Question

IS 3. ChboCKs 2 CH2CH2Br+ 2 Mg+ С-нвосн. CeHsCCH2CH Сн,CH, 109 g/mole 1.47 g/mL 24.3 g/mole 136 g/mole 1.08 g/mL 164 g/mole If 10.0 mL of bromoethane react with 5.0 g Mg. then with 10.0 g of methylbenzoate, and if 9.0 g of 3-phenylpentanol are formed, what is the percent yield? Show your work 4. a) In the reductive amination reaction why is the NaBH4 only added 15 minutes after the amine and the aldehyde have been mixed, and not right away? нви b) What is a rotary evaporator (Rotovap) used for? ame ooems S c) In the Williamson Ether Synthesis, why did we use a reflux assembly and not just boil the reactants in a beaker or flask? Would you expect the melting point of a crude product to be higher or lower than the melting point of a recrystallized product? Why?

Answer 1

® 2CH3 CH2 BY + 2 Ag+ cutter 6-ochy > Etts-e-945 109 g/mol 24:3 136 164 g/mol 1-479/mk 1.089/mel mass of lome Bromo Ethane = dxv =1047 glowy 10ml mass of mg cth CH₂ Briting - 14 79 = 5:0g its ngay Let us find limiting reagent CH₂ CH2 By 8x1099 CH₂ CH₂ Br gives 164g 3-Phenylpentanse .:14.79 - -14.7 x 164 4 2xloq = 11.059g mg 2x24:39 mg forming 164g 3-phenol pertand 59 - ? Exiboh 4.3 = 16.879 Colts doch 3 1368 methyl Benzoate gives 1649 109 - ) 10X1646136 = 126059 As Cats Br toming lower yield & 3-phenyl pentanol Cats By is limiting reagent and field is 11:059g . percent yield = Actual yield x 100 Theoretical yield 3 90 x100 = 81-4). 11:059