Stereochemistry and Chirality

By James Ashenhurst

The Meso Trap

Last updated: January 20th, 2024 |

Recognizing Meso Compounds

  • In chemistry, two molecules that are superimposable mirror images are considered to be identical molecules. They will have identical physical properties in all respects.
  • Two molecules that are non-superimposable mirror images are enantiomers. They will have identical physical properties except they rotate plane-polarized light in equal and opposite directions.
  • Meso compounds contain chiral centers, but these chiral centers are arranged in such a way that the molecule has a mirror plane and is therefore achiral overall. A meso compound is superimposable on its mirror image.
  • Watch out for exam questions where two meso compounds are suggestively drawn as mirror images. They are not enantiomers. They are the same molecule, since they are superimposable!

summary of meso compounds and how to tell if a compound is meso

Table of Contents

    1. The Meso Trap
    2. But Clearly There Are No Mirror Planes In These Molecules…. (Right?)
    3. Fischer Projections, Newman Projections, and Cyclohexane Chairs
    4. Identifying Meso Compounds, Method One: Build A Model
    5. Identifying Meso Compounds, Method Two: Determine (R,S)
    6. Reviewing “Enantiomers, Diastereomers, or The Same” Questions
    7. Notes
    8. Quiz Yourself!
    9. (Advanced) References and Further Reading

1. The Meso Trap

Once you learn about stereochemistry, you’ll start getting asked to identify whether molecules are enantiomers, diastereomers, constitutional isomers or even identical. (See article: Enantiomers vs. Diastereomers vs. The Same – Two Methods for Solving Problems)

Most students by this point have it in their heads that enantiomers are mirror images.

So they are totally ripe for questions like this!

Click to Flip

Ka-pow! These two molecules are drawn like they are mirror images of each other. But they are superimposable mirror images.

If you rotate the whole molecule on the left around, you can lay it right on top of the molecule on the left.

In organic chemistry, we have a name for two molecules that are superimposable. They are identical molecules.

Identical boiling point, freezing point, melting point – whatever physical property you care to name, they are indistinguishable. 

What make the molecule above particularly tricky is that it does have two chiral centers. It’s just that those two chiral centers arranged on the molecule in a way that the molecule itself has an internal mirror plane, which renders the molecule achiral overall.

We call these molecules “meso compounds“.

[Louis Pasteur noted that there were three stereoisomers of tartaric acid. One rotated plane-polarized light to the left (levorotatory), one rotated plane polarized light to the right (dextrorotatory) and the third did not rotate plane-polarized light at all. This third stereoisomer, what we now call (2S, 3R)-tartaric acid, was called the “meso” (for “middle”). [See article – Optical Activity and Optical Rotation]

So I like to call questions like the one above, “Meso Traps”.

Here’s a similar question in this vein.

Click to Flip

Now that you’ve done two, try this one.

Click to Flip

I can’t give you three Meso Traps in a row!  That would make it too easy.

The point is to interleave trick questions with more straightforward ones.

Here’s another one for you.

Click to Flip

You get the idea. If a molecule has a mirror plane – and that mirror plane can cut through a bond or an atom – then it is an achiral molecule.

2. But Clearly There Are No Mirror Planes In These Molecules… (Right?)

Sometimes these questions are asked with a little bit more subtlety.

For example, what about these two molecules? Enantiomers, diastereomers, constitutional isomers, or something else?

Clearly, there’s no mirror plane in this molecule.

Click to Flip

Or is there?

Don’t forget that molecules can adopt many different conformations through rotations about single bonds, and these conformations are all in equilibrium with each other. (See article: Newman Projection of Butane). So it’s very possible that you may be given examples where a bond rotation will reveal a plane of symmetry.

If a bond rotation creates a plane of symmetry in a molecule, that molecule is considered achiral. 

In the case above, if you rotate the molecule along the central C-C bond, you’ll find that these molecules are actually exactly the same as those in the first question. It just takes a little bit more work to see it.

What about this example?

Click to Flip

OK, another change of pace question.  Those two are actually diastereomers.

What about these?

Click to Flip

3. Fischer and Newman Projections

Let’s look at another set of questions. Remember that molecules won’t always be helpfully drawn as line diagrams. You can expect to see the full gamut of Newman, Fischer, Sawhorse, and even cyclohexane chair conformations.

Click to Flip

Sometimes I think that the only reason Fischer projections are still taught is that they give us instructors some variety in asking different stereochemistry questions.

A helpful trick to remember is that exchanging any three groups on a carbon results in doing a bond rotation.

(Swapping any two groups will swap the configuration of a chiral center from R to S or vice-versa – see The Single Swap Rule

Let’s not forget Newman projections!

Click to Flip

And cyclohexane chairs are also fair game.

Click to Flip

In this case, it helps to imagine what the molecule would look like when looking at it from above.

Remember that molecules are three-dimensional objects just like anything else and can be viewed from a variety of perspectives!

4. Identifying Meso Compounds, Method One: Build A Model

Now that most of  the low-hanging fruit of trick questions has been thrown at you, let’s talk about strategies for identifying meso compounds and differentiating different kinds of stereoisomers.

There are two classes of strategies for solving these kinds of problems.

The first involves making a model, and then rotating along C-C bonds to find an axis of symmetry.

Here is an example of doing a bond rotation on a line diagram to reveal a plane of symmetry.


This also works for Fischer projections. Remember that in a Fischer, the arms come out to hug you (or strangle you, if you are feeling dark).


Newman projections can also be rotated, as in the short video below.


You get the idea. Build the two molecules with your model kit, rotate bonds until you can tell if they are superimposable or not.

5.  Identifying Meso Compounds, Method Two: Determining Absolute Configuration (R,S)

Generations of students have been told to use model kits to visualize whether molecules are enantiomers, diastereomers, or the same.

And generations of students have rolled their eyes at this advice like they would at a parent admonishing their children to floss their teeth or eat their vegetables.  They just don’t want to do it.

In my opinion, I think this is actually OK not to build models, provided that you practice like hell to get good at determining whether chiral centers are (R) or (S). Once you master the skill of determining absolute configuration (R/S), making a model is unnecessary.

That’s because there’s One Weird Trick that can help you determine which compounds are meso and which are not. It is also helpful for determining whether molecules are enantiomers, diastereomers, or the same.

Get ready for the One Weird Trick. Here we go.

Look at the R,S designations on these meso compounds, compared to their chiral diastereomers. What do you notice? 

one way to determine if compounds may or may not be meso is to examine connectivity and determine if they are R or S

All the meso compounds are (R,S) or (S,R). Furthermore, the compounds that are (S,S) or (R,R) are never meso. They are chiral.

It makes sense when you think about it!

In a meso compound, the left-hand side of the molecule is the mirror image of the right hand side. In order for this to be true, a molecule with two chiral centers has to have two chiral centers that have opposite configurations – (R,S) or (S,R).

So one way to quickly determine if a compound is meso even if it’s drawn in a strange conformation is to determine (R,S) on each of the chiral centers. If it’s (R,S) or (S,R) your Meso Trap Alarm should be going off. [Note 1]

One word of caution: while it’s a necessary condition that a meso compound with two chiral centers must have (R,S) or (S,R) configurations it’s not a sufficient condition.  The left-hand half must have the same connectivity as the right hand half. 

You should double check that the molecule has the same connectivity on both sides before declaring it to be meso.

(2S, 3R)2-bromo-3-methylpentane is clearly not meso, for example, due to the fact that one chiral center has a bromine atom and the other does not.

just because two chiral centers are R and S doesn't mean the molecule is meso

6. Reviewing “Enantiomers, Diastereomers, or The Same” Questions

In addition to sniffing out meso compounds, determining absolute configuration (R,S) is helpful in determining whether two molecules are enantiomers, diastereomers, or the same.

Just to review what we have covered elsewhere (see Enantiomers, Diastereomers, or the Same – Two Methods For Solving Problems)

For two molecules that have the same connectivity: 

  • If they have identical absolute configuration [e..g both (R,R), (S,S), or (S,R)] then they are considered to be identical.
  • If one is (R,R) and the other is (S,S), they are enantiomers.
  • If one is (R,R) or (S,S) and the other is (R,S) or (S,R), they are diastereomers since at least one of their chiral centers has an identical configuration.
  • If one is (R,S) and the other is (S,R), they may be enantiomers, but they may also be meso compounds. Double check the connectivity to see if a plane of symmetry may be possible. If they are both meso, then they are considered to be identical. 

Good luck – and may you avoid falling into the Meso Trap!


Significantly updated Oct 2023 from a previous version

Note 1. This also applies to meso compounds with more than two chiral centers.

Click to Flip

The largest meso compound I am aware of is nonactinDespite having 16 chiral centers, it has a plane of symmetry and is achiral overall.

the structure of nonactin one of the largest meso compounds

The assignment of chiral centers in nonactin is (R,R,R,R, S, S, S, S, R, R, R, R, S, S, S, S)  – an equal number of (R) and (S) chiral centers, and the left half is the mirror image of the right half. [Source]

Quiz Yourself!


(Advanced) References and Further Reading



Comment section

40 thoughts on “The Meso Trap

  1. Hi James,

    I think your site is great with a lot of high quality material presented in an accessible way. I’m an organic chemistry tutor myself, and I often come by to refresh on reagents or see some of the nuances you point out to flesh out my understanding.

    I encountered a new situation with meso compounds in a session and wanted another perspective. I have always thought of meso compounds as molecules with chiral centers and internal planes of symmetry as you presented. However, this problem involved (1R,2R,4S,5S)-1,4-dibromo-2,5-dichlorocyclohexane, which has no mirror plane but is still identical to its mirror image. It does have an inversion center though. My first impression is that this should be sufficient to make it meso since inversion is still a form of reflective symmetry. I’ve also seen that many molecules with internal mirror planes could be rotated to other conformations with inversion centers, like your example diol drawn in as it is in 2c. So this is my question: is having an inversion center in place of a mirror plane enough to qualify a molecule with chiral centers as meso?

    I’d appreciate your thoughts if you have a moment, and thanks again for this amazing resource.

    1. Hi Philip – Thank you for bringing up inversion centers. That’s another example of molecules that have chiral centers but are achiral due to a symmetry element.

      Pasteur named what we now call (2R, 3S) tartrate “meso” because it had an optical rotation in between those of the levo and dextro isomers.Subsequently we have used the term meso to refer to molecules with chiral centers that have an internal mirror plane.

      Regarding whether a molecule with an inversion center counts as “meso”, I suppose that is really a naming question. I don’t know how much it matters whether it is called meso or not, provided that your students understand that the same type of phenomenon is in play – symmetry giving rise to achirality.

  2. To create a Fisher projection, you must either rotate two ways: 1. counter-clock in 180 degress not in a 120 degress 2. if you must go clockwise, you do so in 90 degress. Therefore 2a is an enantiomer. For an enantiomer to be one in the fisher projection you must make sure that the top and bottom of the molecule are the same which involves rotating one of the molecules in 180 degress. The mirror image shows exactly what it should look like supposing you where holding a mirror. Hope this hasn’t confused anyone

  3. I also ironically ended up here after seeking help for an organic chemistry problem only to realize I fell for this trap before I even realized what it was. Glad I stumbled upon here!

  4. Hi James!
    Could a 2,4 substituted molecule like 2,4 dichloropentane be meso?
    And, on the same thread, if the 3C were substituted as well, would the 3-carbon not have a stereochemistry since the two equal substituents on either side make the center carbon achiral? So therefore we’d only name the 2,4 as R,S and not the 3, even if it had a chlorine atom as well?

    1. Hi Shelby – sorry for ridiculously late response, just back from summer trip.

      Hi James!
      Could a 2,4 substituted molecule like 2,4 dichloropentane be meso?

      —> yes, absolutely. The (2R, 4S) and (2S, 4R) versions turn out to be the same molecule (meso, has a plane of symmetry).

      And, on the same thread, if the 3C were substituted as well, would the 3-carbon not have a stereochemistry
      since the two equal substituents on either side make the center carbon achiral?
      So therefore we’d only name the 2,4 as R,S and not the 3, even if it had a chlorine atom as well?

      —> That is a very interesting question. If 3 is substituted with, say, a methyl group, the answer is that the 3-position on that molecule can be in a situation
      where the 2 position and the 4 position are not exactly the same (e.g. 2 is R, 4 is S) so it is not attached to 4 different substitutents. However, the molecule has a plane of symmetry and does not rotate plane polarized light.

      99.99% of intro level courses will not cover this but the 3 position in that case would be what is known as a pseudoasymmetric center. We use lowercase (r) and (s) to indicate the orientation of the pseudoasymmetric center.

      In the cahn ingold prelog system, (R) is given a higher priority than (S). So for (2R, 4S) 3-methyl dichloropentane, the 3position could be (s) or (r) depending on the orientation of the methyl group.

      1. Thank you for your excellent blog.
        I have a follow up question regarding the lower case r/s designation for pseudo-asymmetric centers. I follow your comments with regard to the compound mentioned above: (2R, 3r/s, 4S) 3-methyl dichloropentane. What I’m finding confusing is why IUPAC prefers naming of compounds that lack any stereogenic centers at all due to symmetry with the r/s designation. For example, (1s,4s)-1,4-dichlorocyclohexane. This is name that is assigned by chemdraw and I believe it is consistent with the IUPAC rules. You probably know it involves a more elaborate application of the CIP rules with bond disconnections, etc.). It seems to me that the main purpose of nomenclature is to be unambiguous and using cis/trans for these types of symmetric molecules is perfectly unambiguous and much easier to assign than the convoluted application of CIP in these cases. I’m I missing something behind the IUPAC recommendation. Thanks for taking my question.

  5. Can you explain why 1,4-disubstituted cyclohexanes are not meso, even though they are achiral and “look” meso (i.e., have a plane of symmetry)? An example would be 1,4-dimethylcyclohexane.

    1. They’re not chiral because there are no chiral centers. You need 4 different “groups”. It’s a bit hard to see, but in each case the 1 and 4 positions have two identical groups – the carbon chains that connect carbons 1 and 4 are identical.

      Note that if it were 1,4 disubstituted cycloheptanes, it WOULD have chiral centers, because in one case you’d have a CH2CH2 group on one side and a CH2CH2CH2 group on the other. Different. Hence, a chiral center.

      Hope this helps – James

  6. In the last figure, cis 1,2-dimethylcyclohexane does not seem to have a plane of symmetry when it is drawn in chair conformation. Can you please elaborate on this? Thanks!

  7. I’m confused about one of your examples above. You say that cis-1,2- dimethyl cyclohexane is meso. However, drawn as a chair, there is no plane of symmetry because one of the substituents is axial, the other equitorial.

    I believe that there exist two separate enantiomers of 1,2 dimethyl cyclohexane, but at room temperature they are in conformational equilibrium, so they don’t have an optical rotation. Am I understanding this correctly?

    1. Drawn as a chair, there is no plane of symmetry. However, if you draw the other chair conformation, you’ll see that it’s the mirror image. Since the molecule spends 50% of its time as either one, you shouldn’t see optical rotation.

      1. True. But it still is not a meso compound. The equilibrium mixture consists of interconverting enantiomers. That is why it is optically inactive, not because it is meso. You can fall into this trap with all 1,2-R,R-disubstituted even-numbered rings.

  8. How to convert a Newman projections with two choral centres into a Fischer projection…..please explain…

  9. Thanks for the info! I am a bit confused though because i thought number 2b was an enantiomer due to the r,s configuration being opposite in both. can you explain?

    1. in 2b, if you rotate each Newman so that the OH groups line up you’ll see that the front carbon completely superimposes on the back – the molecules have mirror planes. They are actually the same molecule, just rotated 180 degrees.

  10. 2)A the fisher projection does not seem to be meso to me. from what I see, the two are mirror images, plus have opposing chiral centers (R,S) (S,R). I also can’t seem to superimpose them. And, they both have been singly substituted at each stereo-center. I also cannot find a plane of symmetry.
    How is 2A a meso compound?

    1. In order to see the mirror plane you’d need to do some bond rotations. Further down the page I show how to do a rotation on the left-hand molecule so that you can clearly see the mirror plane. You can do the same thing (except opposite direction) to the molecule on the right, and you’ll also see that it has a mirror plane.

  11. Great, that finally cleared my doubts. You explained it a gazillion times better than our chem teacher. Thanks a lot.

  12. Indeed a great work……. Thanks a lot for the material and that too presented in an easy way …. You are a genius.

    I have a query… How can a Newman projection be easily converted to Flying Wedge formula.


  13. You ARE a Professor and clearly an excellent one no matter what condition academe is in. Kudos for continuing to teach and doing so accessibly vs. blooming in the desert of some community college–great for the few that go there but not for the rest of us!

  14. very helpful!! quick question though, just to clarify. If a molecule is chiral, is it automatically not meso? It seems like that is true from the first few sentences, but I just want to verify.

  15. thank you so much you saved my life !! thanks but i still dont know how to determine theR and S for newman projection ?

  16. Thank’s a million for all of this great information. I found your site while looking for tips on orgo and I have to say this is a goldmine! I am sure you get this all the time but thank you so much for all of your help. I had just being looking at this one stereochemistry problem for the past 10 minutes and this saved me so much more time.

    Great stuff!

  17. I found this site by accident but what a fortuitous mistake.

    Thank you for this level of detail. I have an upcoming exam and likely would have fallen for this.

    Do you happen to have any more sections on your webpage regarding tricks that professors like to throw into their exams? Your site is unique in that it might as well have been made by a professor (your credentials are more impressive than most professors). Your insight is appreciated tremendously.

    With kind regards


  18. James, I noticed there are no answers posted to these problems on this “meso trap” post. Also there is no explanation to the diagrams where you rotate 120 degrees and in some cases 60 degrees. When can we do this and what are the rules? Thanks! Chris Raabe, UAA

    1. Posted answers, thanks for mentioning this. Regarding bond rotations: If you interchange any three groups on a carbon, you’ve done a bond rotation. Rotating a group 60 degrees is a little trickier, I don’t have a good shortcut for learning how to do this.

  19. hey! thanks for your words of encouragement last month. my blog is kind of just my little secret space where i pout and complain and pour all my frustrations into; stuff i don’t reveal to people in real life cause it would probably make people think i’m self-centered (especially fussing over a B+).

    i guess you could say i have a love-hate relationship with organic chemistry. it’s quite a tough class, but i really do thoroughly enjoy learning the material–at least in comparison with general chemistry. my two apartmentmates (both extremely brilliant curve setters) and i are taking the honors sequence starting this quarter. hopefully i won’t make the same mistake again by doing well in the beginning and underestimating the final exam.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.