The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair. Concept introduction: The pair of Enantiomers is non-superimposable mirror images of each other. The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center). The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as, If the –OH group is located in right side then, the carbohydrate is a D-sugar. If the –OH group is located in left side then, the carbohydrate is an L-sugar.

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Answer 1

Question

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

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Students have asked these similar questions

7. Which one is the major organic product obtained from the following reaction? CHO COOH CH₂OH CHO COOH H- -OH H₂ H- OH H- -OH H- -OH Ni HO -H HO -H HO- -H HO- H HO- H H- -OH H- -OH H- -OH H -OH H- -OH H- -OH H- -OH H- -OH H- -OH H -OH CH₂OH CH₂OH CH₂OH CH₂OH COOH A B C D

4. The compound on the right is- B) a ketopentose C) an aldohexose A) an aldopentose D) a ketohexose E) a ketoheptose OH OH CH2OH OH

12. When glutamic acid is heated to 180°C., it loses a molecule of water to form a lactam. The bond that creates the lactam is an amide bond that forms between the amino nitrogen and the carbon of the side chain -COOH group. What is the most reasonable structure for this lactam? (NOTE: The structure of glutamic acid is shown in Question 1.) COOH COOH ا کچھ جو کچھ C D COOH B COOH E COOH

Answer 2

Question

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

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Answer 3

Question

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Expert Solution & Answer

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Answer 4

Question

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Expert Solution & Answer

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Answer 5

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Answer 6

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Answer 7

Chapter 20.2, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Answer 8

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Answer 9

(b)

Interpretation Introduction

Interpretation:

The enantiomer of the given monosaccharide has to be drawn and D or L-sugar has to be identified from the pair.

Concept introduction:

The pair of Enantiomers is non-superimposable mirror images of each other.

The stereo-descriptor used for carbohydrates is D or L. it is based on the dextrorotatory or levorotatory of smallest carbohydrate glyceraldehyde (1 chiral center).

The stereo-descriptor for other carbohydrates of having more than one chiral center will be evaluated by the location of –OH group (right or left) of farthest chiral carbon (bottom) from the carbonyl group. Such as,

If the –OH group is located in right side then, the carbohydrate is a D-sugar.

If the –OH group is located in left side then, the carbohydrate is an L-sugar.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 20.1 - Classify the following monosaccharides as an...Ch. 20.1 - Prob. 20.2P Ch. 20.2 - Prob. 20.3P Ch. 20.2 - Prob. 20.4P Ch. 20.2 - Prob. 20.6P Ch. 20.3 - D-Talose, a constituent of certain antibiotics,...Ch. 20.3 - Prob. 20.8P Ch. 20.3 - Draw the structure that completes the mutarotation...Ch. 20.4 - Prob. 20.10KCP Ch. 20.4 - Prob. 20.11P

Solution Summary: The author explains that the pair of Enantiomers is non-superimposable mirror images of each other. The stereo-descriptor used for carbohydrates is D or L.

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