Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
Question
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Chapter 23.2, Problem 10SSC
Interpretation Introduction

Interpretation:

The number of possible isomers for galactose, glucose and fructose should be calculated.

Concept introduction:

The compounds having similar chemical formula but different structures are known as isomers.

The compounds having similar chemical or molecular formula but different connectivity is known as constitutional isomers.

Number of stereoisomers is calculated by below expression:

2n

Where, n = number of chiral carbons

Expert Solution & Answer
Check Mark

Answer to Problem 10SSC

Number of isomers of glucose = 16

Number of isomers of galactose = 16

Number of isomers of fructose = 8

Explanation of Solution

Glucose consists of single unit of sugar which can’t be hydrolyzed into simpler molecule.

Chemistry: Matter and Change, Chapter 23.2, Problem 10SSC , additional homework tip  1

When a carbon atom is linked with different groups is known as chiral carbon and center is known as chiral center. Chiral center is represented by a symbol *.

The structure is drawn as with symbol *:

Chemistry: Matter and Change, Chapter 23.2, Problem 10SSC , additional homework tip  2

Thus, four chiral centers are present in the above structure as first four carbon atoms are linked with four different groups.

Number of isomers is calculated as:

Here, n = 4 (3 chiral carbon atoms are present)

Number of isomers = 24

= 16

Therefore, sixteen isomers are present for the structure of glucose.

Galactose consists of single unit of sugar which can’t be hydrolyzed into simpler molecule.

In galactose, the hydroxyl group linked with forth carbon atom of the chain is present on left side whereas in glucose, the hydroxyl group linked with forth carbon atom of the chain is present on right side.

Chemistry: Matter and Change, Chapter 23.2, Problem 10SSC , additional homework tip  3

When a carbon atom is linked with different groups is known as chiral carbon and center is known as chiral center. Chiral center is represented by a symbol *.

The structure is drawn as with symbol *:

Chemistry: Matter and Change, Chapter 23.2, Problem 10SSC , additional homework tip  4

Thus, four chiral centers are present in the above structure as first four carbon atoms are linked with four different groups.

Number of isomers is calculated as:

Here, n = 4 (3 chiral carbon atoms are present)

Number of isomers = 24

= 16

Therefore, sixteen isomers are present for the structure of galactose.

Fructose consists of single unit of sugar which can’t be hydrolyzed into simpler molecule.

Chemistry: Matter and Change, Chapter 23.2, Problem 10SSC , additional homework tip  5

When a carbon atom is linked with different groups is known as chiral carbon and center is known as chiral center. Chiral center is represented by a symbol *.

The structure is drawn as with symbol *:

Chemistry: Matter and Change, Chapter 23.2, Problem 10SSC , additional homework tip  6

Thus, three chiral centers are present in the above structure as first three carbon atoms are linked with four different groups.

Number of isomers is calculated as:

Here, n = 3 (3 chiral carbon atoms are present)

Number of isomers = 23

= 8

Therefore, eight isomers are present for the structure of fructose.

Chapter 23 Solutions

Chemistry: Matter and Change

Ch. 23.2 - Prob. 11SSCCh. 23.3 - Prob. 12SSCCh. 23.3 - Prob. 13SSCCh. 23.3 - Prob. 14SSCCh. 23.3 - Prob. 15SSCCh. 23.3 - Prob. 16SSCCh. 23.3 - Prob. 17SSCCh. 23.3 - Prob. 18SSCCh. 23.3 - Prob. 19SSCCh. 23.4 - Prob. 20SSCCh. 23.4 - Prob. 21SSCCh. 23.4 - Prob. 22SSCCh. 23.4 - Prob. 23SSCCh. 23.4 - Prob. 24SSCCh. 23.4 - Prob. 25SSCCh. 23.5 - Prob. 26SSCCh. 23.5 - Prob. 27SSCCh. 23.5 - Prob. 28SSCCh. 23.5 - Prob. 29SSCCh. 23.5 - Prob. 30SSCCh. 23.5 - Prob. 31SSCCh. 23.5 - Prob. 32SSCCh. 23 - Prob. 33ACh. 23 - Prob. 34ACh. 23 - Prob. 35ACh. 23 - Prob. 36ACh. 23 - Prob. 37ACh. 23 - Prob. 38ACh. 23 - Prob. 39ACh. 23 - Prob. 40ACh. 23 - Prob. 41ACh. 23 - Prob. 42ACh. 23 - Prob. 43ACh. 23 - Prob. 44ACh. 23 - Prob. 45ACh. 23 - Prob. 46ACh. 23 - Prob. 47ACh. 23 - Prob. 48ACh. 23 - Prob. 49ACh. 23 - Prob. 50ACh. 23 - Prob. 51ACh. 23 - Prob. 52ACh. 23 - Prob. 53ACh. 23 - Prob. 54ACh. 23 - Prob. 55ACh. 23 - Prob. 56ACh. 23 - Prob. 57ACh. 23 - Prob. 58ACh. 23 - Prob. 59ACh. 23 - Prob. 60ACh. 23 - Prob. 61ACh. 23 - Prob. 62ACh. 23 - Prob. 63ACh. 23 - Prob. 64ACh. 23 - Prob. 65ACh. 23 - Prob. 66ACh. 23 - Prob. 67ACh. 23 - Prob. 68ACh. 23 - Prob. 69ACh. 23 - Prob. 70ACh. 23 - Prob. 71ACh. 23 - Prob. 72ACh. 23 - Prob. 73ACh. 23 - Prob. 74ACh. 23 - Prob. 75ACh. 23 - Prob. 76ACh. 23 - Prob. 77ACh. 23 - Prob. 78ACh. 23 - Prob. 79ACh. 23 - Prob. 80ACh. 23 - Prob. 81ACh. 23 - Prob. 82ACh. 23 - Prob. 83ACh. 23 - Prob. 84ACh. 23 - Prob. 85ACh. 23 - Prob. 86ACh. 23 - Prob. 87ACh. 23 - Prob. 88ACh. 23 - Prob. 89ACh. 23 - Prob. 90ACh. 23 - Prob. 91ACh. 23 - Prob. 92ACh. 23 - Prob. 93ACh. 23 - Prob. 94ACh. 23 - Prob. 95ACh. 23 - Prob. 96ACh. 23 - Prob. 97ACh. 23 - Prob. 98ACh. 23 - Prob. 99ACh. 23 - Prob. 100ACh. 23 - Prob. 101ACh. 23 - Prob. 102ACh. 23 - Prob. 104ACh. 23 - Prob. 105ACh. 23 - Prob. 106ACh. 23 - Prob. 1STPCh. 23 - Prob. 2STPCh. 23 - Prob. 3STPCh. 23 - Prob. 4STPCh. 23 - Prob. 5STPCh. 23 - Prob. 6STPCh. 23 - Prob. 7STPCh. 23 - Prob. 8STPCh. 23 - Prob. 9STPCh. 23 - Prob. 10STPCh. 23 - Prob. 11STPCh. 23 - Prob. 12STPCh. 23 - Prob. 13STPCh. 23 - Prob. 14STPCh. 23 - Prob. 15STPCh. 23 - Prob. 16STPCh. 23 - Prob. 17STP
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