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Solutions_Manual_for_Organic_Chemistry_6th_Ed 103

Solutions_Manual_for_Organic_Chemistry_6th_Ed 103 - 5-35...

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Unformatted text preview: 5-35 continued (c) This structure is a challenge to visualize. A model helps. One way to approach this problem is to assign R and 5 configurations. Each arrow shows a change at one asymmetric carbon. CH3 H3C CH3 H C H3 H H\S H H �L H H � H 3C H .0. H CH3 H H .0. H CH3 � H3C H H CH3 H H3C H3C H H H Summary R5S5 (K) is the enantiomer of RRR5 (M) 5555 (L) is the enantiomer of RRRR (P) The structures with two R carbons and two 5 carbons (J, N, and 0) have special symmetry. J is a meso structure; it has chirality centers and is superimposable on its mirror image-see Problem 5-20(h). Nand 0 are enantiomers, and are diastereomers o f all of the other structures. Give yourself a gold star if you got this correct! 5-36 s� .Ix. �� S* A meso diastereomers B meso R* C 15,3R equivalent to 1R,35 15,3R equivalent to 1R,35 chiral 1R,3R enantiomers D chiral 15,35 C and Dare enantiomers. All other pairs are diastereomers. *Structures A and B are both meso structures, but they are clearly different from each other. How can they be distinguished? One of the advanced rules of the Cahn-Ingold-Prelog system says: When two groups attached to an asymmetric carbon differ only in their absolute configuration, then the neighboring (R) stereocenter takes priority. Now the configuration of the central asymmetric carbon can be assigned: 5 for structure A, and R for structure B. This rule also applies to problem 5-37. (Thanks to Dr. Kantorowski for this explanation.) 96 ...
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