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Unformatted text preview: 7 - 1 5 continued (b) H "" Ph " Br ':r--f(
Br (CH3 CH2) 3 N : Ph C=C " / H Br only alkene isomer E2 Br
/ " / Ph + m inor substi tution products (c) B Br Ph one enantiomer of the d, l p air ��� (d) NaOH
.. a cetone Cb
HO H C=C " / H Ph only alkene i somer E2 " Ph
+ minor s ubstitution products S N2 i s the only mechanism possi ble because no H can get coplan ar w i th C l ; an eli mination product would violate B redfs Rule ( c) Cl See Appendix 1 i n thi s manual for n umbering and naming bicyc l ic systems. H H Models show that the H on C-3 cannot be anti-coplanar with the Ci on C-2. Thus, this E2 elimi nation must occ ur with a syn-coplanar orientation: the D must be removed as the Cl leave s . 7- 1 6 As shown i n Solved Problem 7 - 3 , the H and the Br must have trans-diaxial orientation for the E 2 reaction t o occur. I n part (a), t h e c i s i somer h a s the methyl i n equatorial position, t h e NaOCH3 can remove a hydrogen from either C - 2 or C-6, giving a mixture of alkenes where the most highly substituted i somer is the maj or product (Zaitsev). In part (b ), t he trans i somer has the methyl in the axial posi tion at C-2 , so no elimination can occur to C-2. The only possible elimination orientation i s toward C-6. (a) .. e Ither H can be removed �l
I 6 Br N aOCH3
.. C H3 0H H c5 c5
+ Z aitsev orientation major minor ( b) o nly this H can be removed cj(.
. 6 Br
1 N aOCH3
.. CH3 0H H c5 o nly alkene formed; stereochemistry of E2 precludes other i somer from forming
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