Problem_Set_1_Answer_Key

Problem_Set_1_Answer_Key - N O B N O NOT a valid resonance...

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H N Me Me O O Cl O H N Me Me O O Cl O Cl R O O H N Me Me R O H N Me Me Cl O R O O H Cl N Me Me H N Me Me O O Cl O Cl H N Me Me O -HCl Cl - Cl - H N Cl Me Me H N Cl Me Me Cl Cl O O R Cl O H N Me Me O R O H O 1. a. + CO 2 + CO + 1. c. 1. b. Cl Cl Organic Chemistry c3046y Problem Set 1 - Answer Key R O O O Cl O Cl Cl Cl O O R O H O R O O O Cl O Cl R O O Cl O O Cl R Cl O + CO 2 + CO + HCl
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Organic Chemistry c3046y Problem Set 1 - Answer Key Page 2 N N O O CH 2 CH 3 N O CH 2 CF 3 A B C 2. Rank the three compounds shown below from lowest frequency to highest frequency for the C=O stretch in the IR spectra. C is a "normal" amide, and we can draw the usual resonance structure: N O CH 2 CH 3 C N O CH 2 CH 3 In A , the nitrogen atom is LESS strongly donating, because the CF 3 group is "pulling" on the N lone pair the other way. Thus, for A we would expect the resonance structure on the right to be making LESS of a contribution (relative to C ) and we'd expect to find the C=O stretch of A at a higher frequency than that of C . So far, so good, but what about B ? Again, start by evaluating the primary resonance structure for any amide:
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Unformatted text preview: N O B N O NOT a valid resonance structure! Bridghead double bond! Here, there is NO donation from the nitrogen to the carbonyl AT ALL. B may technically be an amide, but in a very real sense it is NOT an amide, in that there is no resonance interaction between the N lone pair and the carbonyl pi system. Build a model to convince yourself of the fact that the N lone pair cannot achieve the proper orientation for resonance in this case. Thus, B is really best thought of as a ketone that happens to be attached to a nitrogen atom, and we would expect it to have a ketone-like C=O stretch, and therefore that it would have the highest frequency. C B A lowest frequency highest frequency...
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