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Unformatted text preview: Answers to Hour Examination #3, Chemistry 301X - 2005 1. 1. H 3 O + /H 2 O 2. Na + – N 3 OH (a) (b) 1. H 3 O + /H 2 O (c) 1. CF 3 COOOH 2. KOH/H 2 O OH OH (d) 1. HBr 2. Na + – OCH 2 CH 3 Br (e) 1. Br 2 /H 2 O Br Br OH Br + OH and/or real products can't displace HO – with a nucleophile Markovnikov addition through more substituted carbocation epoxide is opened in S N 2 fashion to give trans no S N 2 at tertiary carbon OH will wind up at the more substituted carbon as the bromonium ion opens Fixes: (a) 1. H 3 O + /H 2 O 2. TsCl 3. Na + – N 3 (b) 1. BH 3 2. HOOH/HO – (c) KMnO 4 (d) 1. H 3 O + /H 2 O 2. NaH 3. CH 3 CH 2 I (e) 1. CF 3 COOOH 2. HBr 2. The data demand that the activation energy for formation of the lower energy 1,4-addition product must be higher than the activation energy for formation of the higher energy product of 1,2-addition. At high temperature (25 °C) the reaction is controlled by thermodynamics - there is enough available energy so that all the barriers can be crossed in both directions. At lower energy (– 78 °C) the reaction is kinetically controlled - controlled by the relative activation barriers for the two possible reactions....
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This note was uploaded on 03/07/2010 for the course ORGO 1 AND V.0243 taught by Professor Jones during the Spring '10 term at NYU.
- Spring '10