ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
2nd Edition
ISBN: 9780393664034
Author: KARTY
Publisher: NORTON
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Chapter 12, Problem 12.56P
Interpretation Introduction

(a)

Interpretation:

The mechanism that accounts for the formation of the mixture of the products in a given reaction is to be drawn.

Concept introduction:

Electrophilic addition of the molecular chlorine into but-2-ene gives anti-addition of the molecular chlorine via the formation of the chloronium ion as an intermediate. Under similar conditions, molecular chlorine undergoes both syn and anti-addition to (E)-1-phenylprop-1-ene that produces the isomer plus enantiomers. In order to show both syn and anti-addition, the carbocation like intermediate must be generated. Hence, this electrophilic addition proceeds via formation of stable benzylic carbocation intermediate instead of the formation of chloronium ion. As this reaction leads to formation of two different benzylic carbocation intermediates, the reaction yields four different products which consists of two pairs of enantiomers.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the electrophilic addition of the molecular chlorine to the 1-phenylprop-1-ene is different from the one for but-2-ene is to be explained.

Concept introduction:

Electrophilic addition of the molecular chlorine into but-2-ene gives anti-addition of the molecular chlorine via the formation of the chloronium ion as an intermediate. Under similar conditions, molecular chlorine undergoes both syn and anti-addition to (E)-1-phenylprop-1-ene that produces the isomer plus enantiomers. In order to show both syn and anti-addition, the carbocation like intermediate must be generated. Hence, this electrophilic addition proceeds via formation of stable benzylic carbocation intermediate instead of the formation of chloronium ion. As this reaction leads to formation of two different benzylic carbocation intermediates, the reaction yields four different products which consists of two pairs of enantiomers.

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Chapter 12 Solutions

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