Stereocenters

In Chirality, the carbon of CHBrClF was bonded to four different atoms, and that carbon is why the molecule cannot be laid on its mirror image. Most chiral molecules owe their chirality to one or more carbons like it. This section shows how to find those carbons quickly and reliably.

Finding them is the first practical skill of stereochemistry. It is also where careless mistakes start. A carbon that looks like a stereocenter often is not one, and a carbon that is easy to overlook often is.

The definition

Start by comparing two alcohols. In butan-2-ol, CH₃–CH(OH)–CH₂CH₃, the OH sits on the second carbon, C2. That carbon is bonded to four groups: OH, H, a methyl group (CH₃) and an ethyl group (CH₂CH₃). No two of them are the same.

Propan-2-ol, CH₃–CH(OH)–CH₃, has one carbon fewer. Its C2 is bonded to OH, H and two methyl groups. Two of its four groups are the same.

C2 of butan-2-ol is a stereocenter, and C2 of propan-2-ol is not. A stereocenter is most often an sp³ carbon, one with four single bonds, that is bonded to four different groups. Many textbooks call it a chirality center. Butan-2-ol is chiral, because its mirror image is a different molecule. Propan-2-ol is achiral, because its mirror image is the same molecule.

The four groups matter because of what happens when two of them trade places. Swap the OH and the H on C2 of butan-2-ol, and no amount of turning makes the result match the molecule you started with. You have made a different stereoisomer, a molecule with the same bonds but a different arrangement in space. Here the new stereoisomer is the mirror image of the first.

Swap the OH and the H on C2 of propan-2-ol, and you get the same molecule back. A half turn brings every group back to where it started, because the two methyl groups can stand in for each other.

Different means different all the way out

Look at C3 of 3-methylhexane. It carries an H, a methyl group, an ethyl group (two carbons) and a propyl group (three carbons). The ethyl and the propyl group both begin with a CH₂, so at first they look alike. One carbon farther out, they differ: the ethyl group ends in a CH₃, while the propyl group has a second CH₂.

So C3 is a stereocenter. Two groups count as the same only if they match as whole groups, right to their ends. When two groups begin with the same atom, compare them one atom at a time, moving outward along each, until you find a difference. If you reach the ends of both without finding one, they are the same group.

How to find them fast

The same two passes work on any molecule. The first pass rules out every carbon that cannot qualify:

• any carbon in a double bond, which has only three groups, or in a triple bond, which has two;
• any CH₃, which has three identical hydrogens;
• any CH₂, which has two, whether it sits in a chain or in a ring;
• any carbon that plainly carries two identical groups, such as two methyl groups.

That usually removes most of the molecule. The second pass takes each carbon that is left and compares its four groups, moving outward along any two that begin the same way. A carbon whose four groups all differ is a stereocenter.

Ring carbons: walk both ways

A ring carbon puts two of its four bonds into the ring. Those two bonds look alike, since both are C–C bonds into the same ring. Treat them as two groups and compare them the way you compared the ethyl and propyl groups: walk outward along each, one ring carbon at a time, until the two walks differ.

In 3-methylcyclohexan-1-ol, C1 carries OH, H and two ring bonds. Walk one way round the ring from C1, and the methyl-bearing carbon is the second carbon you reach. Walk the other way, and it is the fourth. The two walks differ, so C1 is a stereocenter.

In 4-methylcyclohexan-1-ol, the methyl-bearing carbon is the third carbon you reach either way. The ring reads the same from either side, so C1 is not a stereocenter.

Now walk from C3 of 3-methylcyclohexan-1-ol. C1, the carbon with the OH, is the second carbon you reach one way and the fourth the other, so C3 is a stereocenter too, and the molecule has two. In 4-methylcyclohexan-1-ol, C4 fails for the same reason as C1.

Stereocenters and other stereogenic units

But-2-ene, CH₃–CH=CH–CH₃, has no stereocenter. Its two middle carbons share a double bond, and its end carbons are CH₃ groups. Yet it has two stereoisomers. In cis-but-2-ene the two methyl groups sit on the same side of the double bond, and in trans-but-2-ene they sit on opposite sides. A double bond cannot turn the way a single bond does, so neither can become the other.

The double bond passes the same test a stereocenter passes. Swap the H and the CH₃ on one end of cis-but-2-ene, and you get trans-but-2-ene, a different compound rather than the same one turned around.

A ring can also lock groups into two arrangements. In 4-methylcyclohexan-1-ol, the compound still has two stereoisomers: one with the OH and the CH₃ on the same face of the ring, and one with the OH and CH₃ on opposite faces. No ring flip moves a group from one face to the other, so the two never interconvert.

Any part of a molecule that gives rise to stereoisomers is a stereogenic unit. A stereocenter is the most common kind, and a double bond like the one in but-2-ene is another.

C1 and C4 of 4-methylcyclohexan-1-ol are stereogenic in this wider sense: swapping the OH and the H on C1 turns cis into trans. Neither is a stereocenter, though, since the two ring bonds on each one lead to the same groups. In this course, stereocenter always means a single tetrahedral atom with four different groups, so when a question asks you to count stereocenters, count only atoms that pass the four-different-groups test. Cis/trans and E/Z covers double bonds and rings.

Counting stereoisomers: the 2ⁿ rule

3-Chlorobutan-2-ol, CH₃–CH(OH)–CHCl–CH₃, has two stereocenters, C2 and C3. At C2 the OH can sit on a wedge or on a hash. At C3 the Cl can too, whichever way the OH points. That makes 2 × 2 = 4 stereoisomers.

Each of the two arrangements at a stereocenter is a configuration. A molecule with n stereocenters therefore has at most 2ⁿ stereoisomers, since each center can take either configuration. Two stereocenters give up to 4 stereoisomers, three give up to 8, and four give up to 16.

The four stereoisomers of 3-chlorobutan-2-ol pair up. Reflecting a drawing through the page turns every wedge into a hash, so each one has exactly one mirror image among the other three, called its enantiomer. The other two are its diastereomers, stereoisomers that are not mirror images of each other. Enantiomers and Diastereomers work through these relationships in full.

The same holds in general. A chiral stereoisomer has exactly one enantiomer, and every other stereoisomer of it is a diastereomer. With three stereocenters and no internal symmetry, each of the eight stereoisomers has one enantiomer and six diastereomers.

The words "at most" matter. Symmetry inside a molecule can make two of the 2ⁿ arrangements the same molecule, and then the real count is lower. Tartaric acid, HOOC–CH(OH)–CH(OH)–COOH, has two stereocenters and so up to four stereoisomers, but only three exist: a pair of enantiomers and one achiral form. An achiral stereoisomer that has stereocenters is a meso compound, the subject of Meso compounds.

The rule counts only the stereoisomers that stereocenters create. Other stereogenic units add more. Each stereogenic double bond can double the count again, and so can a ring like the one in 4-methylcyclohexan-1-ol.

Stereocenters that are not carbon

Any tetrahedral atom with four different groups can be a stereocenter. Phosphorus is one in a phosphine oxide, R₃P=O, when its three R groups differ.

Sulfur is one in a sulfoxide, R–S(=O)–R′, where a lone pair counts as the fourth group. The sulfur is pyramidal, and it turns inside out only very slowly, so its two mirror-image forms can be kept apart. Esomeprazole, a widely prescribed drug that lowers stomach acid, is the single-enantiomer form of the sulfoxide omeprazole.

An amine nitrogen with three different groups looks like the same case. Its lone pair would be the fourth group, and the nitrogen really is pyramidal. But the nitrogen turns inside out like an umbrella in a gust of wind. On the way it passes through a flat transition state, the highest-energy arrangement along the path, where the three groups lie in one plane with the nitrogen and the lone pair sits in a p orbital. It then comes out as the mirror-image pyramid. This flip is called nitrogen inversion.

The energy barrier to the flip is only about 6 kcal/mol. At room temperature each molecule therefore flips roughly 10⁸ to 10⁹ times a second, about once every few nanoseconds. Cooling slows the flip enough to see it by NMR spectroscopy, which NMR Spectroscopy covers. For an ordinary amine, though, no practical temperature slows it enough to separate the two forms. So the nitrogen of an ordinary amine does not count as a stereocenter, and an amine whose only possible stereocenter is its nitrogen behaves as an achiral compound.

One exception is a nitrogen locked into a rigid ring system. In Tröger’s base, a rigid amine built from four fused rings, each nitrogen sits at a bridgehead, where two rings join, so it cannot flatten out and flip. Its two mirror-image forms can be separated.

A quaternary ammonium ion, an N⁺ bonded to four different carbon groups, has no lone pair to move into a p orbital, so it has no flat transition state to pass through. It cannot invert, and its nitrogen is a genuine stereocenter.

What carries forward

Locating stereocenters is the first step in assigning R/S configuration, in sorting out how stereoisomers are related, and in tracking what a reaction does to stereochemistry. From Substitution & Elimination onward, a standard part of predicting a product is asking what happened at each stereocenter: whether it was inverted, kept, destroyed or newly created. You cannot answer that without first knowing where the stereocenters are.