Steroisomerism in cyclic compounds [closed]








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What about stereochemistry of 1,2,3,4-tetrahdroxy cyclobutane 
How will we decide the stereocentre it's eating me can anybody help.


Answer is 4 stereoisomers




In the [RULES FOR THE NOMENCLATURE OF ORGANIC CHEMISTRY
SECTION E: STEREOCHEMISTRY](https://www.iupac.org/publications/pac/1976/pdf/4501x0011.pdf), published in *Pure Appl. Chem.*, **1976**, *45*, 11-30, you find:



> 
> **E—2.3.3.** When one substituent and one hydrogen atom
>  are attached at each of more than two positions of a
>  monocycle, the steric relations of the substituents are
>  expressed by adding *r* (for reference substituent), followed
>  by a hyphen, before the locant of the lowestnumbered
>  of these substituents and *c* or *t* (as appropriate),
>  followed by a hyphen, before the locants of the
>  other substituents to express their relation to the reference
>  substituent.
> 
> 
> 


This means: pick one centre as a reference (*r*-1) and express the relative stereochemistry of the other centres, that is the relative orientation of the $\ce{OH}$ groups with respect to the reference, using *c* (if the substituent is on the same face of the ring) or *t* (if the substituent is on the opposite face of the ring).


 2

From what i see...at a basic point of view...there is no carbon with 4 different functional groups...hence i guess its optically inactive...as in each carbon is bonded to 2 identical carbons which are further bonded to the same carbon ...so the functional group dosent matter...Hopefully thats it


 0

To solve this first we number the carbons as
[![enter image description here](https://i.stack.imgur.com/RYnjz.jpg)](https://i.stack.imgur.com/RYnjz.jpg)


Now when we consider cis form of 1 and 2 carbon there will be two stereoisomers cis and trans for 3 and 4 carbon


Now when we consider trans form of 1 and 2 carbon there will be two stereoisomers cis and trans for 3 and 4 carbon


Hence total 4 stereoisomers


[![enter image description here](https://i.stack.imgur.com/j2tcw.jpg)](https://i.stack.imgur.com/j2tcw.jpg)


 0