Consider the following reaction: 0.269 g of phenol (MW 94.11 g/mol) is reacted w
ID: 1067676 • Letter: C
Question
Consider the following reaction: 0.269 g of phenol (MW 94.11 g/mol) is reacted with 0.491 g of methyl iodide (MW 141.94 g/mol) in aqueous sodium hydroxide at 60 degree C using tetrabutylammonium bromide as a phase transfer catalyst, the corresponding ether, known as anisole (MW=108.14) is obtained. What is the theoretical yield of the product anisole in grams? Do NOT include units (g or grams) as part of your answer. For example, if your answer is 0.318 grams, only enter 0.318 in the answer box.Explanation / Answer
Given
Phenol = 0.269 g
No. of moles of phenol = (weight of phenol/M.Wt of phenol)
= 0.269/94 = 0.00286 moles
Methyl iodide = 0.491 g
No. of moles of methyl iodide = wt/M.wt = 0.491/142 = 0.00345
From the given reaction to get the product both the starting materials should be used in 1:1 mole ratio
But methyl iodide was used in excess. This indicates that phenol is the limiting reagent.
As 0.00286 moles of phenol was used we should get same moles of anisole (as phenol is the limiting reagent)
Theorital yield of anisole is
No. of moles of anisole = weight/M.wt
0.00286 = weight/108
Weight = 0.00286 x 108 = 0.3088 g
Theoritical yiels of anisole is 0.3088