Solutions_Manual_for_Organic_Chemistry_6th_Ed 130

Solutions_Manual_for_Organic_Chemistry_6th_Ed 130 - 6-57...

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Unformatted text preview: 6-57 substitution Br H + + 25,35 B, H + B' "--- -----y---� 2R,3R CH 3 H + CH 3 + H Br ) racemic mixture KOH HO Br + + 2R,35 CH 3 H H + H H + 25,3R + CH 3 Br OH Each of the substitution products has one chiral center inverted from the starting material. The mechanism that accounts for inversion is S N2 . If an S N 1 process were occurring, the product mixture would also contain 2R,3R and 25,35 diastereomers. Their absence argues agai nst an SN 1 process occurring here. Regardless of which bromine is substituted on each molecule, the same mixture of products results. H Br -:. elimination '-Y / '" Br ::: H racemic mixture B rotate.. The other enantiomer gives the same product (you should prove this to yourself). H H CH 3 H and Br anti -coplanar � :M S=H3 B, H'-') � HO ' trans 6-58 (a) The absence of cis product is evidence that only the E2 elimination is occurring in one step through an anti-coplanar transition state with no chance of rotation. If El had been occuring, rotation around the carbocation intermediate would have been possible, leading to both the c is and trans products. + 15.58° +15.90° (b) The 1% of radioactive iodide has produced exactly 1 % of the R enantiomer. Each substitution must occur with inversion, a classic S N2 mechanism. 6-59 Thus, 98% of the 5 enantiomer and 2% racemic mixture gives an overall composition of 99% 5 and 1% R. x 100% = .. . 98% of ongmal optIcal activity = 98% e.e. (a) An S N 2 mechanism with inversion will convert R to its enantiomer, 5. An accumulation of excess 5 does not occur because it can also react with bromide, regenerating R. The system approaches a racemic mixture at equilibrium. f\ :: Br -C-H CH2CH3 R ��e B + inversion S N2 .. 123 H -C-Br CH2CH 3 5 �H 3 + Br- ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.

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