26 Reaction Scheme transformation 1 CH CH I CH3 CH,CH20 CH3 Williamson Ether phe

Question

26 Reaction Scheme transformation 1 CH CH I CH3 CH,CH20 CH3 Williamson Ether phenacetin acetamidophenol Tylenol) transformation 2 amide hydrolysis NH2 transformation 3 CH,CH20 NH2 CH3CHzo nucleophilic acyl NH2 NH2 phenitidine substitution dulcin Procedure wear safety glasses during the entire laboratory period and use gloves when handling any chemicals. All chemicals, reactions and procedures should be used or performed in the hood. All organic wastes should be put in the hazardous waste bottle unless otherwise stated The following procedures are literature descriptions for the formation of Dulcin from Tylenol. The Preparation of Phenacetin from Tylenol Part I): Grind 4 tablets of Tylenol mg of acetaminophen per tablet or equivalent amount) using a mortar and pestle and place the in a round bottom with a magnetic stir Note: NaOH is skin irritant and ethyl iodide is toxic. Wear gloves and goggles and use these chemicals in the hood. wastes should be disposed of immediately in the hazardous bottle. To the powdered tablets add 15.5-mL of a M ethanolic NaoH for Fit bottom flask with a water condenser and bring to reflux. Maintain 2.1 15 minutes and then remove the flask from its heat source. To the hot mL of and return to reflux an additional 15 minutes. Filter the hot ution under vacuum through a Buchner funnel starches on flask containing 40-mL mixture of ice and water. Co 300-mL side arm from the the filter paper. The phenacetin, upon contact the vacuum filtrate as a white solid While still cold, with the cold water, precipitates filtration using a Buchner the solid phenacetin is funnel and washpd with

Explanation / Answer

From the above shown reaction,

In step 1, 1 mole of acetaminophen starting material after first step would produce 1 mole of phenacetin

moles of acetaminophen = 4 x 0.5 g/151.163 g/mol

                                        = 0.0132 mol

moles of phenacetin produced = 0.0132 mol

theoretical yield in grams of phenacetin = 0.0132 mol x 179.216 g/mol

                                                                = 2.366 g

Now, in step 2, 1 mole of phenacetin would give 1 mole of phenitidine

moles of phenacetin starting material = 0.0132 mol

theoretical yield in grams of phenitidine = 0.0132 mol x 137.182 g/mol

                                                               = 1.811 g

In step 3, 1 mole of phenitidine would produce 1 mole of dulcin

moles of phenitidine = 0.0132 mol

So,

moles of dulcin formed = 0.0132 mol

theoretical yield in grams of dulcin = 0.0132 mol x 180.2 g/mol

                                                       = 2.384 g

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