Stereochemistry and Chirality
Enantiomers vs Diastereomers vs The Same? Two Methods For Solving Problems
Last updated: December 6th, 2022 |
Enantiomers vs. Diastereomers vs. The Same
This post on how to solve “Enantiomers vs. Diastereomers vs. the Same?” problems, a classic and important type of exam question.
Table of Contents
- The key distinctions: isomers vs. non-isomers, constitutional isomers vs. stereoisomers, and enantiomers vs. diastereomers.
- How to distinguish isomers from non-isomers
- How to distinguish constitutional isomers from stereoisomers
- Enantiomers, Diastereomers or The Same? Ground rules & easy examples
- Examples with multiple chiral centers: Technique #1 – Using a model kit
- Examples with multiple chiral centers: Technique #2 – The “R/S” Method
- Summary: Enantiomers vs Diastereomers vs The Same
- Quiz Yourself!
Last post we described the most important classes of isomers. See post: Classification of Isomers
The three important distinctions are:
- isomers (same molecular formula) versus non-isomers (different molecular formula)
- constitutional isomers (different connectivity) versus stereoisomers (different arrangement in space)
- enantiomers (stereoisomers that are non-superimposable mirror images) versus diastereomers (stereoisomers that are not non-superimposable mirror images)
Among these distinctions, it’s the third (enantiomers versus diastereomers) that gives students the most headaches, and we’re going to focus on it nearly exclusively.
There’s also an important “non-distinction”:
two molecules that are superimposable on each other, through rotation of bonds or of the whole molecule, are considered to be “identical molecules”.
But before we get there, let’s quickly discuss the first two distinctions.
In theory, identifying isomers is simple. It’s like being able to tell that the words “SLIME” , “MILES”, “SMILE” are made up of the exact same letters but differ in their arrangement.
What makes it trickier in organic chemistry is that you need to be able to interpret line diagrams, and interpret them quickly.
Here’s an exercise: find the three pairs of constitutional isomers below (and one decoy). Shouldn’t take more than a minute.
If this is tough for you, I don’t have any short cuts, but I do have a prescription. Hunt down some exercises that will teach you these skills: look for a chapter in a textbook entitled “alkanes” , usually chapter 3 or so, and start working the exercises.
Going into an exam without being able to quickly and accurately interpret line diagrams will be fatal to your grade.
If two molecules have the same molecular formula, the next question is whether or not they have the same connectivity or not.
- Constitutional isomers have the same molecular formula and different connectivity
- Stereoisomers have the same molecular formula but the same connectivity
How can we tell if molecules have the same connectivity?
So if you can quickly tell if two given molecules have the same IUPAC name (or not), you’re all set for recognizing if they’re constitutional isomers.
Darnit. Does this mean that you have to do a full IUPAC name before you can answer an exam question?
Usually not. If naming molecules takes a long time for you, take heart. A few questions will usually get you at least 80% of the way:
- are the same functional groups present?
- is the length of the main chain the same in both molecules?
- what substituents are present in both molecules? are they identical?
- are the substituents located in the same place relative to the main chain?
If they are still identical at this point, then do a nitpicky double-check (with a pen or pencil, ideally) to make sure the connectivity is identical. If it is, and if the molecules only differ in their R/S, cis/trans, or E/Z designations, then they are stereoisomers.
Here’s a few (simple) examples:
If you found this easy, don’t pat yourself on the back just yet. When drawn in line/wedge form, this question is a little too easy to be exam material. Expect to see some curveballs.
The classic way to make this type of question more challenging is to change the type of depiction, using Fischer, Newman, or a mixture of multiple projections:
When preparing for an exam on stereochemistry, I advise trying to do pretty much every “classify these isomers” question you can until it gets boring.
The Stereochemistry Practice Quizzes have a lot of these kinds of problems (must be an MOC Member for full access). The point is to get used to solving problems for molecules in different renderings.
Now we get to the third, and most challenging case, where we will spend the bulk of our time. “Are these molecules enantiomers, diastereomers, or the same?”.
Let’s cross off three quick wins to start:
First: Superimposable Molecules Are Identical
Molecules that are superimposable are considered to be identical molecules (i.e. “the same”).
Note that “superimposable” includes molecules that are
1. “superimposable through bond-rotations” (conformational isomers), as well as
2.molecules that are superimposable through rotation of the whole molecule.
It also includes some molecules that are drawn as having stereochemistry… but don’t. Look closely ↓
Second: Recall The Key Differences Between Enantiomers and Diastereomers:
Stereoisomers always have the same connectivity. Among molecules with the same connectivity:
- Molecules that are mirror images but non-superimposable are enantiomers.
- If they aren’t superimposable, and they aren’t mirror images, then they’re diastereomers.
Third: cis-trans isomers and double-bond isomers are always diastereomers
Now for the not-so-easy cases.
5. “Enantiomers vs Diastereomers vs The Same” On Molecules With Multiple Chiral Centers: Using A Model Kit
What about molecules with two or more chiral centers? Like these two examples?
Here, I have good news and bad news.
The bad news is that there is no feature on any of these molecules (like different arrangement on a double bond or ring) that allows for quick and easy determination of the relationship in < 5 seconds.
Therefore, the only way to solve these problems is to compare the configurations on each of the chiral centers and see how they relate to each other.
The good news is that these problems can be readily solved using one of two key techniques.
- One method is easy to learn but tends to require a long time to solve problems.
- The second method takes some time to get good at, but allows problems to be solved extremely quickly.
Technique #1 : Use A Model Kit
The traditional method to solve “enantiomers, diastereomers, or the same” is to build models, and see if the two molecules are superimposable or not.
By all means, start with this! Especially in a non-exam situation.
I recommend using this method the first few times you do the exercise because you have to start somewhere, and the model kit will serve as a set of training wheels to help you build confidence.
Models also help you get used to the fact that molecules are 3-dimensional objects. In this respect they are fundamentally no different from cars, cats, or any other commonplace object you care to name.
So let’s go to our “enantiomers, diastereomers, or the same” question, and try to solve each of them using models.
The first set: 2,3-dibromobutanes.
Let’s build the model of the 2,3-dibromobutane isomer using the Fischer projection. (Remember that “the arms come out to hug you”). That gives us the following model:
Next, we build the second model, based on the Newman projection. Here’s what the model would look like:
Finally, once we’ve built these two models, we can then try and move them around to see how they relate to each other.
Here, we take the model on the right hand side, and see that if we perform a rotation about the C2-C3 bond, the two molecules are actually completely superimposable!
Being superimposable, they are therefore are the same!
Let’s look at the second set.
The second set: pentan-2,3,4-triol
When we build the model for the first molecule of the set, we get the following:
When we build the model for the second molecule of the set as a Fischer projection (remembering again that the “arms come out to hug you”) we get this:
Now, let’s try to move around the molecule on the right so that we can compare it to the one on the left.
Again, these two models turn out to be the same!
While there’s nothing wrong with making models, there is a significant drawback.
It takes time to build the darn things, and then, it takes even more time to compare the two models. While this might be OK for molecules with a stereocenter or two, model-building becomes a huge bottleneck as molecular complexity increases.
For instance, say you were asked to compare the following two molecules:
Are you really going to take ~ 5 minutes to make the models, rotate each of them around, and then compare them? Probably not.
Thankfully there is an easier way. And when you get good, you can answer these questions in less than 60 seconds.
Step 1: Start by identifying all of the chiral centers in each molecule, and determining each of their configurations as R or S.
Why is this helpful?
Take the question above. Say that you were given the full names of the two molecules, without their structures.
Could you figure out how these molecules are related based on the names alone?
Because once you know the (R,S) configurations of a molecule, you can easily figure out what the configurations of its enantiomer and its diastereomers should be!
Step 2: Compare The Configurations of The Chiral Centers To Obtain Their Stereochemical Relationship
For two molecules with the same connectivity:
- The enantiomer of a molecule will always have an opposite R/S configuration. So to get the enantiomer of (2R, 3R, 4R)-2,3,4,5-tetrahydroxypentanal all we need to do is flip all the stereocenters: (2S, 3S, 4S)
- Diastereomers arise when at least two molecules share at least one (but not all) chiral center(s) with identical (R/S) configuration. So to find the diastereomers of (2R,3R,4R)-2,3,4,5-tetrahydroxypentanal , all we need to do is keep at least one stereocenter the same, and flip any or all of the rest.
- (obviously, if the two molecules have the same connectivity with all the R/S designations are the same, the two molecules are also the same)
Our molecule (2R, 3R, 4R)-2,3,4,5-tetrahydroxypentanal goes by a more pronounceable name: D-ribose.
- The enantiomer is (2S, 3S, 4S)
- There are six possible diastereomers (see below) which possess at least one (but not all) identical configuration(s) at their chiral centers.
The other bonus of this method is that since it doesn’t matter what conformation (or what projection) a molecule is drawn in. Since all you’re really doing is comparing 1) connectivity and 2) configuration of stereocenters, you don’t need to worry about figuring out if two models are non-superimposable mirror images.
Let’s do a few more examples just based on the name alone.
- (S)-2-butanol and (R)-2-butanol. enantiomers, since they have opposite R,S.
- (S,R)-cyclopentane-1,2-diol and (R,R)-cyclopentane-1,2-diol. diastereomers since they share the same configuration of one of the stereocenters
- To be really ambitious, let’s do: methyl (1R,2R,3S,5S)-3- (benzoyloxy)-8-methyl-8-azabicyclo[3.2.1] octane-2-carboxylate and methyl (1S,2S,3R,5R)-3- (benzoyloxy)-8-methyl-8-azabicyclo[3.2.1] octane-2-carboxylate.
Same IUPAC descriptor, opposite (R,S)-descriptors: enantiomers.
Step 3: Double Check That They’re Not Meso
What about (S,R)-2,3-dibromobutane and (R,S)-2,3-dibromobutane? Opposite R/S, therefore enan… oh, snap. They’re actually the same.
We can’t forget to mention the one little fly that sometimes lands in the ointment: meso compounds.
If a molecule with chiral centers has a plane of symmetry, then it can be written two equivalent ways that have the appearance of being enantiomers, but are in fact the same:
- (S,R)-2,3-dibromobutane and (R,S)-2,3-dibromobutane
- (R,S)-tartaric acid and (S,R)-tartaric acid
- (R,S)-cyclohexane-1,2-diol and (S,R)-cyclohexane-1,2-diol
So before declaring that two molecules with opposite (R,S) designations are enantiomers, double check that the molecule doesn’t have a plane of symmetry.
One way to do this is to try to name the molecule in two directions. If you obtain the same IUPAC name going left-to-right as you do going right-to-left (or clockwise / counterclockwise in the case of cyclic molecules) then you are looking at a meso compound.
The drawback of this final method is that it takes lots of practice to get good at quickly assigning R and S.
To that end, I recommend doing lots of exercises where you work on molecules drawn as line diagrams, Newman projections, Fischer projections, Sawhorse projections, and so on.
That’s probably enough for now. In the next post I’ll provide several examples and applications of using the R/S method to determine if molecules are enantiomers, diastereomers, or the same.
In the meantime, I recommend the following videos, since they apply these principles:
Thanks to Chlöe and the staff at IndustriousHQ for lending their stable hands towards filming the GIFs used in this post.