CARBOHYDRATE ESTERIFICATION kcohols react readily with acid anhydrides to form e

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CARBOHYDRATE ESTERIFICATION kcohols react readily with acid anhydrides to form esters. Glucose contai to form esters. Glucose contains five molecule (including the hemiacetal hydroxyl group) and can ester by reaction with acetic anhydride. A proper choice of catalyst. not only gpends the rate of formation of the product, but also influences the a- or B- configuratio the hemiacetal ester group. For example, when the reaction is based-catalyzed with a wea ase, anbydrous sodium acetate the E-D.glucase pentaacetate is formed, whereas the use of an acidic catalyst, zinc chloride, yields a-D-glucose pentaacetate. base, Preparation of B-D-glucose pentaacetate CHy CH Using a mortar and pestle, grind together a homogeneous mixture of 2.7 g moles) of D-O-glucose and 2.0 g of anhydrous sodium acetate. NOTE: Be sure the reagent bottle specifies "anhydrous" or the reaction will not be effectively catalyzed and the reaction will fail. Transfer the ground mixture into a 100 mL round bottom flask; add 14.0 mL (15.3 g moles) of acetic anhydride and a magnetic stir bar. Set up a reflux condenser being sure to use a small amount of stopcock grease on the joint. Reflux for 1 hour using a pan of boiling water on a magnetic stirrerhot plate. After 1 hour, slowly pour the hot reaction mixture while stirring, into a 250 mL beaker which is 1/3 filled with crushed ice and a little water. WAIT and allow the product to solidify. Filter the crude crystalline product on the Buchner funnel and proceed directly to recrystallize the entire amount of product. To obtain a maximum recrystallized yield, you must correctly estimate by progressive addition of small portions of solvent (in this case, water) which, when boiling, will totally

Explanation / Answer

Beta D Glucose = A

Acetic Anhydride = B

Beta D Glucose Pentaacetate = C

We can see from the reaction that

A + 2.5 B ------> C + 2.5H2O

Given that A is 2.7g and B is 15.3g

We know that

Molar Mass of A =180g and Molar Mass of B =102g

Therefore No of Moles of A =2.7g / 180g = 0.015

No of Moles of B = 15.3g / 102g = 0.15

We can easily see that A is the limiting molar reagent [ as B is in large excess as compared to A also we need only 0.0375 moles of B to completely react the 0.015 M of A]

Therefore the theoritical yield of the reaction is the weight of the product formed in the reaction

Now theoritical yield = No of Moles of C formed * Molar Mass of C

[No of moles of C formed = No of moles of A ( due to limiting molar reagent)]

yield = 0.015 * 208g

yield = 3.12g

  

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