EBK ORGANIC CHEMISTRY
EBK ORGANIC CHEMISTRY
12th Edition
ISBN: 9781119233664
Author: Snyder
Publisher: VST
bartleby

Concept explainers

Question
Book Icon
Chapter 22, Problem 1PP
Interpretation Introduction

Interpretation:

The chirality centers in aldotetrose and ketopentose are to be calculated and the stereoisomers for each general case are to be determined.

Concept Introduction:

Carbohydrates are categorized mainly as monosaccharides, disaccharides, and polysaccharides. Monosaccharides are single sugar units, mainly glucose and fructose, disaccharides are two sugar units, such as sucrose, and polysaccharides are more than two sugar units, such as starch and cellulose.

Monosaccharides containing 3-carbon atoms are called triose, 4-carbon atoms called tetrose, 5-carbon atoms called pentose, and so on.

In chiral molecules, carbon atom having four nonidentical substituent groups is called the chirality center of that molecule. Chirality center may also be called stereocenter, which signifies any point in the molecule where the interchanging of any two groups may lead to stereoisomers. The carbon of a carbohydrate can be considered as chiral when the carbon has all four different substituents attached to it.

The stereoisomers are calculated as follows:

Number of stereoisomers=2n

Here, n is the number of chiral centers.

Expert Solution & Answer
Check Mark

Answer to Problem 1PP

Solution:

a) Two

b) Two

c) Four

Explanation of Solution

a) The aldotetrose

A monosaccharide containing four carbon atoms is called a tetrose. An aldotetrose is a monosaccharide that contains aldehyde group.

The structure of aldotetrose is as follows:

EBK ORGANIC CHEMISTRY, Chapter 22, Problem 1PP , additional homework tip  1

The carbon atom attached to four different groups is chiral carbon. The chiral center in ketopentose is marked by * as follows:

EBK ORGANIC CHEMISTRY, Chapter 22, Problem 1PP , additional homework tip  2

Hence, an aldotetrose has two chirality centers.

b) The ketopentose

A monosaccharide containing five carbon atoms is called a pentose. A pentose containing a keto group is called a ketopentose.

The structure of ketopentose is as follows:

EBK ORGANIC CHEMISTRY, Chapter 22, Problem 1PP , additional homework tip  3

The carbon atom attached to four different groups is chiral carbon. The chiral center in ketopentose is marked by * as follows:

EBK ORGANIC CHEMISTRY, Chapter 22, Problem 1PP , additional homework tip  4

Hence, a ketopentose has two chirality centers.

c) The number of stereoisomers that will be expected from each general structure

Stereoisomers of a molecule have the same molecular formula, but different arrangement of atoms in space. Stereoisomers are different from enantiomers as they are not mirror images of each other, while enantiomers are mirror images of one another.

The compounds aldotetrose and ketopentose have two sets of enantiomers. The number of stereoisomers is calculated as:

Number of stereoisomers=2n

Substitute 2 for n in the above expression as follows:

Number of stereoisomers =22=4

Hence, they will have four stereoisomers for each general structure.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Photochemical smog is formed in part by the action of light on nitrogen dioxide. The wavelength of radiation absorbed by NO2 in this reaction is 197 nm.(a) Draw the Lewis structure of NO2 and sketch its π molecular orbitals.(b) When 1.56 mJ of energy is absorbed by 3.0 L of air at 20 °C and 0.91 atm, all the NO2 molecules in this sample dissociate by the reaction shown. Assume that each absorbed photon leads to the dissociation (into NO and O) of one NO2 molecule. What is the proportion, in parts per million, of NO2 molecules in this sample? Assume that the sample behaves ideally.
Correct each molecule in the drawing area below so that it has the skeletal ("line") structure it would have if it were dissolved in a 0.1 M aqueous solution of HCI. If there are no changes to be made, check the No changes box under the drawing area. No changes. HO Explanation Check NH, 2 W O :□ G ©2025 M unter Accessibility
An expression for the root mean square velocity, vrms, of a gas was derived. Using Maxwell’s velocity distribution, one can also calculate the mean velocity and the most probable velocity (mp) of a collection of molecules. The equations used for these two quantities are vmean=(8RT/πM)1/2 and vmp=(2RT/M)1/2 These values ​​have a fixed relationship to each other.(a) Arrange these three quantities in order of increasing magnitude.(b) Show that the relative magnitudes are independent of the molar mass of the gas.(c) Use the smallest velocity as a reference for establishing the order of magnitude and determine the relationship between the larger and smaller values.
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
  • Text book image
    Organic Chemistry
    Chemistry
    ISBN:9781305580350
    Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
    Publisher:Cengage Learning
    Text book image
    Organic Chemistry
    Chemistry
    ISBN:9781305080485
    Author:John E. McMurry
    Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9781305080485
Author:John E. McMurry
Publisher:Cengage Learning