Organic Chemistry
Organic Chemistry
11th Edition
ISBN: 9781118133576
Author: T. W. Graham Solomons, Craig Fryhle
Publisher: Wiley, John & Sons, Incorporated
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Chapter 5, Problem 32PP
Interpretation Introduction

Interpretation:

The Fischer projections for the given compounds are to be drawn and (R) or (S) configuration at each chirality center is to specified.

Concept Introduction:

The molecules which are non-superimposable or not identical with its mirror image are known as chiral molecules.

The pair of two mirror images which are non-identical are known as enantiomers and these are optically active.

The enantiomers, in which the path traced from the highest atomic number to the lowest atomic number is in an anticlockwise direction, are designated as S.

The enantiomers, in which the path traced from the highest atomic number to the lowest atomic number is in the clockwise direction, are designated as R.

The objects or molecules which are superimposable with its mirror images are achiral objects or molecules and these objects have a centre of symmetry or plane of symmetry.

The achiral compounds in which the plane of symmetry is present internally and consists of chiral centres are known as meso compounds, but they are optically inactive.

The stereoformula which is depicted in two dimensions, in which stereochemical information is not destroyed, is determined by the Fisher Projection formula.

The stereoisomers which are non-superimposable on each other and not mirror images of each other are known as diastereomers.

Chiral molecules are capable of rotating plane polarized light

The molecules which are superimposable or identical with their mirror images are known as achiral molecules, and achiral molecules are not capable of rotating the plane-polarised light.

Plane of symmetry is the plane that bisects the molecule in two equal halves, such that they are mirror images of each other.

Compounds having plane of symmetry are usually achiral as they do not have different atoms around the central carbon atom.

Glyceraldehyde consists of one chiral center and due to which it exists in a pair of enantiomers.

Tartaric acid has two enantiomers and two diastereomers. It has two chiral centres.

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Chapter 5 Solutions

Organic Chemistry

Ch. 5 - Practice Problem 5.11 List the substituents in...Ch. 5 - Prob. 12PPCh. 5 - Practice Problem 5.13 Tell whether the two...Ch. 5 - Prob. 14PPCh. 5 - Prob. 15PPCh. 5 - Prob. 16PPCh. 5 - Prob. 17PPCh. 5 - Prob. 18PPCh. 5 - Prob. 19PPCh. 5 - Prob. 20PPCh. 5 - Practice Problem 5.21 The following are formulas...Ch. 5 - Practice Problem 5.22 Write three-dimensional...Ch. 5 - Prob. 23PPCh. 5 - Practice Problem 5.24 Give names chat include (R)...Ch. 5 - Prob. 25PPCh. 5 - Prob. 26PPCh. 5 - Prob. 27PPCh. 5 - Prob. 28PPCh. 5 - Practice Problem 5.29 Write formulas for all of...Ch. 5 - Prob. 30PPCh. 5 - Prob. 31PPCh. 5 - Prob. 32PPCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - 5.35 Designate the (R) or (S) configuration at...Ch. 5 - Prob. 36PCh. 5 - (a) Write the structure of...Ch. 5 - Shown below are Newman projection formulas for...Ch. 5 - 5.39 Write appropriate structural formulas...Ch. 5 - Discuss whether each of the following compounds...Ch. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Compound F has the molecular formula C5H8 and is...Ch. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - For the following molecule, draw its enantiomer as...Ch. 5 - 5.49 (Use models to solve this...Ch. 5 - 5.50 (Use models to solve this...Ch. 5 - (Use models co solve this problem.) Write...Ch. 5 - 5.52 Tartaric acid was an important compound in...Ch. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 1LGPCh. 5 - Prob. 2LGPCh. 5 - Prob. 3LGPCh. 5 - Prob. 2QCh. 5 - Prob. 3QCh. 5 - Prob. 4QCh. 5 - Select the words that best describe what happens...Ch. 5 - Prob. 6QCh. 5 - Prob. 7QCh. 5 - Prob. 8QCh. 5 - 5.9 Which is untrue about the following...Ch. 5 - Prob. 10Q
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