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The shape of molecules is super important to life as we know it. In this episode of Crash Course Organic Chemistry we’re learning about stereochemistry and how to identify molecules as chiral or achiral. And as always, we’ll be doing a lot of practice!

Episode Sources:
“THINK BIG! Must the molecules of life always be Left-Handed or Right-Handed?” Smithsonian Magazine.
Spinoff 2004 - “A NATURAL WAY TO STAY SWEET”, NASA.

Series Sources:
Brown, W. H., Iverson, B. L., Ansyln, E. V., Foote, C., Organic Chemistry; 8th ed.; Cengage Learning, Boston, 2018.
Bruice, P. Y., Organic Chemistry, 7th ed.; Pearson Education, Inc., United States, 2014.
Clayden, J., Greeves, N., Warren., S., Organic Chemistry, 2nd ed.; Oxford University Press, New York, 2012.
Jones Jr., M.; Fleming, S. A., Organic Chemistry, 5th ed.; W. W. Norton & Company, New York, 2014.
Klein., D., Organic Chemistry; 1st ed.; John Wiley & Sons, United States, 2012.
Louden M., Organic Chemistry; 5th ed.; Roberts and Company Publishers, Colorado, 2009.
McMurry, J., Organic Chemistry, 9th ed.; Cengage Learning, Boston, 2016.
Smith, J. G., Organic chemistry; 6th ed.; McGraw-Hill Education, New York, 2020.
Wade., L. G., Organic Chemistry; 8th ed.; Pearson Education, Inc., United States, 2013.

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Hi! I’m Deboki Chakravarti and welcome back to Crash Course Organic Chemistry!

In 1990, while trying to detect life on Mars, the American Engineer Dr. Gilbert Levin wondered if Martian life would eat the same food as life on Earth. So in the samples he sent to Mars on the Viking lander, he included some D-glucose, a “right-handed” sugar that most life on Earth uses for energy.

But he also sent some of the “left-handed” version of that compound, L-glucose, in case Martians preferred the opposite food. By the way, because the IUPAC nomenclature for right-handed and left-handed glucose is really a mouthful, we’re going to just use their common names here! When Dr.

Levin was getting his samples ready, he found that the taste of L-glucose was the same as D-glucose. And he found that humans can’t digest L-glucose because our enzymes don’t recognize it. If we can’t digest L-glucose, that means it won’t be converted into energy or stored as fat.

So, theoretically, we could eat as much of this sweet molecule as we wanted. Dr. Levin thought he stumbled upon a sweet surprise that was better than any sugar substitute that already existed, with a less artificial taste.

But it turns out that it’s just too dang expensive to make L-glucose, so this sugar substitute unfortunately never made it to market. This story reinforces what we’ve been saying about how structure affects what chemicals can do: the right-handed version of glucose provides energy while the left-handed version gets ignored by our bodies. It’s all because of tiny geometric differences, and specifically stereochemistry. [Theme Music].

Biology, chemistry, and biochemistry have all established that the shape of molecules is important to life as we know it. And in this series, we’ve learned the term isomer to describe molecules that have the same parts as each other with small differences. Two constitutional isomers have the same molecular formula, but the atoms are connected in different ways.

And the geometric isomers we’ve talked about are in molecules with double bonds, because the attached groups can be arranged differently around the bond. If they’re on the same side, we call it the cis isomer. Or if they’re on the opposite sides, it’s the trans isomer.

A geometric isomer is just one kind of stereoisomer. Stereoisomers have the same atoms connected in the same bond, but they differ in the spatial relationships between those atoms. Something that isn’t a stereoisomer is two identical pairs of sunglasses.

I can set them across from each other so they’re mirror images. Then, I can pick up one pair and turn it around so they would melt into each other, if they could. That means they’re superimposable and have the same spatial relationship.

If I took out one lens on each pair, they’re still mirror images. But no matter how I flip them around, the dark lenses won’t line up, so they’re non-superimposable mirror images. You don’t even need cheap sunglasses to see this!

Let’s take a look at our hands. If my palms are facing each other, I can line them up so the thumb and pinky meet, so they’re mirror images. But if I place one on top of the other, they won’t perfectly line up, so they’re non-superimposable.

In chemistry, we say that anything with a non-superimposable mirror image has chirality. The simplest example of chirality is a carbon with four different groups attached to it. This kind of central carbon goes by many names, but we’ll call it the chiral center or chirality center.

We can check whether this whole molecule is chiral by building it and doing the same test we did with our hands or those half-sunglasses. First, we’ll draw the molecule’s mirror image. We can see that the reflection can’t be superimposed on the original molecule.

No matter how you turn them, the four different groups will never line up. These two molecules are a pair of non-superimposable mirror images. And because that phrase is kind of a mouthful, we call a pair of these chiral molecules enantiomers.

Enantio- is from the Greek for opposite, because they’re opposite images. By now, you’ve probably noticed that we do a lot of drawing in organic chemistry. So, surprise!

We’ll need to draw pairs of enantiomers too. For example, in the molecule butan-2-ol, carbon 2 is a chiral center. It has a methyl group, an ethyl group, an alcohol group, and a hydrogen atom.

Four different groups attached to one carbon, like our model from a second ago. To draw the enantiomer, we have to start by drawing a line to represent our mirror! Next, we’ll draw the reflection of the group that’s closest to the mirror.

In this case, the hydrogen atom. Then, we’ll continue building the molecule and draw the methyl group, the alcohol group, and the ethyl group. When all of the atoms are there, we’re done!

Another way to draw a mirror image is to place the mirror behind the molecule, so the solid wedges become dashed wedges, and the dashes become solid. Remember that solid wedges mean things are coming out towards us and dashes are pointing away from us. To practice this kind of drawing, let’s look at a common asthma medication, albuterol.

Even though it looks like a pretty complicated molecule, albuterol only has one carbon surrounded by four different groups. We’ll highlight this chiral center with a red dot just so we’re on the same page! So if we place the mirror behind albuterol and draw its enantiomer, we only need to make one change: the solid wedge connecting to the alcohol group needs to become dashed.

Now, when we draw enantiomers, it’s also important that we give each structure a name that communicates which one it is. After all, one enantiomer of albuterol is an asthma medication, but the other might have a different effect, or do nothing at all. To name enantiomers, we adopt the Cahn-Ingold-Prelog convention, where the chiral centers of a molecule are labeled R for right-handed and S for left-handed.

Like all puzzles in organic chemistry, we have rules we can follow and a pattern we can learn. This convention says that the first step is assigning priority to the four groups around a chiral carbon. And, just to make it fun... enantiomer priority is slightly different than functional group or substituent priority when we name molecules.

To figure out priority for enantiomers, we have to pay attention to the atomic number of different atoms. For example, let’s go back to butan-2-ol. The chiral center has a methyl group, an ethyl group, an alcohol group, and a hydrogen atom attached to it.

The lowest priority group has the lowest atomic number. It’s almost always a hydrogen atom, because their atomic number is 1. So here, the hydrogen atom has lowest priority, and we’ll label that as 4.

When assigning the chiral center, we have to make sure the lowest priority group is pointing away from us because that’s the rule for consistency. Otherwise the naming will get messed up. The hydrogen is pointing away from us, because there’s a dashed wedge.

Easy! Next, the highest priority group has the element with the highest atomic number. In this case, the alcohol group has highest priority, because oxygen’s atomic number is 8.

So let’s label that as 1. Then, we have to prioritize the middle groups: the methyl and the ethyl. To do this, we’ll look at the atoms attached to the first carbons away from the chiral center.

The methyl carbon has three hydrogens attached to it, so the highest atomic number is 1. The ethyl carbon has two hydrogens and another methyl group attached, so it has a carbon, which has an atomic number of 6. So congratulations ethyl carbon, you’re higher priority!

When it comes to priority, we’ll give the methyl group a 3 and the ethyl group a 2. For both enantiomers, the priority numbering will be the same. From highest to lowest, it’s: the alcohol group, the ethyl group, the methyl group, and the hydrogen atom.

Now we have to decide which one is R, or right-handed, and which one is S, or left-handed. To do that, let’s draw an arrow around both molecules from highest priority, 1, to lowest priority, or 4. On one enantiomer, the arrow points in a clockwise arc, which means it’s the R-enantiomer.

So the molecule is (2R)-butan-2-ol. On the other, we have a counterclockwise arc. So it’s the S-enantiomer and we call it (2S)-butan-2-ol.

If you have a 3D molecule set on hand, you can figure out enantiomer names in 3D if you want! Just hold onto the lowest priority group, in this case, the hydrogen, like you’re admiring a sunflower and the hydrogen is the stem. Then, you can twist the molecule from highest to lowest priority.

If you’re twisting it right, it’s the R-enantiomer, and if you’re twisting it left, it’s the S-enantiomer. Like all things orgo, the only way to get better at assigning chirality is practice, so let’s assign R and S to the albuterol enantiomers too. One group bonded to the chiral center is an alcohol group with a solid wedge, so it’s coming out toward us.

The other two groups around the chiral center are in the same plane. So the fourth group is an implied hydrogen atom pointing away from us. If it helps, we can always draw that hydrogen ourselves with a dashed wedge.

Now that we have our four groups, we can assign priority to them. The hydrogen is the lowest priority again, and it’s pointing away from us, so we’re good. We can label it with a 4.

And, again, we have an alcohol group and oxygen’s atomic number is 8. So that’s the highest priority and we can label it with a 1. To assign the middle priorities, we’ll look one step away from the chiral center.

The carbon to the left of the chiral center is bonded to two carbons. And the carbon to the right of the chiral center is bonded to two hydrogens and a nitrogen. Because nitrogen has an atomic number of 7 compared to carbon’s 6, the right group wins priority.

Sorry, carbon. So we’ll label the winner 2 and the loser 3. If we draw the arc again, or imagine spinning a 3D molecule, we can see that it goes clockwise.

So this molecule is the R-enantiomer or (R) albuterol. So the other one is (S)-albuterol. But if we label its groups with priority and draw the arc, it’s not counterclockwise like the S-enantiomer should be.

You know how I keep saying that we want the lowest priority group pointing away from us? It’s for this exact situation. Let’s draw in the fourth group again, which is the hydrogen with a solid wedge.

The hydrogen atom is coming out of the page at us, so we need to redraw the molecule with the hydrogen pointing away from us to make our enantiomer-naming rules work. Imagine grabbing that hydrogen bond like a flower stem! The arc goes counterclockwise, and the naming makes sense!

Sometimes, we’re faced with an even more brain-bending molecule where the lowest-priority group is in the same plane as the chiral center. So a quick trick is to invert the hydrogen with whatever is pointing back away from us and label it INVERTED to remember we did this flip. Next, we can use our toolkit to assign R or S to the chiral center of the inverted structure.

In this case, it’s an R. But then we need to switch the inverted structure back. So the original chiral center is an S.

So, I don’t know, but maybe you’re thinking, “Deboki. This is too easy. Please, show me something a little MORE complicated.” Well, let me introduce you to cyclic compounds, and one more thing to think about when it comes to chirality.

For example, methylcyclopentane has superimposable mirror images, so we say that it’s achiral. Even though one has a methyl group pointing away from us and one has a methyl group pointing toward us, we can overlap the mirror images and get the same arrangement of atoms. This dotted blue line is like a mirror within the molecule, which we call an internal plane of symmetry.

Any molecule with an internal plane of symmetry is achiral and doesn’t have enantiomers. But a molecule like 2-methylcyclopent-1-ene doesn’t have an internal plane of symmetry, so it is chiral. The mirror image is different and we can’t overlap them to get the same arrangements of bonds and atoms.

To name these enantiomers, we just need to assign priority around the chiral center again. We’re basically pros at it now, so we got this! The double-bonded carbon in the ring is highest priority because it’s like being bonded to two carbons.

The CH2 in the ring is next, because it’s bonded to two hydrogens and one carbon. The methyl group is third because it’s bonded to three low priority hydrogens. And the invisible hydrogen is last, like always.

That’s just the way it is, hydrogen. Using these priorities, we can name (3R)-3-methylcyclopent-1-ene and (3S)-3-methylcyclopent-1-ene. Okay, so up until this point, we’ve been dealing with molecules with one chiral center.

But there are some molecules with two chiral centers that have enantiomers too. For example, look at (3R,4S)-4 bromohexan-3-ol. We use numbered carbons to specify which chiral center is R and which chiral center is S, after doing the whole assigning-priority-to-groups thing.

But some molecules with two chiral centers are actually achiral, which we can tell with two checks: first, is there a superimposable mirror image, and second, is there an internal plane of symmetry. Let’s consider cis- and trans-1,2 dibromocyclohexane. Both molecules have two chiral centers, which we’ve marked with red dots!

Just look at them. Trans-1,2-dibromocyclohexane is chiral, with non-superimposable mirror images. But cis-1,2-dibromocyclohexane is achiral.

And since we’ve covered so much so quickly, here’s a handy flow chart to help you with your chirality gut checks. By following these steps, it can help you tell whether a molecule has enantiomers or not. Okay.

Time for some rapid fire problems! Using that handy dandy flow chart if you want, decide if the following examples are chiral or achiral. We’re going to put four examples on screen and then tell you the answers, and you can pause this if you want to take a second to solve them yourself.

Ready? The first problem is… chiral. See, the red-marked carbon has four different groups.

The second problem is… achiral. It has chiral centers, but it also has an internal plane of symmetry. The third problem is… chiral.

It has two chiral centers but no internal plane of symmetry. And the last problem is… also chiral. It just has one chiral center, which we marked with a red dot.

Stereochemistry is a really important part of our organic chemistry toolbox. And even just 8 episodes in, we’ve got a lot of tools for understanding all the exciting and weird and fascinating reactions to come. In this episode we learned about:.

How to name chiral centers R or S,. And how to determine if a molecule is chiral or achiral. Next time, we’ll look more closely at the properties of enantiomers, how they interact with light, and how to separate them so that drugs with Dr.

Jekyll and Mr. Hyde personalities don’t harm us! Thanks for watching this episode of Crash Course Organic Chemistry.

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