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Solutions_Manual_for_Organic_Chemistry_6th_Ed 148

Solutions_Manual_for_Organic_Chemistry_6th_Ed 148 - 7 1 5...

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7- 1 5 continued (b) (c) (d) (c) H ':r--f( j Ph H " Ph " " Br Br meso B Ph Br one enantiomer of the d,l pair Cl H Ph Ph (CH 3 CH2) 3 N : " / C = C NaOH .. acetone / " Br H only alkene isomer E2 Br Ph " / + minor substitution products C = C + minor substitution products Ph / " H only alkene isomer E2 HO H Cb SN2 is the only mechanism possible because no H can get coplanar with Cl; an elimination product would violate Bredfs Rule H See Appendix 1 in this manual for numbering and naming bicyclic systems. Models show that the H on C-3 cannot be anti-coplanar with the Ci on C-2. Thus, this E2 elimination must occur with a syn-coplanar orientation: the D must be removed as the Cl leaves. 7- 1 6 As shown in Solved Problem 7-3, the H and the Br must have trans-diaxial orientation for the E2
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