I know there are many similar questions to this but none could clear my confusion over how exactly the mirrored molecules is rotated i.e on which axis it is rotated to superpose with the original molecule (a) 2-Propanol (V) and its mirror image (VI). (b) When either one is rotated, the two structures are
superposable and so do not represent enantiomers. They represent two molecules of the same compound.
2-Propanol does not have a chirality center.

Here the molecules are superimposible but I cannot figure out how the molecule is rotated to acheive this. If the molecule is rotated from y-axis (↑) then the molecule is again mirrored defeating the purpose.

  • $\begingroup$ If having troubles with 3D imagination, try to make yourself physical ball-stick/pin model, each stick pointing to a vertex of a thought regular tetraedr. Make 2 mutually mirror images and then try to manipulate by whatever rotation to look identical. $\endgroup$
    – Poutnik
    Mar 16 at 21:49
  • $\begingroup$ You can use whatever method for checking it, as long as it works ;) BTW maybe get better source than what you used for this picture. $\endgroup$
    – Mithoron
    Mar 16 at 22:12
  • $\begingroup$ The molecule in your example is not rotated at all. As it is drawn, VI is exactly the same as V, no rotation required. $\endgroup$
    – Loong
    Mar 17 at 9:57
  • $\begingroup$ The example is taken from Solomons by Wiley Publication $\endgroup$
    Mar 20 at 6:36

1 Answer 1


TL;DR Combined rotation and reflection bring this particular achiral molecule into a form indistinguishable from the original, in appearance.

If the molecule is rotated from y-axis (↑) then the molecule is again mirrored defeating the purpose.

No, it isn't mirrored again, it looks as if, that's the whole point. Defeat is not defeat, it is the whole purpose of the exercise, what you sought to prove. The initial reflection transformed the molecule but a rotation about a vertical axis that goes through the chiral center is enough to bring it back into an orientation indistinguishable in appearance from the original. In a chiral molecule a similar rotation cannot bring the reflected molecule into alignment no matter how hard you try. A simple exercise is to swap one of the methyl groups with a third type of substituent to see the effect of the same operations.

The question is similar to another which also deals with 2-propanol. It is worth noting that using these symmetry operations to determine whether a molecule has an enantiomer is one algorithmic test but one can define enantiomers in other ways. A chiral molecule lacks reflection (Cs) and rotoreflection symmetries (S2n). The atomic composition, connectivity, and bond order are unchanged (as are thermodynamic properties of the pure substance), but their interaction with polarized light differs. An equivalent test is to invert the sign of all dihedral angles and attempt to superimpose the molecules:

Newman projection of monosubstituted center Here the molecules are shown using a Newman projection. If the blue atoms are of the same type (including isotope, etc) the two molecules are indistinguishable.

On a final note, isopropanol might not be an ideal example for such an exercise. The gas phase molecule in its ground state is classified as having C1 and not Cs symmetry, so strictly speaking its mirror images are not superimposable (that H on the hydroxyl breaks the symmetry).


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