Solutions_Manual_for_Organic_Chemistry_6th_Ed 178

Solutions_Manual_for - 8-35(a � HO OH HO OH meso(b)� � � HO OH HO OH racemic(d,l(c � � C H3C03H H2 O Os04 H20 2 rotate meso(d rotate

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Unformatted text preview: 8-35 (a) � HO OH HO OH meso (b)� � � HO OH HO OH racemic (d,l) (c) � � C H3C03H .. H2 O Os04 H20 2 rotate .. meso (d) .. rotate .. � racemic (d,l) Have you noticed yet? For symmetric alkenes and symmetric reagents (addition of two identical X groups): cis-alkene + syn addition ----7 ----7 meso racemic racemic meso ----7 ----7 cis-alkene + anti addition trans-alkene + anti addition trans-alkene + syn addition Assume that cis/synlmeso are "same", and trans/anti! racemic are "opposite". Then any combination can be predicted, just like math! +1 -1 -1 x x +1 +1 - 1 = = +1 x x -1 +1 = = -1 -1 +1 8-36 Solve these ozonolysis problems by working backwards, that is, by "reattaching" the two carbons of the new carbonyl groups into alkenes. Here's a hint. When you cut a circular piece of string, you still have only one piece. When you cut a linear piece of string, you have two pieces. Same with molecules. If ozonolysis forms only one product with two carbonyls, the alkene had to have been in a ring. If ozonolysis gives two molecules, the alkene had to have been in a chain. (a) two carbonyls from ozonolysis are in a chain, so alkene had to have been in a nng (b) two carbonyls from ozonolysis are in two different products, so alkene had to have been in a chain, not a ring (c) two carbonyls from ozonolysis are in two different products, so alkene had to have been in a chain, not a ring E or Z of alkene cannot be determined from products 8 - 37 (a) � 03 � Mc,� �o + o� H OH O (b) � 17 1 ...
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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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