(a) Interpretation: The glycosidic linkage in cellobiose should be located. Concept Introduction: Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Question

Want to see the full answer?

Answer 1

Question

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Convert 1.38 eV into wavelength (nm) and wavenumber (cm-1) (c = 2.998 x 108 m/s; h = 6.626 x 10-34 J*s).

Can you help me understand the CBC method on metal bridging by looking at this problem?

A partir de Aluminio y Co(NO3)2ꞏ6H2O, indicar las reacciones a realizar para obtener Azul de Thenard (Al2CoO4).

Answer 2

Question

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Answer 6

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Answer 7

Chapter 20.4, Problem 20.17P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage in cellobiose should be located.

Concept Introduction:

Disaccharides are carbohydrate produced by the combination of two monosaccharides. Disaccharides are acetals. Monosaccharide contains hemiacetal. When a hemiacetal hydroxyl group is reacted with a hydroxyl group of another monosaccharide, a disaccharide is yielded.

Answer 8

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Answer 9

Interpretation Introduction

(b)

Interpretation:

Carbon atoms in both rings should be numbered.

Concept Introduction:

When numbering carbons in a cyclic carbohydrate, the numbering starts at acetal or hemiacetal carbon that means at the anomeric carbon, giving it the number 1. And the numbering continues clockwise. O atom in the ring is not given a number.

Answer 10

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Answer 11

Interpretation Introduction

(c)

Interpretation:

The glycosidic linkage should be classified as a and β and the location of the linkage should be designated using the numbers.

Concept Introduction:

If the glycosidic linkage is oriented down or below the plane of the ring then it is an a glycoside. If the glycosidic linkage is oriented up, above the plane of the ring, then it is a β-glycoside.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 15

Ch. 20.1 - Label the hemiacetal carbonthe carbon bonded to...Ch. 20.1 - Prob. 20.2P Ch. 20.1 - Prob. 20.3P Ch. 20.1 - Prob. 20.4P Ch. 20.1 - Prob. 20.5P Ch. 20.2 - Prob. 20.6P Ch. 20.2 - Prob. 20.7P Ch. 20.2 - Prob. 20.8P Ch. 20.2 - Prob. 20.9P Ch. 20.2 - Prob. 20.10P

Solution Summary: The author explains that the glycosidic linkage in cellobiose should be located.

Concept explainers

Want to see the full answer?

Chapter 20 Solutions