(a) Interpretation: The names of the two monosaccharides that would form on hydrolysis of melibiose are 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.

Question

Explanation Check F1 H₂O H₂ Pd 1) MCPBA 2) H3O+ 1) Hg(OAc)2, H₂O 2) NaBH4 OH CI OH OH OH hydration halohydrin formation addition halogenation hydrogenation inhalation hydrogenation hydration ☐ halohydrin formation addition halogenation formation chelation hydrogenation halohydrin formation substitution hydration halogenation addition Ohalohydrin formation subtraction halogenation addition hydrogenation hydration F2 80 F3 σ F4 F5 F6 1 ! 2 # 3 $ 4 % 05 Q W & Å © 2025 McGraw Hill LLC. All Rights Reserved. F7 F8 ( 6 7 8 9 LU E R T Y U A F9

Answer 1

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 17.52E

Interpretation Introduction

(a)

Interpretation:

The names of the two monosaccharides that would form on hydrolysis of melibiose are 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:

Whether melibiose is a reducing sugar or not is to be stated. The explanation for the corresponding answer 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 melibiose 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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Explanation Check F1 H₂O H₂ Pd 1) MCPBA 2) H3O+ 1) Hg(OAc)2, H₂O 2) NaBH4 OH CI OH OH OH hydration halohydrin formation addition halogenation hydrogenation inhalation hydrogenation hydration ☐ halohydrin formation addition halogenation formation chelation hydrogenation halohydrin formation substitution hydration halogenation addition Ohalohydrin formation subtraction halogenation addition hydrogenation hydration F2 80 F3 σ F4 F5 F6 1 ! 2 # 3 $ 4 % 05 Q W & Å © 2025 McGraw Hill LLC. All Rights Reserved. F7 F8 ( 6 7 8 9 LU E R T Y U A F9

Show the mechanism steps to obtain the lowerenergy intermediate: *see image

Soap is made by the previous reaction *see image. The main difference between one soap and another soap isthe length (number of carbons) of the carboxylic acid. However, if a soap irritates your skin, they mostlikely used too much lye.Detergents have the same chemical structure as soaps except for the functional group. Detergentshave sulfate (R-SO4H) and phosphate (R-PO4H2) functional groups. Draw the above carboxylic acidcarbon chain but as the two variants of detergents. *see image

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 17.52E

Interpretation Introduction

(a)

Interpretation:

The names of the two monosaccharides that would form on hydrolysis of melibiose are 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:

Whether melibiose is a reducing sugar or not is to be stated. The explanation for the corresponding answer 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 melibiose 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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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 17.52E

Interpretation Introduction

(a)

Interpretation:

The names of the two monosaccharides that would form on hydrolysis of melibiose are 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:

Whether melibiose is a reducing sugar or not is to be stated. The explanation for the corresponding answer 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 melibiose 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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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 17.52E

Interpretation Introduction