize 1-bromobutane. NoBr 3.0g/S.0mlbutarol S.0mLbn) Sulhicand

Answer 1

OH - NaBr Hşc B r нс 1-bromopentane pentan-1-ol 1) Molar mass of 1-pentanol = 88.15g/mol Mass of 1-pentanol = 4.05 g number of moles of 1-pentanol = mass/molar mass = 4.05/88.15 = 0.0459 moles Moles of NaBr = 5.50g/102.89g/mol = 0.0535 moles Thus NaBr is in excess and 1-pentanol is the limiting reagent. Theoretical #moles of 1-bromopentane = 0.0459 moles Mass of 1-bromopentane = 0.0459mol*151.05g/mol = 6.933 g Theoretical yield = 6.933 g Actual yield = 2.01 g % yield = (actual/theoretical)*100 = (2.0176.933)*100 = 28.99% 2) a) H2SO4 is used for the protonation of hydroxyl group of alcohol which leads to a good leaving group. b) NaOH is to neutralize any excess of acid remaining in the reaction mixture c) MgSO4 is used as drying agent, to absorb traces of water present in the organic layer.

3) As the reactant is a primary alcohol, the reaction is SN2 reaction. Because it is less likely to give primary carbocation and hence less likely to follow SN2. Instead SN2 is more likely as Nucloephile Br- can attacks directly to break C-O bond. 4) Molar mass of 1-butanol = 74.12g/mol Density of 1-butanol = 0.810g/mL Mass of 1-butanol = Volume * density = 5.0mL*0.810g/mL = 4.05g number of moles of 1-butanol = mass/molar mass = 4.05/74.12 = 0.0546 moles Moles of NaBr = 3g/102.89g/mol = 0.0292 moles Thus NaBr is limiting reagent and 1-butanol is in excess. Theoretical #moles of 1-bromobutane = 0.0292 moles Mass of 1-bromobutane = 0.0292mol*137.02g/mol = 4.00 g Theoretical yield = 4.00 g H2SO4 is used as a catalyst hence may not influance the yield.