(a) Interpretation: The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide. Concept Introduction: When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group . Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

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Answer 1

Question

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

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Answer 2

Question

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

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Answer 3

Question

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

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Answer 4

Question

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

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Answer 5

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Answer 6

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Answer 7

Chapter 20, Problem 80CP

Interpretation Introduction

(a)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Answer 8

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Answer 9

Interpretation Introduction

(b)

Interpretation:

The cyclic form of monosaccharide given should be converted to an acyclic monosaccharide.

Concept Introduction:

When an aldehyde or a ketone is reacted with one equivalent of alcohol, a hemiacetal is formed. The aldoses and ketoses contains both hydroxyl groups and aldehyde or ketone functional group. Therefore, aldoses and ketoses undergo intramolecular cyclization to form stable cyclic hemiacetals. The O atom on the chirality center farthest from the carbonyl is involved in the cyclization.

Answer 10

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Answer 12

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Answer 13

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

Solution Summary: The author explains that when an aldehyde or a ketone is reacted with one equivalent of alcohol, the hemiacetal is formed.

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