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Calculate the composition of a mixture composed of α-D- glucose (specific rotationH2O = 112°) and β-D-glucose (specific rotationH2O of 19°), which has a specific rotation of 94°
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- The following compound has two asymmetric centers and four stereoisomers. Two of these are d-erythrose and d-threose, which are naturally occurring sugars. The configuration of d-erythrose is (2R,3R), and the configuration of d-threose is (2S,3R). a. Which structure represents d-erythrose? b. Which represents d-threose?The specific rotation of α-d-galactose is 150.7 and that of β-d-galactose is 52.8. When an aqueous mixture that was initially 70% α-d-galactose and 30%β-d-galactose reaches equilibrium, the specific rotation is 80.2. What is the percentage of α-d-galactose and β-d galactose at equilibrium?The specific rotation of a-d-glucose is +112.2. Q.) When a-d-glucose is dissolved in water, the specific rotation of the solution changes from +112.2 to +52.7. Does the specific rotation of a-l-glucose also change when it is dissolved in water? If so, to what value?
- Like glucose, galactose mutarotates when it dissolves in water. The specific rotation ofa-d-galactopyranose is +150.7°, and that of the b anomer is +52.8°. When either ofthe pure anomers dissolves in water, the specific rotation gradually changes to +80.2°.Determine the percentages of the two anomers present at equilibrium.Identify the configuration of the stereogenic center (R or S)A mixture of the stereoisomers of glutamic acid has an observed rotation of +6°. The specific rotation of pure (R)-(–)-glutamic acid is –24°. Calculate (a) the enantiomeric excess, (b) the composition of the mixture, and (c) draw the 3D structure of the excess glutamic acid.
- Aswer all parts of the question. A freshly prepared solution of αα ‑D‑glucose shows a specific rotation of +112°.+112°. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to [?]25 °CD=+52.5°.[α]D25 °C=+52.5°. In contrast, a freshly prepared solution of ?β‑D‑glucose has a specific rotation of +19°.+19°. The rotation of this solution increases over time to the same equilibrium value as that shown by the ?α anomer. A solution of one enantiomer of a given monosaccharide rotates plane‑polarized light to the left (counterclockwise) and is the levorotatory isomer, designated (−). The other enantiomer rotates plane‑polarized light to the same extent but to the right (clockwise) and is the dextrorotatory isomer, designated (+). An equimolar mixture of the (+) and (−) forms does not rotate plane‑polarized light. The optical rotation, the number of degrees by which plane‑polarized light rotates on passage through a given path length of a…1. Draw the Fischer projection for a) D-Glucose and b) D-Galactose and their corresponding enantiomers. 2. Assign the absolute configuration for each chiral carbon by writing the carbon number followed by R or S enclosed in parenthesis (e.g. 2R, 3S, 4R, 5S). Answers must be written on the right side of the Fischer projection aligned with the corresponding chiral carbon. 3. Underline the absolute configuration, you have to show that it is italicized by underlining the absolute configuration.Pure cholesterol has a specific rotation of −32. A sample of cholesterolprepared in the lab had a specific rotation of −16. What is the enantiomericexcess of this sample of cholesterol?
- Draw the Fischer projections for the following compounds and define the stereochemistry at each site using R/S notation. You must report the configuration in the form of carbon number and notation e.g. (1S,2R,3R,4S). a) L-glucose b) D-mannose c) L-erythrose d) D-riboseThe specific rotation of a-d-glucose is +112.2. Q.) What is the specific rotation of a-l-glucose?Identify pairs of molecules that represent enantiomers and diastereomers and identify each center stereogenic by writing R or S next to it.