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Given Fischer's projections of a few stereoisomers of some compound, how can we find out if they are enantiomers, diastereomers, etc.? From what I understand, we can rotate the projections to have the same groups on the top and bottom and then compare (non-superimposable mirror images --> enantiomers, non-mirror image --> diastereomers) and we can convert to eclipsed sawhorse projections and vice-versa, could anyone explain further?

EDIT: I have a doubt in the following MCQ question from VMC's JEE archive -->

Q. Which of the following forms a pair of enantiomers? enter image description here

(A) I and II

(B) III and IV

(C) I and IV

(D) I and III

The answer is given as (D), that is, I and III. I also got this as an answer, but I also got III and IV, which is (B) as enantiomers, as I and IV were identical in my solution

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    $\begingroup$ Assign Cahn-Ingold-Prelog designations to each stereocenter and compare the related centers. $\endgroup$
    – Zhe
    Mar 31 at 13:50
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    $\begingroup$ You don't need the eclipsed sawhorse. As for the rest, you explained it yourself pretty well; what's to explain further? $\endgroup$ Mar 31 at 15:09
  • $\begingroup$ @IvanNeretin I just edited the question...my doubt is actually in an MCQ question, which I have now included $\endgroup$ Apr 4 at 8:54
  • $\begingroup$ On my end, the question shows no signs of having been edited. $\endgroup$ Apr 4 at 9:01
  • $\begingroup$ Maybe its some connectivity issues or something on my side, please check again in some time maybe $\endgroup$ Apr 4 at 9:08

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