Stereochemistry and Chirality

By James Ashenhurst

How To Draw The Enantiomer Of A Chiral Molecule

Last updated: November 6th, 2023 |

How To Draw The Enantiomer Of A Compound – The Single Swap Rule

  • Exchanging any two groups on a chiral center will flip the R/S designation from R to S or vice versa. (A subsequent flip of two groups will give you back the original configuration). This is sometimes referred to as the “single swap rule“.
  • For a chiral molecule with only one chiral center, a single swap on that chiral center will convert it into its enantiomer
  • To draw the enantiomer of a chiral molecule, 1) identify all the chiral centers and 2) do a single swap on each chiral center (Note – it’s equivalent to just convert all wedged bonds to dashed bonds and vice versa)
  • The single swap rule is also useful for determining R/S in cases where the #4 priority group points out of the page (“wedged”). First, do a single swap such that the #4 group is dashed. Then, determine R/S normally. Since a single swap was performed, whatever value you get will be the opposite of the true configuration.  (With enough practice you will just do this in your head)
  • This trick also works for when the #4 group is in the plane of the page. Just swap the #4 group (in the plane) with whichever group is dashed, then determine R/S normally. Whatever value you get will also be the opposite.
  • To draw the diastereomer of a molecule, do a single swap on at least one but not all chiral centers.

how to draw the enantiomer of a chiral compound summary single swap rule

Table of Contents

  1. The Single Swap Rule
  2. Switching Any Two Groups Interconverts (R/S) (It doesn’t matter which two!)
  3. One Swap Gives Inversion of Configuration. Two Swaps Gives Retention
  4. Two Ways To Draw The Enantiomer Of A Molecule With 1 Chiral Center
  5. Drawing The Enantiomer Of A Molecule With 2 or More Chiral Centers
  6. Just Watch Out For Meso Compounds
  7. Applications of The Single Swap Rule – Determining R/S When #4 Group Is In The Front
  8. Applications of The Single Swap Rule (2) – Determining R/S When #4 Group Is In The Plane of The Page
  9. Drawing Diastereomers
  10. Flipping 3 Groups Gives A Bond Rotation
  11. Summary
  12. Notes
  13. Quiz Yourself!
  14. (Advanced) References and Further Reading

1. The Single Swap Rule

Here’s a neat trick.

  • Swapping any two groups on an asymmetric (“chiral”) carbon atom transforms the configuration around carbon to its mirror image.
  • For a molecule with a single chiral center, this means that doing a single swap on that chiral center will result in the enantiomer.
  • second swap of two groups will restore the original configuration.

A nice description of this principle (not my invention) is to call it the “single swap rule”



For chemistry purposes, two molecules that are superimposable will have identical chemical and physical properties and are considered to be identical molecules.

Two molecules that are non-superimposable [Note 1 ] are not identical and will have at least some differences in their physical and chemical properties (like melting points, boiling points, solubility properties, and more)

The special case where two non-superimposable molecules are also mirror images is called optical isomerism and the two molecules are referred to as enantiomers. [See article: Types of Isomers]

Enantiomers have identical physical properties except for the direction in which they rotate plane-polarized light. [See article: Optical Rotation, Optical Activity and Specific Rotation]

Optical isomerism is most commonly encountered in the case of molecules containing a carbon attached to four different groups (an asymmetric carbon center or “chiral carbon”) . [Note 2]

2. Yes, Really – Switch Any Two Groups – (It doesn’t matter which two)

If you do the math, a tetrahedral carbon with 4 different substituents will have six possible ways to swap two groups.

Are they all really equivalent?

Yes! You can verify this yourself by using a model kit. Starting with two identical molecules, swap any two groups and see what happens. You should find that you have created the enantiomer.

Alternatively…. here is a video of flipping all 6, so you don’t have to do it



3. Doing a single swap inverts R/S.

You may recall that one way to distinguish enantiomers is that they have the same connectivity but opposite (R/S) values.

For example the enantiomer of (R)-AwesomeMoleculene will naturally be (S)-AwesomeMoleculene

Since doing a single swap converts a chiral molecule with one asymmetric carbon to its enantiomer, it is also true is that doing a single swap on a chiral carbon will flip the configuration of that carbon from (R) to (S)  (or from S to R).

[For more on (R) and (S) see this article on The Cahn-Ingold-Prelog Rules (CIP).] Briefly, the four atoms directly attached to the chiral carbon are assigned priorities 1-2-3-4 based on atomic weight (with rules for breaking ties,  if necessary). Then with the #4 substituent in the back, we trace the path of the 1-2-3 substituents to see if they follow a clockwise (R) or counterclockwise (S) direction.]

Let’s say for the molecule in the videos above, these 4 atoms have priorities

  • B = 1,
  • G = 2,
  • R = 3,
  • W = 4

(I’ve done this based on alphabetical order, but we could assign B = bromine G = chlorine R = fluorine and W = hydrogen if we want to) 

Our starting molecule has the (R) configuration. Flipping any two groups (e.g. blue and red) gives the enantiomer, which has the (S) configuration.

[Note that in this example we are only exchanging the groups and otherwise leaving the bonds alone –  much like what one would do when plucking atoms off a model kit and reattaching in a different order.  There’s a second way to do a single swap where the groups are left alone and the dashed/wedged bonds are swapped that we will discuss shortly]

the single swap rule states that swapping the position of any two groups will result in flipping the configuration from R to S or vice versa

A second swap (e.g. of blue and green) gives the enantiomer of (S), which now has the (R) configuration – the configuration of our original molecule!

single swap rule - two successive single swaps results in retention of configuration about a chiral center

So every time we do a single swap we toggle between (S) and (R) – an inversion of configuration.

It doesn’t matter which two groups you pick, you will get the same result.

Two single swaps will lead to retention of configuration.

4. How To Draw The Enantiomer Of Any Molecule With A Single Chiral Center (Two Ways)

All right. Let’s say we have a chiral molecule with a single chiral center and we are asked to draw the enantiomer.

How would we do that?

When you just have a single chiral center and a linear molecule, it’s pretty straightforward – even if the molecule is drawn in a Newman, Fischer or sawhorse projection. Here’s how.

While keeping the bonds from the asymmetric carbon constant (i.e. don’t move the bonds, don’t convert wedges to dashes or vice-versa, etc.)  all we have to do is swap the position of two groups. Just like you would when interconverting two groups on a model kit.

Click to Flip

This can work pretty well. However,  there are some situations where swapping the position of two groups will give ridiculous-looking results, especially when we have chiral centers that are part of a ring – hover here or click the link.

Thankfully there is a second,  equivalent method of doing a single swap that works even better for most cases.

The second method of doing a single swap is to keep all groups in the same place, and then swap a wedged bond with a dashed bond.

Like this:

interconverting a wedge and a dash results in flipping a chiral center from r to s or vice versa

Note that we didn’t “move” the position of the atoms, we just interconverted which bond was wedged and which was dashed. But it still results in flipping (R) to (S) !

There are many times when you will encounter an asymmetric carbon atom drawn such that only one group is drawn as a dash or a wedge.

This is often the case when the chiral carbon has an implied (“hidden”) hydrogen  hover here (link) or when specifying the stereochemistry of just one group is sufficient to determine (R/Shover here (link).

In these cases, just toggle that one bond between wedge/dash  and you will have performed a single swap.

cases where there is only one dash or one wedge include the case where there is an implicit hydrogen or a second dash-wedge would be superfluous

Here are some exercises on drawing enantiomers using the single swap principle:

Click to Flip

5. Drawing The Enantiomer Of Molecules With 2 Or More Chiral Centers

What if a molecule has more than one chiral center? Does this still work for drawing enantiomers? Yes! *

(* see section 6)

Here is (2R, 3R) tartaric acid. How would you draw the enantiomer?

Click to Flip

If you said, “just make all the wedges into dashes, and all the dashes into wedges”, then good job. This is sufficient to convert the molecule into its enantiomer.

Now that you’ve done that, draw the enantiomers of these molecules each with two chiral centers  (some of which you may have heard of).

Click to Flip

Let’s stretch it out a bit.  What if our molecule has three (or even four!) chiral centers?

Still works! Just find the chiral centers and do single swaps on each:

Click to Flip

Let’s get even more ambitious. Here are two famous molecules with multiple chiral centers. See if you can draw their enantiomers.

Click to Flip

Note that once your molecule reaches a certain size, interconverting dashes and wedges becomes much more convenient than moving atoms around.

6. Just Watch Out For Meso Compounds

Now that we’re feeling confident, let’s try another fun example.

Convert (2R, 3S) tartaric acid to its mirror image by doing two single swaps.

Click to Flip

Sure enough, this gives us (2S, 3R) tartaric acid.

But is this the enantiomer? No!

Why not?  Because (R,S) tartaric acid is a meso compound – a molecule with chiral centers that contains an internal mirror plane. It’s an achiral molecule. (See article – Meso Compounds)

(This is easier to see once you do a bond rotation to reveal the mirror plane.)

So the “single swap rule” has a catch. Flipping all the chiral centers will only result in the enantiomer if the compound is chiral.

If you flip all the chiral centers on a meso compound, you’ll still have an achiral compound.

7. Using the single swap rule for determining R/S when #4 is not in front.

The single swap rule is also helpful for determining R/S.

As mentioned above, R/S is determined by placing the priority #4 substituent in the back and then tracing the path of the 1-2-3 substituents.

(See article – Introduction to Assigning (R) and (S) – the Cahn Ingold Prelog Priority Rules).

The trouble is that we will often encounter situations where  #4 isn’t always helpfully drawn in the back (i.e. as a “dash”).

One way to solve this is to get out your model kit, build the molecule and then rotate it around so that #4 is in the back.

You know, like in this molecule.

example of trying to determine r s when the fourth ranked substituent is in the front

As one of my students once said, “when I hear my instructor say “make a model” it’s like hearing my mom say, “clean your room“.



OK, OK! You don’t need to make a model.

Single swap rule to the rescue!

Just flip the #4 priority group with the group that is a dash. When you do this, you will flip R to S (or vice versa).

Now you can determine R/S like you normally would.

-example of using the single swap rule to determine r s when the number 4 ranked substituent is in the front

The thing is, whatever absolute configuration you get (R or S), the true configuration will be the opposite since you previously did a single swap. 

Knowing this, just “flip” the value you got to the opposite, and you’ll have the true absolute configuration.

Note that although I’ve drawn this entire process out, I think you’ll find that with a little bit of practice, you’ll just end up doing this in your head when you see that the hydrogen (or whatever other #4 priority group you encounter) is pointing out of the page.

It’s a very useful trick.

8. Using The Single Swap Rule When #4 Is In The Plane of the Page

This trick also works when the #4 group is in the plane of the page.

As before, just do a single swap to put the group in the back. Then determine R/S.

example of determining r s when the number 4 substituent is in the plane of the page

Whatever value of R/S you obtain will the the opposite of the “true” absolute configuration. So toggle the R/S value to get the true absolute configuration.

9. Drawing Diastereomers

The single swap rule is also useful for drawing diastereomers.

Recall that diastereomers are stereoisomers that aren’t enantiomers. (See article – Types of Isomers)

Since not all chiral centers have to have opposite values from each other, finding a diastereomer of a given molecule isn’t too difficult.

For example, a molecule with 5 chiral centers will have 25 possible stereoisomers (32).

Testosterone (below) represents one of those 32 possibilities. Its enantiomer is another.

The other 30 possibilities are all diastereomers of testosterone.

how to draw the diastereomer of a molecule with multiple chiral centers - just flip at least one but not all chiral centers

So to draw a diastereomer of a molecule you just need to draw a stereoisomer that isn’t an enantiomer.

  • That means that for a molecule with N chiral centers, do at least 1 but less than N single swaps.
  • Alternatively, flip a double bond from (E) to (Z)

10. Flipping 3 is a bond rotation

When doing a single swap, make sure you stop at interconverting just two groups.


In the example below, three groups have been interchanged. What happens to the absolute configuration (R/S) ?

Click to Flip

That’s right. Interchanging 3 groups results in a bond rotation. I’ll have more to say about that in a future post.

11. Summary

  • Swapping any two groups on an asymmetric carbon will flip its configuration from (R) to (S) or vice versa
  • A second swap will result in retention of configuration
  • For a chiral molecule with a single chiral center, this means that performing a single swap will result in its enantiomer.
  • An alternative way to perform a single swap is to interchange a wedged and dashed bond on a chiral carbon.
  • To draw the enantiomer of a molecule with two or more chiral centers, swap dashed/wedged bonds on all chiral carbons. Just watch out for meso compounds (which are achiral and don’t have an enantiomer).
  • To determine (R/S) on a molecule where the #4 priority substituent is in the front (or in the plane of the page!) , perform a single swap to put the #4 group in the back. Then determine (R/S). The true absolute configuration will be the opposite of this value.
  • Interchanging 3 groups results in a bond rotation (with retention of absolute configuration!)


Note 1 – Determining whether two molecules are superimposable also implies a certain time frame. For example, it’s possible for a molecule to have two conformations which are non-superimposable mirror images of each other. However, if these two conformers interconvert on a fast-enough time frame, then any effects of these enantiomeric conformers – such as equal and opposite optical rotation – will be cancelled out.

One fairly arbitrary (but practical!) test is, “Can you store the two enantiomers in separate bottles without them interconverting”. Certain conformers with hindered rotation – known as atropisomers – fit this description. One can purchase the enantiomers of 2-binaphthol.

Note 2 – Other forms of chirality exist besides the situation where there are four different groups on tetrahedral carbon. Another prominent example often encountered in undergraduate organic chemistry is molecules that include an axis of chirality, such as certain substituted allenes. [See article – Chiral Allenes and Chiral Axes]


Quiz Yourself!

Click to Flip

Click to Flip

Click to Flip

(Advanced) References and Further Reading



Comment section

25 thoughts on “How To Draw The Enantiomer Of A Chiral Molecule

  1. I want to start off by saying thank you so much for your amazing website! It has been so helpful and I use it to prepare before every test! Is it valid to keep the “positions” of the substituents the same but change the wedges and dashes? So for example, if we had (S)-2-butanol could we make (R)-2-butanol by making the wedge of the OH group a dash and the dash of the H a wedge without doing any other swapping?

  2. In determining r and s configuration, with respect to what are we defining it clockwise or anticlockwise?I mean there could be many representations to the same substituents attack d to a carbon atom

  3. Fantastic post as always James!! Just a few corrections needed in my opinions.
    I think your assignments of the stereocenters for the two tartaric acid molecules were wrong. I have tried to convert the drawings to line-wedge diagrams and found out that your (R,R) molecule should be (S,S) instead!
    The Reusch textbook also seems to imply the same:
    Correct me if I am wrong. Thanks for your great content !!

      1. I also just noticed that the Newman projection molecule on the left (just to the right of the two tartaric acid molecules) should have S configuration instead! When I converted the Newman projection to line-wedge drawing, the H is on a wedge, while OH is on a dashed bond. So I think that the real configuration should be reversed to S!

  4. hi! thanks for this.
    Hmm please correct me if im wrong but i dont think this works for substitution reactions. because what there is a change in priority order when naming? Ex) original was R but depending on whether the LG is larger/smaller than the Nu, the priority will change and hence naming it will be R, even though there was yes, a physical inversion of config.

    1. It will accurately show the change in configuration. However it will not accurately predict whether R is converted into S. I recommend doing this yourself on the substitution product.

  5. Thanks!
    but this doesn’t work for SN2 reactions. The reason is, if there’s a change in priority order, then the R –> is still R for example.
    it’s not S. yes, there is an actual physical inversion but the R is still R depending on whether the Nu is bigger or smaller than the LG.

    Feel free to correct me if i’m wrong. i really want to use the single swap rule even for SN2.

    1. Obviously some thought has to go into it when applying it for SN2 reactions. The SN2 reaction will always invert stereochemistry, but it will not always convert R to S (or vice versa).

  6. I just learned another trick – by using my hands. Apparently, you can point your thumb of one hand in the direction of the 4th priority group (that means point it into the page if that’s where the 4th group is going), and whichever hand’s fingers roll in the direction of the 1,2,3 priorities (left hand is S, right hand is R) is the winner. This seems to work even if the 4th priority is pointing out of the page, or across it. Hope that’s clear!

    1. Interesting. I’m getting it to work in most situations, but not all. For instance check the first diagram, second row, example on the left. Using that method I get S. What about you?

      1. Actually, I get R using the hand. Just remember to always point your thumb toward the 4th priority group. Then, you MUST turn from 1 to 2 to 3. If it is 3 to 2 to 1, it is the other hand. R R Right. S S that’s all that’sleft…

      2. The hand rule is also what I learned, and I’m pretty sure it always works. After reading your comment I went back to all of the molecules on this page and checked…they all agree with the hand trick, so I’m not sure of the problem you’re having. The hand rule requires you to picture the molecule in 3D and rotate your fingers that way, rather than just looking at clockwise/counterclockwise. It’s never failed me. It’s how I know if I’ve rotated molecules correctly or not when doing problems. Super helpful, in my opinion.

      3. This is because you used the other hand. The thumb rule works for only one hand – either left or right. If you use the other hand, the REVERSE is true. Hope you get what I’m saying.

        I’m yet to go through this topic so couldn’t really tell you which hand is this particular direction true for. [This is similar to determining the direction of cross product of two vectors]


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