Stereochemistry · Section 36 of 121

Prochirality: which hydrogens are the same?

Practice this — interactive lesson

A CH₂ group has two hydrogens drawn identically, attached to the same carbon, made of the same atoms. It is natural to assume they are interchangeable. Often they are not — and the difference is not a technicality, because it decides how many signals a molecule gives in an NMR spectrum, which hydrogen an enzyme removes, and which face of a carbonyl a nucleophile attacks. This section gives you the one test that settles it and the vocabulary that reports the answer.

The question, and the test that answers it

Two groups on a molecule are equivalent if the molecule has a symmetry operation that exchanges them — a rotation that carries one onto the other and leaves everything else looking the same. That definition is exact and almost useless at the bench, because spotting a rotation axis in a drawing is hard. So chemists use an operational stand-in, and it is the only tool you need here.

The substitution test. Take the two groups in question. Replace the first one with some test group — a deuterium, or an imaginary group written Q — and write down the product. Put the molecule back the way it was, replace the second group with the same test group, and write down that product. Now ask what relationship the two products have. Three answers are possible, and each has a name:

Nothing new is needed to run it. You already decide same-versus-enantiomer-versus-diastereomer from the previous sections, and you already assign R and S. The test just points those skills at a new question. The test group does not have to be deuterium and is not a real reagent you would buy — it is a thought experiment whose only job is to break the tie between two things that currently look alike.

Two identical-looking groups on one atomReplace one with D. Undo it. Replace the other.Now compare the two products.the SAME compoundhomotopicthe CH₂ of propaneone NMR signal,and nothing can tell them apartENANTIOMERSenantiotopicthe CH₂ of ethanolone NMR signal, but a chiralenvironment picks one outDIASTEREOMERSdiastereotopicthe C3 H₂ of 2-bromobutaneTWO NMR signals, and theysplit each otherTwo questions decide it: does a stereocenter appear, and was anything already stereogenic?
The whole section as one procedure. Everything else here — the pro-R and pro-S labels, the Re and Si faces, the NMR consequences — is vocabulary hung on these three branches.The bottom row is what the classification buys you. Homotopic and enantiotopic groups are indistinguishable to an ordinary spectrometer; only diastereotopic ones are guaranteed to show up separately, which is why that branch is the one an NMR problem cares about.
Worked example 1 — the CH₂ of propane: homotopic

Step 1. Propane is CH₃–CH₂–CH₃. Pick C2, which carries two hydrogens plus two methyl groups.

Step 2. Replace the first hydrogen with D. The product is CH₃–CHD–CH₃, 2-deuteriopropane.

Step 3. Go back and replace the other hydrogen with D. The product is CH₃–CHD–CH₃ again.

Step 4. Compare. Is that carbon a stereocenter in either product? Its four groups are CH₃, CH₃, H and D — two of them identical, so no. With no stereocenter anywhere, there is only one 2-deuteriopropane, and the two products are the same bottle of the same substance.

Verdict: homotopic. Propane does have the symmetry to prove it directly — a 180° rotation about the axis through C2 swaps the two methyl groups and swaps the two hydrogens at the same time — but you did not need to find that axis to get the answer.

CH₃CH₃HaHbCpropane — C2 carries two hydrogens and two methylsreplace Hareplace HbCH₃CH₃HDCCH₃CH₃HDC=identicalCH₃–CHD–CH₃CH₃–CHD–CH₃
Why the labeled carbon never becomes a stereocenter: it is left holding two identical methyl groups, so there is only one 2-deuteriopropane and both routes arrive at it. That is what homotopic means.A methyl group behaves the same way for the same reason, which is why a CH₃ in a spectrum is a clean three-hydrogen signal and never argues with itself. Homotopic is the easy branch — the trouble starts when substitution does create a stereocenter.
Worked example 2 — the CH₂ of ethanol: enantiotopic

Step 1. Ethanol is CH₃–CH₂–OH. The carbon of interest is C1, carrying OH, CH₃ and two hydrogens.

Step 2. Replace one hydrogen with D. C1 now holds OH, CH₃, H and D — four different groups, so it is a stereocenter, and the product is one enantiomer of 1-deuterioethanol.

Step 3. Replace the other hydrogen instead. Same four groups at C1, but the arrangement in space is the opposite one, so this is the other enantiomer.

Step 4. Check it with the CIP rules rather than trusting the drawing: OH outranks CH₃, which outranks D, which outranks H by mass number. Run the assignment on each product and one comes out R while the other comes out S. Two compounds, identical connectivity, mirror-image configurations.

Verdict: enantiotopic. Ethanol has no stereocenter at all, and that changed nothing — enantiotopic hydrogens live on perfectly ordinary achiral molecules.

OHCH₃HaHbCethanol — C1 carries OH, CH₃ and two hydrogensreplace Hareplace HbOHCH₃DHCOHCH₃DHCmirror(S)-1-deuterioethanol(R)-1-deuterioethanolmirror images ⇒ the two hydrogens are enantiotopic
Substitution turns C1 into a stereocenter, because OH, CH₃, D and H are four different groups — deuterium outranks ordinary hydrogen on mass number, which is the rule that makes the test work at all. The two products are mirror images, so the hydrogens are enantiotopic.Ethanol itself has no stereocenter and is not chiral, and that is the point worth taking away: a molecule does not need a stereocenter to carry enantiotopic groups. It needs one only for the diastereotopic case.
Worked example 3 — C3 of 2-bromobutane: diastereotopic

Step 1. Take (R)-2-bromobutane, CH₃–CHBr–CH₂–CH₃. C2 is already a stereocenter and stays R throughout; the pair under test is the two hydrogens on C3.

Step 2. Replace one C3 hydrogen with D. C3 now carries H, D, CH₃ and the CHBrCH₃ branch — four different groups, so C3 becomes a stereocenter too. Call the product (2R,3R).

Step 3. Replace the other C3 hydrogen instead. C2 is untouched and still R; C3 has the opposite arrangement, so this product is (2R,3S).

Step 4. Compare (2R,3R) with (2R,3S). One center matches and one is inverted — some but not all — which is the definition of a diastereomer pair, not enantiomers.

Verdict: diastereotopic. Diastereomers are different compounds with different physical properties, and that difference is the whole reason this classification earns a section.

BrHaHbC2C3(R)-2-bromobutane — C2 is already a stereocenterreplace Hareplace HbBrD(2R,3R)BrD(2R,3S)C2 the same in both, C3 opposite⇒ the two products are diastereomers
The stereocenter that was already there is what changes the answer. C2 is untouched by either substitution and stays R; C3 becomes a new stereocenter with opposite configurations in the two products. One center matching and one inverted is the definition of diastereomers, so the C3 hydrogens are diastereotopic.Diastereomers are different compounds — different melting points, different spectra, different reaction rates. That is the whole reason this branch of the flowchart matters, and it is why these two hydrogens can and usually do appear as two separate signals in a proton spectrum.
Run the test on a methyl group and it comes out homotopic every time: the three hydrogens of a freely rotating CH₃ are exchanged by the C3 rotation about the C–C bond, and replacing any one of them gives the identical product. This is why a methyl always integrates as a clean 3H and never splits itself. A CH₂ has no such guarantee, and that asymmetry between CH₃ and CH₂ is what trips people up.

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

A carbon that becomes a stereocenter when you swap one of two identical groups for something new is called a prochiral center. C1 of ethanol is the standard example: it is not a stereocenter, but it is one substitution away from being one. C2 of propane is not prochiral, because the substitution leaves two methyls behind and no stereocenter appears. So "prochiral" and "has enantiotopic groups" describe the same situation from two directions. Read it as a statement about the carbon rather than about the whole molecule, because that is the form that keeps working: a molecule that is already chiral has no enantiotopic groups anywhere, yet its CH₂ carbons can still be prochiral centers whose two hydrogens take pro-R and pro-S labels.

Because the two hydrogens of a prochiral CH₂ are genuinely distinguishable, they can be named individually. The rule is a small extension of CIP: give the hydrogen you are naming a priority just above its twin, leave every other priority alone, and assign R or S to the center as usual. Whichever descriptor comes out is the hydrogen's name — pro-R if the center reads R, pro-S if it reads S. The hydrogens are then written HR and HS.

Worked example — naming the two hydrogens of ethanol

Step 1. Draw C1 with OH pointing up and CH₃ to the lower left, both in the plane of the page, and the two hydrogens to the lower right — one on a bold wedge coming at you, one on a hash going back.

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

Step 3. Priority 4 is on a wedge, pointing at you, so trace 1 → 2 → 3 as drawn and then reverse. As drawn the path runs counterclockwise; the flip makes the center R.

Step 4. So the hashed hydrogen is pro-R and its partner, by elimination, is pro-S. You never have to run the assignment twice — naming one hydrogen names both.

OHCH₃HHCpro-Spro-Rethanol, C1the wedged H comes at you;the hashed H goes backHOW THE LABEL IS ASSIGNED1.Pick one hydrogen — here the hashed one.2.Promote it just above its twin in priority.3.Priorities: OH > CH₃ > this H > the other H.4.The other H is on the wedge, pointing at you,so trace 1→2→3 as drawn, then reverse it.5.It reads counterclockwise, so the center is R:this hydrogen is pro-R, its partner pro-S.
The rule in full: promote the hydrogen you are naming just above its twin, change nothing else, and assign the center the way you always do. The descriptor that comes out is the hydrogen’s name.You never run the assignment twice. Promoting the other hydrogen instead reverses exactly one priority relationship, which reverses the descriptor — so one calculation labels both. Note also that the molecule is achiral: pro-R and pro-S name hydrogens, not compounds.

Most exam questions stop one step short of this and only ask whether a pair is diastereotopic or not, because that is the part NMR needs. Learn the labels anyway: enzymology is written in them, and a question that asks which hydrogen an enzyme removes has no other vocabulary to answer in.

Faces: Re and Si

The same idea applies to a flat trigonal carbon — the carbon of a C=O or a C=C — except that there the two things being compared are not groups but the two faces you could approach it from. A reagent arriving from the top gives one product; the same reagent arriving from the bottom gives another, and the relationship between those two products is exactly the question the substitution test asks.

Faces get their own labels. Look at the trigonal carbon from one side, rank its three attached groups by CIP priority, and trace 1 → 2 → 3. Clockwise means you are looking at the Re face; counterclockwise means the Si face. For acetaldehyde the ranking is O, then CH₃, then H, and the two faces sit on opposite sides of the molecular plane, so labeling one labels the other.

NAMING THE FACE YOU ARE LOOKING ATOCH₃HC1231→2→3 runs clockwise from this side,so you are looking at the Re face.The far side is the Si face.WHY A FLAT CARBON GIVES A MIXTUREthe sp² plane, seen edge-onNu⁻ from the Re faceNu⁻ from the Si face→ one enantiomer→ the mirror imageNothing achiral prefers one side, so bothhappen equally and the product is racemic.Put a stereocenter elsewhere and the two faces stop being equivalent —the two products then come out in unequal amounts.
Left: the face label is just a CIP trace done from one side — clockwise for 1 → 2 → 3 is Re, counterclockwise is Si, and the opposite side automatically gets the other name. Right: what the labels are for.The right-hand panel is also the picture behind racemization at a carbocation. A trigonal carbon is flat, a nucleophile reaches it from either side with equal ease when nothing biases the approach, and a 50:50 mixture is the unavoidable result — not an experimental accident but a consequence of the geometry.

What matters is not the label but the relationship. In an achiral molecule such as acetaldehyde the two faces are enantiotopic: attack on one gives one enantiomer, attack on the other gives its mirror image, and because an achiral reagent in an achiral solvent has no way to prefer either, the two happen at exactly the same rate and the product is a racemate. Hang a stereocenter elsewhere on the molecule and the two faces become diastereotopic: the two transition states are now diastereomeric, they differ in energy, and the two diastereomeric products form in unequal amounts. That is the structural reason a nucleophile adding to a chiral aldehyde does not give 50:50 — a result you will meet as a rule of thumb in Carbonyl Chemistry (a preview).

The same picture explains a result you will see even sooner. A carbocation is trigonal and flat, so a nucleophile can capture it from either face; when the rest of the cation is achiral those faces are enantiotopic, capture is equally likely on both, and the product is a 50:50 mixture. That is what "racemization" means in the SN1 mechanism, taught in Substitution & Elimination (a preview) — the flatness is not an incidental detail of the intermediate, it is the reason for the stereochemical outcome.

Why this matters

Proton NMR — the main reason this section exists. An NMR spectrum reports chemical environments, not atoms, and diastereotopic hydrogens sit in genuinely different environments: no symmetry operation exchanges them, so nothing forces their shifts to match. They therefore give separate signals, and because they are not equivalent they also couple to each other, adding splitting that a naive n + 1 count does not predict. This is why the CH₂ next to a stereocenter so often appears as two multiplets where a beginner expected one clean peak, and why the C3 hydrogens of 2-bromobutane make a mess of an otherwise simple spectrum. You will do this properly in Spectroscopy (a preview); the test you just learned is where the diagnosis comes from.

Enzymes. An enzyme's active site is built from single enantiomers of amino acids, so it is a chiral environment, and a chiral environment can distinguish enantiotopic groups — the two hydrogens present it with two different fits. Alcohol dehydrogenase, which oxidizes ethanol to acetaldehyde, removes the pro-R hydrogen from C1 and leaves the pro-S hydrogen alone. The molecule itself is achiral and the two hydrogens are chemically indistinguishable to any achiral reagent; the enzyme still picks the same one every time.

Rings. A ring does the same job a stereocenter does, because it gives the molecule a top and a bottom. In methylcyclohexane the two hydrogens on C2 are diastereotopic: replacing one puts the label cis to the methyl group and replacing the other puts it trans, and cis and trans isomers are diastereomers. Note that methylcyclohexane contains no stereocenter at all, which is worth holding on to — the ring alone was enough.

METHYLCYCLOHEXANE — NO STEREOCENTERCH₃HaHbC1C2Ha → D gives the D cis to the methylHb → D gives the D trans to itcis and trans are diastereomers,so Ha and Hb are DIASTEREOTOPICA PREVIEW: WHAT A SPECTRUM SHOWSone signalif the two were equivalentHaHbdiastereotopic: two shifts, splitting each otherTwo different environments need not match.A ring gives a molecule a top and a bottom — that is all the test ever needs.
The two claims this section makes that a drawing settles faster than a sentence. Left: methylcyclohexane has no stereocenter, yet the two hydrogens on C2 are diastereotopic, because the ring gives the molecule a face to be on — one substitution puts the label cis to the methyl and the other puts it trans, and those are diastereomers. Right: the consequence, sketched.The right-hand panel is a preview of Spectroscopy and is drawn only to the level this section needs: diastereotopic hydrogens may give two signals rather than one, and because they are inequivalent they also couple to each other. How much they are separated is not predictable from topicity — sometimes the two shifts happen to coincide and the pair looks equivalent. Topicity tells you they are allowed to differ, never by how much.

Mistakes to avoid

"They're on the same carbon, so they're equivalent." Being attached to the same atom is not a symmetry argument and proves nothing. Run the test. A CH₃ passes it by rotation, a CH₂ has to earn it.

Mixing up the two -topic words. Enantiotopic groups behave identically toward everything achiral — same NMR shift, same reaction rate — and differ only in a chiral environment such as an enzyme. Diastereotopic groups are inequivalent to everything, all the time, including a plain NMR spectrometer. If you find yourself predicting two NMR signals for a pair, you had better have concluded diastereotopic, because enantiotopic hydrogens give one signal.

Assuming a stereocenter is required. Enantiotopic groups need no stereocenter anywhere in the molecule; ethanol has none. Diastereotopic groups do need some other stereogenic feature, but a stereocenter is only the most common one — a ring or a C=C creates the same top-and-bottom distinction, which is why methylcyclohexane and the terminal CH₂ of an alkene both qualify.

Reading "prochiral" as "chiral". A prochiral molecule is achiral. The prefix says it stands one substitution away from chirality, not that it has arrived — and a molecule that is already chiral, like 2-bromobutane, can still carry prochiral-type CH₂ groups whose hydrogens get pro-R and pro-S labels.

What carries forward

Topicity is the bridge between the stereochemistry you have just finished and the two places it gets used hardest. It is what makes a proton NMR spectrum readable rather than a puzzle with more signals than you counted, and it is the language every stereoselective reaction is described in — which face was attacked, which hydrogen was removed, which of two diastereomeric transition states won. Keep the substitution test in your hands: replace one, replace the other, compare the products. Every question in this area reduces to that.