Question

[No Subject]

Me to Aisha OsanyinpejuSent

The synthesis of 1,1-diphenylethanol via the grignard reaction is given by the following equation (below):

A student decides to perform the above reaction. She starts with 0.32ml of bromobenzene (M.W. = 157) and uses 0.0072g of Mg (M.W. = 24.3) and 0.23ml of acetophenone (M.W. = 120.15).

a. Calculate the limiting reagent

b. Calculate the theoretical yield of 1,1-diphenyethanol

c. She obtains 0.172 grams of of the product. What is her actual yield?

d. Explain how the student could use IR to determine if she had obtained the desired product.

(By the way, here is the structure meant to be placed above) -

Me to Aisha Osanyinpeju

The synthesis of 1,1-diphenylethanol via the grignard reaction is given by the following equation (below):

A student decides to perform the above reaction. She starts with 0.32ml of bromobenzene (M.W. = 157) and uses 0.0072g of Mg (M.W. = 24.3) and 0.23ml of acetophenone (M.W. = 120.15).

a. Calculate the limiting reagent

b. Calculate the theoretical yield of 1,1-diphenyethanol

c. She obtains 0.0036grams of of the product. What is her actual yield?

d. Explain how the student could use IR to determine if she had obtained the desired product.

(By the way, here is the structure meant to be placed above) -

Mg MgBr ether MgBr OMgBr othor ether H20, H PROCEDURE: Synthesis of bromide

Answer 1

Here mg, bromobenzene and actophenone are used in equivalents, so, one which is less in md is 1 ka limiting agent Mg = 0.0072 g/24.3 = 2.963 times 10^-4 mol. Rightarrow loss and limiting agent bromobenzene = 0.32 ml/1.5 g/mL times 157 = 1.359 times 10^-3 mol. autophenone = 0.23ml/1.03 g/ml times 120.15 = 1.859 times 10^-3 mol theoretical liquid: 24.3 g of mg will give 198.26 g of therefore 0.0072 g of mg = 0.0072 times 198.26/24.3 = 0.059 g actual yield: obtained yield = 0.0036 g yield = 0.0036 (obtained)/0.059(calculated times 100 = 6.13% IR determination: rightarrow shows alcoholic o - H stretching band around 3400 cm^-1 in IR spectrum