(a) Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location. Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

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Answer 1

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Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

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LABORATORY REPORT FORM Part I. Determination of the Formula of a Known Hydrate 1. Mass of empty evaporating dish 3. Mass of hydrate Using subtraction or mass by difference, find the mass of the hydrate 76.96 -75.40 75.40g 76.968 1.568 01.56 76.90 g 2. Mass of evaporating dish + hydrate 4. Mass of evaporating dish + hydrate (after heating) First 76.98 g Third 76.66g Second Fourth (if necessary) 76.60g 5. Mass of anhydrate 6. Mass of water lost by the hydrate 7. Percent of water of hydration (Show Calculations) 8. Moles of water (Show Calculations) mol mass of water = MM of water (g/m) 9. Moles of anhydrate (Show Calculations) 10. Ratio of moles of water to moles of anhydrate 11 F(Show Calculations) 11. Formula of hydrate - Mass of water (g) x 100 % water hydration g g % Mass of hydrate (9) x IC % = (Mass of hydrate- mass of an) mass of hydrate (g) % = (1.569- × 100= mol 1.569 mol Mol Mass of anhydrate/MM of anhydrate 12. What was the color of the hydrate? blue What was the color of the…

compared t-critical with t-calculated and 95% confidence interval to answer this question

Comparing two means. Horvat and co-workers used atomic absorption spectroscopy to determine the concentration of Hg in coal fly ash. Of particular interest to the authors was developing an appropriate procedure for digesting samples and releasing the Hg for analysis. As part of their study they tested several reagents for digesting samples. Their results using HNO3 and using a 1+3 mixture of HNO3 and HCl are shown here. All concentrations are given as ppb Hg sample. HNO3: 161, 165, 160, 167, 166 1+3 HNO3–HCl: 159, 145, 140, 147, 143, 156 Determine whether there is a significant difference between these methods at the 95% confidence interval.

Answer 2

Question

Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

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Answer 3

Question

Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

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Answer 4

Question

Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

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Answer 5

Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

Answer 6

Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Answer 7

Chapter 26, Problem 68P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkage in disaccharide C is to be identified and the glycosidic linkage is to be classified as α or β and the numbers are to be used to designate the location.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Answer 8

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

Answer 9

Interpretation Introduction

(b)

Interpretation: The compounds E and F are to be identified.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 26.2 - Prob. 1P Ch. 26.2 - Prob. 2P Ch. 26.2 - Label each stereogenic center as R or S. a. b. c....Ch. 26.2 - Convert the ball-and-stick model to a Fischer...Ch. 26.2 - Prob. 5P Ch. 26.2 - Prob. 6P Ch. 26.3 - Prob. 7P Ch. 26.3 - Prob. 8P Ch. 26.4 - Prob. 9P Ch. 26.4 - Prob. 10P

Solution Summary: The author explains that the glycosidic linkage in disaccharide C is to be identified and the numbers are to designate the location.

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Chapter 26 Solutions