Organic And Biological Chemistry
Organic And Biological Chemistry
7th Edition
ISBN: 9781305081079
Author: STOKER, H. Stephen (howard Stephen)
Publisher: Cengage Learning,
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Question
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Chapter 7, Problem 7.93EP

(a)

Interpretation Introduction

Interpretation: The Haworth projection formula for the given monosaccharide has to be drawn.

Concept introduction: The Haworth projection formula can be drawn by using the following rules.

  • Ø For α and β configurations, if hydroxyl group is drawn at the right position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn below the ring.
  • Ø If hydroxyl group is drawn at the left position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn above the ring.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction.
  • Ø For α. configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the opposite direction.
  • Ø The hydroxyl group that is obtained from the carbonyl group can be placed above or below the ring which is dependent upon the ring closure of the cyclic form.
  • Ø For D configuration, CH2OH group is always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group is always placed below the cyclic ring.

(a)

Expert Solution
Check Mark

Answer to Problem 7.93EP

The Haworth projection formula for the given monosaccharide is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  1

Explanation of Solution

The given monosaccharide is αDgalactose. The open-chain structure of αDgalactose is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  2

The Haworth projection formula for the given monosaccharide, αDgalactose is shown as,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  3

Thus, in this Haworth projection formula of the given monosaccharide, CH2OH group is placed above the cyclic ring.  The position of OH group at the first carbon with respect to CH2OH group is in the opposite direction.  Thus, this Haworth projection formula is a αDmonosaccharide.

(b)

Interpretation Introduction

Interpretation: The Haworth projection formula for the given monosaccharide has to be drawn.

Concept introduction: The Haworth projection formula can be drawn by using the following rules.

  • Ø For α and β configurations, if hydroxyl group is drawn at the right position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn below the ring.
  • Ø If hydroxyl group is drawn at the left position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn above the ring.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction.
  • Ø For α. configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the opposite direction.
  • Ø The hydroxyl group that is obtained from the carbonyl group can be placed above or below the ring which is dependent upon the ring closure of the cyclic form.
  • Ø For D configuration, CH2OH group is always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group is always placed below the cyclic ring.

(b)

Expert Solution
Check Mark

Answer to Problem 7.93EP

The Haworth projection formula for the given monosaccharide is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  4

Explanation of Solution

The given monosaccharide is βDgalactose. The open-chain structure of βDgalactose is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  5

The Haworth projection formula for the given monosaccharide, βDgalactose is shown as,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  6

Thus, in this Haworth projection formula of the given monosaccharide, CH2OH group is placed above the cyclic ring.  The position of OH group at the first carbon with respect to CH2OH group is in the same direction.  Thus, this Haworth projection formula is a βDmonosaccharide.

(c)

Interpretation Introduction

Interpretation: The Haworth projection formula for the given monosaccharide has to be drawn.

Concept introduction: The Haworth projection formula can be drawn by using the following rules.

  • Ø For α and β configurations, if hydroxyl group is drawn at the right position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn below the ring.
  • Ø If hydroxyl group is drawn at the left position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn above the ring.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction.
  • Ø For α. configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the opposite direction.
  • Ø The hydroxyl group that is obtained from the carbonyl group can be placed above or below the ring which is dependent upon the ring closure of the cyclic form.
  • Ø For D configuration, CH2OH group is always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group is always placed below the cyclic ring.

(c)

Expert Solution
Check Mark

Answer to Problem 7.93EP

The Haworth projection formula for the given monosaccharide is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  7

Explanation of Solution

The given monosaccharide is αLgalactose. The open-chain structure of αLgalactose is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  8

The Haworth projection formula for the given monosaccharide, αLgalactose is shown as,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  9

Thus, in this Haworth projection formula of the given monosaccharide, CH2OH group is placed below the cyclic ring.  The position of. OH. group at the first carbon with respect to CH2OH group is in the opposite direction.  Thus, this Haworth projection formula is a αLmonosaccharide.

(d)

Interpretation Introduction

Interpretation: The Haworth projection formula for the given monosaccharide has to be drawn.

Concept introduction: The Haworth projection formula can be drawn by using the following rules.

  • Ø For α and β configurations, if hydroxyl group is drawn at the right position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn below the ring.
  • Ø If hydroxyl group is drawn at the left position in Fischer projection formula then in the cyclic form of Haworth projection, the same OH group should be drawn above the ring.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction.
  • Ø For α. configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the opposite direction.
  • Ø The hydroxyl group that is obtained from the carbonyl group can be placed above or below the ring which is dependent upon the ring closure of the cyclic form.
  • Ø For D configuration, CH2OH group is always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group is always placed below the cyclic ring.

(d)

Expert Solution
Check Mark

Answer to Problem 7.93EP

The Haworth projection formula for the given monosaccharide is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  10

Explanation of Solution

The given monosaccharide is βLgalactose. The open-chain structure of βLgalactose is,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  11

The Haworth projection formula for the given monosaccharide, βLgalactose is shown as,

Organic And Biological Chemistry, Chapter 7, Problem 7.93EP , additional homework tip  12

Thus, in this Haworth projection formula of the given monosaccharide, CH2OH group is placed below the cyclic ring.  The position of OH group at the first carbon with respect to CH2OH group is in the same direction.  Thus, this Haworth projection formula is a βLmonosaccharide.

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Chapter 7 Solutions

Organic And Biological Chemistry

Ch. 7.4 - Prob. 3QQCh. 7.4 - Prob. 4QQCh. 7.5 - Prob. 1QQCh. 7.5 - Prob. 2QQCh. 7.6 - Prob. 1QQCh. 7.6 - Which of the following Fischer projection formulas...Ch. 7.6 - Prob. 3QQCh. 7.6 - Prob. 4QQCh. 7.7 - Prob. 1QQCh. 7.7 - Prob. 2QQCh. 7.8 - Prob. 1QQCh. 7.8 - Which of the following statements about...Ch. 7.8 - Prob. 3QQCh. 7.9 - Prob. 1QQCh. 7.9 - Prob. 2QQCh. 7.9 - Prob. 3QQCh. 7.9 - In which of the following pairs of monosaccharides...Ch. 7.9 - In which of the following pairs of monosaccharides...Ch. 7.10 - Prob. 1QQCh. 7.10 - Which of the following structures represents a...Ch. 7.10 - Prob. 3QQCh. 7.10 - Prob. 4QQCh. 7.10 - Prob. 5QQCh. 7.11 - Prob. 1QQCh. 7.11 - Which of the following is the correct Haworth...Ch. 7.12 - Prob. 1QQCh. 7.12 - Prob. 2QQCh. 7.12 - Prob. 3QQCh. 7.12 - Prob. 4QQCh. 7.12 - Prob. 5QQCh. 7.13 - Which of the following disaccharides contains...Ch. 7.13 - Which of the following disaccharides will produce...Ch. 7.13 - In which of the following disaccharides is the...Ch. 7.13 - In which of the following pairs of disaccharides...Ch. 7.13 - Which of the following disaccharides is not a...Ch. 7.13 - The terms milk sugar and table sugar apply,...Ch. 7.14 - Prob. 1QQCh. 7.14 - Prob. 2QQCh. 7.15 - Which of the following statements about...Ch. 7.15 - Prob. 2QQCh. 7.16 - Which of the following storage polysaccharides has...Ch. 7.16 - Prob. 2QQCh. 7.16 - Prob. 3QQCh. 7.16 - Prob. 4QQCh. 7.17 - Prob. 1QQCh. 7.17 - Which of the following statements about cellulose...Ch. 7.17 - Chitin is a polysaccharide in which the...Ch. 7.18 - Which of the following statements about the...Ch. 7.18 - Which of the following statements about the...Ch. 7.19 - Which of the following is not classified as a...Ch. 7.19 - Prob. 2QQCh. 7.20 - Which of the following types of compounds are...Ch. 7.20 - Which of the following is not a biochemical...Ch. 7 - Prob. 7.1EPCh. 7 - Prob. 7.2EPCh. 7 - Prob. 7.3EPCh. 7 - Prob. 7.4EPCh. 7 - Prob. 7.5EPCh. 7 - Prob. 7.6EPCh. 7 - Prob. 7.7EPCh. 7 - Prob. 7.8EPCh. 7 - Prob. 7.9EPCh. 7 - Prob. 7.10EPCh. 7 - Prob. 7.11EPCh. 7 - Prob. 7.12EPCh. 7 - Prob. 7.13EPCh. 7 - Prob. 7.14EPCh. 7 - Prob. 7.15EPCh. 7 - Prob. 7.16EPCh. 7 - Prob. 7.17EPCh. 7 - Prob. 7.18EPCh. 7 - Prob. 7.19EPCh. 7 - Prob. 7.20EPCh. 7 - Prob. 7.21EPCh. 7 - Prob. 7.22EPCh. 7 - Prob. 7.23EPCh. 7 - Prob. 7.24EPCh. 7 - Prob. 7.25EPCh. 7 - Indicate whether or not each of the molecules in...Ch. 7 - Prob. 7.27EPCh. 7 - Prob. 7.28EPCh. 7 - Prob. 7.29EPCh. 7 - Prob. 7.30EPCh. 7 - Prob. 7.31EPCh. 7 - Prob. 7.32EPCh. 7 - Prob. 7.33EPCh. 7 - Prob. 7.34EPCh. 7 - Prob. 7.35EPCh. 7 - Draw the Fischer projection formula for each of...Ch. 7 - Prob. 7.37EPCh. 7 - Prob. 7.38EPCh. 7 - Prob. 7.39EPCh. 7 - Prob. 7.40EPCh. 7 - Prob. 7.41EPCh. 7 - Prob. 7.42EPCh. 7 - Prob. 7.43EPCh. 7 - Prob. 7.44EPCh. 7 - Prob. 7.45EPCh. 7 - Prob. 7.46EPCh. 7 - Prob. 7.47EPCh. 7 - Prob. 7.48EPCh. 7 - Prob. 7.49EPCh. 7 - Prob. 7.50EPCh. 7 - Prob. 7.51EPCh. 7 - Prob. 7.52EPCh. 7 - Prob. 7.53EPCh. 7 - Prob. 7.54EPCh. 7 - Prob. 7.55EPCh. 7 - Prob. 7.56EPCh. 7 - Prob. 7.57EPCh. 7 - Prob. 7.58EPCh. 7 - Prob. 7.59EPCh. 7 - Prob. 7.60EPCh. 7 - Prob. 7.61EPCh. 7 - Prob. 7.62EPCh. 7 - Prob. 7.63EPCh. 7 - Prob. 7.64EPCh. 7 - Prob. 7.65EPCh. 7 - Prob. 7.66EPCh. 7 - Prob. 7.67EPCh. 7 - Prob. 7.68EPCh. 7 - Prob. 7.69EPCh. 7 - Prob. 7.70EPCh. 7 - Prob. 7.71EPCh. 7 - Prob. 7.72EPCh. 7 - Prob. 7.73EPCh. 7 - Prob. 7.74EPCh. 7 - Prob. 7.75EPCh. 7 - Prob. 7.76EPCh. 7 - Prob. 7.77EPCh. 7 - Prob. 7.78EPCh. 7 - Prob. 7.79EPCh. 7 - Prob. 7.80EPCh. 7 - Prob. 7.81EPCh. 7 - Prob. 7.82EPCh. 7 - Prob. 7.83EPCh. 7 - Prob. 7.84EPCh. 7 - Prob. 7.85EPCh. 7 - Prob. 7.86EPCh. 7 - Prob. 7.87EPCh. 7 - Prob. 7.88EPCh. 7 - Prob. 7.89EPCh. 7 - Prob. 7.90EPCh. 7 - Prob. 7.91EPCh. 7 - Prob. 7.92EPCh. 7 - Prob. 7.93EPCh. 7 - Prob. 7.94EPCh. 7 - Prob. 7.95EPCh. 7 - Prob. 7.96EPCh. 7 - Prob. 7.97EPCh. 7 - Classify each of the glucose derivatives in...Ch. 7 - Prob. 7.99EPCh. 7 - Prob. 7.100EPCh. 7 - Prob. 7.101EPCh. 7 - Prob. 7.102EPCh. 7 - Prob. 7.103EPCh. 7 - Prob. 7.104EPCh. 7 - Prob. 7.105EPCh. 7 - Prob. 7.106EPCh. 7 - Prob. 7.107EPCh. 7 - Prob. 7.108EPCh. 7 - Prob. 7.109EPCh. 7 - Prob. 7.110EPCh. 7 - Prob. 7.111EPCh. 7 - Prob. 7.112EPCh. 7 - Prob. 7.113EPCh. 7 - Prob. 7.114EPCh. 7 - Prob. 7.115EPCh. 7 - Prob. 7.116EPCh. 7 - Prob. 7.117EPCh. 7 - Prob. 7.118EPCh. 7 - Prob. 7.119EPCh. 7 - Prob. 7.120EPCh. 7 - Prob. 7.121EPCh. 7 - Prob. 7.122EPCh. 7 - Prob. 7.123EPCh. 7 - Prob. 7.124EPCh. 7 - Prob. 7.125EPCh. 7 - Prob. 7.126EPCh. 7 - Prob. 7.127EPCh. 7 - Prob. 7.128EPCh. 7 - Prob. 7.129EPCh. 7 - Prob. 7.130EPCh. 7 - Prob. 7.131EPCh. 7 - Prob. 7.132EPCh. 7 - Prob. 7.133EPCh. 7 - Prob. 7.134EPCh. 7 - Prob. 7.135EPCh. 7 - Prob. 7.136EPCh. 7 - Prob. 7.137EPCh. 7 - Prob. 7.138EPCh. 7 - Prob. 7.139EPCh. 7 - Prob. 7.140EPCh. 7 - Prob. 7.141EPCh. 7 - Prob. 7.142EPCh. 7 - Prob. 7.143EPCh. 7 - Prob. 7.144EPCh. 7 - Prob. 7.145EPCh. 7 - Prob. 7.146EPCh. 7 - Prob. 7.147EPCh. 7 - Prob. 7.148EPCh. 7 - Prob. 7.149EPCh. 7 - Prob. 7.150EPCh. 7 - Prob. 7.151EPCh. 7 - Prob. 7.152EPCh. 7 - Prob. 7.153EPCh. 7 - Prob. 7.154EPCh. 7 - Prob. 7.155EPCh. 7 - Prob. 7.156EPCh. 7 - Prob. 7.157EPCh. 7 - Prob. 7.158EPCh. 7 - Prob. 7.159EPCh. 7 - Prob. 7.160EPCh. 7 - Prob. 7.161EPCh. 7 - Prob. 7.162EPCh. 7 - Prob. 7.163EPCh. 7 - Prob. 7.164EPCh. 7 - Prob. 7.165EPCh. 7 - Prob. 7.166EPCh. 7 - Prob. 7.167EPCh. 7 - Prob. 7.168EPCh. 7 - Prob. 7.169EPCh. 7 - Prob. 7.170EPCh. 7 - Prob. 7.171EPCh. 7 - Prob. 7.172EP
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