R and S Configuration Explorer

What is R and S configuration?

R and S name the two mirror-image ways four different groups can be arranged around a stereocentre, which is usually a carbon atom. Rank the groups 1 to 4 by the Cahn-Ingold-Prelog rules, turn the molecule so that group 4 points away from you, and follow 1 to 2 to 3: clockwise = R and anticlockwise = S. R comes from the Latin rectus, right, and S from sinister, left. Written with vectors, if r₁, r₂ and r₃ are the bond directions to groups 1, 2 and 3, the centre is R when (r₁ × r₂) · r₃ < 0 and S when that product is positive, and that is how the explorer decides it.

The two arrangements are enantiomers: the same atoms joined in the same order, which cannot be laid on top of each other however they are turned, like a left hand and a right hand. A carbon carrying four different groups, called a chiral centre or stereocentre, is the commonest source of this handedness. Its four bonds point to the corners of a tetrahedron, 109.5° apart, the shape the VSEPR Molecular Geometry visualiser builds for four electron domains.

Ranking the groups: the Cahn-Ingold-Prelog rules

The ranking comes from the sequence rules of Robert Cahn, Christopher Ingold and Vladimir Prelog, first published in 1956, set out in full in 1966, revised by Prelog and Günther Helmchen in 1982 and restated as rule P-92 of the IUPAC 2013 recommendations. For the molecules most courses use, one rule does all the work: the higher atomic number ranks higher.

• Start at the atoms bonded to the stereocentre. Compare their atomic numbers, which the periodic table gives: bromine 35 beats chlorine 17, which beats oxygen 8, nitrogen 7, carbon 6 and hydrogen 1.
• Break a tie by moving out one bond. For each tied atom, list the three atoms it carries, highest first, as a set such as (O, H, H), and compare the sets atom by atom.
• Stop at the first point of difference. The first atom that differs decides, and nothing further out counts. Only if the sets match do you move out another bond, following the higher-ranked branch first.

The first-point rule is the one that catches people out. CH₂OH carries (O, H, H) and C(CH₃)₃ carries (C, C, C), so CH₂OH ranks higher: its single oxygen beats the first carbon, and the two hydrogens after it never come into play. Adding the atomic numbers instead, 8 + 1 + 1 = 10 against 6 + 6 + 6 = 18, gives the opposite answer, and it is wrong.

Some ties last longer. The stereocentre of 3-methylhexane carries H, CH₃, an ethyl group and a propyl group. The three carbons tie at the first sphere. At the second, CH₃ carries (H, H, H) and drops to 3, but ethyl and propyl both carry (C, H, H). At the third sphere the CH₃ at the end of ethyl carries (H, H, H) while the next CH₂ of propyl carries (C, H, H), so propyl ranks 1 and ethyl 2. Load 3-methylhexane to see the step list need all three spheres.

Double bonds, duplicate atoms and phantom atoms

A double bond is counted as two single bonds. Each end of it is given a duplicate of the atom at the other end, so the carbon of a C=O carries the oxygen twice, and a triple bond gives each end two duplicates. That is why COOH carries (O, O, O): its carbonyl oxygen counts twice and its OH oxygen once. CHO carries (O, O, H) and CH₂OH carries (O, H, H), so the three rank COOH, then CHO, then CH₂OH, which is what settles glyceraldehyde.

A duplicate atom carries nothing further: its empty places are filled with phantom atoms of atomic number zero, and so are the places that lone pairs take on an oxygen or a nitrogen. The explorer’s step list writes a phantom atom as 0. Phenyl and ethynyl both carry (C, C, C) at their first carbon. One bond further out, phenyl’s two ring carbons carry (C, C, H) each, while ethynyl has one (C, C, H) and two duplicates carrying (0, 0, 0), so phenyl ranks higher. Neither group is bigger in any sense that matters; the rule simply reads the next sets.

Using the explorer

The explorer opens on (S)-lactic acid and plays the assignment once: the molecule turns until the hydrogen, priority 4, is at the back, and an arrow traces 1 to 2 to 3. With reduced motion set, it opens on that finished view instead. Press Play to run the assignment again from any angle, drag the molecule or use the arrow keys to turn it yourself, and press Fischer view to see it as a Fischer projection.

• Molecule loads a real molecule in its natural form: lactic acid, alanine, glyceraldehyde, butan-2-ol, bromochlorofluoromethane, cysteine, serine, phenylalanine, 1-phenylethanol or 3-methylhexane.
• Top, right, bottom and left group place the four groups by their positions in the Fischer projection, so you can copy a projection from a worksheet. There are 28 groups to choose from, from H to iodine. Give two places the same group to see a carbon that is not a stereocentre.
• Swap two groups exchanges any pair, which always flips R and S.
• Ranking, step by step lists every comparison: the atomic numbers at sphere 1, what each tied group carries further out, and the first difference that settles it.

Each group’s first atom is coloured by element and numbered by priority. Bonds are drawn the way you would draw them on paper from the current angle: a solid wedge toward you, a hashed wedge away from you and a plain line across. The caption under the molecule says how to read the arrow from where it is turned, directly when 4 is at the back and reversed when 4 points at you, and asks you to turn it when 4 is side-on. The readouts give the configuration, the molecule’s full name, the four priorities, the sphere at which the last tie was settled, and the triple product.

Worked example: (S)-lactic acid

The explorer opens with COOH at the top of the Fischer projection, H on the right, CH₃ at the bottom and OH on the left.

• Atoms on the stereocentre: O (8) > C (6) = C (6) > H (1). OH ranks 1 and H ranks 4, and COOH and CH₃ tie on carbon.
• One bond further out, COOH carries (O, O, O), counting the C=O oxygen twice, and CH₃ carries (H, H, H). O beats H at the first atom, so COOH ranks 2 and CH₃ ranks 3.
• Turned with H at the back, OH sits at 3 o’clock, COOH at 11 o’clock and CH₃ at 7 o’clock, with H peeping out behind the carbon at 5, opposite COOH, where the arrow never goes. From 3 up to 11 and round to 7 is anticlockwise, so the configuration is S.
• From the Fischer projection instead: OH at 9 o’clock, COOH at 12 and CH₃ at 6 are drawn clockwise, but H is on a horizontal bond, pointing at you, so the answer is the reverse, S again.
• As vectors, r₁ = (−0.816, 0, 0.577) points to OH, r₂ = (0, 0.816, −0.577) to COOH and r₃ = (0, −0.816, −0.577) to CH₃. Then r₁ × r₂ = (−0.471, −0.471, −0.667) and (r₁ × r₂) · r₃ = +0.770, which is positive, so S.

This is (S)-lactic acid, the L-(+) form made in muscle, and the readout names it. Swap OH and H and every ranking step stays the same; only the reading changes, because the arrow now runs clockwise, and the molecule becomes (R)-lactic acid. The triple product is the same size, 0.770, at every arrangement of a perfect tetrahedron, so its sign is all that carries information.

Reading R and S from a Fischer projection

Emil Fischer’s projection, from 1891, draws a stereocentre as a cross: the horizontal bonds come toward you and the vertical bonds go away. Press Fischer view and the explorer turns to exactly that view, with the horizontal bonds drawn as wedges and the vertical ones as dashes.

• Lowest priority on a vertical bond: it already points away, so read 1 to 2 to 3 as drawn.
• Lowest priority on a horizontal bond: it points at you, so read the turn and reverse it.
• Moving the drawing: turning a Fischer projection through 180° in the plane of the page keeps its configuration, while turning it through 90° gives the enantiomer, because the horizontal and vertical bonds trade places.

D-glyceraldehyde shows the reversal. CHO is at the top, OH on the right, CH₂OH at the bottom and H on the left. OH at 3 o’clock, CHO at 12 and CH₂OH at 6 are drawn anticlockwise, but H is on a horizontal bond, so the answer is clockwise: R. Load glyceraldehyde, press Fischer view, and the caption makes the same reversal in words.

Why swapping two groups flips R and S

Exchanging any two groups turns a stereocentre into its mirror image. In the vector rule a swap either exchanges two of the vectors in the triple product or puts the bond to group 4 in place of one of them, and either way the sign changes. So one swap turns R into S, a second swap puts it back, and rotating three groups round, which is two swaps, leaves it unchanged.

That gives a shortcut when the lowest priority sits in an awkward place on paper: swap it with whichever group points away, read the configuration, and then reverse your answer, because you made one swap. Turning the whole molecule never changes R or S, since no bond is broken, which is why the explorer gives one answer at every angle. A ring in the Cyclohexane Chair Conformation Explorer makes the same point: flipping one chair into the other moves groups between axial and equatorial positions without changing the configuration at any carbon.

R and S, D and L, and the sign of rotation

R and S describe the arrangement. Whether a sample turns plane-polarised light clockwise, written (+), or anticlockwise, written (−), is measured, and it does not follow from the letter: (R)-glyceraldehyde is (+), while (R)-butan-2-ol is (−). The older D and L labels compare a molecule with glyceraldehyde in a Fischer projection, and they do not map onto R and S either. They began with a guess: Fischer could not tell which arrangement (+)-glucose had, so he chose one, and it was not until 1951 that Johannes Bijvoet’s X-ray study of sodium rubidium tartrate showed that the guess had been right.

The amino acids show this. The chiral amino acids in proteins are all L, and nearly all of them are S, as alanine, serine and phenylalanine are in the explorer. Cysteine is the familiar exception. Its CH₂SH carries sulfur, atomic number 16, which outranks the oxygens of COOH at 8, so CH₂SH ranks 2 and COOH 3, and L-cysteine is R. Nothing about its shape differs from L-alanine’s; only the ranking has changed.

Enantiomers are hard to tell apart in most experiments. They have the same melting point, the same solubility and the same NMR spectrum in an ordinary solvent, so the NMR Splitting Pattern Visualiser would draw identical multiplets for both. A stereocentre does leave a mark nearby, though: the two hydrogens of a neighbouring CH₂ are diastereotopic, and they can have different chemical shifts.

What this model leaves out

• More than one stereocentre. There is a single centre here, so there are no diastereomers and no meso compounds such as meso-tartaric acid, and none of the rules that rank groups by their own configuration.
• The later sequence rules. Only atomic number is used. Isotopes, ranked by mass number so that deuterium beats hydrogen, the cis and trans geometry of double bonds, and stereocentres inside the groups all need rules beyond it.
• Other kinds of stereocentre. Only carbon is offered. A sulfoxide, a phosphine or a quaternary ammonium ion can be chiral too, and allenes and biaryls are chiral without any stereocentre at all.
• Rings, apart from phenyl. Phenyl is written as one Kekulé structure, which is enough for every comparison here. Stereocentres in rings, such as those of cyclohexanes or of sugars in their ring forms, are not offered.
• Real geometry. The tetrahedron is regular, every bond is the same length, and each group is drawn as the one atom bonded to the centre.

Common mistakes

• Adding atomic numbers. Compare sets atom by atom from the highest. The (O, H, H) of CH₂OH beats the (C, C, C) of C(CH₃)₃ at the first atom, although its total is lower.
• Ranking by size or mass. A tert-butyl group is large, but a single bromine outranks it, and so does CH₂OH.
• Forgetting the duplicate atoms. COOH carries (O, O, O), not (O, O), because the C=O oxygen counts twice.
• Reading with 4 pointing at you. The turn you see is backwards. Turn the molecule round, or read it and reverse your answer.
• Treating R as (+). The letter comes from a naming rule and the sign from a polarimeter, and neither predicts the other.
• Turning a Fischer projection through 90°. That changes which bonds come forward and gives the other enantiomer. Only a 180° turn in the page is safe.