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MCAT Stereochemistry Practice Questions

By StudyStore · Published 2026-10-20 · 10 original questions · about 15 min

Ten free MCAT-style stereochemistry practice questions on R and S assignment, enantiomers versus diastereomers, meso compounds, E and Z alkenes, specific rotation, enantiomeric excess and resolution, with full explanations.

Quick concept primer

A carbon with four different groups is a chiral center. To name it, rank the groups by the Cahn-Ingold-Prelog rules: higher atomic number at the first point of difference wins, and a double bond counts as two single bonds to the same atom. With the lowest-priority group pointing away, a clockwise path from 1 to 2 to 3 is R and counterclockwise is S; if the lowest group points toward you, reverse the answer. The same priorities name alkenes: E when the two higher-priority groups sit on opposite sides, Z when they share a side.

Stereoisomers that are nonsuperimposable mirror images are enantiomers, and they invert every chiral center. Stereoisomers that differ at some but not all centers are diastereomers. A meso compound has chiral centers but an internal mirror plane, so it is achiral and optically inactive. Enantiomers share physical properties except the direction of optical rotation and their behavior toward other chiral substances; diastereomers differ in ordinary physical properties, which is why resolution converts enantiomers into separable diastereomers.

Traps: the R or S label is a naming convention and does not predict whether rotation is (+) or (-), and cis-trans wording can disagree with E-Z labels on substituted alkenes.

10 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A newly isolated compound is assigned the (R) configuration at its only chiral center by X-ray analysis. Which statement about its optical rotation is justified without further measurement?

  1. It must be dextrorotatory, because R compounds rotate light clockwise
  2. The direction of its rotation cannot be predicted from the R label; it must be measured with a polarimeter
  3. It must be levorotatory, because R corresponds to the L series
  4. It is optically inactive, because a single chiral center cannot rotate light
Show answer and explanation

Answer: B. The direction of its rotation cannot be predicted from the R label; it must be measured with a polarimeter

R and S describe absolute configuration from a set of priority rules, while (+) and (-) are experimental observations of how the sample rotates plane-polarized light. There is no general relationship between them, so the sign must be measured. Choices A and C wrongly link the labels, and any pure sample with one chiral center is optically active, so choice D is false.

Question 2

What is the relationship between (2R,3S)-2-bromo-3-chlorobutane and (2R,3R)-2-bromo-3-chlorobutane?

  1. Identical compounds
  2. Enantiomers
  3. Diastereomers
  4. Constitutional isomers
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Answer: C. Diastereomers

Both have the same connectivity, so they are stereoisomers, not constitutional isomers. Enantiomers would differ at every chiral center, but these share the R configuration at C2 and differ only at C3, which makes them diastereomers. Because the two ends carry different halogens, no symmetry lets the two sets of labels describe the same molecule, so they are not identical.

Question 3

Which of the following compounds is meso?

  1. (2R,3R)-2,3-dibromobutane
  2. (2R,3S)-2,3-dibromobutane
  3. (2R,3S)-2-bromo-3-chlorobutane
  4. (R)-2-bromobutane
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Answer: B. (2R,3S)-2,3-dibromobutane

A meso compound needs at least two chiral centers with identical sets of substituents and opposite configurations, so an internal mirror plane makes the molecule achiral. (2R,3S)-2,3-dibromobutane meets this. The (2R,3R) isomer is chiral, the bromo-chloro compound has unlike ends so no mirror plane is possible, and 2-bromobutane has only one chiral center.

Question 4

A student dissolves 0.50 g of a pure chiral compound in enough ethanol to make 10.0 mL of solution and places it in a 2.0 dm polarimeter tube. The observed rotation is +3.0 degrees. What is the specific rotation?

  1. +30
  2. +3.0
  3. +15
  4. +60
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Answer: A. +30

Specific rotation equals observed rotation divided by (path length in dm times concentration in g/mL). The concentration is 0.50 g / 10.0 mL = 0.050 g/mL, so the value is 3.0 / (2.0 × 0.050) = +30. Choice D forgets to divide by the path length, and choice B treats the observed rotation as the specific rotation.

Question 5

A pure sample of the S enantiomer of a compound has a specific rotation of +40. A mixture of the R and S enantiomers of the same compound shows a specific rotation of +12 under the same conditions. What is its composition?

  1. 30 percent S and 70 percent R
  2. 70 percent S and 30 percent R
  3. 60 percent S and 40 percent R
  4. 65 percent S and 35 percent R
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Answer: D. 65 percent S and 35 percent R

Enantiomeric excess is 12/40 = 30 percent, meaning the S enantiomer exceeds the R enantiomer by 30 percentage points. Solving S - R = 30 and S + R = 100 gives 65 percent S and 35 percent R. Choice A reverses the major enantiomer, and choice B treats the excess as the percentage of one enantiomer rather than the difference.

Question 6

A carbon atom bears four groups: -OH, -CHO, -CH2OH and -CH3. What is the correct Cahn-Ingold-Prelog priority order, highest first?

  1. -OH > -CHO > -CH2OH > -CH3
  2. -OH > -CH2OH > -CHO > -CH3
  3. -CHO > -OH > -CH2OH > -CH3
  4. -CH2OH > -CHO > -OH > -CH3
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Answer: A. -OH > -CHO > -CH2OH > -CH3

The first atoms are O, C, C and C, so -OH ranks first. Among the carbons, compare the sets of attached atoms: -CHO has (O, O, H) because the double bond counts twice, -CH2OH has (O, H, H), and -CH3 has (H, H, H). So the aldehyde outranks the hydroxymethyl group. Choice B misses the duplicated oxygen, and choices C and D ignore that oxygen outranks carbon at the first atom.

Question 7

In a drawing of a chiral center, the lowest-priority group is on a wedge pointing toward the viewer, and the path from priority 1 to 2 to 3 runs clockwise. What is the configuration?

  1. R
  2. Cannot be assigned unless the lowest-priority group points away
  3. S
  4. Meso
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Answer: C. S

The clockwise-equals-R rule assumes the lowest-priority group points away from the viewer. When it points toward the viewer, the apparent direction is reversed, so a clockwise path means S. The configuration can still be assigned from any view by applying this reversal, and meso describes a whole molecule, not one center.

Question 8

A drug is sold as two separate products, each containing one pure enantiomer of the same compound. All of the following are expected to be true of the two products EXCEPT:

  1. Their boiling points are identical
  2. Their specific rotations have the same magnitude but opposite signs
  3. Their densities are identical
  4. They bind a protein receptor in the body with the same affinity
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Answer: D. They bind a protein receptor in the body with the same affinity

Enantiomers have identical physical properties measured in an achiral setting, such as boiling point and density, and their specific rotations are equal in size and opposite in sign. A protein receptor is built from L-amino acids and is itself chiral, so each enantiomer forms a different diastereomeric complex with it, and binding affinities usually differ, sometimes greatly.

Question 9

2-Bromopent-2-ene has the structure CH3C(Br)=CHCH2CH3. In one of its stereoisomers, the bromine on C2 and the ethyl group on C3 point toward the same face of the C=C. What is the correct descriptor for this isomer?

  1. Z, even though carbons C1 and C4 of the main chain are trans to each other
  2. E, because carbons C1 and C4 of the main chain are trans to each other
  3. Z, because the two vinyl hydrogens are cis to each other
  4. E, because bromine outranks the ethyl group
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Answer: A. Z, even though carbons C1 and C4 of the main chain are trans to each other

On C2, bromine (atomic number 35) outranks the methyl carbon (C1); on C3, the ethyl carbon outranks hydrogen. The two higher-priority groups, Br and ethyl, are together, so the isomer is Z. That places C1 across from the ethyl group, so the chain itself has a trans shape, which is exactly why E-Z labels replace cis-trans for such alkenes. The molecule has only one vinyl hydrogen, so choice C is impossible, and choice D compares groups on different carbons.

Question 10

A chemist has a racemic mixture of a chiral amine and wants to isolate each enantiomer. Which approach works?

  1. Fractional distillation of the racemic mixture
  2. React the amine with a single enantiomer of a chiral acid to form diastereomeric salts, separate them by crystallization, then free each amine with base
  3. Chromatography on ordinary achiral silica gel
  4. React the amine with racemic tartaric acid, then recrystallize
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Answer: B. React the amine with a single enantiomer of a chiral acid to form diastereomeric salts, separate them by crystallization, then free each amine with base

Enantiomers have identical boiling points and identical interactions with achiral media, so distillation and achiral chromatography cannot separate them. Pairing each with one enantiomer of a chiral acid gives diastereomers, which differ in solubility and can be separated, after which base regenerates the free amines. Using racemic tartaric acid gives a mixture of all combinations and defeats the purpose.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.