Glencoe Chemistry: Matter and Change, Student Edition
Glencoe Chemistry: Matter and Change, Student Edition
1st Edition
ISBN: 9780076774609
Author: McGraw-Hill Education
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 21, Problem 89A
Interpretation Introduction

Interpretation:

The meaning of dextrorotatory should be explained.

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.

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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.

On passing a polarized light through a sample cell containing optical isomer, then one isomer will rotate the light in one direction whereas the second isomer will rotate the light in other direction. The rotation of light is clockwise when the isomer is D-isomer whereas for L-isomer the rotation of light is counter-clockwise, the D and L represents dextrorotatory and levorotatory respectively.

No rotation in polarized light is observed when there is an equal or racemic mixture of optical isomers are present that results in cancelling the rotation out.

For example: glucose is referred as dextrose because it rotates polarized light in clockwise direction.

Chapter 21 Solutions

Glencoe Chemistry: Matter and Change, Student Edition

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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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY