Textbook Question
Name four important monosaccharides and tell where each occurs in nature.
Ch.20 Carbohydrates
McMurry8th EditionFundamentals of General, Organic, and Biological ChemistryISBN: 9780134015187
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Table of contents
- Ch.1 Matter and Measurements27
- Ch.2 Atoms and the Periodic Table20
- Ch.3 Ionic Compounds8
- Ch.4 Molecular Compounds23
- Ch.5 Classification & Balancing of Chemical Reactions12
- Ch.6 Chemical Reactions: Mole and Mass Relationships28
- Ch.7 Chemical Reactions: Energy, Rate and Equilibrium64
- Ch.8 Gases, Liquids and Solids24
- Ch.9 Solutions52
- Ch.10 Acids and Bases25
- Ch.11 Nuclear Chemistry22
- Ch.12 Introduction to Organic Chemistry: Alkanes57
- Ch.13 Alkenes, Alkynes, and Aromatic Compounds47
- Ch.14 Some Compounds with Oxygen, Sulfur, or a Halogen50
- Ch.15 Aldehydes and Ketones45
- Ch.16 Amines42
- Ch.17 Carboxylic Acids and Their Derivatives57
- Ch.18 Amino Acids and Proteins82
- Ch.19 Enzymes and Vitamins63
- Ch.20 Carbohydrates44
- Ch.21 The Generation of Biochemical Energy65
- Ch.22 Carbohydrate Metabolism45
- Ch.23 Lipids42
- Ch.24 Lipid Metabolism33
- Ch.25 Protein and Amino Acid Metabolism24
- Ch.26 Nucleic Acids and Protein Synthesis45
McMurry8th EditionFundamentals of General, Organic, and Biological ChemistryISBN: 9780134015187Not the one you use?Change textbook
Chapter 20, Problem 40
Only three stereoisomers are possible for 2,3-dibromo-2, 3-dichlorobutane. Draw them, indicating which pair are enantiomers (optical isomers). Why does the other isomer not have an enantiomer?
Verified step by step guidance1
Step 1: Understand the structure of 2,3-dibromo-2,3-dichlorobutane. The molecule has a butane backbone with bromine (Br) and chlorine (Cl) atoms attached to the second and third carbon atoms. These carbons are chiral centers because they are bonded to four different groups.
Step 2: Draw the possible stereoisomers. Since there are two chiral centers, each can have two configurations (R or S). This gives a maximum of 2^2 = 4 stereoisomers. However, due to symmetry in the molecule, only three unique stereoisomers exist.
Step 3: Identify the enantiomers. Enantiomers are non-superimposable mirror images of each other. Draw the pair of stereoisomers where the configurations of the two chiral centers are opposite (e.g., (R,R) and (S,S)). These are the enantiomers.
Step 4: Identify the meso compound. The third stereoisomer is a meso compound, which has an internal plane of symmetry. This symmetry makes it superimposable on its mirror image, meaning it does not have an enantiomer.
Step 5: Explain why the meso compound lacks an enantiomer. The internal symmetry of the meso compound cancels out optical activity, making it achiral. As a result, it is identical to its mirror image and does not have a distinct enantiomer.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Stereoisomers
Stereoisomers are compounds that have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of those atoms. This category includes enantiomers, which are non-superimposable mirror images of each other, and diastereomers, which are not mirror images. Understanding stereoisomers is crucial for analyzing the optical activity and reactivity of molecules.
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Stereoisomers Concept 2
Enantiomers
Enantiomers are a specific type of stereoisomer that are mirror images of each other and cannot be superimposed. They typically arise in chiral molecules, which have at least one carbon atom bonded to four different substituents. Enantiomers often exhibit different optical activities, meaning they rotate plane-polarized light in opposite directions, which is significant in fields like pharmaceuticals.
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3:33Enantiomers vs Diastereomers Concept 1
Chirality
Chirality refers to the geometric property of a molecule that makes it non-superimposable on its mirror image, akin to how left and right hands are mirror images but not identical. A chiral center, usually a carbon atom with four different substituents, is essential for the formation of enantiomers. In the case of 2,3-dibromo-2,3-dichlorobutane, the presence of two chiral centers leads to the formation of stereoisomers, but not all configurations will yield enantiomers.
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Related Practice
Textbook Question
In its open-chain form, D-mannose, an aldohexose found in orange peels, has the structure shown here. Coil mannose around and draw it in the cyclic hemiacetal ⍺ and β forms.
Textbook Question
How many chiral carbon atoms are present in each of the molecules shown in Problem 20.31?
a.
b.
c.
d.
Textbook Question
Sucrose and D-glucose rotate plane-polarized light to the right; D-fructose rotates light to the left. When sucrose is hydrolyzed, the glucose–fructose mixture rotates light to the left.
b. Why do you think the mixture is called “invert sugar”?
Textbook Question
In Section 15.6, you saw that aldehydes react with reducing agents to yield primary alcohols (RCH=O → RCH2OH). The structures of two D-aldotetroses are shown. One of them can be reduced to yield a chiral product, but the other yields an achiral product. Explain.
Textbook Question
Draw the open-chain structure of a 4-carbon deoxy sugar.