Organic Chemistry As a Second Language, 3e: First Semester Topics

Published by John Wiley & Sons
ISBN 10: 111801040X
ISBN 13: 978-1-11801-040-2

Chapter 8 - Mechanisms - 8.3 Drawing Intermediates - Problems - Page 176: 8.18

Answer

The intermediate is on the left of the image below:
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Work Step by Step

1. Analyze the arrows. First arrow (Top): - Tail: It is breaking a bond... - Head: to give the electrons pair to the top "Br". Second arrow (Middle): - Tail: It is taking 2 electrons from the bottom "Br"... - Head: to form a bond between this "Br" and a carbon. Third arrow (Bottom): - Tail: It is breaking a pi bond... - Head: to form other bond between the bottom "Br" and another carbon. 2. Now, let's find the formal charges: The top "Br" had neutral charge, then it lost a bond (+1), and gained 2 electrons (-2) = -1 formal charge. The bottom "Br" had neutral charge, then it lost 2 electrons (+2); it lost a bond (Br-Br) (+1), and formed 2 bonds (-2) = (+1) formal charge.
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