Question

please provide theoretical yield of the reaction

MgBr E120 Ph+ Ph MgBr C13H100 182.21 Form M. Wt Conc Vol Gms HCI Ph- OH Ph CigH 80 260.32 Th. Yld.=.001 Moles 2.5mL = 260gms 3 Molar .5mL .182 2mL Moles .001 .0015 To a 5-ml reaction vial equipped with a spin vane and a serum cap was added benzophenone (.182gms,.001 moles) and anhydrous diethyl ether (2mL). The mixture was stirred until all solids had dissolved, then a 3 molar ether solution of phenyl magnesium bromide (.5mL, .0015 moles) was injected into the stirred reaction mixture over a 1-minute period. An exothermic reaction ensued and a thick white precipitate separated from solution. This precipitate soon made magnetic stirring impossible. After 5 min., the almost solid reaction mixture was mixed thoroughly using a small spatula, then allowed to stand for an additional 10 minutes. Subsequently, 6-molar HCI (2.5mL) was carefully added to the solid mixture. After stirring for a few minutes, all solids had dissolved yielding a two-phase solution. The top (organic) phase was removed and added to a 10mL Erlenmeyer flask. The aqueous layer was again extracted with ether (3mL). The ether extracts were combified, dried (Na2SO4), and the solvent evaporated to give an oily yellow/white solid. This solid was sequentially triturated with petroleum ether (30°- 60', 3mL), vacuum filtered, washed with pet ether (3 X ImL) and air dried. The resulting white solid was recrystallized from 2-propanol to give the desired triphenylmethanol as a white solid (.161 gms, 61.9%): m.p.= 162-163° C; ir 3450 cm (OH).

Answer 1

One mole of benzophenone reacts with one mole phenyl magnesium bromide to give one mole of triphenylmethanol after acidic workup.

In the given reaction, 0.182 g (0.001 mol) of benzophenone reacts with excess of phenyl magnesium bromide (0.5 mL of 3 molar which corresponds to 0.0015 mol). Thus, benzophenone is limiting reagent. As per the balance equation, theoretically 0.001 moles of benzophenone will give 0.001 moles of triphenylmethanol.

Therefore, theoretical yield will be 0.001 mole of triphenylmethanol. Calculate weight of 0.001 mole of triphenylmethanol by multiplying with molar mass of triphenylmethanol (260.32 g/mol)

Theoretical yield = mole of triphenylmethanol x molar mass

Theoretical yield = (0.001 mol ) x (260.32 g/mol)

Theoretical yield = 0.260 g

Therefore, theoretical yield of triphenylmethanol is 0.260 g