Solutions_Manual_for_Organic_Chemistry_6th_Ed 102

Solutions_Manual_for_Organic_Chemistry_6th_Ed 102 - 5Å35...

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5-34 (b) Rotation Of the enantIomer will be equal in magnitude, Opposite in sign: - 15.90°± (c± The rotatIon -7±95° ²s what percent oF - 15²90³? ´µ9 5 ° ³´' µ x 1¶0% = 50% e.e. · 15¸90° ¶here Is 50% excess of (·¸¹2ºiOdobutane over the racem²c m»xture; that is¼ anOther 25% Must be ½ aND 25% ¾us¿ be S. Àhe total cOmposItIon Is 75% (½¸º(Á¸Â2ÃIodoButane anD 25% (S¸Ä(+¸Å2ÅIOdOButane.
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Unformatted text preview: 5Å35 All structures ²n this problem are ch²ral. C al H + Oh H + OH cH20H A chO HO ± H HO ² h ch20H Æ enan¿²oÇersÈ A and ÆÉ C anD D cHO h ³ Oh hO ´ H cH2±H C D²astereOMers: A and C; A anD D; ÆaÊd C; Æand D (b¸ E F G enantIOmers: E anD F± G anD H DIastereoMersË E and G± E anD H± F and G± F and H 95 cHo HO µ h h ¶ oh ch2±H D H...
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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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