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Total number of stereoisomers of the compound will be:

truxillic acid

This was a question asked in our mock test. I've tried by considering pseudo-chirality on the carbon atoms. But I don't know where to start.

I need help finding the different stereoisomers. It would be appreciated if you could draw the different isomers.

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    $\begingroup$ Wikipedia already has the answer. However, I strongly suggest you to try one more time before looking up the answer. $\endgroup$
    – andselisk
    Commented May 19, 2020 at 12:43

2 Answers 2

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At first glance, not only does it seem like there are no stereogenic centres in truxillic acid, it is also highly symmetrical. However, if we consider different isomers of the chemically identical units (H, COOH or Ph), which give rise to different isomers, some carbons at the 4-membered ring can possibly become stereogenic.

Stereogenic centres are labelled with an asterisk: ε-truxillic acid and peri-truxillic acid do not have stereogenic carbon centres as their 4-membered ring carbons have chemically identical neighbours.

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I offer a correction to the answer by @Anand Maneesh regarding the truxillic acids.

From the IUPAC Blue Book: P-92.1.1
"A stereogenic unit (i.e. a unit generating stereoisomerism) is a grouping within a molecular entity that may be considered to generate stereoisomerism."

Thus, inversion of the stereochemistry at any cyclobutane carbon will lead to one of the other stereoisomers. All cyclobutane carbons are stereogenic, not just some of them. The carbons having R/S descriptors are both stereogenic and chirotopic. Those carbons bearing r/s descriptors are stereogenic and achirotopic and describe planes of symmetry. All five truxillic acids are achiral. α-Truxillic acid (1) has a center of symmetry while the other four have planes of symmetry. They are not meso compounds.

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