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Organic Chemistry I · Stereochemistry

R and S configuration practice problems, with answers

Every chirality center is either R or S. To tell which, you rank its four groups with the Cahn–Ingold–Prelog (CIP) rules, point the lowest one away from you, and see which way 1 → 2 → 3 turns. Nearly all mistakes happen in one of two places: ranking the groups, or forgetting to reverse when the lowest priority points toward you.

Already know the basics? Jump to the 8 practice questions or the worked examples.

Step 1: rank the four groups (CIP rules)

Every R/S problem starts with the Cahn–Ingold–Prelog (CIP) sequence rules. You give each of the four groups on the chirality center a priority from 1 (highest) to 4 (lowest). Work through these rules in order and stop as soon as there's a winner.

  1. Atomic number of the attached atom. Look only at the four atoms bonded directly to the chirality center. Higher atomic number wins: I > Br > Cl > S > F > O > N > C > H. The size of the whole group doesn't matter. –Cl beats –CH₂Br, because the comparison is Cl (17) against C (6).
  2. Ties: go to the first point of difference. If two attached atoms are the same, list the three atoms each one carries, highest first, and compare the lists item by item. –CH₂OH carries (O, H, H) and –CH(CH₃)₂ carries (C, C, H). Compare the first items: O beats C, so –CH₂OH wins. One higher atom beats any number of lower ones. If the lists are identical, move one bond further out along the highest-ranked branch.
  3. Double and triple bonds count as duplicated atoms. A C=O carbon is treated as if it carries two oxygens. So –CHO counts as (O, O, H), –COOH as (O, O, O), and –CH=CH₂ as (C, C, H).
  4. Isotopes. If the atoms are the same element, the heavier isotope ranks higher. Deuterium (²H) beats ordinary hydrogen.
Common groups, highest priority first
GroupWhat you compareWhy it lands here
–I, –Br, –ClI, Br, ClThe heaviest attached atoms you'll usually see
–SHS (16)Sulfur outranks fluorine and oxygen
–FF (9)Fluorine beats oxygen, 9 vs 8
–OH, –OCH₃O (8)Any oxygen-bonded group beats any nitrogen or carbon group
–NH₂N (7)Nitrogen beats every carbon group
–COOHC (O, O, O)Three oxygens once C=O is duplicated
–CHOC (O, O, H)Two oxygens after duplication
–CH₂OHC (O, H, H)One oxygen still beats any all-carbon set
–C(CH₃)₃C (C, C, C)Three carbons
–CH=CH₂C (C, C, H), then (C, H, H)Ties with isopropyl at the first carbon, but wins one bond further out
–CH(CH₃)₂C (C, C, H), then (H, H, H)Loses to vinyl at the second sphere
–CH₂CH₃C (C, H, H)Beats methyl
–CH₃C (H, H, H)Lowest carbon group
–HH (1)Almost always priority 4

Step 2: point the lowest priority away, then read 1 → 2 → 3

Once you have the priorities, look at the molecule with group 4 pointing away from you. Then trace 1 → 2 → 3, skipping 4. Clockwise is R (rectus, right). Counterclockwise is S (sinister, left). On paper you rarely need to rotate anything in your head. Check where group 4 is drawn and use the matching shortcut.

Reading R or S from a wedge-and-dash drawing
Group 4 is onWhat to do
A dash (pointing away)Read 1 → 2 → 3 as drawn. Clockwise = R, counterclockwise = S.
A wedge (pointing toward you)Read 1 → 2 → 3 as drawn, then reverse the answer. Clockwise means S, counterclockwise means R.
A plain line (in the plane)Swap group 4 with whatever group is on the dash. Assign the new drawing, then reverse the answer, because one swap flips R and S.
A Fischer projectionHorizontal bonds point toward you and vertical bonds point away. If group 4 is vertical, read the answer directly. If it's horizontal, reverse it.

The swap rule. Swapping any two groups on a chirality center turns R into S and S into R. Swap twice and you're back where you started. That's why the swap shortcut works: you make one swap to get group 4 onto the dash, read the answer, and then undo it by reversing.

One more trap. The R/S label has nothing to do with the sign of optical rotation. An R compound can be (+) or (−); you only find out by measuring it.

Worked examples

Example 1: rank groups that tie at the first atom

Rank these four groups: –CH₂OH, –CHO, –CH=CH₂, and –CH(CH₃)₂.

  1. Every group is attached through carbon, so rule 1 gives a four-way tie.
  2. List what each carbon carries, with double bonds duplicated. –CHO: (O, O, H). –CH₂OH: (O, H, H). –CH=CH₂: (C, C, H). –CH(CH₃)₂: (C, C, H).
  3. Compare the lists one item at a time. Both oxygen groups beat both carbon groups at the first item. Between them, –CHO wins at the second item (O beats H).
  4. Vinyl and isopropyl still tie, so go one bond further out. Vinyl's CH₂ carbon carries (C, H, H), where the C is the duplicate from the double bond. Each isopropyl CH₃ carries only (H, H, H). Vinyl wins.

Answer: –CHO > –CH₂OH > –CH=CH₂ > –CH(CH₃)₂.

Example 2: H on a dash (read it directly)

2-Bromobutane is drawn with Br straight up, CH₂CH₃ to the lower left, CH₃ to the lower right, and H on a dash pointing down. Is it R or S?

2-Bromobutane with H on a dashA chirality center labeled C. Br is bonded straight up. CH2CH3 is bonded to the lower left and CH3 to the lower right, both drawn as plain lines. H is bonded straight down on a hashed (dashed) wedge, so it points away from the viewer. C Br CH₂CH₃ CH₃ H 1 = Br 2 = CH₂CH₃ 3 = CH₃ 4 = H (dash)
2-Bromobutane drawn with H on a dash. Br (top) to ethyl (lower left) to methyl (lower right) runs counterclockwise, so this drawing is S.
  1. Priorities: Br (35) is 1. Ethyl beats methyl at the first point of difference, so CH₂CH₃ is 2 and CH₃ is 3. H is 4.
  2. H is on the dash, so it already points away. Read the drawing as it is.
  3. Br (top) → CH₂CH₃ (lower left) → CH₃ (lower right). Top, then lower left, then lower right is counterclockwise.

Answer: S. This drawing is (S)-2-bromobutane.

Example 3: H on a wedge (reverse it)

Lactic acid, CH₃CH(OH)COOH, is drawn with OH up, COOH to the lower right, CH₃ to the lower left, and H on a wedge pointing toward you. R or S?

  1. Priorities: OH is 1 (oxygen). COOH (O, O, O) beats CH₃ (H, H, H), so COOH is 2 and CH₃ is 3. H is 4.
  2. Trace as drawn: OH (top) → COOH (lower right) → CH₃ (lower left). That's clockwise, which would mean R.
  3. But H points toward you, so you're looking at the center from the wrong side. Reverse the answer.

Answer: S. This drawing is (S)-lactic acid.

Example 4: H in the plane (use one swap)

2-Chlorobutane is drawn in a common zigzag style. H is a plain line to the upper left and CH₂CH₃ a plain line to the upper right. CH₃ is on a wedge to the lower left, and Cl is on a dash to the lower right. R or S?

  1. Priorities: Cl is 1, CH₂CH₃ is 2, CH₃ is 3, H is 4.
  2. H is in the plane, not on the dash. Swap H with the dashed group (Cl). Now Cl is upper left, H is on the dash at the lower right, and the others stay put.
  3. Read the swapped drawing: Cl (upper left) → CH₂CH₃ (upper right) → CH₃ (lower left). That's clockwise, so the swapped drawing is R.
  4. One swap flipped the configuration, so the original is the opposite.

Answer: S. The original drawing is (S)-2-chlorobutane.

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Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 1Rank these groups from highest to lowest CIP priority: –CH₂Br, –OH, –Cl, –SH.

Show solution

Rule 1 looks only at the attached atoms: Cl (17), S (16), O (8), C (6). The bromine in –CH₂Br is one bond too far away to count. It would only matter if there were a tie at carbon.

Question 2Which group has the highest CIP priority: –C(CH₃)₃, –CH₂OH, –CH=CH₂, or –CH(CH₃)₂?

Show solution

All four attach through carbon, so compare what each carbon carries: –CH₂OH (O, H, H), –C(CH₃)₃ (C, C, C), –CH=CH₂ (C, C, H), –CH(CH₃)₂ (C, C, H). The first items are O, C, C, C, so –CH₂OH wins right away. The full order is –CH₂OH > –C(CH₃)₃ > –CH=CH₂ > –CH(CH₃)₂.

Question 3Rank these from highest to lowest priority: –CH₂OH, –COOH, –CH₂CH₃, –CHO.

Show solution

All four attach through carbon. Duplicate each C=O oxygen: –COOH is (O, O, O), –CHO is (O, O, H), –CH₂OH is (O, H, H), and –CH₂CH₃ is (C, H, H). Compare item by item to get –COOH > –CHO > –CH₂OH > –CH₂CH₃.

Question 42-Butanol is drawn with H on a dash, OH up, CH₃ to the lower left, and CH₂CH₃ to the lower right. What is the configuration?

Show solution

Priorities: OH (1), CH₂CH₃ (2), CH₃ (3), H (4). H is on the dash, so no correction is needed. OH at the top → CH₂CH₃ at the lower right → CH₃ at the lower left runs clockwise. The drawing is (R)-2-butanol.

Question 51-Bromo-1-chloroethane is drawn with Br up, Cl to the lower right, CH₃ to the lower left, and H on a wedge pointing toward you. What is the configuration?

Show solution

Priorities: Br (1), Cl (2), CH₃ (3), H (4). As drawn, top → lower right → lower left is clockwise. Because H points toward you, reverse the answer: the drawing is S.

Question 6Alanine, CH₃CH(NH₂)COOH, is drawn with H on a dash, NH₂ on a wedge pointing straight down, COOH in the plane to the upper right, and CH₃ in the plane to the upper left. What is the configuration?

Show solution

Priorities: NH₂ (1, nitrogen), COOH (2, carbon carrying O, O, O), CH₃ (3), H (4). H is on the dash, so read directly. Bottom → upper right → upper left is counterclockwise, so this is (S)-alanine, the form found in proteins (L-alanine).

Question 7In a Fischer projection of glyceraldehyde, CHO is at the top, CH₂OH at the bottom, OH on the right, and H on the left. What is the configuration?

Show solution

Priorities: OH (1), CHO (2, carries O, O, H), CH₂OH (3, carries O, H, H), H (4). In a Fischer projection, horizontal bonds come toward you. Right → top → bottom is counterclockwise, which would be S. H is horizontal, so reverse it: R. This is (R)-glyceraldehyde, also called D-glyceraldehyde.

Question 82-Butanol is drawn zigzag style. H is a plain line to the upper left and CH₃ a plain line to the upper right. OH is on a wedge to the lower left, and CH₂CH₃ is on a dash to the lower right. What is the configuration?

Show solution

Priorities: OH (1), CH₂CH₃ (2), CH₃ (3), H (4). Swap H with the dashed CH₂CH₃. Now ethyl is upper left and H is on the dash. OH (lower left) → CH₂CH₃ (upper left) → CH₃ (upper right) is clockwise, so the swapped drawing is R. One swap flips the label, so the original is S.

Answer key

  1. B, –Cl > –SH > –OH > –CH₂Br: rank by the atomic number of the attached atom.
  2. C, –CH₂OH: (O, H, H) beats every all-carbon set at the first point of difference.
  3. D, –COOH > –CHO > –CH₂OH > –CH₂CH₃: duplicate the C=O oxygens, then compare item by item.
  4. A, R: H is on the dash and 1 → 2 → 3 runs clockwise.
  5. B, S: clockwise as drawn, reversed because H is on a wedge.
  6. A, S: NH₂ > COOH > CH₃, counterclockwise with H in back.
  7. C, R: counterclockwise, reversed because H is on a horizontal bond.
  8. B, S: swap H onto the dash, read R, then reverse.

Frequently asked questions

Is R and S configuration on the MCAT?

Yes. Stereochemistry, including absolute configuration and the R/S convention, appears in the organic chemistry part of the AAMC MCAT content outline. Expect to assign R or S quickly from a wedge-and-dash drawing or a Fischer projection, often as one step in a longer passage question.

What if the lowest priority group is pointing toward me?

Read 1 to 2 to 3 as drawn, then reverse the answer. Clockwise becomes S and counterclockwise becomes R.

Does R mean the compound rotates light to the right?

No. R and S describe the arrangement of groups in space. The direction of optical rotation, (+) or (-), has to be measured and does not follow from the R/S label.

How do I break a tie between two carbon groups?

List the three atoms each carbon carries, highest first, counting double bonds as duplicated atoms. Compare the lists one item at a time, and the first difference decides. If the lists are identical, move one bond further out along the highest-ranked branch.

Why does swapping two groups flip R and S?

Exchanging any two groups on a chirality center turns the molecule into its mirror image at that center. Two swaps cancel out, which is a quick way to check your work.

Sources

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