If we add substituents R to this structure, we can have two diastereoisomers with the two R groups on the same side (syn) of the flat ring or on opposite (anti) sides. Although the plane of the paper is no longer a plane of symmetry, neither isomer is chiral as the other plane bisects the substituents and is still a plane of symmetry.
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4$\begingroup$ The plane of symmetry contains both R's and both H's, so they don't move when reflected. $\endgroup$– orthocresolCommented Jun 4, 2023 at 21:13
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2$\begingroup$ Exactly like it is drawn in the picture... $\endgroup$– MithoronCommented Jun 4, 2023 at 23:22
2 Answers
The two $\ce{CHR}$ groups at opposite corners of the ring are co-planar, and both isomers are symmetric about this plane.
Yes, seeing a molecule in three dimensions when you have a paper drawing is difficult. In my day at school the recommendation was to use a kit to build a molecular model, which you could rotate to get different views. Today we can use computer images.
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$\begingroup$ So, what is wrong this time? (Other than getting burned by Autocorrect, which forced an early-morning edit?) $\endgroup$ Commented Jun 5, 2023 at 9:35
Both molecules have a plane of symmetry. The $\ce{R}$ groups lie on the plane of symmetry. Try visualizing it this way:
Unless the $\ce{R}$ groups are chiral, both the molecules can exist in an achiral conformation.
Also, likely, the ring is not planer because $\pu{120 ^\circ}$ is not the natural angle for the $\ce{N}$ atoms in (somewhat) $\mathrm{sp^3}$ hybridization.
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3$\begingroup$ Amide nitrogens are very much planar. $\endgroup$ Commented Jun 5, 2023 at 12:19
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$\begingroup$ @orthocresol 'very much' being the keywords, the six-membered ring is strained in the planer configuration. Based on geometrical optimization, non-planarity adds stability. $\endgroup$– anantaCommented Jun 5, 2023 at 13:24