14D_f11_e01_key

14D_f11_e01_key - Chemistry 14D Fall 2011 Exam 1 Solutions...

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Chemistry 14D Fall 2011 Exam 1 Solutions Page 1 Statistics : High score, average and low score will be posted on the course web site after exam grading is complete. The exam is ready to be picked up when these numbers are posted. A note about exam keys : The answers presented here may be significantly longer than expected from a student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as well. To see the projected course grade cutoffs, consult the grading scale on the Chem 14D course web page. 1. (a) I Cl Cl In an aprotic solvent, Cl - is a better nucleophile than Br - . (b) Ph I I Ph I rds OH 2 Ph I O H OH 2 H Ph I OH Ionization of the tertiary iodide is favored because it leads to the more stable carbocation. The oxonium is then deprotonated by the strongest base present . (c) CH 3 Cl H 3 C Br H H OCH 3 CH 3 Cl H 3 C Bromide ion is a better leaving group than chloride. The E2 periplanarity requirement prevents the more stable trisubstituted alkene to be formed, so Hofmann elimination occurs in this case. (d) CH 3 Ph OSO 2 CF 3 CH 3 Ph Ph CH 3 H HOCH 3 Ph CH 3 rds Carbocation rearrangement (2 o to 3 o with resonance) results in a significant increase in stability. Methyl migration yields a carbocation that is 3 o with resonance whereas phenyl group migration yields a carbocation that is 3 o without resonance stabilization. The carbocation is then deprotonated by the strongest base present. The more stable alkene product (tetrasubstituted and conjugated versus trisubstituted and conjugated) is favored . 2.
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This note was uploaded on 03/12/2012 for the course CHEM 14D taught by Professor Hardinger during the Winter '08 term at UCLA.

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14D_f11_e01_key - Chemistry 14D Fall 2011 Exam 1 Solutions...

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