?LUC Sakai : CHEM 361 E Untitled document-G LUC Sakai : CHEM 226 ? ?Acylation-Pr

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?LUC Sakai : CHEM 361 E Untitled document-G LUC Sakai : CHEM 226 ? ?Acylation-Procedure+ ? i MasteringGenetics D13: The Genet c Code-X L Raw x × C Kami PDF and Document Markup chrome-extension://ecnphlgnajanjnkcmbpancdjoidceilk/ r.html?source extension ile:https%3A% 0 Kami Upgrade 2015.pdf NH2 HN CH3 CH3 CH Procedure In a 125 mL Erlenmeyer flask, gently warm 70 mL of a 0.334 M solution of p-toluidine hydrochloride using a hot plate. Do not let the temperature go above 50 °C In a 50 mL beaker, dissolve 3.75 g of sodium acetate trihydrate in 8 mL of deionized water Again, use a warm hot plate to gently speed up the process if necessary. The purpose of the sodium acetate is to buffer the solution. The pH optimum of the reaction is around 5 While swirling rapidly, add 2.5 mL of acetic anhydride. Then, as quickly as possible, add the sodium acetate solution prepared above. A white precipitate should form. Once the solution is thoroughly mixed, cool the flask in an ice-water bath. Collect the solid product via vacuum filtration. While the solid is on the filter paper, wash it three times with a little cold water. After washing the solid, allow the aspirator to continue to run for a few minutes. This will help to remove some of the water from the product. Once the product is fully dry, determine the mass, measure the melting point and calculate the percent yield. Let your Teaching Assistant evaluate your product and dispose of it properly Pre-Lab Exercises Why is it necessary to buffer the reaction mixture for this reaction?

Explanation / Answer

Ans. Step 1: Determine Limiting Reactant:

# Following stoichiometry, the theoretical molar ratio of reactants

                        p-toluidine : acetic anhydride = 1 : 1

# Moles of p-toluidine hydrochloride = Molarity x Volume of soln. in liters

                                                            = 0.334 M x 0.070 L

                                                            = 0.02338 mol

Since 1 mol p-toluidine hydrochloride yields 1 mol p-toluidine, the moles of these two chemical species must be the same.

            So, moles of p-toluidine taken = 0.02338 mol

# Moles acetic anhydride = Mass / MW = (Vol x Density) / MW

                                                = (2.5 mL x 1.08 g mL-1) / 102.09 g mol-1

                                                = 0.02645 mol

# The experimental molar ratio of reactants

                        p-toluidine : acetic anhydride = 0.02338 mol : 0.02645 mol = 0.9 : 1

Comparing the theoretical and experimental molar ration, the experimental moles of p-toluidine is less than its theoretical value whereas that of acetic anhydride is kept constant at 1 mol. Therefore, p-toluidine is the limiting reactant.

# Step 2: The formation of product follows the stoichiometry of limiting reactant.

Following stoichiometry, 1 mol p-toluidine forms 1 mol product (N-(m-tolyl)acetamide).

So, moles of N-(m-tolyl)acetamide formed = 0.02338 mol x

Now,

            Theoretical yield = Moles of N-(m-tolyl)acetamide x MW

                                                = 0.02338 mol x 149.19 g mol-1

                                                = 3.490 g

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