Question

ATOM ECONOMY: The traditional way of evaluating the efficiency of an organic reaction is to determine the percent yield of the reaction. Most organic chemists would be quite pleased with a reaction with >90% isolated yield of the target molecule. On the other hand, a yield of 30% is typically considered to be a disappointment. As a review of the concept of theoretical yield and percent yield, let's consider the following Sy2 reaction CH3CH2CH2CH2OH + NaBr + H2SO4 → CH CH CH CH Br + NaHSO4 + H20 The procedure says "1.33 g of NaBr is dissolved in 1.5 mL of water, followed by the addition of 1.0 mL of l-butanol and 1.1 mL of concentrated sulfuric acid." The student doing this reaction isolated 1.2 g of pure l-bromobutane as the product of this reaction. QUESTION 1. Fill in the rest of this table. Quantity (mol) Reagent Molecular Mass Volume Density Weight (g) (mL) (g/mL) IL/mol) 1-butanol 74.121 0 11. 0 .810 NaBr 82. 133 XXXXXX XXXXXX H SOCO 1.84 QUESTION 20 Calculate the theoretical yield of this reaction QUESTION 2b. Calculate the percent yield if 1.2g of product are isolated
Can someone please help? Thank you.

Answer 1

Page No. Date moiz Griven S. dreaction _+ H2O CH3CH2CH2CH201_+_ NaBr + H2SOn → CH2CH2CH/Ben + NOHSOU Procedure says labr dissolved in 1.5ml water followed by <Ime of l- butonal eisd. lelme of conc. H₂SO4 7 Student isolated 1.2 gm of Pure - -bromobuleme we know that Theoretical yield - (males of limiting creureht); _xlStoichiepis tenie brulio 2 - *' Thearetical gield (Malas ob climaiting Reugene) (Stoichiometerie oratio; desired Product Mlimining Rea gest ) cmw of desired product) - Teacher's Signature...

Page No. Date Sercentage _{Сллах чреу-100-) the arctical Yield 1% field) Bmw of Culgoh = 76.12 d = 0.810 gm me Coiven) Volume = Ime (Given) se mass = dx valume 30-310 qm .no. of males = 0-8100-otog - 742 1 1 1 2 1 1 M.Wof HeSon – 98.08 - d = 1.84 gr/ml V- lolme Corivon) Griven) 1 1 mass = 1.84 X11 - 2.024 gry 1 1 0. 206 mol moles = 2.024 - 98.08 Teacher's Signature......

l van DI Mw of Naber = 102-91 muss 1.33 sm No. of males = 1.33 102:32 0.0129 moles la (hm) (mu) (grlands Sel Tables 7 Reagent M.W muss 1) CuHgOH 74.12 (6.810 2) NaBr 102.91 1.33 2 Holboy 98.08 2.024 -M val. Density Moles 1 0.810 (0-009 0.0129 II 184 1020k/ Teacher's Signature.

Page No. Date Ol>OR: CHICH₂ CH₂ CH₂OH at nabr t 4₂son > CH₂ CH2 CH₂ CH₂ Ber ANASOu1_1120. o ho 9 moh 0.0129 1 0.206 mol - MOI I so gimiting Reagent will be cu Hg OH then theoreticaly 0.0109 Mol of CuHgOH 0.0109 mal of Naber and o- olog mal of Hason is consumed. Theoretical yield 1 (males of br) (Stoichiometric Ratio) x Mwofi desired Product A Goolof 7 The anipical 0.0109 x (H) x 137-03 1.49 gm y fm.wo Suttg Bet : 137-03 Teacher's Signature.

Page No. Date Sal :-K0:3) actual yield 1. Peruentage gield theoretical yield 12xtoo -full ---- %field - 80.53/ 7 actual yiell I k Given in Problem