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Chapter 13.9, Problem 17P
Interpretation Introduction

Interpretation:

The steps how the two products formed from reaction of methylenecyclohehane with NBS has to be given.  Stereoisomers has to be disregarded.

Concept introduction:

Bromination of Allylic Carbons:

Organic Chemistry; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card;  Study Guide and Student Solutions Manual for Organic Chemistry, Books a la Carte Edition (7th Edition), Chapter 13.9, Problem 17P , additional homework tip  1

N-bromosuccinimide (NBS) is used for the allylic bromination through radical reaction.  Bromination of allylicc carbon requires low concentration of bromine and low concentration of hydrobromic acid. If high concentration of bromine and high concentration of hydrobromic acid which leads to the formation of bromonation in the double bond.

Organic Chemistry; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card;  Study Guide and Student Solutions Manual for Organic Chemistry, Books a la Carte Edition (7th Edition), Chapter 13.9, Problem 17P , additional homework tip  2

Bromination reaction starts with the homolytic cleavage of N-Br bond in N-bromosuccinimide (NBS) which creates bromine radical to initiate the radical bromination reaction.

NBS bromine radical removes the allylic hydrogen which forms hydrogen bromide and allylic radical in the first propagation step, the allylic radical is stabilized by the double bond in ring. This allylic radical reaction with bromine molecule and forms allylic bromide in the second propagation step which are shown above.

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Chapter 13 Solutions

Organic Chemistry; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card; Study Guide and Student Solutions Manual for Organic Chemistry, Books a la Carte Edition (7th Edition)

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