Portmess Homework Q&S 67

Portmess Homework Q&S 67 - hexane rings that are...

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3.31 Both structures have approximately the same degree of angle strain and of torsional strain. Structure B has more van der Waals strain than A because two pairs of hydrogens (shown here) approach each other at distances that are rather close. 3.32 Five bond cleavages are required to convert cubane to a noncyclic skeleton; cubane is pentacyclic. 3.33 Conformational representations of the two different forms of glucose are drawn in the usual way. An oxygen atom is present in the six-membered ring, and we are told in the problem that the ring exists in a chair conformation. The two structures are not interconvertible by ring f ipping; therefore they are not different confor- mations of the same molecule. Remember, ring f ipping transforms all axial substituents to equato- rial ones and vice versa. The two structures differ with respect to only one substituent; they are stereoisomers of each other. 3.34 This problem is primarily an exercise in correctly locating equatorial and axial positions in cyclo-
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Unformatted text preview: hexane rings that are joined together into a steroid skeleton. Parts ( a ) through ( e ) are concerned with positions 1, 4, 7, 11, and 12 in that order. The following diagram shows the orientation of axial and equatorial bonds at each of those positions. ( a ) At C-1 the bond that is cis to the methyl groups is equatorial (up). ( b ) At C-4 the bond that is cis to the methyl groups is axial (up). Both methyl groups are up. CH 3 CH 3 e c a b d 4 1 11 12 7 O HOCH 2 OH HO OH HO One axial OH substituent written in its most stable conformation as O HOCH 2 OH HO HO OH written in its most stable conformation as All substituents equatorial O CH 2 OH OH HO HO HO OH OH O CH 2 OH HO HO A: More stable stereoisomer Van der Waals strain destabilizes B H H H H CONFORMATIONS OF ALKANES AND CYCLOALKANES 61...
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