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 21.4, Problem 30SSC
Interpretation Introduction

Interpretation:

Two different optical isomers needs to be drawn by attaching the following atoms or groups to the carbon atom:

-H,  -CH3,  -CH2CH3,  -CH2CH2CH3

Concept introduction:

Two mirror image representations are possible for a molecule having chiral center in its structure. If the mirror images are non-superimposable to each other then, they are said to be optical isomers.

Expert Solution & Answer
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Explanation of Solution

A molecule is said to be chiral molecule when all the attached groups to central carbon atom are different and lacks in plane of symmetry in the molecule. The molecules having chiral centres show optical isomerism as there are two mirror image representations possible for a molecule having chiral center in its structure and if the mirror images are non-superimposable to each other then, they are said to be optical isomers.

Now, starting from a single carbon atom, two different optical isomers can be drawn by attaching -H,  -CH3,  -CH2CH3,  -CH2CH2CH3 to the carbon atom as:

The single compound formed will be:

Chemistry: Matter and Change, Chapter 21.4, Problem 30SSC , additional homework tip  1

Since the central carbon atom is attached to four different atoms and groups so, it is chiral in nature shown by * as:

Chemistry: Matter and Change, Chapter 21.4, Problem 30SSC , additional homework tip  2

Now, drawing the optical isomers using mirror as:

Chemistry: Matter and Change, Chapter 21.4, Problem 30SSC , additional homework tip  3

Since, the mirror image is not superimposable to each other so, they are optical isomers of each other.

Chapter 21 Solutions

Chemistry: Matter and Change

Ch. 21.2 - Prob. 11PPCh. 21.2 - Prob. 12SSCCh. 21.2 - Prob. 13SSCCh. 21.2 - Prob. 14SSCCh. 21.2 - Prob. 15SSCCh. 21.2 - Prob. 16SSCCh. 21.3 - Prob. 17PPCh. 21.3 - Prob. 18PPCh. 21.3 - Prob. 19SSCCh. 21.3 - Prob. 20SSCCh. 21.3 - Prob. 21SSCCh. 21.3 - Prob. 22SSCCh. 21.3 - Prob. 23SSCCh. 21.3 - Prob. 24SSCCh. 21.4 - Prob. 25SSCCh. 21.4 - Prob. 26SSCCh. 21.4 - Prob. 27SSCCh. 21.4 - Prob. 28SSCCh. 21.4 - Prob. 29SSCCh. 21.4 - Prob. 30SSCCh. 21.5 - Prob. 31PPCh. 21.5 - Prob. 32PPCh. 21.5 - Prob. 33SSCCh. 21.5 - Prob. 34SSCCh. 21.5 - Prob. 35SSCCh. 21.5 - Prob. 36SSCCh. 21.5 - Prob. 37SSCCh. 21 - Prob. 38ACh. 21 - Prob. 39ACh. 21 - Prob. 40ACh. 21 - Prob. 41ACh. 21 - Prob. 42ACh. 21 - Prob. 43ACh. 21 - Prob. 44ACh. 21 - Prob. 45ACh. 21 - Prob. 46ACh. 21 - Prob. 47ACh. 21 - Prob. 48ACh. 21 - Prob. 49ACh. 21 - Prob. 50ACh. 21 - Prob. 51ACh. 21 - Prob. 52ACh. 21 - How does the structure of a cycloalkane differ...Ch. 21 - Prob. 54ACh. 21 - Prob. 55ACh. 21 - Prob. 56ACh. 21 - Prob. 57ACh. 21 - Prob. 58ACh. 21 - Prob. 59ACh. 21 - Prob. 60ACh. 21 - Prob. 61ACh. 21 - Prob. 62ACh. 21 - Prob. 63ACh. 21 - Prob. 64ACh. 21 - Prob. 65ACh. 21 - Prob. 66ACh. 21 - Prob. 67ACh. 21 - Prob. 68ACh. 21 - Prob. 69ACh. 21 - Prob. 70ACh. 21 - Prob. 71ACh. 21 - Prob. 72ACh. 21 - Prob. 73ACh. 21 - Prob. 74ACh. 21 - Prob. 75ACh. 21 - Prob. 76ACh. 21 - Prob. 77ACh. 21 - Prob. 78ACh. 21 - Prob. 79ACh. 21 - Prob. 80ACh. 21 - Prob. 81ACh. 21 - Prob. 82ACh. 21 - Prob. 83ACh. 21 - Prob. 84ACh. 21 - Prob. 85ACh. 21 - Prob. 86ACh. 21 - Prob. 87ACh. 21 - Prob. 88ACh. 21 - Prob. 89ACh. 21 - Prob. 90ACh. 21 - Prob. 91ACh. 21 - Prob. 92ACh. 21 - Prob. 93ACh. 21 - Prob. 94ACh. 21 - Prob. 95ACh. 21 - Prob. 96ACh. 21 - Prob. 97ACh. 21 - Prob. 98ACh. 21 - Prob. 99ACh. 21 - Prob. 100ACh. 21 - Prob. 101ACh. 21 - Prob. 1STPCh. 21 - Prob. 2STPCh. 21 - Prob. 3STPCh. 21 - Prob. 4STPCh. 21 - Prob. 5STPCh. 21 - Prob. 6STPCh. 21 - Prob. 7STPCh. 21 - Prob. 8STPCh. 21 - Prob. 9STPCh. 21 - Prob. 10STPCh. 21 - Prob. 11STPCh. 21 - Prob. 12STPCh. 21 - Prob. 13STPCh. 21 - Prob. 14STPCh. 21 - Prob. 15STPCh. 21 - Prob. 16STPCh. 21 - Prob. 17STP
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