In an examination I was asked to determine the molecules which are chiral as well as dissymmetric. There were four options, and among them, two were achiral, as they had an improper axis of symmetry. The other two options are 2-bromobutan-2-ol and trans-1,2-dimethylcyclobutane:

enter image description here

These two compounds are chiral, as they do not have an improper axis of symmetry. I couldn't find any axis of symmetry as well. The question has only one correct answer, but I don't know which one is correct. Please help me with the answer.

  • $\begingroup$ Chiral molecules are always dissymmetric $\endgroup$
    – Lalit
    Commented Aug 5, 2021 at 18:44
  • $\begingroup$ (c) has an axis of symmetry. $\endgroup$ Commented Aug 5, 2021 at 21:20
  • 2
    $\begingroup$ To be honest, at another time of day, this probably would not have been closed. And I don't think it should, while it could still have used some improvements. Anyway, here is something related: chemistry.stackexchange.com/q/118985/4945 $\endgroup$ Commented Aug 6, 2021 at 0:29

1 Answer 1


I assume you already studied molecular symmetry and point groups. So, first of all, among the existing point groups, you can have chiral point groups (point groups C1, Cn, Dn) and achiral point groups (all the others). As a matter of fact, chirality is unchanged when it comes to operations such as E or Cn (that equal an even number of reflections). On the contrary, this is not the case when it comes to operations that equal an odd number of reflections (σ, Sn, inversion), as they can convert a right-handed object into a left-handed object. So an improper rotation axis (Sn) implies that the molecule cannot be chiral. Both molecules that you show here are chiral, as you correctly pointed out.

2-bromobutan-2-ol belongs to point group C1 as it has no symmetry other than the identity (which is the C1 axis that every molecule has and that corresponds to a rotation by 360° that of course brings the molecule back to the same initial position). Being the C1 axis the only element, C1 molecules are defined asymmetric.
The other two chiral point groups (Cn and Dn) are dissymmetric.

trans-1,2-dimethylcyclobutane has a C2 axis that allows you to rotate the molecule by 180° (which is 360°/2, that's why it corresponds to a 2-fold rotation axis) obtaining a situation that is equivalent to the initial one:

C2 axis in trans-1,2-dimethylcyclobutane

Let's see this with colors, so it's easier:

Rotation in trans-1,2-dimethylcyclobutane

So, in conclusion, since (1R,2R)-1,2-dimethylcyclobutane belongs to the C2 point group, it is "chiral as well as dissymmetric".

Note: The above is only true in a first order approximation. A more detailed look will reveal that the cyclobutane ring is puckered and the C2 axis no longer exists.

  • 1
    $\begingroup$ In first approximation yes, but the cyclobutane ring is puckered, so: no. $\endgroup$ Commented Aug 5, 2021 at 23:04
  • 1
    $\begingroup$ @Martin-マーチン Sure, but I think the examination under consideration here assumes the ring to be planar for simplicity. $\endgroup$
    – sasam
    Commented Aug 5, 2021 at 23:36
  • 4
    $\begingroup$ Yes, I merely stated that it isn't that simple. If some more advanced people come here, then this fact should be included. As a side note: examiners shouldn't pick examples that are only correct at a glance; I don't think they are doing a good job then. I find it very important to not think in schematic drawings, and questions like these trends to reinforce the view that these drawings are somewhat accurate. $\endgroup$ Commented Aug 5, 2021 at 23:42
  • 1
    $\begingroup$ I think that this should be re-opened because this answer togheter with the comments can be useful to "teach" a new student or whenever a false simple exercise of this kind is administered. The answer will benefit if the point raised by the comment is integrated. $\endgroup$
    – Alchimista
    Commented Aug 6, 2021 at 7:29
  • 1
    $\begingroup$ I took the liberty to do some reformatting and I've included my initial comment as a note. The other one is too personal to be included into a technical answer in my opinion. I'll also reopen the question. cc @Alchimista $\endgroup$ Commented Aug 6, 2021 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.