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Solutions to Problems 93 (g) With ring compounds there are some simplifying "tricks" that you can employ. For example, a ring may be treated as if it were flat-planar-for the purpose of determining whether or not it contains a plane of symmetry. These two structures are identical. The compound pictured has two stereocenters but contains a plane of symmetry, making it meso: Plane of symmetry (h) Harder. Not meso. The mirror plane that works for the cis compound in part (g) doesn't in the trans, because it does not reflect one atom into the other. So the trans compound is chiral. Are these structures both of the same enantiomer or of the mirror images? Rotate each by 90° CI and put them side by side: Enantiomers. Mirror (i) Different connectivity: constitutional isomers. (j) Same thing. Constitutional isomers. (k) Identical. (I) Diastereomers (cis/trans isomer pairs are stereoisomers but not mirror images-the definition of diastereomer). Did you notice that the molecules in (k) and (1) do not have stereocenters? The substituted carbons do not have four different groups on them. (m) Look to see if groups have been switched. Switching a pair of groups around a stereocenter changes the stereochemistry at that stereocenter. If both stereocenters are switched, you have two enantiomers. If only one is switched, you have diastereomers. These are diastereomers: Only the Cl-containing center is switched. H Br Br H (n) Less difficult than it looks: Enantiomers. H H CI Mirror (o) This one requires some work. One approach: Determine R/S for each stereocenter and compare the structures. Easier: Notice that rotation of either one by 180° in the plane of the paper turns it into the other. H Br H becomes H H Br (p) Just as in (n), the structures are mirror images of each other. But now each stereocenter has the same groups, so we could be dealing either with two enantiomers or with two images of a single meso compound. Options: Either determine R/S for each stereocenter, or (better) rotate into an eclipsed conformation to see if an internal plane of symmetry is present.