Organic Chemistry
Organic Chemistry
8th Edition
ISBN: 9781305580350
Author: William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher: Cengage Learning
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Chapter 29, Problem 29.28P
Interpretation Introduction

Interpretation:

A series of contributing structures for the less reactive radical has to be drawn and account for its stability.

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Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction.(a) Using a generic radical R• as a reactant with propylene, draw the mechanism and products for the two competing steps.(b) Which step produces the more stable product?(c) How do your results explain propylene’s poor reactivity in free radical polymerization?
For the radical polymerization of 2-methyl-1-propene shown below, give the initiation step, the first propagation step, and another propagation step in a mechanism for the reaction. Indicate electron flow with arrows. Show the structures of intermediate products.
show the mechanism for the condensation polymerisation reaction shown below. draw the FULL electron pushing mechanism with ALL intermediates, lone pairs, and formal charges clearly labelled. Show ALL electron pushing arrows and label the nucleophile and electrophile where appropriate for non-radical mechanisms.
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