Choose the correct relationship for \(\alpha\) -D-glucose (a) and \(\beta\) -D-glucose (B). (a) \(\mathrm{A}\) and \(\mathrm{B}\) are crystal modification (b) \(\mathrm{A}\) and \(\mathrm{B}\) are epimers (c) \(\mathrm{A}\) is an aldose and \(\mathrm{B}\) is a ketose (d) \(\mathrm{A}\) is a pyranose sugar and \(\mathrm{B}\) is a furanose sugar

Short Answer

Expert verified
\(\alpha\)-D-glucose (A) and \(\beta\)-D-glucose (B) are epimers.

Step by step solution

01

Understand the significance of alpha and beta in glucose nomenclature

The prefixes \(\alpha\) and \(\beta\) in glucose nomenclature refer to the stereoisomers of D-glucose that differ in the configuration around the anomeric carbon (carbon-1). In \(\alpha\)-D-glucose, the hydroxyl group attached to the anomeric carbon is on the opposite side of the ring compared to the CH2OH group at carbon-5. In contrast, in \(\beta\)-D-glucose, these groups are on the same side.
02

Eliminate incorrect relationships based on chemical properties

Option (a) is incorrect as crystal modifications refer to different crystal structures of the same molecule, which is not the case with \(\alpha\) and \(\beta\) stereoisomers. Option (c) is incorrect because both \(\alpha\) and \(\beta\) forms of D-glucose are aldoses, not ketoses. Option (d) is incorrect because both \(\alpha\) and \(\beta\) forms of D-glucose are pyranose sugars, owing to their six-membered ring structure.
03

Identify the correct relationship between (A) \(\alpha\)-D-glucose and (B) \(\beta\)-D-glucose

Epimers are stereoisomers that differ in configuration at only one specific carbon. Since \(\alpha\)-D-glucose and \(\beta\)-D-glucose differ only in their configuration around the anomeric carbon, they are epimers. Therefore, the correct answer is (b) \(\mathrm{A}\) and \(\mathrm{B}\) are epimers.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbohydrate Chemistry
Carbohydrate chemistry focuses on the study of carbohydrates, a diverse group of biomolecules that serve as energy sources, structural components, and participate in various biological processes. Simple sugars, or monosaccharides, such as glucose, fructose, and galactose, are the building blocks of more complex carbohydrates.

Monosaccharides consist of carbon, hydrogen, and oxygen atoms with the generic formula \( C_{n}(H_2O)_{n} \). They are classified based on the number of carbon atoms they contain, such as trioses for three carbons, pentoses for five, and hexoses for six. Glucose, for example, is a hexose.

These simple sugars can be found in ring-shaped structures called pyranoses, resembling a six-membered ring, or furanoses, which are five-membered rings. The ring forms when the hydroxyl group on one of the carbon atoms reacts with the aldehyde or ketone group. In this cyclic form, the carbon atom that was part of the aldehyde or ketone group becomes a new chiral center, called the anomeric carbon, which is the site where \(\alpha\) and \(\beta\) isomers differ.
Stereoisomerism
Stereoisomerism refers to the spatial arrangement of atoms in molecules that have the same sequence of bonded atoms but differ in the three-dimensional orientations of those atoms. It plays a critical role in carbohydrate chemistry because the specific arrangement of atoms determines the properties and functions of these biological molecules.

There are several types of stereoisomerism, one of which is optical isomerism. Optical isomers, or enantiomers, are non-superimposable mirror images of each other, like left and right hands. Another form is diastereomerism, which includes epimerism. Epimers are a subset of diastereomers that differ in configuration at only one chiral center among many. In glucose, an important type of diastereomerism is the distinction between \(\alpha\)-D-glucose and \(\beta\)-D-glucose, which are two epimers that differ only at the anomeric carbon.

The configuration of the hydroxyl group on the anomeric carbon (up for \(\alpha\), down for \(\beta\)) gives rise to distinct physical and chemical properties. These differences are essential for the functionality of carbohydrates in biological systems, like the differing roles of \(\alpha\) and \(\beta\) glucose in human metabolism.
Monosaccharides
Monosaccharides are the simplest form of carbohydrates and are the fundamental units from which all larger carbohydrate molecules are constructed. These single-sugar molecules cannot be hydrolyzed into simpler sugars. Glucose is one of the most common monosaccharides and plays a central role in energy metabolism.

The structure of monosaccharides can significantly affect their behavior and interaction with other biological molecules. The \(\alpha\) or \(\beta\) designation in glucose refers to the stereoisomers that form due to the creation of the cyclic structure involving the anomeric carbon. The \(\alpha\)-anomer has the OH group at the anomeric carbon facing opposite to the CH2OH group at carbon-5, while the \(\beta\)-anomer has these groups on the same side, which influences the way these molecules interact with enzymes and other elements within the body.

Understanding the unique characteristics of monosaccharides like glucose is crucial in fields ranging from biochemistry to nutrition, as these molecules are not only energy sources but also intermediates in various metabolic pathways. The difference between \(\alpha\)-D-glucose and \(\beta\)-D-glucose is a prime example of how slight variations in structure lead to different physiological functions and nutritional properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following statements about amino acids: 1\. Nitrous acid liberates nitrous oxide from amino acids 2\. An important sensitive test for the detection of amino acids is the ninhydrin colour test 3\. Glycine is the only amino acid which does not possess a chiral centre. 4\. All the natural amino acids that occur as constituents of proteins belong to the L-series Which of these statements are correct? (a) 1,2 and 3 (b) 1,2 and 4 (c) 2,3 and 4 (d) 2 and 3 only

An example for a saturated fatty acid, present in nature is: (a) Oleic acid (b) Linoleic acid (c) Linolenic acid (d) Palmitic acid

Consider the following statements: 1\. Monosaccharides are optically active polyhydroxy carbonyl compounds 2\. Fructose does not reduce Fehling's solutio because it is a ketose 3\. \(\alpha-\mathrm{D}(+)\) -glucose and \(\beta-\mathrm{D}(+)\) -glucose are anomers 4\. D-glucose and D-mannose are C-2 epimers Which of the following statements are correct? (a) 1,2 and 3 (b) 1,3 and 4 (c) 1,2 and 4 (d) 2,3 and 4

Consider the following statements about sucrose. 1\. Hydrolysis of sucrose with dilute acid yields an equimolar mixture of D-glucose and D-fructose 2\. Acid hydrolysis of sucrose is accompanied by a change in optical rotation 3\. In sucrose, the glycosidic linkage is between \(\mathrm{C}-1\) of glucose and \(\mathrm{C}-2\) of fructose 4\. Aqueous solution of sucrose exhibits mutarotation Which of the statements are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1,2 and 3 (d) 1,3 and 4

Sucrose itself is a non reduction sugar. It is due to: (a) The linkage of both the anomeric carbon atoms in an acetal form (b) The presence of \(\alpha\) -hydroxy keto group in its structure (c) The presence of equal amount of \(\mathrm{D}(+)\) glucose and \(\mathrm{D}(-)\) fructose (d) The easy conversion of sucrose into invert sugar on hydrolysis

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free