Question

Assume that we ran this reaction with 1.0 g Mg, 4.5 mL bromobenzene, and 2.3 mL methylbenzoate. If you got 2.15 g of product calculate your percent yield for this reaction.

Answer 1

Mg 7 O CH3 Mg Br E 1.08 gut! Br -Br Mg El Long Br only Methylbenzoate Mgor D final product molar massa 260 gmot! Methyl benzoate = 2.3 ml • Bromobeu zene =4.5ml density density = 1.5 gb Total weight = 6.75 g. Total weight دان -Omgbr Again Attack ✓ Hook up or 6 2,48 g

Now, calculating no. of moles of all the given species: Maceday weight (given) no. of moles of Mg = Atomic weight 1.0 g 12oul mol 24.31 ĝmolt = 2041 mol moles of Bromobenzene a 6.75g 157.01 gmolt 0.042 moles 0.018 moles moles of Methylbenzoate = 2484 136. 15 gmol C 0.02. Mgbr OCH3 (Ph); C-OH + مو can all [Myer you see here that alt moles of given species are balanced approx.. Mgbr no. of moles of grignard reagent Should be two times that of methyl benzoate. And given moles are approx. double accordingly. You can calculate the oritical yield by any of given reactant. So from methylbenzoate of methyl benzocite gives 260 g. of product And 2.48 g of *2.48= 4.757. Actual yield & 2.15 g. Theoritical Yield. 136.15 g a 260 136.15 2:15 Percentage yield Actual Theoritical x100 Xloo 4:25 % yidd 45.26% Aus.
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