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I had this question in my book which asked us to comment whether 2,3-dihydroxybutanedioic acid is optically active or not. I have been taught that molecules with a plane of symmetry or center of symmetry or alternate axis of symmetry are achiral.

So here's the compound:

enter image description here

I can see a horizontal plane of symmetry passing through the middle of the compound, so it should be optically inactive but the answer says: it's optically active.

So can someone explain why the molecule is optically active?

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    $\begingroup$ It isn't.$\mathstrut$ $\endgroup$ Sep 13 at 6:37
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    $\begingroup$ @DivyanshVerma thanks just need to confirm one thing : the question was whether the compound shows optical isomerism. So aren't the two same ? Showing optical isomerism and being chiral are equivalent right ? $\endgroup$
    – Ankit
    Sep 13 at 6:43
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    $\begingroup$ This formula has something unusual. There is a symbol $\ce{(C)}$ beside $\ce{H}$. This has no obvious meaning. When drawing a structure, symbols are never in parentheses. $\endgroup$
    – Maurice
    Sep 13 at 9:52
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    $\begingroup$ "Showing optical isomerism and being chiral are equivalent right ?"; but in the OP "whether this molecule is optically active or not". To clarify: optical isomerism and optical activity are related but different concepts, and 'optical isomerism' is a historical term that is best replaced by 'stereoisomerism' (see en.wikipedia.org/wiki/Chirality_(chemistry)#History ). A chiral molecule may display optical activity, but it does not necessarily do so. On the other hand, an achiral molecule will definitely not display optical activity. The molecule discussed here is achiral. $\endgroup$ Sep 13 at 16:34
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    $\begingroup$ @Ankit To avoid confusion, always make sure the information you present satisfy necessity and sufficiency criteria. "Given compound" is meaningless; use unambiguous chemical name instead. Crop the image properly so that unrelated (and poorly placed) label "(C)" and some watermark that has nothing to do with the question don't confuse readers and don't waste their time. $\endgroup$
    – andselisk
    2 days ago

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