John added 0.5 g of sodium borohydride to 50 methanol solution containing 5 g of

Question

John added 0.5 g of sodium borohydride to 50 methanol solution containing 5 g of benzophenone, He boiled solution for 10 min, add 10 mL of sulfuric acid (4 mol/l), and add solution to a 100 mL beaker full of ice. He collectd product, recrystallized it and, after drying, obtained 2 g of product (diphenylmethanol) according to the reaction below. What was the yield (in %) of his reaction? John measured IR and NMR spectra of benzophenone, diphenylmethanol and ethoxybenzene. Assign spectra below to each substance. Assign IR spectra to benzophenone, diphenylmethanol and ethoxybenzene

Explanation / Answer

a).amount of benzophenone taken=5g

molar mass of benzophenone = 182.22g/mol

No of moles=mass/molar mass=5g/ 182.22g/mol=0.0274mol

if you take 4 moles of benzophenone you will get 4 moles of product (from equation given)

in the same way 0.0274mol of benzophenone give 0.0274mol of product

theoritical yield=No of molesxmolar mass=0.0274molx184.24=5.048g

%yield=actual yield/theoritical yieldx100=2g/5.048gx100=39.61%

b.

1-diphenylmethanol sharp peak for-OH around 3000 cm^-1

2.benzophenone

3.ethoxy benzene-strong peak around 1100 cm^-1

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