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Give the stereochemical relationships between each pair of

Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L.G. Wade Jr ISBN: 9780321768414 33

Solution for problem 30SP Chapter 5

Organic Chemistry | 8th Edition

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Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L.G. Wade Jr

Organic Chemistry | 8th Edition

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Problem 30SP

Give the stereochemical relationships between each pair of structures. Examples are same compound, structural isomers, enantiomers, diastereomers. Which pairs could you (theoretically) separate by distillation or recrystallization?

Step-by-Step Solution:
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Solution 30SP

(a)

The first two compounds are both glycols.

These two compounds both have a mirror plane of symmetry running horizontally through the center of the compound. Rotation of either compound by 180o leads to a structure that can be superimposed with its partner. This indicates that these two structures are actually identical compounds. By definition, recrystallization or distillation cannot be used to separate the same compound from itself.

(b)

The second pair of compounds is both triols.

Because the configuration of both asymmetric carbon atoms is reversed, these two compounds are enantiomers. Enantiomers have identical chemical and physical properties (except for their behavior towards plane-polarized light), and so they could not be separated by distillation or recrystallization.

(c)

The third pair of compounds is both glycols.

Because the configuration of both asymmetric carbon atoms is reversed, these two compounds are enantiomers. Enantiomers have identical chemical and physical properties (except for their behavior towards plane-polarized light), and so they could not be separated by distillation or recrystallization.

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(d)

The fourth pair of compounds is halogenated hydrocarbons.

By rotating the first structure so that the bromine substituent is pointing upwards, it’s easy to see that these two compounds are mirror images of each other that are non-superimposable. This makes them enantiomers. Enantiomers have identical chemical and physical properties (except for their behavior towards plane-polarized light), and so they could not be separated by distillation or recrystallization.

(e)

The fifth pair of compounds is Newman projections.

These two compounds are diastereomers. Rotate the front carbon of the first compound to the left by one position, and it matches the front carbon of the other compound. However, the back carbons are mirror images of each other. The entire molecule is not a mirror image of its partner; that would make this pair diastereomers. Because diastereomers have different chemical and physical properties, it should be (theoretically) possible to separate these two molecules using recrystallization and distillation.

(f)

The sixth pair of compounds is methylated cyclohexanes.

Because only one of the asymmetric carbons has an inverted configuration, these two compounds are diastereomers. Because diastereomers have different chemical and physical properties, it should be (theoretically) possible to separate these two molecules using recrystallization and distillation.

(g)

The seventh pair of compounds is also methylated cyclohexanes.

Note that each compound has two asymmetric centers, and the configuration is inverted at both of these asymmetric carbons. These compounds are non-superimposable mirror images, making them enantiomers. Enantiomers have identical chemical and physical properties (except for their behavior towards plane-polarized light), and so they could not be separated by distillation or recrystallization.

(h)

The eighth pair of compounds is a set of fused cyclopentanes.

These two compounds are same so they could be separated by distillation or recrystallization.

(i)

The ninth and last pair of compounds in this exercise set are bicyclic ketones.

These compounds are enantiomers because they have different configurations at each chiral center in both compounds.. By definition, recrystallization or distillation cannot be used to separate the compound.

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Chapter 5, Problem 30SP is Solved
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Textbook: Organic Chemistry
Edition: 8
Author: L.G. Wade Jr
ISBN: 9780321768414

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