1559T_ch14_259-27511/3/059:27Pa9e265 EQA Solutions to Problems265 (e)CH,CH,C-CH-CH2 (racemic)CH,CH,C-CHCH,Br Br (dICH,CHCH=CHCH,CH←一CH,CH=CHCHCH,CH (e)CH,CHCH-CHCHCH,+CH.CH-CHCHCH-CH (CH.C-CS CH3-C-CH CH.CH-CHCHL.TMED THF (Grignard reagent isalso okay.) 40.(a)cis-2-trans-5-Heptadiene.or (2E.5Z)-2.5-heptadiene (b)2.4-Pentadicn-1-ol (c)(5S.6S)-5.6-Dibromo-1.3-cyclooctadicne (d)4-Ethenylcyclohexene H 41.CH,=CH-CH-CH=CH2 1,4-Pentadiene has weak C- o长品一 Ci2二ci屵cH-cH芒Cii2=ICH2-CH=CH-CH=CH2←一 CH=CH-CH-CH=CH←一CH=CH-CH=CH-CH] Solutions to Problems • 265 CH3 CH3 A A (c) CH3CH2COCHPCH2 (racemic) CH3CH2CPCHCH2Br A Br (d) CH3CHOH CH3CHOH (All possible A A stereoisomers for (e) CH3CHCHPCHCH2CH3  CH3CHPCHCHCH2CH3 each structure) (f) 38. 39. 40. (a) cis-2-trans-5-Heptadiene, or (2E, 5Z)-2,5-heptadiene (b) 2,4-Pentadien-1-ol (c) (5S, 6S)-5,6-Dibromo-1,3-cyclooctadiene (d) 4-Ethenylcyclohexene H A m 41. CH2PCHOCHOCHPCH2 1,4-Pentadiene has the weakest COH bond (arrow), a bond that is doubly allylic (DH 77 kcal mol1 ); this isomer will therefore be brominated fastest. Because only a very weak COH bond needs to be broken, its first propagation step has a much smaller Ea relative to 1,3-pentadiene, where a stronger methyl COH bond needs to be broken. However, both will give identical product mixtures because identical radicals are formed from each: CH2 CH CH CH CH2 [CH2 CH CH CH CH2 CH2 CH CH CH CH2 CH2 CH CH CH CH2] or CH3CH2CH2CH2Li, TMEDA NBS, h , CCl4 Li, THF (Grignard reagent is also okay.) THF Li Br CH3 C CH3 OH 2. H, H2O 1. CH3CCH3, THF O CH3 MgCl CH3 D and CH3 MgCl D OD DO D OD DO CH3 D (CH3)2C SCH3 C CH3 H CH [CH3CHCH CHCH2CH3 CH3CH CHCHCH2CH3] Li 1559T_ch14_259-275 11/3/05 9:27 Page 265