Butan-2-ol has one stereocenter, the carbon that carries OH, H, CH₃ and CH₂CH₃. Because of that carbon, butan-2-ol exists as two molecules that are mirror images of each other, as Stereocenters showed. A name has to say which of the two you mean, and "the left-handed one" means nothing on paper.
The Cahn–Ingold–Prelog system, or CIP system, solves this. It is named after the three chemists who devised it. It gives each stereocenter a label, R or S, worked out from the structure alone, with no measurement. That label is the stereocenter's configuration: the way its four groups are arranged in space. Assigning it follows a fixed routine.
The four-step method
Start with bromochlorofluoromethane, CHFClBr. Its one carbon carries four different atoms: Br, Cl, F and H. The drawing uses wedges and hashes as in Bonding: a plain line lies in the page, a solid wedge comes toward you, and a hashed wedge goes away from you. Four steps give its label.
- Rank the four groups. Each group gets a priority, a number from 1 (highest) to 4 (lowest), set by the CIP rules in the next section. For CHFClBr the order is Br, Cl, F, H.
- Orient the molecule so that priority 4, usually H, points straight away from you, behind the page. Here H already sits on the hash, so this step is done.
- Trace a path from priority 1 to 2 to 3, around the three groups that face you. Here that path runs Br → Cl → F.
- Read the direction of the path. Clockwise means R, from the Latin rectus, right. Counterclockwise means S, from sinister, left. Br → Cl → F turns clockwise, so this is (R)-bromochlorofluoromethane.
Step 2 matters because R and S are defined as seen from the side opposite priority 4. Drawings that put priority 4 anywhere else are handled in The flip trick, further down this page.
CIP priority rules
Rule 1: the higher atomic number wins. Compare the four atoms bonded directly to the stereocenter. In CHFClBr those are Br (atomic number 35), Cl (17), F (9) and H (1), so the order is Br > Cl > F > H. The atoms you will meet most often rank I > Br > Cl > S > O > N > C > H.
Atomic number decides, not electronegativity. Iodine outranks fluorine, even though fluorine is far more electronegative.
Isotopes are the one place where mass counts. Two atoms of the same element are ranked by mass number, the number of protons plus neutrons. Deuterium (²H, written D) therefore outranks ordinary hydrogen, and ¹³C outranks ¹²C. This matters mostly for molecules labeled with isotopes, and it often appears on exams.
Butan-2-ol shows what happens when two attached atoms are the same element. Its stereocenter carries O, H and two carbons. OH is priority 1 and H is priority 4, but the methyl and the ethyl both attach through carbon, so rule 1 cannot separate them.
So look one atom further out. For each tied carbon, list the three atoms bonded to it other than the stereocenter. The ethyl carbon carries (C, H, H): the next carbon and two hydrogens. The methyl carbon carries (H, H, H). In the first position C beats H, so ethyl is priority 2 and methyl is priority 3.
Rule 2: on a tie, compare the next atoms out. List the three atoms attached to each tied atom as a set, highest atomic number first. Then compare the two sets one position at a time. The first position where the sets disagree, the first point of difference, decides the ranking. Nothing after it counts.
The comparison also stops at the first difference, however heavy an atom further out may be. Compare an isopropyl group, CH(CH₃)₂, with a 2-hydroxyethyl group, CH₂CH₂OH. Their first carbons carry (C, C, H) and (C, H, H). The sets differ in the second position, so isopropyl wins there. The oxygen of CH₂CH₂OH sits one atom further out, and the comparison never reaches it.
If two sets tie in every position, move out one more atom. Go first along the highest-ranked branch of each group, then along the next, and compare those sets in the same way.
C3 carries H, CH₃, CH₂CH₂OH and CH(CH₃)₂. H is priority 4. The other three all attach through carbon, so compare their sets: CH(CH₃)₂ carries (C, C, H), CH₂CH₂OH carries (C, H, H) and CH₃ carries (H, H, H).
CH(CH₃)₂ wins in the second position, so it is priority 1. CH₂CH₂OH beats CH₃ in the first position, so it is 2, and CH₃ is 3. The OH at the end of the chain plays no part.
In the drawing, H is on the hash and 1 → 2 → 3 turns counterclockwise. This is (S)-3,4-dimethylpentan-1-ol.
Glyceraldehyde, HOCH₂–CH(OH)–CHO, brings in double bonds. Its stereocenter carries OH, H, CHO and CH₂OH. The CHO carbon and the CH₂OH carbon each bond to one oxygen, but in CHO that bond is a double bond.
Rule 3: treat a double bond as two single bonds, and a triple bond as three. For a double bond, count the atom at the far end twice: once for real and once as a duplicate atom. A duplicate is a placeholder, written in brackets. It has the atomic number of the atom it copies, but nothing is attached to it. A triple-bonded partner counts three times: once for real and twice as duplicates.
So the CHO carbon carries (O, O, H) and the CH₂OH carbon carries (O, H, H). The sets agree in the first position and differ in the second, where O beats H. CHO outranks CH₂OH.
The same count puts a carboxylic acid carbon, COOH, at (O, O, O), which beats almost any other group attached through carbon. A nitrile carbon, C≡N, counts as (N, N, N).
Both ends of a multiple bond get a duplicate. The oxygen of CHO carries a duplicate carbon in return. That second duplicate only matters when a comparison reaches that far out.
Duplicate atoms also settle a comparison that comes up often: a phenyl group against a vinyl group. A phenyl group is a benzene ring attached through one of its carbons. A vinyl group is CH=CH₂, attached through the CH.
Draw the phenyl ring with alternating double bonds. Its attachment carbon bonds to two ring carbons, and one of those bonds is double, so that ring carbon counts twice: (C, C, C). The vinyl carbon has its CH₂ partner, a duplicate of that partner and one hydrogen, so it carries (C, C, H). The two sets agree in the first two positions, and phenyl wins in the third.
A tert-butyl group also carries (C, C, C), so it outranks vinyl too. An isopropyl group carries (C, C, H), the same set as vinyl, and that tie breaks one atom further out. There the vinyl CH₂ carries (C, H, H), where the C is the duplicate of the first carbon, against a methyl's (H, H, H), so vinyl wins. Cis/trans and E/Z works through this comparison with a figure.
A ring raises a different problem. Walk outward along a ring and you come back to an atom you have already passed. When that happens, stop and write a duplicate of that atom. Like the duplicates for double bonds, it has nothing attached, so the walk ends there.
Compare a cyclopropyl group, a three-membered ring, with a pentan-3-yl group, CH(CH₂CH₃)₂. Both attachment carbons carry (C, C, H). At the next atom out, each group has two CH₂ carbons, and all four carry (C, H, H), so the groups still tie. One atom further, the chain reaches its CH₃ carbons, which carry (H, H, H). The ring instead reaches its other CH₂, whose other carbon neighbor is the attachment carbon you started from. That carbon goes in as a duplicate, so the set is (C, H, H), and cyclopropyl wins.
The flip trick
OH is 1, ethyl is 2, methyl is 3 and H is 4, as ranked above. The drawing puts OH at the top, ethyl lower left, methyl lower right and H on a hash.
OH → ethyl → methyl turns counterclockwise. H already points away, so no flip is needed: this is (S)-butan-2-ol.
OH is 1, CHO is 2 and CH₂OH is 3, from the duplicate-atom comparison above. H is 4. The drawing uses the same pose as butan-2-ol: OH at the top, CHO lower left and CH₂OH lower right.
OH → CHO → CH₂OH again turns counterclockwise. This time, though, H sits on a wedge and points toward you. Reverse the answer: this is (R)-glyceraldehyde.
Those two drawings trace the same direction on the page and have opposite configurations. The only difference is which bond the hydrogen sits on.
That is the flip trick. When priority 4 points toward you, trace 1 → 2 → 3 exactly as drawn, then reverse the answer: R becomes S, and S becomes R. There is no need to redraw the molecule.
It works because a wedge to priority 4 means you are viewing the center from the wrong side. Viewing it from the other side swaps left and right, so every turn you see runs the other way.
Some drawings put priority 4 on a plain bond, in the plane of the page. Take the drawing of 2-bromobutane below. Br is 1, ethyl is 2, CH₃ is 3 and H is 4, and H is drawn in the page, so you can neither read the turn directly nor use the flip trick.
Swap priority 4 with the group that points away. Here that means exchanging H and the hashed CH₃. The swapped drawing has H on the hash, and Br → ethyl → CH₃ turns clockwise, so the swapped drawing is R. Reverse that for the original: the drawing shows (S)-2-bromobutane.
The reversal is needed because swapping any two groups on a stereocenter turns it into its mirror image, as Stereocenters showed. The swapped drawing therefore has the opposite label, and reversing it recovers the original. The swap rules for Fischer projections, later in this chapter, rest on the same fact.
R/S is not (+)/(−), and R/S is not D/L
Two other labels also describe chiral molecules, and they answer different questions. R/S comes from the structure, by the CIP rules. (+)/(−) records which way a sample rotates plane-polarized light, a measurement that Enantiomers describes. No rule reads it off a drawing.
D/L is an older system, still standard for sugars and amino acids. It compares a molecule with glyceraldehyde. You draw the molecule in a fixed pose, a Fischer projection (taught later in this chapter) with its carbon chain vertical and its most oxidized carbon at the top. Then you check which side one key group sits on: in a sugar, the OH on the stereocenter farthest from the top; in an amino acid, the NH₂. On the right, as in D-glyceraldehyde, the compound is D. On the left, it is L.
The three systems do not line up. (S)-glyceraldehyde and (S)-alanine rotate light in opposite directions, though both are S. Most L-amino acids are S, but L-cysteine is R. Its groups sit in the same places as in the other L-amino acids; only the ranking changes. Its CH₂SH carbon carries (S, H, H) and the carboxyl carbon carries (O, O, O). S beats O in the first position, so CH₂SH outranks COOH. One label never tells you the others.
Naming whole molecules
3-Bromobutan-2-ol has two stereocenters, C2 and C3. Each one is ranked and read on its own, and each gets its own label with its locant in front. The drawing below is (2R,3S)-3-bromobutan-2-ol. The rankings are C2: OH > CHBrCH₃ > CH₃ > H, and C3: Br > CH(OH)CH₃ > CH₃ > H. At C2, H is on the hash and the turn is clockwise, so C2 is R. At C3, H is on the wedge, so the clockwise answer flips: C3 is S.
The mirror image of (2R,3S)-3-bromobutan-2-ol is (2S,3R), with every label reversed. The other two stereoisomers, (2R,3R) and (2S,3S), are not its mirror images. They are its diastereomers: stereoisomers that are not mirror images of each other. Diastereomers covers them. Once each center has a label, sorting stereoisomers into these relationships is bookkeeping.
What carries forward
R/S labels are how the rest of the course reports what a reaction does to a stereocenter. In Substitution & Elimination they show that SN2 inverts a stereocenter and that SN1 gives a racemic mixture, an equal mix of both mirror images. They also record whether an addition put its two new groups on the same face of an alkene (syn) or on opposite faces (anti). The CIP ranking itself comes back unchanged in Cis/trans and E/Z, where it labels the geometry of double bonds.