With the data given for the reaction of the acetylation of ferrocene using aceti

Question

With the data given for the reaction of the acetylation of ferrocene using acetic anhydride and an acid catalyst, H3PO4 whay is the expected yield of Acetylferrocene?

Materials Table and Reactions Density (g/mol) g/mL) Formula MP/BP (C) MW Amt used Expected Comment mol) 0.00269 0.0211 0.00576 Name Yield 172.5/249 186.04 111 73.1/139.8 102.09 1.08 42.35/158 97.994 1.88 Ferrocene C10H1oFe Acetic Anhydride C4H603 H3PO4 (14.7M) Phosphoric Acid 50/851 84.007 2.2 0.0298 Sodium Bicarbonate NaHCO3 Acetylferrocene Fe(C H4COCH 81 to 228.07 1.014 0.00055 83/161 116.3/34.6 74.12 0.7134 -95/69 -63.5/61.2 119.38 149 1713/2950 60.08 0.700 0.024 0.05698 0.0624 0.0652 mol Dietyl ether zHs)zo Hexane 86.18 0.6548 CsH14 CHCl3 Sio2 Chloroform Silica gel All properties obtained from: Chemical reactions using ChemDraw H,PO4 Fe

Explanation / Answer

128 g of ferrocene will give 228 g of acetyl ferrocene is 100 % yield

128 g of ferrocene is = 1 mole

Weight for 0.00269 mole = 0.00269 x 128 = 0.500 g

Similarly for acetyl ferrocene

228 g = 1 mole

Then for 0.00055 mole = 228 x 0.00055 = 0.125 g

Theoretical 100 % yield for 0.500 g of ferrocene is calculated from known data

186 g of Ferrocene will produce 228 of acetyl ferrocene is 100 % yield

Then for 0.500 g of ferrocene is obtained by cross multiplying

228 x 0.500 / 186 = 0.612 g

but acetyl ferrocene obtained is 0.125 g

Therefore percentage yield obtained

Real yield gms / theoretical yield gms x 100

0.125 / 0.612 x 100 = 20.5 % is yield percentage of acetyl ferrocene

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