Friday Session-1-sol - Chem 3580 Spring 2010 Friday Session...

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Chem 3580 Spring 2010 Friday Session - Week #1- Solutions 1. Here, consider both planar and non-planar cases. Select the scenario that gives the more stable structure. Order of decreasing stability: aromatic > nonaromatic > antiaromatic N H Total ! e - 's: 10 (4n + 2; n = 2) => Aromatic Assume Planar (N (sp 2 )) Total e - 's: 10 But N has no p-orbital => Nonaromatic Assume Non-Planar (N (sp 3 )) Preferred structure: The more stable planar aromatic N H B CH 3 Total ! e - 's: 14 (4n + 2; n = 3) => Aromatic C is sp 2 hybridized Whole structure will be planar B and C's are sp 2 hybridized. Structure has to be planar 4 ! e - 's => antiaromatic N H Total ! e - 's: 8 (4n; n = 2) => Antiaromatic Assume Planar (N (sp 2 )) Total e - 's: 8 But N has no p-orbital => Nonaromatic Assume Non-Planar (N (sp 3 )) N H Preferred structure: The more stable nonaromatic Experiment suggests that the compound has some antiaromatic character. The distortion from planarity in this case is not enough to completely avoid antiaromaticity. O
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Friday Session-1-sol - Chem 3580 Spring 2010 Friday Session...

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