Chemistry for Today: General, Organic, and Biochemistry
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN: 9781305960060
Author: Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher: Cengage Learning
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Chapter 17, Problem 17.51E
Interpretation Introduction

(a)

Interpretation:

The type of linkage present in the lactose is to be stated.

Concept introduction:

A monosaccharide is the simplest form of carbohydrate. It possesses either ketone or aldehyde with one or more than one hydroxyl (OH) groups. Disaccharides, trisaccharides, oligosaccharides, and polysaccharides all involve monosaccharide as subunits.

Interpretation Introduction

(b)

Interpretation:

The type of linkage present in the maltose is to be stated.

Concept introduction:

Monosaccharide is the simplest form of carbohydrate. It possesses either ketone or aldehyde with one or more than one hydroxyl (OH) groups. Disaccharides, trisaccharides, oligosaccharides, and polysaccharides all involve monosaccharide as subunits.

Interpretation Introduction

(c)

Interpretation:

The type of linkage present in the sucrose is to be stated.

Concept introduction:

Monosaccharide is the simplest form of carbohydrate. It possesses either ketone or aldehyde with one or more than one hydroxyl (OH) groups. Disaccharides, trisaccharides, oligosaccharides, and polysaccharides all involve monosaccharide as subunits.

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First image: I have to show the mecanism for the reaction on the left, where the alcohol A is added fast in one portion Second image: I have to show the mecanism of the reaction at the bottom. Also I have to show by mecanism why the reaction wouldn't work if the alcohol was primary
First image: I have to explain why the molecule C is never formed in those conditions. Second image: I have to propose a synthesis for the lactone A
First image: I have to explain why the molecule C is never formed in these conditions Second image: I have to propose a synthesis for the lactone A

Chapter 17 Solutions

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