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Free-radical bromination of the following compound

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

Solution for problem 36SP 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 36SP

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

Propose a mechanism to show why free-radical halogenation occurs almost exclusively at the benzylic position.

Draw the two stereoisomers that result from monobromination at the benzylic position.

Assign R and S configurations to the asymmetric carbon atoms in the products.

What is the relationship between the two isomeric products?

Will these two products be produced in identical amounts? That is, will the product mixture be exactly \(50: 50\)?

Will these two stereoisomers have identical physical properties like boiling point, melting point, solubility, etc.?

Could they be separated (theoretically, at least) by distillation or recrystallization?

Equation Transcription:

Text Transcription:

CH3

HO

Br2

hv

50:50

Step-by-Step Solution:
Step 1 of 6

Solution 36SP:

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring.

Step 2 of 6

Chapter 5, Problem 36SP is Solved
Step 3 of 6

Textbook: Organic Chemistry
Edition: 8
Author: L.G. Wade Jr
ISBN: 9780321768414

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Free-radical bromination of the following compound