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Unformatted text preview: Answers to Exam 3, Chemistry 0243 – 2008 1. a. H H H H H H H H H H H (CH 3 ) 3 C b. CH 3 CH 3 Cl Cl CH 3 CH 3 Cl + Cl CH 3 CH 3 Cl Cl 2. (a) CH 2 N 2 (diazomethane) and light (b) CF 3 COOOH (c) KMnO 4 or (b) and either HO – /H 2 O or H 3 O + /H 2 O (d) Br 2 /H 2 O (e) 1. BH 3 2. HOOH/HO – (f) (e) and HBr or HBr/ROOR (g) (e) and NaH followed by (f) (h) H 2 /Pd (i) 1. HBr (makes the tertiary bromide) 2. base (E2) tert-butoxide can be made from tert-butyl alcohol and NaH. (j) (i) and Br 2 followed by base (double E2) (k) 1. O 3 2. HOOH 3. Protonation of 1 can take place in two ways to give carbocations that can lead to the 1,2- and 1,3-dibromo compounds. Br HBr Br Br Br Br 1 2 3 Br Br Br Br cis-1,3-Dibromocyclopentane is a meso compound, so it must be optically inactive; there is no mechanistic information here. trans-1,3 Dibromocyclopentane is formed by capture of the open cation, it must be optically active, as it is. So, this observation is consistent with the mechanism outlined above in which an open cation is the critical intermediate. outlined above in which an open cation is the critical intermediate....
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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