Prochirality: which… — Organic Chemistry Notes

Look at the CH₂ group of ethanol, CH₃–CH₂–OH. Its two hydrogens sit on the same carbon and are drawn the same way. It is natural to assume they are interchangeable, but they are not: the liver enzyme that breaks down ethanol removes one of them and never the other.

Whether two look-alike groups are really interchangeable matters all through the course. It decides which hydrogens count as equivalent, which NMR Spectroscopy relies on. It also decides what happens when a reagent attacks a flat C=O group from one side or the other. This page gives you one test that settles the question, and the names for its three possible answers.

The question, and the test that answers it

The test is easiest to see on a case where the answer is obvious, so start with propane.

Worked example 1 — the CH₂ of propane: homotopic

Step 1. Propane is CH₃–CH₂–CH₃. Its middle carbon, C2, carries two methyl groups and two hydrogens. Call the hydrogens Ha and Hb.

Step 2. Replace Ha with deuterium, D. Deuterium is hydrogen with an extra neutron in its nucleus, so it counts as a different group on paper while behaving almost exactly like hydrogen. The product is CH₃–CHD–CH₃.

Step 3. Go back to propane and replace Hb instead. The product is CH₃–CHD–CH₃ again.

Step 4. Compare the two. In each one, C2 carries CH₃, CH₃, H and D. Two of those groups are identical, so C2 is not a stereocenter, and there is only one 2-deuteriopropane. Both routes lead to it.

Verdict: homotopic. Replacing either hydrogen gives the same compound, so nothing can tell Ha and Hb apart.

The half turn swaps the two identical methyl groups and swaps the wedge and the hash, so the second product becomes the first exactly.

That is the substitution test, and it works on any two groups that look alike. Replace the first with a test group and write down the product. Then go back to the starting molecule, replace the second group with the same test group, and write down that product. Finally, decide how the two products are related. There are three possible answers, and each one gives the pair of groups a name:

The same compound → the two groups are homotopic.
Enantiomers → the two groups are enantiotopic.
Diastereomers → the two groups are diastereotopic.

Together, these three labels describe the pair's topicity.

You already have every skill the test needs. The earlier sections of this chapter showed you how to tell enantiomers from diastereomers, and how to assign R/S configuration. Nobody runs this reaction. The substitution is a thought experiment whose only job is to break the tie between two groups that look alike. Deuterium is the usual choice, and any other new achiral group gives the same answer.

Behind the test sits an idea about symmetry. A symmetry operation of a molecule, such as a rotation or a reflection in a mirror plane, carries it onto itself so that it looks the same afterward. If a rotation swaps the two groups, they are homotopic. If only a mirror swaps them, they are enantiotopic. If nothing swaps them, they are diastereotopic. In propane, a half turn about an axis through C2 swaps the two methyl groups and the two hydrogens at the same time. Spotting an axis or a mirror plane in a drawing is hard, which is why chemists run the substitution test instead.

The substitution test as one decision. The bottom row previews what each answer means for a spectrum and for reactions.

Worked example 2 — the CH₂ of ethanol: enantiotopic

Step 1. Ethanol is CH₃–CH₂–OH. The carbon to test is C1, which carries OH, CH₃ and two hydrogens. Draw it with OH up, CH₃ to the lower left, Ha on a wedge (toward you) and Hb on a hash (away from you).

Step 2. Replace Ha with D. C1 now carries OH, CH₃, D and H, four different groups, so it has become a stereocenter. Rank them: OH is 1, CH₃ is 2, D is 3 and H is 4. D outranks H because, when two atoms are the same element, the heavier isotope wins. H is on the hash, pointing away, and 1 → 2 → 3 runs counterclockwise, so this product is (S)-1-deuterioethanol.

Step 3. Start again from ethanol, and this time replace Hb with D. The ranking is the same, but now H is on the wedge, pointing at you. The trace still runs counterclockwise as drawn, so you flip the answer: this product is (R)-1-deuterioethanol.

Step 4. Compare. The two products have the same atoms joined the same way, and one is S while the other is R. They are enantiomers.

Verdict: enantiotopic. Ethanol has no stereocenter, and it did not need one to have enantiotopic hydrogens. Enantiotopic hydrogens sit on perfectly ordinary achiral molecules.

The two products differ only in which bond D and H sit on. Swapping a wedge and a hash is a reflection through the page, so the products are mirror images.

Enantiotopic groups behave identically toward anything achiral. An achiral reagent removes either one at the same rate, and in an ordinary (achiral) solvent the two hydrogens share one NMR signal. This is why, in Radical halogenation (a preview), two hydrogens count as equivalent when replacing either one gives the same compound or its mirror image. Only a chiral environment, such as the inside of an enzyme, can tell enantiotopic groups apart.

Worked example 3 — C3 of 2-bromobutane: diastereotopic

Step 1. Take (R)-2-bromobutane, CH₃–CHBr–CH₂–CH₃. C2 is already a stereocenter. The pair to test is the two hydrogens on C3, Ha on a wedge and Hb on a hash.

Step 2. Replace Ha with D. C3 now carries the C2 branch, the C4 methyl, D and H: four different groups, so C3 becomes a second stereocenter. Rank them at C3. The C2 carbon (Br, C, H) beats the C4 carbon (H, H, H), so C2 is 1 and C4 is 2; D is 3 and H is 4. H points away and the trace runs clockwise, so C3 is R. C2 was not touched, so the product is (2R,3R).

Step 3. Replace Hb instead. The ranking is the same, but H now points at you, so the clockwise trace flips and C3 is S. C2 is still R, so this product is (2R,3S).

Step 4. Compare (2R,3R) with (2R,3S). One center matches and the other is inverted. Stereoisomers that differ at some centers but not all are diastereomers, not enantiomers.

Verdict: diastereotopic. The stereocenter that was already there is what made the difference.

The numbers are the CIP priorities at C3, the new stereocenter. C2 is untouched by either substitution and stays R.

Diastereomers are different compounds, with different melting points, different spectra and different reaction rates. So diastereotopic groups differ toward everything, including an achiral reagent.

A methyl group always comes out homotopic. Replace any one of its three hydrogens and you get CH₂D. Turning the methyl about its C–C bond carries each hydrogen into the next one's place, so all three replacements give the same compound. That is why a CH₃ always gives one signal for all three of its hydrogens. A CH₂ has no such guarantee.

A ring or a double bond can do the same job

Methylcyclohexane has no stereocenter, yet the two hydrogens on C2, the carbon next to the methyl, are diastereotopic. Draw the ring flat, with the methyl on a wedge. Put one C2 hydrogen, Ha, on a wedge, on the same face as the methyl, and the other, Hb, on a hash.

Replace Ha with D and the D ends up cis to the methyl, on the same face. Replace Hb and it ends up trans, on the opposite face. Cis and trans ring isomers are diastereomers, so Ha and Hb are diastereotopic. Each substitution creates two stereocenters at once, C1 and C2, and the ring's two faces are what make the two products differ.

Wedges point toward you and hashes away, so two wedged groups sit on the same face of the flat ring.

A double bond can do the same. In propene, CH₂=CH–CH₃, one hydrogen of the CH₂ is on the same side of the double bond as the methyl group, and the other is on the opposite side. Replace the first with D. D is now the higher-priority group on its carbon and CH₃ is the higher one on the other, and they sit on the same side: that is the Z isomer. Replace the second instead and they sit on opposite sides: the E isomer. E and Z isomers are diastereomers (see Cis/trans and E/Z), so the two =CH₂ hydrogens are diastereotopic. No stereocenter is involved at all.

On each carbon of the double bond the higher-priority group is tagged: D outranks H on the left carbon, and CH₃ outranks H on the right one.

Here is a quick check to try before the full test. If the substituted carbon cannot become a stereocenter and no cis/trans or E/Z pair can form, the groups are homotopic. Otherwise, if one new stereocenter forms and the molecule has nothing else stereogenic, they are enantiotopic. Otherwise (the molecule is already chiral, or the substitution creates two stereocenters at once, or a cis/trans or E/Z pair forms) they are diastereotopic. Apply the three rules in that order. The C4 hydrogens of methylcyclohexane are an example of the last kind: C4 never becomes a stereocenter, yet one substitution gives a cis product and the other a trans one. When in doubt, draw both products.

Prochiral centers, and the pro-R / pro-S labels

C1 of ethanol is not a stereocenter, because it carries two identical hydrogens. Replace either hydrogen with D, though, and it becomes one. A carbon like this, with two identical groups and two different ones, is called a prochiral center: it is one substitution away from being a stereocenter.

C2 of propane is not a prochiral center. It also carries two identical hydrogens, but its two methyl groups are identical too, so replacing a hydrogen still leaves it without four different groups.

In an achiral molecule such as ethanol, the two identical groups on a prochiral center are enantiotopic. A prochiral center can also sit in a chiral molecule. C3 of (R)-2-bromobutane is one, and there the two hydrogens are diastereotopic. (A chiral molecule never has enantiotopic groups, because it has no mirror symmetry that could swap them.) Either way, the two hydrogens can be told apart, so each one can be given a name.

The naming rule adds one step to the CIP rules. To name one hydrogen, pretend it outranks its twin: give it a priority just above the other hydrogen, and leave every other priority as it is. Then assign R or S to the carbon in the usual way. If the carbon comes out R, that hydrogen is pro-R, written H_R; if it comes out S, the hydrogen is pro-S, written H_S. These labels name hydrogens, not compounds; ethanol itself stays achiral.

Worked example — naming the two hydrogens of ethanol

Step 1. Draw C1 as before: OH up, CH₃ to the lower left, one hydrogen on a wedge and one on a hash.

Step 2. Name the hashed hydrogen first. Promote it above its twin, so the priorities are OH (1), CH₃ (2), the hashed H (3) and the wedged H (4).

Step 3. Priority 4 is on the wedge, pointing at you. Trace 1 → 2 → 3 as drawn: it runs counterclockwise. Flip it, and the center reads R.

Step 4. So the hashed hydrogen is pro-R, and the wedged one is pro-S. You do not need a second assignment. Promoting the other hydrogen instead swaps priorities 3 and 4, and swapping two priorities always reverses the descriptor. The lower drawing in the figure runs that check anyway.

The highlighted hydrogen in each drawing is the one being named. Both drawings show the same molecule in the same orientation.

The rule ties straight back to the substitution test. Deuterium outranks hydrogen, so replacing a hydrogen with D promotes it in exactly this way. Replacing the pro-R hydrogen of ethanol with D therefore gives (R)-1-deuterioethanol. Worked example 2 shows it: Hb, on the hash, is the pro-R hydrogen, and replacing it gave the R product.

Many exam questions stop one step short of these labels and ask only whether a pair is diastereotopic, because that is the distinction used most. Learn the labels anyway. Biochemistry is written in them, and a question about which hydrogen an enzyme removes can only be answered in these terms.

Faces: Re and Si

Acetaldehyde, CH₃CHO, has a flat carbonyl carbon bonded to O, CH₃ and H. All four atoms lie in one plane, so a reagent can reach that carbon from in front of the plane or from behind it. Those two sides are the carbon's two faces.

Each face gets a name from the CIP priorities of the three groups. For acetaldehyde, O is 1, CH₃ is 2 and H is 3. Look at the carbon from one face and trace 1 → 2 → 3. If the trace runs clockwise, you are looking at the Re face; if it runs counterclockwise, you are looking at the Si face. A C=C carbon is flat too, and its faces are named the same way. The figure shows why the side opposite a Re face is always a Si face.

Turning the molecule over to look at the back face moves CH₃ and H to opposite sides of the drawing, so the same three priorities trace the other way.

Now let a nucleophile add to acetaldehyde. A nucleophile is an electron-rich reagent that forms a new bond to a carbon; How Reactions Happen, the next chapter, covers nucleophiles in full. If the group it brings is anything other than H or CH₃, the carbonyl carbon ends up with four different groups, a new stereocenter. Attack on the Re face puts the new group on one side of the old plane, and attack on the Si face puts it on the other. The two products are mirror images, so the two faces of acetaldehyde are enantiotopic, just like the two hydrogens of ethanol.

Nu stands for the group the nucleophile brings.

An achiral nucleophile in an achiral solvent has no way to prefer either face. It attacks both at the same rate, and the product is a racemic mixture, a 50:50 mix of the two enantiomers. Carbonyl Chemistry (a preview) shows the same result for additions to aldehydes and ketones.

Put a stereocenter elsewhere in the molecule, and the two faces become diastereotopic. Attack on each face passes through its own transition state, the highest-energy arrangement along the way from starting material to product. Those two transition states are now diastereomers of each other, so they differ in energy. The two products are diastereomers too, and they form in unequal amounts. That is why a nucleophile adding to a chiral aldehyde does not give a 50:50 mixture.

The same picture explains a result you will meet sooner. A carbocation, a carbon with only three bonds and a positive charge, is flat, so a nucleophile can reach it from either face. When the rest of the ion is achiral, those faces are enantiotopic, capture on each is equally likely, and the product is racemic. That is what racemization means in the S_N1 mechanism in Substitution & Elimination (a preview). The flat shape is not a side detail of that mechanism; it is the reason for the outcome.

Why this matters

Spectroscopy. Proton NMR, taught in NMR Spectroscopy, gives one signal per kind of hydrogen: homotopic and enantiotopic hydrogens share a signal, while diastereotopic ones, such as the two on C3 of 2-bromobutane, can give two.

Enzymes. An enzyme is built from single enantiomers of amino acids, so the pocket where it holds its target, the active site, is chiral. A chiral pocket can tell enantiotopic groups apart, because the two hydrogens fit it in two different ways. Alcohol dehydrogenase, the enzyme that oxidizes ethanol to acetaldehyde, removes the pro-R hydrogen from C1 and leaves the pro-S hydrogen alone. To any achiral reagent the two hydrogens are indistinguishable, yet the enzyme picks the same one every time.

Mistakes to avoid

"They're on the same carbon, so they're equivalent." Being attached to the same atom proves nothing. Run the test. A CH₃ always passes it, but a CH₂ has to earn it.

Mixing up the two -topic words. Enantiotopic groups differ only in a chiral environment. Diastereotopic groups differ everywhere.

Assuming a stereocenter is required. Enantiotopic groups need no stereocenter at all, as ethanol shows. Diastereotopic groups need something else stereogenic, but a ring or a C=C can supply it, as methylcyclohexane and propene show.

Reading "prochiral" as "chiral". A prochiral center is not a stereocenter; it is one substitution away from becoming one. Many books also call a whole achiral molecule prochiral when one substitution would make it chiral, so ethanol is a prochiral molecule. Either way, the prefix means "one substitution short": of a stereocenter for a center, or of chirality for a whole molecule.

Common questions

What does prochiral mean? It describes a carbon that becomes a stereocenter after one substitution, because two of its four groups match and the other two differ. C1 of ethanol is the classic case: swap either hydrogen for deuterium and you have a stereocenter.

What are pro-R and pro-S hydrogens? They are names for the two look-alike hydrogens on a prochiral carbon. Rank the hydrogen you are naming just above its twin, assign the carbon as usual, and the descriptor you get becomes that hydrogen's label. The labels name hydrogens, not compounds.

What carries forward

Topicity links the stereochemistry you have just finished to the two places it is used most. It is how NMR spectroscopy decides which hydrogens give separate signals. It is also the language every stereoselective reaction is described in: which face was attacked, which hydrogen was removed, and which of two diastereomeric transition states won. Keep the substitution test ready. Replace one, replace the other, and compare the products; every question in this area comes down to that.