I am facing trouble with the following problem.

Is 2,6-dimethyl-2',6'-dinitro-1,1'-biphenyl chiral?

enter image description here

My idea is that there is a axis of symmetry along the line joining the two benzene molecules. Hence it should be achiral.

However my book states it otherwise.


Ortho-Ortho-tetrasubstituted biphenyls become non-planar at room temperature in order to have minimum electronic repulsion among substituents.In this orientation(phenyl planes perpendicular to each other) the free rotation of C-C single bond is restricted and molecule shows optical activity due to molecular disymmetry.

Actually looks like this:

enter image description here

The two phenyl rings become perpendicular to each other thus leading to disymmetry which in turn leads to optical activity(chirality).

Conditions for biphenyls to be chiral: enter image description here

enter image description here


But here since two substituted groups are the same that is 2 methyl groups on the same phenyl ring and 2 nitro groups on the other phenyl rings, this according to the second condition leads to the presence of a line of symmetry or a plane of symmetry thus making it achiral.

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Yes, the molecule should exhibit chirality below room temperature.

In the optimised1 molecular configuration the molecule cannot be superimposed on its mirror image. This is because of the out-of-plane rotation of the nitro groups.

molecular rotation of 2,6-dimethyl-2',6'-dinitro-1,1'-biphenyl

The thermodynamic most favourable structure is of $C_2$ symmetry and has no mirror plane (see above).
Contrary to the previously presented arguments, an analysis of the Lewis-structure is misleading as it is massively misjudging the space requirement of the nitro moieties. A feature that is already present in 2,4,6-trinitrotoluene.2 From the abstract of Clarkson et.al.:3

Two genuine energy minimum structures were found. In both structures the 4-nitro group is planar to the phenyl ring, while the 2,6-nitro groups are slightly out of plane with the phenyl ring due to steric interaction with the methyl group.

They later describe:

The two stable molecular structures of TNT, Fig. 1, are related by internal rotations of the 2 and 6-nitro groups and the methyl group. [...] Structure A is the more stable geometry, though the difference in energy is small ($0.650~\pu{kcal mol^−1}$ at B3LYP/6-311+G**, Table 4). Structure A displays near ideal $C_\mathrm{s}$ symmetry with one of the methyl hydrogen atoms perpendicular to the phenyl ring in the plane of $\sigma_\mathrm{h}$. The 2,6-nitro groups of A are non-planar and rotated into the same face of the phenyl ring, maximizing the number of van der Waals interactions to the methyl group.

The 2,6-phenyl moiety however is much larger than a methyl group, which is the reason, why the symmetry of the nitro groups is broken. Forcing this $C_\mathrm{s}$ symmetry will lead to a first order saddle point (transition state) on the potential energy surface which is only about $1.6~\pu{kJ mol^-1}$ higher in energy than the minimum structure.
At room temperature the inter-conversion between the enantiomers is likely to happen through this structure. I have not checked whether this state actually connects via an intrinsic reaction co-ordinate calculation, but from the visual investigation of the imaginary mode this seems very likely.

imaginary mode of the cs structure

If you cool the system enough you should be able to identify two enantiomers, however, at room temperature inter-conversion should be too fast to observe.


It would be good to know which book you have used and in which context the statement was issued. From a purely theoretical point of view, I would agree that the molecule is chiral. From practical point of view I would not asses this property, since it is probably too hard to measure.

The suggested $C_\mathrm{2v}$ structures are either second or fourth order saddle points, depending on the rotation of the methyl groups. These structures are at least $15~\pu{kJ mol^-1}$ higher in energy than the former and no viable pathway for inter-conversion.

high energy symmetric structures


  1. All structures have been optimised on the DF-M06L/def2-SVP level of theory with Gaussian 09 Rev. D01. Stationary points have been characterised by calculating the vibrational frequencies at the same level of theory.
  2. W. Robert Carper, Larry P. Davis, Michael W. Extine, J. Phys. Chem. 1982, 86 (4), 459–462.
  3. John Clarkson, W. Ewen Smith, David N. Batchelder, D. Alastair Smith, Alison M. Coats, J. Mol. Struc. 2003, 648 (3), 203-214.
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The simplified point of view in a way that students should be expected to answer the question

It is not the axis of symmetry (which is a $C_2$ axis in this case) that is important but rather that the molecule additionally features planes of symmetry ($\sigma_\mathrm{v}$). There are two perpendicular planes of symmetry in your example compound which both contain the $C_2$ axis. Thus, the molecule’s point group is $C_\mathrm{2v}$ and it must be achiral.

Obviously, your textbook contains an error.

The practical point of view

See Mart’s answer. In a nutshell: the molecule would be chiral but racemic (rapidly interconverting) at room temperature due to the specific most stable orientation of the nitro groups.

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    $\begingroup$ I think the textbook is correct. The minimum energy structure only has $C_2$ symmetry, see my answer for details. I think for biphenyls and related compounds point groups are never obvious, tiny changes can easily break symmetry. $\endgroup$ – Martin - マーチン Jan 6 '17 at 10:24
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    $\begingroup$ @Martin-マーチン Ah, the pitfalls in simple looking molecules. $\endgroup$ – Jan Jan 6 '17 at 20:45
  • $\begingroup$ I don't know about point groups and the axis of symmetry.can you please share a resource from where to know about them $\endgroup$ – user471651 Nov 10 '17 at 8:40
  • $\begingroup$ @user471651 chemistry.stackexchange.com/q/43073 $\endgroup$ – Jan Nov 10 '17 at 11:03

I do not believe the molecule to be chiral. While other answers correctly note that rotation about the central phenyl-phenyl bond is hindered (under reasonable conditions), the substituents on each individual ring are identical. For this reason, one cannot unambigously assign a chirality designator, such as $S$ or $P$.

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  • $\begingroup$ you are right my buddy! i didnt have a proper look at the compound $\endgroup$ – Prakhar Dec 24 '16 at 11:39

This question wants you to check for axial chirality. However, this molecule doesn't possess axial chirality (even though in my haste to answer, I did fall into the trap of saying it did, my apologies. I did not make not of the fact that the substitution pattern here is symmetrical, and because of the existence of this element of symmetry, the given compound should be achiral.

Axial chirality, which is a special case of chirality in which a molecule does not possess a "stereogenic center" (which is what is most commonly taught in introudctory organic chemistry courses) but an axis of chirality, i.e an axis about which a set of substituents is held in a spatial arrangement that is not superposable on its mirror image.

It is commonly observed in atropisomeric biaryl compounds wherein the rotation about the aryl-aryl bond is restricted. Owing to restricted rotation around the biaryl bond, the two stereoisomers cannot interconvert under ambient conditions. I wrote a lengthy post about chirality in the past; you might find instructive.

Whatever, I said about axial chirality (in my previous post) still holds true, it just isn't applicable in this case. My bad! sorry

Additionally, since OP asked about how to assign stereochemistry in the case of compounds possessing axial chirality, I'll put down a few words on that.

First draw a Newman projection along the axis of hindered rotation. Now assign the ortho, and meta substituents priorities based on Cahn–Ingold–Prelog rules. Now, starting with the substituent of highest priority in the closest ring and moving along the shortest path to the substituent of highest priority in the other ring (as illustrated below), the absolute configuration is assigned P for clockwise and M for counterclockwise. In the example shown, A has priority over B.

enter image description here

Now, try doing this for your given compound, and you run into trouble. You cannot unambiguously assign a configuration.

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I would check the simplest definition of chirality before looking at potential chirality centers or point group representations:

  • Is the molecule's mirror image superimposable on the original?

If you redraw the biphenyl along any mirror plane, you will get the exact same image. That means it's mirror is superimposable on the original, and the molecule is achiral.

Even in difficult chem classes, I always check the simplest definition first. As long as you're confident of the stereochem, it's the quickest to test and hardest to mistake. I would only worry about elements of symmetry if you're in an upper level inorganic class or into that sort of thing.

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    $\begingroup$ You're assuming that biphenyl's twist is $90^{\circ}$. If it's less than that, the molecule is axially chiral... Chirality has a simple definition but that doesn't always translate to a simple scenario. That's why Elial's book is so damn thick. $\endgroup$ – Zhe Dec 29 '16 at 3:12
  • $\begingroup$ In the trivial example of two perpendicular and symmetric biphenyls, simple works quite well. I wasn't suggesting point group theory isn't useful, but in my opinion the poster should lock down what chiral means before learning what a C2v axis is. $\endgroup$ – Max Hoffman Dec 29 '16 at 3:22
  • $\begingroup$ That, I agree with, especially since $C_{2v}$ is not the important symmetry element to look at. $\endgroup$ – Zhe Dec 29 '16 at 13:33
  • $\begingroup$ C2v point group*, not axis. Don't have enough points to edit. Also meant phenyls* instead of biphenyls. $\endgroup$ – Max Hoffman Dec 29 '16 at 13:48

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