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 18.81EP

(a)

Interpretation Introduction

Interpretation: The given Haworth projection formula is whether an αDmonosaccharide or a βDmonosaccharide has to be identified.

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 always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group always placed below the cyclic ring.

(a)

Expert Solution
Check Mark

Answer to Problem 18.81EP

The given Haworth projection formula is an αDmonosaccharide.

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 18.81EP , additional homework tip  1

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 an αDmonosaccharide.

(b)

Interpretation Introduction

Interpretation: The given Haworth projection formula is whether an αDmonosaccharide or a βDmonosaccharide has to be identified.

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 always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group always placed below the cyclic ring.

(b)

Expert Solution
Check Mark

Answer to Problem 18.81EP

The given Haworth projection formula is an αDmonosaccharide.

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 18.81EP , additional homework tip  2

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 an αDmonosaccharide.

(c)

Interpretation Introduction

Interpretation: The given Haworth projection formula is whether an αDmonosaccharide or a βDmonosaccharide has to be identified.

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 always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group always placed below the cyclic ring.

(c)

Expert Solution
Check Mark

Answer to Problem 18.81EP

The given Haworth projection formula is a βDmonosaccharide.

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 18.81EP , 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 same direction.  Thus, this Haworth projection formula is a βDmonosaccharide.

(d)

Interpretation Introduction

Interpretation: The given Haworth projection formula is whether an αDmonosaccharide or a βDmonosaccharide has to be identified.

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 always placed above the cyclic ring.
  • Ø For L configuration, CH2OH group always placed below the cyclic ring.

(d)

Expert Solution
Check Mark

Answer to Problem 18.81EP

The given Haworth projection formula is a βDmonosaccharide.

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 18.81EP , additional homework tip  4

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.

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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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