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I tried writing all the geometrical isomers for 1,2-dibromo-3,4-dichlorocyclobutane, and then saw if they were optical or not. And I am getting the answer as 10:

list of 1,2-dibromo-3,4-dichlorocyclobutane isomers

However, the given answer is 8. I can't really figure out where I went wrong. I am not sure whether the last one is a stereoisomer or not.

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  • $\begingroup$ The last two are both optically inactive ( they have axis of symmetry..).Hope that helps. $\endgroup$
    – user115396
    Aug 31, 2021 at 8:15
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    $\begingroup$ @Abhinav Can you please once say where the axis will be? $\endgroup$
    – Perseus
    Aug 31, 2021 at 9:15
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    $\begingroup$ The two Cl atoms are not symmetric right? One is above the plane and one is below... $\endgroup$
    – Perseus
    Aug 31, 2021 at 9:17
  • $\begingroup$ Huh? It won't be even 8 as at least two of them are identical. $\endgroup$
    – Mithoron
    Aug 31, 2021 at 14:04

1 Answer 1

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I think if the question is asking for number of stereoisomers, OP is correct about 10. If it is about optically active isomers, then answer is 8 as shown in following image:

optically active isomers

Since the given molecule contains four chiral centers, theoretically it should have maximum of $2^4 = 16$ stereoisomers. The best way to find the correct number of isomers is to assign the $R/S$ (Cahn-Ingold-Prelog) configuration for each chiral center. Since the molecule has at least one plane of symmetry based on the stereoisomer as shown for the structure 1, the structure can name two different ways: For example, for the structure 1, it could be $(1S,2R,3R,4S)$ going counter-clockwise (red-numbering) or $(1R,2S,3S,4R)$ by going clockwise. Thus, the corresponding mirror images would have $(1R,2S,3S,4R)$ going counter-clockwise (red-numbering) or $(1S,2R,3R,4S)$ by going clockwise configurations. As shown in the image, these for configurations are identical. Thus, structure 1 is not optically active (meaning its mirror image is superimposable because it has plane of symmetry as indicated) and called meso-isomer. Yet, it is a stereoisomer.

The same is true for the structure 4 as well and hence optically inactive (the second meso-isomer). However, structures 2, 3, 5, and 6 have non-imposable mirror images as indicated in above image (show the assigned $R/S$ configurations). As a consequence, these four and their mirror images are optically active (total of eight structures). Thus, 1,2-dibromo-3,4-dichlorocyclobutane has 10 stereoisomers: 8 optically active isomers and 2 meso-isomers.

OP's doubt about last structure can be solved by assigning corresponding $R/S$-configurations as well: For the given structure it is $(1S,2S,3R,4R)$ (clockwise) or $(1S,2S,3R,4R)$ (counter-clockwise; same as clockwise). The assigned mirror image is $(1R,2R,3S,4S)$ (clockwise) or $(1R,2R,3S,4S)$ (counter-clockwise; same as clockwise). There are no coinciding names. Keep in mind that there is no plane of symmetry because 1,2-dibromo versus 1,2-dichloro here. Thus, structure 6 is optically active.

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    $\begingroup$ @Mather Mahindaratne Thnx for the solution. In the question they have mentioned to find the number of stereoisomers and not optically active stereoisomers, but seeing as you too got 10 for stereoisomers, the answer must be wrong... Again thnx for clarifying my doubt in this much detail! $\endgroup$
    – Perseus
    Sep 1, 2021 at 3:47
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    $\begingroup$ @MathewMahindaratne downvoted because you assign two different names to each structure, although only one is correct $\endgroup$
    – Loong
    Sep 5, 2021 at 9:26
  • $\begingroup$ @Loong: Not necessarily. The two names ($R/S$-configurations) are based on numbering. Rest of the name is the same. $\endgroup$ Sep 5, 2021 at 13:41
  • $\begingroup$ Exactly. One numbering is correct and the other one is wrong. $\endgroup$
    – Loong
    Sep 5, 2021 at 16:59
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    $\begingroup$ @Loong I dont think he numbered/labeled the whole compound as either R or S, he just numbered all the chiral carbons individually. $\endgroup$
    – Perseus
    Sep 8, 2021 at 9:43

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