The process by which the hemiacetal group of a lactose molecule is able to react with Benedict’s reagent is to be explained. Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. The carbohydrates with free aldehyde or ketone groups can act as reducing sugars. Under basic conditions, weak oxidizing agents such as Tollen’s reagent and Benedict’s solution oxidize the aldehyde or ketone group of the monosaccharide.

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Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 17, Problem 17.46E

Interpretation Introduction

Interpretation:

The process by which the hemiacetal group of a lactose molecule is able to react with Benedict’s reagent is to be explained.

Concept introduction:

Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. The carbohydrates with free aldehyde or ketone groups can act as reducing sugars. Under basic conditions, weak oxidizing agents such as Tollen’s reagent and Benedict’s solution oxidize the aldehyde or ketone group of the monosaccharide.

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Answer 2

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 17, Problem 17.46E

Interpretation Introduction

Interpretation:

The process by which the hemiacetal group of a lactose molecule is able to react with Benedict’s reagent is to be explained.

Concept introduction:

Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. The carbohydrates with free aldehyde or ketone groups can act as reducing sugars. Under basic conditions, weak oxidizing agents such as Tollen’s reagent and Benedict’s solution oxidize the aldehyde or ketone group of the monosaccharide.

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 3

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 17, Problem 17.46E

Interpretation Introduction

Interpretation:

The process by which the hemiacetal group of a lactose molecule is able to react with Benedict’s reagent is to be explained.

Concept introduction:

Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. The carbohydrates with free aldehyde or ketone groups can act as reducing sugars. Under basic conditions, weak oxidizing agents such as Tollen’s reagent and Benedict’s solution oxidize the aldehyde or ketone group of the monosaccharide.

Expert Solution & Answer

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Answer 4

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 17, Problem 17.46E

Interpretation Introduction