Solutions_Manual_for_Organic_Chemistry_6th_Ed 137

Solutions_Manual_for_Organic_Chemistry_6th_Ed 137 - 6 -72...

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Unformatted text preview: 6 -72 (a) " Ph " m ajor Br __ (b) H and Br must be anti -coplanar in the transition state ,H H Ph I CH3CHCHCH3 I = p henyl NaOCH3 E2 ... CH3-C = CHCH3 + I Ph Ph I CH3CHCH = CHz rru nor 2 R,3R Ph CH3 \ \\ ( c) CH3 Ph th � Br H Ph H f K \\ \\ CH3 Na OCH3 � � Br H Br II C" C CH3""- - . 'CH3 "1 \\ \\ Ph I H methyls C IS CH3 CH3 CH3 ---{ � Ph H ';( CH3 CH3 H CH3 ---- Ph V -+ r.. 'r H Ph " \\ CH3 II := C C" CH3 CH3....... 'H '1 ,\ 2 S,3 R m ethyls trans ( d) The 2S, 3S is the mirror i mage of 2 R, 3R; it would give the mirror image of the alkene that 2 R, 3R p roduced (with two methyl groups c is). T he alkene product i s planar, not chiral , so i ts mirror image i s t he same : the 2S, 3S a nd the 2R, 3 R g i ve the same alkene. 1 30 ...
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