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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:
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.
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:
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.
