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 6 - Conformations - 6.2 Ranking The Stability Of Newman Projections - Problems - Page 111: 6.9

Answer

In both structures, the back carbon will have: Up: Methyl (Me) Right: Methyl (Me) Left: Methyl (Me) In the most stable structure, the front carbon will have: Down: Methyl (Me) Right: Methyl (Me) Left: Methyl (Me) In the less stable structure, the front carbon will have: Up: Methyl (Me) Right: Methyl (Me) Left: Methyl (Me)
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Work Step by Step

1. For both most and less stable conformations, draw the atoms connected to the back carbon. (Represented by the circle.) In this case: Up: Methyl (Me) Right: Methyl (Me) Left: Methyl (Me) 2. Identify the largest group connected to the front carbon, and for the back carbons: In this case: Front carbon: Methyl (Me) Back carbon: Methyl (Me) 3. Now, for the most stable one, we will rotate only the groups/atoms on the front carbon (The point on the center.), trying to put the largest groups we identified as distant as possible. In this case, the front carbon will have: Down: Methyl (Me) Right: Methyl (Me) Left: Methyl (Me) 4. Now, for the most stable one, we will rotate only the groups/atoms on the front carbon (The point on the center.), trying to put the largest groups we identified as close as possible. In this case, the front carbon will have: Up: Methyl (Me) Right: Methyl (Me) Left: Methyl (Me)
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