Question

please help on questions 2 ans 3

EX AA = = = = Es Styles Style X ASA abr - Pane 2. Given the following experimental data calculate the percent yield for the product of the So- dium Borohydride Reduction portion of the experiment. Compound Amount Ethyl Acetoacetate 2.45 mL NaBH 0.4327 (9) Product + Round Bottom Flask 60.335 Round Bottom Flask 59.660 3. Given the following experimental data calculate the percent yield for the product of the En- zymatic Reduction portion of the experiment Compound Amount Ethyl Acetoacetate Na2HPO4 Sucrose Yeast Celite Product Round Bottom Flask Round Bottom Flask 2.45 mL 0.254 g 40.035 g 7.050 g 10.130 g 59.910 g 59.660 g Page 2 of 5 558 words O Focus *

Answer 1

2 thawer and The Sodium reaction between Ethyl acctoacetate Boro hydride can be given as OH thon O + I mol NaBity Solvent ou Revende nusture of Ethyl-3-hydroxybutyrate. of Ethylace to acetate gives I mol of Imol. I mol Ellup -3 hydroxybutyrate. Here can number of calculated moles of as Ethyl acetoacetate be number of moles - mass of substance molar mass of substance Here mass = density & volume of Excyl acetoacetate = 2.45 X 1.021 g / ml = 2.50145 g number of mole = 2.50145 20.04922 130.14 give Thus 0.01922 moles 2019 22 moles of of Ethyl aceto acetate will product.

0.01922 moles of product = 1.04922 x molar mass I of product lg product = (0.019 22 X 132.14 g : mass of = 2:5397308 g Now yield obtained after reaction = (Product & Round bottom flask - Round bottom flack) = (60.335 - 59.660) = 0.675 g. % yield = x 100 Obtained Theoretical yield Yield = 50.675 x 100 = 26.5 77 %. 2. 539731 :(% yield) of product = 26.5 79.7.

The reduction by enzymatic method reaction can be of Ethyl acetoacetate given as- I mol Na, IPO, Celite Stereo cherical hacemic minture of Ethyl-3-hydroxy butyrate (Ind) Ethyl acetoacetate will give & mol l racemic minture of Exluyt-3 hydrony I mol of product butyrate) 2.45ml Ethyl acetoacetate = 2.50 145 g = 2.50145 nel = 0.01922 mol 130.14 0.0 1922 mol Ethyl acetoacetate should glue 0.01922 mol of product. = (0.01922 X 132.14 g = 2.5397301 g of Product obtained yield = ( Product & Round bottom flask - Round bettom Flack) (59.910 - 59.6600g = 0.25 g

% yield = ostorned yield x 100 Theoretical Yield ro.25 - x 100) = (0.0984 36x 100) %. 2.53 973 9.90 = 9.8437 Percentage yield = 9.843 %.