Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group. Concept Introduction: Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses. The number of carbon atoms present in the chain characterize the monosaccharide. They are given below. Carbon chain with three carbon atoms is triose. Carbon chain with four carbon atoms is tetrose. Carbon chain with five carbon atoms is as pentose. Carbon chain with six carbon atoms is as hexose.

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Answer 1

Question

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

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Answer 2

Question

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

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Answer 3

Question

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

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Answer 4

Question

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

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Answer 5

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

Answer 6

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

Answer 7

Chapter 14, Problem 14.22UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

Answer 8

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

Answer 9

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

Answer 10

Expert Solution & Answer

Answer 11

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Answer 12

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Answer 13

Ch. 14.1 - Draw a Lewis structure for glucose that clearly...Ch. 14.2 - Prob. 14.2P Ch. 14.2 - Prob. 14.3P Ch. 14.2 - Prob. 14.4P Ch. 14.2 - Rank the following compounds in order of...Ch. 14.2 - Prob. 14.6P Ch. 14.2 - Prob. 14.7P Ch. 14.2 - Prob. 14.8P Ch. 14.2 - Prob. 14.9P Ch. 14.3 - Prob. 14.10P

Solution Summary: The author explains how monosaccharides are classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept explainers

Interpretation:

Chiral centers has to be located in the given compounds.

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Chapter 14 Solutions

Solution Summary: The author explains how monosaccharides are classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept explainers

Interpretation: Chiral centers has to be located in the given compounds.

Want to see the full answer?

Chapter 14 Solutions

Interpretation:

Chiral centers has to be located in the given compounds.