What is the relationship between the following pair of molecules? Is the underlined answer correct? Explain.
I'm getting mixed answers. Some say these are identical molecules. Other say these are enantiomers. Which is correct and why?
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2 Answers
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What is the relationship between given pair of molecules?
They are enantiomers as given underlined answer.
How did you know that?
You know that by assigning $R/S$ configuration (following Cahn-Ingold-Prelog rules) on each chiral carbon. That is the easiest way to identify the enantiomers. If you have one chiral center in the molecule and its CIP-rotation is $(S)$, then its enantiomer (mirror image) has CIP-rotation of $(R)$ and vise versa. If you have two chiral centers in your molecule of interest and its CIP-rotation is $(S,S)$, then its enantiomer (mirror image) has CIP-rotation of $(R,R)$. If your comparing molecule has CIP-rotation of $(R,S)$ or $(S,R)$ instead of $(R,R)$, then they are not enantiomers. They are called diastereomers. In your given two molecules (1-bromo-3-methylcyclohexane) where $\ce{C}$1 and $\ce{C}$3 are chiral carbons (two chiral carbon system). First assign $R/S$ configuration on each molecule:
For convenience to visualize, I avoid the chair conformation but put the relevant chair conformation below each molecule (e.g., the planar molecule to the left is the right structure in the question). The priorities of groups are given in red for the $\ce{C}$1 center and in powder blue for that of $\ce{C}$3. The molecule to the left has $(1S,3R)$-configuration while the molecule to the right has that of $(1R,3S)$. Since there is no plane of symmetry in each molecule, they are enantiomers (mirror images of each other).
answered Apr 14, 2021 at 16:13
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1$\begingroup$ Is it possible to draw the chair conformation from the planar drawing directly without seeing the R,S configuration (or vice versa)? Because for me, it takes time to first find out their configurations individually, then draw the chair form, and then again verifying their configurations... basically what should I take care of, to not draw the planar drawing of the left molecule as the chair form of the right molecule? Or in other words, how do I show that the structures you have drawn on left/right are indeed equivalent? $\endgroup$– V.GCommented Apr 14, 2021 at 16:48
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$\begingroup$ I don't have that capability at home. Visualize from top view of the molecule to the left (chair conformation) and rotate your molecule $180^\circ$ clockwise. Do the same for the molecule to the right and rotate it $180^\circ$ anti-clockwise. You get two chair confirmation of mirror images. $\endgroup$ Commented Apr 14, 2021 at 16:57
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Enatiomers are "optical isomers which are mirror images of each other" For any compound to show optical isomerism disymmetry is the thumb rule. For checking disymmetry compound should have dissymetric center(s). If we consider the given structures they have 2 chiral centres each and now we need to draw the fisher diagram and find R/S configuration of each chiral carbon and then compare like for example if Compound 1 has the configuration (1R,3S) and compound 2 has (1S,3R) then they are enantiomers.
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2$\begingroup$ What is a "fisher" diagram? You don't even need a Fischer structure! Just do this: chemistry.stackexchange.com/questions/99721/… $\endgroup$ Commented Apr 13, 2021 at 19:30
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$\begingroup$ I'd appreciate whoever downvote this answer give the reason for downvoting. $\endgroup$ Commented Apr 14, 2021 at 17:13
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2$\begingroup$ @Mathew I didn't downvote, but I can give you a couple of reasons why some would. 1) The definition of enantiomer (in quotes without source) is not quite correct. 2) Some are quite allergic to rules of the thumb. 3) Applying a Fischer projection is completely unnecessary to find the configuration of the stereocenters. 4) Spelling and grammar. | meta: It's a 'fastest gun in the west' situation. | I still think it could have been helpful with some cleanup. With your answer that hardly seems worth it. $\endgroup$– Martin - マーチン ♦Commented Apr 14, 2021 at 21:31
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$\begingroup$ @Martin: Actually this is not my post. I thought this shouldn't get downvoted but should have given good reason to correct as you did. I wasn't agree with this answer and post my own yesterday. I appreciate your reasoning. $\endgroup$ Commented Apr 15, 2021 at 6:07
C[C@H]1CCC[C@@H](Br)C1on the left, bromine on R, methyl on S; vsBr[C@H]1CCC[C@@H](C)C1right, bromine on S, methyl on R configurated carbon centre. $\endgroup$