Organic Chemistry, Books a la Carte Edition and Study Guide and Student's Solutions Manual for Organic Chemistry (7th Edition)
7th Edition
ISBN: 9780133903652
Author: Paula Yurkanis Bruice
Publisher: PEARSON
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Chapter 21, Problem 60P
Interpretation Introduction
Interpretation:
The percentage of
Concept Introduction:
The chiral compound shows the property of specific rotation. Specific rotation is intensive property.
The slow change in optical rotation to an equilibrium value is called mutarotation
Equation for calculating the percentage ,
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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?
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.
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?
Chapter 21 Solutions
Organic Chemistry, Books a la Carte Edition and Study Guide and Student's Solutions Manual for Organic Chemistry (7th Edition)
Ch. 21.1 - Prob. 1PCh. 21.2 - Prob. 2PCh. 21.2 - Prob. 3PCh. 21.3 - Prob. 4PCh. 21.3 - Prob. 5PCh. 21.3 - Prob. 6PCh. 21.4 - Prob. 7PCh. 21.4 - Prob. 8PCh. 21.5 - Prob. 9PCh. 21.5 - Prob. 10P
Ch. 21.5 - Prob. 11PCh. 21.6 - Prob. 12PCh. 21.6 - Prob. 13PCh. 21.6 - Prob. 14PCh. 21.7 - Prob. 15PCh. 21.8 - Prob. 16PCh. 21.9 - Prob. 18PCh. 21.10 - Prob. 20PCh. 21.10 - Prob. 21PCh. 21.10 - Prob. 22PCh. 21.11 - Prob. 24PCh. 21.11 - Prob. 25PCh. 21.15 - Prob. 27PCh. 21.16 - Prob. 28PCh. 21.17 - Prob. 29PCh. 21.18 - Refer to Figure 20.5 to answer the following...Ch. 21 - Prob. 31PCh. 21 - Prob. 32PCh. 21 - Prob. 33PCh. 21 - Prob. 34PCh. 21 - Prob. 35PCh. 21 - Prob. 36PCh. 21 - Prob. 37PCh. 21 - Prob. 38PCh. 21 - Prob. 39PCh. 21 - Prob. 40PCh. 21 - Prob. 41PCh. 21 - Prob. 42PCh. 21 - Prob. 43PCh. 21 - Prob. 44PCh. 21 - The 1H NMR spectrum of D-glucose in D2O exhibits...Ch. 21 - Prob. 46PCh. 21 - Prob. 47PCh. 21 - Prob. 48PCh. 21 - Prob. 49PCh. 21 - Prob. 50PCh. 21 - Prob. 51PCh. 21 - Prob. 52PCh. 21 - Prob. 53PCh. 21 - Prob. 54PCh. 21 - Prob. 55PCh. 21 - A hexose is obtained when the residue of a shrub...Ch. 21 - Prob. 57PCh. 21 - Prob. 58PCh. 21 - Prob. 59PCh. 21 - Prob. 60PCh. 21 - Prob. 61PCh. 21 - Prob. 62PCh. 21 - Prob. 63PCh. 21 - Prob. 64PCh. 21 - Prob. 65PCh. 21 - Prob. 66P
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- When a pyranose is in the chair conformation in which the CH2OH group and the C-1 OH group are both in axial positions, the two groups can react to form an acetal. This is called the anhydro form of the sugar (it has “lost water”). The anhydro form of D-idose is shown here. Explain why about 80% of d-idose exists in the anhydro form in an aqueous solution at 100 °C, but only about 0.1% of D-glucose exists in the anhydro form under the same conditions.arrow_forwardAswer 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…arrow_forwardThe 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?arrow_forward
- When a pyranose is in the chair conformation in which the CH2OH group and the C-1 OH group are both in axial positions, the two groups can react toform an acetal. This is called the anhydro form of the sugar (it has “lost water”). The anhydro form of d-idose is shown here. Explain why about 80% of d-idose exists in the anhydro form in an aqueous solution at 100 °C, but only about 0.1% of d-glucose exists in the anhydro form under the same conditions.arrow_forwardThe specific rotation of a-d-glucose is +112.2. Q.) What is the specific rotation of a-l-glucose?arrow_forwardCalculate the percentages of α-d-glucose and b-d-glucose present at equilibrium from the specific rotations of a-d-glucose, β-d-glucose, and the equilibrium mixture. Compare your values with those given in Section 20.10. (Hint: The specific rotation of the mixture equals the specific rotation of α-d-glucose times the fraction of glucose present in the a-form plus the specific rotation of β-d-glucose times the fraction of glucose present in the β-form.)arrow_forward
- When a crystalline sample of α-glucopyranose is dissolved in water and the specific rotation is measured immediately, the observed rotation is +112°. When the same amount of βglucopyranose is dissolved, the observed rotation is +18.7°. However, over time, at room temperature, both solutions reach a specific rotation of +52.7°. Explain why these two solutions reach the same specific rotation over time.arrow_forwardA second category of six-carbon carbohydrates, called ketohexoses, has the constitution shown. How many stereoisomeric 2-ketohexoses are possible?arrow_forwardCalculate 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°arrow_forward
- Carbohydrates. Determination of the extent of branching in glycogen. The amount of branching (number of α 1⟶ 6 glycosidic bonds) in glycogen can be determined by the following procedure. A sample of glycogen is exhaustively methylated-treated with a methylating agent (methyl iodide) that replaces the hydrogen of every sugar hydroxyl with a methyl group, converting –OH to –OCH3. All the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amounts of α -D-2,3-di-O-methylglucopyranose and α -D-2,3,6-tri-O-methylglucopyranose formed are measured. 1- Draw the cyclic structure of α-D-glucopyranose, α-D-2,3-di-O-methylglucopyranose and α-D-2,3,6-tri-O-methylglucopyranose, including the numbering of carbon atoms. 2- Which of α-D-2,3-di-O-methylglucopyranose or α-D-2,3,6-tri-O-methylglucopyranose represents a glucose unit in glycogen which was originally carrying a α 1⟶ 6 glycosidic bond? Explain.arrow_forwardHoney bees collect nectar which is approximately 10% sucrose and convert it into honey (a concentrated solution of about 40% each of glucose and fructose). They do this conversion by mixing the nectar with the salivary enzyme invertase then busily aerating and fanning the solution to drive off water. Given the specific rotation of sucrose (+66.5°), glucose (+52.7°) and fructose (-92°). Calculate the rotation that accomplishes honey productionarrow_forwardD-Glucose most often exists as a pyranose, but it can also exist as a furanose. Draw the Haworth projection of α-d-glucofuranose.arrow_forward
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