Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
7th Edition
ISBN: 9781305081086
Author: STOKER, H. Stephen
Publisher: Brooks Cole
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Chapter 18, Problem 18.84EP

(a)

Interpretation Introduction

Interpretation: The open-chain form of the given βDmonosaccharide has to be drawn.

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

  • Ø For α and β configurations, if in the cyclic form of Haworth projection, OH group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
  • Ø If in the cyclic form of Haworth projection, OH group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

(a)

Expert Solution
Check Mark

Answer to Problem 18.84EP

The open-chain form of the given βDmonosaccharide is,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  1

Explanation of Solution

The given Haworth projection formula is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  2

The open-chain form of the given Haworth projection formula of an βDmonosaccharide is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  3

Thus, in this open-chain form of the given βDmonosaccharide, CH2OH group is placed at the last carbon.  The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide.  But as it is a βDmonosaccharide, thus OH group at the first carbon with respect to CH2OH group is drawn at the left side of the open-chain structure.  The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

(b)

Interpretation Introduction

Interpretation: The open-chain form of the given αDmonosaccharide has to be drawn.

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

  • Ø For α and β configurations, if in the cyclic form of Haworth projection, OH group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
  • Ø If in the cyclic form of Haworth projection, OH group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

(b)

Expert Solution
Check Mark

Answer to Problem 18.84EP

The open-chain form of the given αDmonosaccharide is,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  4

Explanation of Solution

The given Haworth projection formula is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  5

The open-chain form of the given Haworth projection formula of an αDmonosaccharide is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  6

Thus, in this open-chain form of the given αDmonosaccharide, CH2OH group is placed at the last carbon.  The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide.  The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

(c)

Interpretation Introduction

Interpretation: The open-chain form of the given αDmonosaccharide has to be drawn.

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

  • Ø For α and β configurations, if in the cyclic form of Haworth projection, OH group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
  • Ø If in the cyclic form of Haworth projection, OH group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

(c)

Expert Solution
Check Mark

Answer to Problem 18.84EP

The open-chain form of the given αDmonosaccharide is,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  7

Explanation of Solution

The given Haworth projection formula is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  8

The open-chain form of the given Haworth projection formula of an αDmonosaccharide is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  9

Thus, in this open-chain form of the given αDmonosaccharide, CH2OH group is placed at the last carbon.  The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide.  The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

(d)

Interpretation Introduction

Interpretation: The open-chain form of the given αDmonosaccharide has to be drawn.

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

  • Ø For α and β configurations, if in the cyclic form of Haworth projection, OH group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
  • Ø If in the cyclic form of Haworth projection, OH group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
  • Ø For β configuration, the position of OH group at the first carbon with respect to CH2OH group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

(d)

Expert Solution
Check Mark

Answer to Problem 18.84EP

The open-chain form of the given αDmonosaccharide is,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  10

Explanation of Solution

The given Haworth projection formula is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  11

The open-chain form of the given Haworth projection formula of an αDmonosaccharide is shown as,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 18, Problem 18.84EP , additional homework tip  12

Thus, in this open-chain form of the given αDmonosaccharide, CH2OH group is placed at the last carbon.  The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide.  The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

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Chapter 18 Solutions

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th

Ch. 18.4 - Prob. 3QQCh. 18.4 - Prob. 4QQCh. 18.5 - Prob. 1QQCh. 18.5 - Prob. 2QQCh. 18.6 - Prob. 1QQCh. 18.6 - Which of the following Fischer projection formulas...Ch. 18.6 - Prob. 3QQCh. 18.6 - Prob. 4QQCh. 18.7 - Prob. 1QQCh. 18.7 - Prob. 2QQCh. 18.8 - Prob. 1QQCh. 18.8 - Which of the following statements about...Ch. 18.8 - The smallest monosaccharides that can exist are a....Ch. 18.9 - Prob. 1QQCh. 18.9 - Prob. 2QQCh. 18.9 - Prob. 3QQCh. 18.9 - In which of the following pairs of monosaccharides...Ch. 18.9 - In which of the following pairs of monosaccharides...Ch. 18.10 - Prob. 1QQCh. 18.10 - Which of the following structures represents a...Ch. 18.10 - Prob. 3QQCh. 18.10 - Prob. 4QQCh. 18.10 - Prob. 5QQCh. 18.11 - Prob. 1QQCh. 18.11 - Which of the following is the correct Haworth...Ch. 18.12 - Prob. 1QQCh. 18.12 - Prob. 2QQCh. 18.12 - Prob. 3QQCh. 18.12 - Prob. 4QQCh. 18.12 - Prob. 5QQCh. 18.13 - Which of the following disaccharides contains...Ch. 18.13 - Which of the following disaccharides will produce...Ch. 18.13 - In which of the following disaccharides is the...Ch. 18.13 - In which of the following pairs of disaccharides...Ch. 18.13 - Which of the following disaccharides is not a...Ch. 18.13 - The terms milk sugar and table sugar apply,...Ch. 18.14 - Prob. 1QQCh. 18.14 - Prob. 2QQCh. 18.15 - Which of the following statements about...Ch. 18.15 - Prob. 2QQCh. 18.16 - Prob. 1QQCh. 18.16 - Which of the following storage polysaccharides is...Ch. 18.16 - Prob. 3QQCh. 18.16 - Which of the following statements about storage...Ch. 18.17 - Prob. 1QQCh. 18.17 - Which of the following statements about cellulose...Ch. 18.17 - Chitin is a polysaccharide in which the...Ch. 18.18 - Which of the following statements about the...Ch. 18.18 - Prob. 2QQCh. 18.19 - Which of the following is not classified as a...Ch. 18.19 - Prob. 2QQCh. 18.20 - Prob. 1QQCh. 18.20 - Which of the following is not a biochemical...Ch. 18 - Prob. 18.1EPCh. 18 - Prob. 18.2EPCh. 18 - Prob. 18.3EPCh. 18 - Prob. 18.4EPCh. 18 - Prob. 18.5EPCh. 18 - Prob. 18.6EPCh. 18 - Prob. 18.7EPCh. 18 - Prob. 18.8EPCh. 18 - Prob. 18.9EPCh. 18 - Prob. 18.10EPCh. 18 - Prob. 18.11EPCh. 18 - Prob. 18.12EPCh. 18 - Prob. 18.13EPCh. 18 - Prob. 18.14EPCh. 18 - Prob. 18.15EPCh. 18 - Prob. 18.16EPCh. 18 - Prob. 18.17EPCh. 18 - Prob. 18.18EPCh. 18 - Prob. 18.19EPCh. 18 - Prob. 18.20EPCh. 18 - Prob. 18.21EPCh. 18 - Prob. 18.22EPCh. 18 - Prob. 18.23EPCh. 18 - Prob. 18.24EPCh. 18 - Prob. 18.25EPCh. 18 - Prob. 18.26EPCh. 18 - Prob. 18.27EPCh. 18 - Prob. 18.28EPCh. 18 - Prob. 18.29EPCh. 18 - Prob. 18.30EPCh. 18 - Prob. 18.31EPCh. 18 - Prob. 18.32EPCh. 18 - Prob. 18.33EPCh. 18 - Prob. 18.34EPCh. 18 - Draw the Fischer projection formula for each of...Ch. 18 - Prob. 18.36EPCh. 18 - Prob. 18.37EPCh. 18 - Prob. 18.38EPCh. 18 - Prob. 18.39EPCh. 18 - Prob. 18.40EPCh. 18 - Prob. 18.41EPCh. 18 - Prob. 18.42EPCh. 18 - Prob. 18.43EPCh. 18 - Prob. 18.44EPCh. 18 - Prob. 18.45EPCh. 18 - Prob. 18.46EPCh. 18 - Prob. 18.47EPCh. 18 - Prob. 18.48EPCh. 18 - Prob. 18.49EPCh. 18 - Prob. 18.50EPCh. 18 - Prob. 18.51EPCh. 18 - Prob. 18.52EPCh. 18 - Prob. 18.53EPCh. 18 - Prob. 18.54EPCh. 18 - Prob. 18.55EPCh. 18 - Prob. 18.56EPCh. 18 - Prob. 18.57EPCh. 18 - Prob. 18.58EPCh. 18 - Prob. 18.59EPCh. 18 - Prob. 18.60EPCh. 18 - Prob. 18.61EPCh. 18 - Prob. 18.62EPCh. 18 - Prob. 18.63EPCh. 18 - Prob. 18.64EPCh. 18 - Prob. 18.65EPCh. 18 - Prob. 18.66EPCh. 18 - Prob. 18.67EPCh. 18 - Prob. 18.68EPCh. 18 - Prob. 18.69EPCh. 18 - Prob. 18.70EPCh. 18 - Prob. 18.71EPCh. 18 - Prob. 18.72EPCh. 18 - Prob. 18.73EPCh. 18 - Prob. 18.74EPCh. 18 - Prob. 18.75EPCh. 18 - Prob. 18.76EPCh. 18 - Prob. 18.77EPCh. 18 - Prob. 18.78EPCh. 18 - Prob. 18.79EPCh. 18 - Prob. 18.80EPCh. 18 - Prob. 18.81EPCh. 18 - Prob. 18.82EPCh. 18 - Prob. 18.83EPCh. 18 - Prob. 18.84EPCh. 18 - Prob. 18.85EPCh. 18 - Prob. 18.86EPCh. 18 - Prob. 18.87EPCh. 18 - Prob. 18.88EPCh. 18 - Prob. 18.89EPCh. 18 - Prob. 18.90EPCh. 18 - Prob. 18.91EPCh. 18 - Prob. 18.92EPCh. 18 - Prob. 18.93EPCh. 18 - Prob. 18.94EPCh. 18 - Prob. 18.95EPCh. 18 - Prob. 18.96EPCh. 18 - Prob. 18.97EPCh. 18 - Prob. 18.98EPCh. 18 - Prob. 18.99EPCh. 18 - Prob. 18.100EPCh. 18 - Prob. 18.101EPCh. 18 - Prob. 18.102EPCh. 18 - Prob. 18.103EPCh. 18 - Prob. 18.104EPCh. 18 - For each structure in Problem 18-103, identify the...Ch. 18 - For each structure in Problem 18-104, identify the...Ch. 18 - Prob. 18.107EPCh. 18 - Prob. 18.108EPCh. 18 - Prob. 18.109EPCh. 18 - Prob. 18.110EPCh. 18 - Prob. 18.111EPCh. 18 - Prob. 18.112EPCh. 18 - Prob. 18.113EPCh. 18 - Prob. 18.114EPCh. 18 - Prob. 18.115EPCh. 18 - Prob. 18.116EPCh. 18 - Prob. 18.117EPCh. 18 - Prob. 18.118EPCh. 18 - Prob. 18.119EPCh. 18 - Prob. 18.120EPCh. 18 - Prob. 18.121EPCh. 18 - Prob. 18.122EPCh. 18 - Prob. 18.123EPCh. 18 - Prob. 18.124EPCh. 18 - Prob. 18.125EPCh. 18 - Prob. 18.126EPCh. 18 - Prob. 18.127EPCh. 18 - Prob. 18.128EPCh. 18 - Prob. 18.129EPCh. 18 - Prob. 18.130EPCh. 18 - Prob. 18.131EPCh. 18 - Prob. 18.132EPCh. 18 - Prob. 18.133EPCh. 18 - Prob. 18.134EPCh. 18 - Prob. 18.135EPCh. 18 - Prob. 18.136EPCh. 18 - 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