
Concept explainers
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:
Here,

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
The structure of aldotetrose is as follows:
The carbon atom attached to four different groups is chiral carbon. The chiral center in ketopentose is marked by * as follows:
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:
The carbon atom attached to four different groups is chiral carbon. The chiral center in ketopentose is marked by * as follows:
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:
Substitute 2 for
Hence, they will have four stereoisomers for each general structure.
Want to see more full solutions like this?
Chapter 22 Solutions
ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
- OA. For the structure shown, rank the bond lengths (labeled a, b and c) from shortest to longest. Place your answer in the box. Only the answer in the box will be graded. (2 points) H -CH3 THe b Нarrow_forwardDon't used hand raitingarrow_forwardQuizzes - Gen Organic & Biological Che... ☆ myd21.lcc.edu + O G screenshot on mac - Google Search savings hulu youtube google disney+ HBO zlib Homework Hel...s | bartleby cell bio book Yuzu Reader: Chemistry G periodic table - Google Search b Home | bartleby 0:33:26 remaining CHEM 120 Chapter 5_Quiz 3 Page 1: 1 > 2 > 3 > 6 ¦ 5 > 4 > 7 ¦ 1 1 10 8 ¦ 9 a ¦ -- Quiz Information silicon-27 A doctor gives a patient 0.01 mC i of beta radiation. How many beta particles would the patient receive in I minute? (1 Ci = 3.7 x 10 10 d/s) Question 5 (1 point) Saved Listen 2.22 x 107 222 x 108 3.7 x 108 2.22 x 108 none of the above Question 6 (1 point) Listen The recommended dosage of 1-131 for a test is 4.2 μCi per kg of body mass. How many millicuries should be given to a 55 kg patient? (1 mCi = 1000 μСi)? 230 mCiarrow_forward
- Q4: Rank the relative nucleophilicity of halide ions in water solution and DMF solution, respectively. F CI Br | Q5: Determine which of the substrates will and will not react with NaSCH3 in an SN2 reaction to have a reasonable yield of product. NH2 Br Br Br .OH Brarrow_forwardClassify each molecule as optically active or inactive. Determine the configuration at each H соон Chirality center OH 애 He OH H3C Ноос H H COOH A K B.arrow_forwardQ1: Rank the relative nucleophilicity of the following species in ethanol. CH3O¯, CH3OH, CH3COO, CH3COOH, CH3S Q2: Group these solvents into either protic solvents or aprotic solvents. Acetonitrile (CH3CN), H₂O, Acetic acid (CH3COOH), Acetone (CH3COCH3), CH3CH2OH, DMSO (CH3SOCH3), DMF (HCON(CH3)2), CH3OHarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
