Question

IS 3. OH C H5CCH2CH3 CH2CH3 eros 164 g/mole uto er 2 CH2CH2Br + 2 Mg Cон,восн. evdy 109 g/mole 1.47 g/mL 24.3 g/mole 136 g/mole 1.08 g/mL 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. SCH CHAL VOH 13 O0409 O M hy is the NaBH only added 15

Answer 1

mass of CH3CH2Br = vomume * dssnity
                  = 10*1.47 = 14.7g
2CH3CH2Br + 2Mg + C6H5COOCH3 ------> C6H5C(OH)(CH2CH3)2
2 mole of CH3CH2Br react with Mg and C6H5COOCH3 to gives 1 mole of C6H5C(OH)(CH2CH3)2
2*10.9g of CH3CH2Br react with Mg and C6H5COOCH3 to gives 164g of C6H5C(OH)(CH2CH3)2
14.7g of CH3CH2Br react with Mg and C6H5COOCH3 to gives = 164*14.7/(2*10.9) = 110.6g of C6H5C(OH)(CH2CH3)2
2 mole of Mg react with CH3CH2Br and C6H5COOCH3 to gives 1 mole of C6H5C(OH)(CH2CH3)2
2*24.3g of Mg react with CH3CH2Br and C6H5COOCH3 to gives 164g of C6H5C(OH)(CH2CH3)2
5g of Mg react with CH3CH2Br and C6H5COOCH3 to gives = 164*5/(2*24.3) = 16.87g of C6H5C(OH)(CH2CH3)2
1 mole of C6H5COOCH3 react with Mg and CH3CH2Br to gives 1 mole of C6H5C(OH)(CH2CH3)2
136g of C6H5COOCH3 react with Mg and CH3CH2Br to gives 164g of C6H5C(OH)(CH2CH3)2
10g of C6H5COOCH3 react with Mg and CH3CH2Br to gives 164*10/136 = 12.05g of C6H5C(OH)(CH2CH3)2
The theoretical yield of product = 12.05g
actual yield of product = 9g
percent yield   = actual yield *100/theoretical yield
                = 9*100/12.05   = 74.68% >>>>answer