Selectivity_SUPP_ANS - Problem S.1 HO H(a Ph(b Ph(c H CH3...

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1 Problem S.1 HO H H (a) (b) Ph H CH 3 CH 3 H (c) CH 3 H (d) (e) Ph H H CH 3 Ph H O H H Br OCH 3 CH 3 H Br H H although both are secondary, this is less hindered path OH H CH 3 H (f) OH Ph same rationale as in part (d) (a) N O O CH(CH 3 ) 2 O Ph (b) N O O CH 3 O Ph Ph (c) N O O CH 2 Ph O (d) N O O CH 2 Ph O Problem S.2 Ph Ph Ph Problem S.3 Cl O A B N O O CH(CH 3 ) 2 O C N O O CH(CH 3 ) 2 O Li D N O O CH(CH 3 ) 2 O HO E O F O O N G N H O I H 3 CH 2 CO OH J
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2 Problem S.4 HO O O (a) (b) H 2 N O HO O H 2 O H 2 SO 4 HO O SOCl 2 Cl O NLi O O CH(CH 3 ) 2 N O O CH(CH 3 ) 2 O LDA N O O CH(CH 3 ) 2 O Li Br N O O CH(CH 3 ) 2 O NaOH H 2 O SOCl 2 Cl O NLi O O CH(CH 3 ) 2 N O O CH(CH 3 ) 2 O LDA N O O CH(CH 3 ) 2 O Li Ph Br N O O CH(CH 3 ) 2 O Ph CH 3 NHOCH 3 N O OCH 3 1) CH 3 CH 2 Li (low temp) 2) H 3 O 1) 2) H 3 O
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3 Problem S.5 (a) B H 3 Li is a “large” hydride reducting agent, and the preference is for equatorial attack, however the fraction of equatorial in this case (79%) is less than when the nearby
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