1,4-Dichloro-2,5-dimethylbenzene is symmetric about the plane of paper.

I have never seen a plane of symmetry positioned like this. Would it be correct to call it a plane of symmetry?


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    $\begingroup$ The "plane of the paper" depends on how you draw the molecule; the molecule cannot know what plane that is, there is nothing special about it. $\endgroup$ – orthocresol Sep 21 '19 at 16:00
  • $\begingroup$ This question seems a bit weird. Any halogenated benzene does not change if you use the paper as a mirror plane. Also the unstubstituted benzene. $\endgroup$ – Karl Sep 21 '19 at 19:39
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    $\begingroup$ The short answer is "yes" - a planar molecule like this can have the plane of the paper as a symmetric mirror plane. (See the answer below for more context about methyl groups, though. $\endgroup$ – Geoff Hutchison Sep 21 '19 at 19:56

In short: The presence of the $\sigma_h$ plane you see depends on the conformation of the methyl groups.

If your example were either benzene or 1,4-dichlorobenzene, then you were be right, the molecule would contain a plane of symmetry identical to the plane defined by the benzene ring. This $\sigma_h$ plane then could be described as in the following example by the yellow disk:

enter image description here

(credit to symmetry@otterbein, example symmetry $C_{6h}$, hexakis(Me2N)benzene).

Your example however is about a methyl-substituted benzene. While the carbon atoms of the methyl groups are in the plane of the symmetry, you no longer can apply the same operation (reflection on this mirror plane) on the hydrogens of these methyl groups to yield corresponding, symmetry equivalent hydrogens on the other side of the mirror. This is only possible if both the methyl groups share a special conformation: for each exactly one hydrogen must reside in this $\sigma_h$ mirror plane; then, the other two hydrogens of the methyl group in question are symmetry equivalent.

In all other conformations, the three hydrogens of one methyl group were no longer symmetry equivalent in respect to this $\sigma_h$ plane. And it would require only one methyl group to be «off» to erase the $\sigma_h$ plane for the whole molecule.

I would suggest to venture out the interactive tutorials around symmetry@otterbein. All you need is an internet browser with javascript enabled. Later, if interested in applying the concepts of symmetry into crystallography (including the transition of Schoenflies notation used in IR-spectroscopy by the Hermann-Maguin about point groups), you may find its sibling crystals@otterbein a valuable complement.

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    $\begingroup$ +1 Generally since the methyl rotation is fast at room temperature, a methyl group is often considered a sphere for the case of symmetry perception. (On exercises I give, I usually clarify this point.) $\endgroup$ – Geoff Hutchison Sep 21 '19 at 19:55
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    $\begingroup$ @GeoffHutchison I agree that a methyl (or trifluoromethyl) group may rotate almost at «no additional energetic cost» for molecules in solution / gas phase at room temperature and above, and by consequence, this perspective. Of course not so for the restraints and constraints of packed, crystalline state. Speaking of such, I assume your exercises equally discern between ferrorcene in either symmetry $D_{5d}$ or $D_{5h}$, too. $\endgroup$ – Buttonwood Sep 21 '19 at 21:57
  • $\begingroup$ Yes - those are good thought exercises for the difference between those two point group categories - "staggered ferrocene" vs. "eclipsed ferrocene" - even though at room temperature, there's interconversion. $\endgroup$ – Geoff Hutchison Sep 21 '19 at 22:52

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