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.

Question

Want to see the full answer?

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

The beta-lactamase hydrolyzes the lactam-ring in penicillin. Describe the mechanism of hydrolysis, insuring to include the involvement of S, D, & K in the reaction sequence. Please help

To map the active site of beta-lactamase, the enzyme was hydrolyzed with trypsin to yield a hexapeptide (P1) with the following amino acids. Glu, Lys, Leu, Phe, Met, and Ser. Treatment of P1 with phenyl isothiocyanate yielded a PTH derivative of phenylalanine and a peptide (P2). Treatment of P1 with cyanogenbromide gave an acidic tetrapeptide (P3) and a dipeptide (P4).Treatment of P2 with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis, yields N-2,4-dinitrophenyl-Glu. P1, P2, and P3 contain the active site serine. Why doesn't D in this hexapeptide not participate in the hydrolysis of the beta-lactam ring even though S, K, and D are involved in the catalyst?

To map the active site of -lactamase, the enzyme was hydrolyzed with trypsin to yield a hexapeptide (P1) with the following amino acids. Glu, Lys, Leu, Phe, Met, and Ser. Treatment of P1 with phenyl isothiocyanate yielded a PTH derivative of phenylalanine and a peptide (P2). Treatment of P1 with cyanogenbromide gave an acidic tetrapeptide (P3) and a dipeptide (P4).Treatment of P2 with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis, yields N-2,4-dinitrophenyl-Glu. P1, P2, and P3 contain the active site serine. Using the experimental results described above derive the primary sequence of the active site hexapeptide. Please help!

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

Want to see the full answer?

Check out a sample textbook solution

See solution

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.

Concept explainers

Want to see the full answer?

Chapter 20 Solutions