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Organic Chemistry Some Basic Principles and Techniques · TOPIC LESSON

Stereoisomerism

Read connectivity first, then use three-dimensional geometry, priority rules and symmetry to distinguish genuinely different molecules.

JEE Main · JEE Advanced · NEET UG48 min study guide10 indexed sections

BY THE END, YOU CAN

  • Classify structural isomers, conformers, enantiomers and diastereomers with a reliable decision sequence.
  • Assign R/S and E/Z from explicitly defined structures rather than from apparent drawing direction.
  • Use Fischer projections without accidentally inverting a stereocenter.
  • Recognize meso compounds and count unique configurations after symmetry.
  • Calculate optical rotation and enantiomeric excess under stated measurement assumptions.

BEFORE YOU START

  • Bond-line and condensed organic structures.
  • Tetrahedral carbon, sigma and pi bonds, atomic numbers and isotopes.
  • Basic molecular formula and functional-group identification.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. A drawing is a representation, not automatically a new molecule

Two structures are isomers only if they have the same molecular formula. First count the atoms and charges; then compare which atoms are connected to which. If connectivity differs, the pair is constitutional or structural isomers. If connectivity is the same, ask whether the three-dimensional arrangements are identical, merely different conformations, or distinct configurations.

Rotating the entire molecule in space changes the viewpoint, not the molecule. Rotating around an ordinary single bond can change conformation while preserving each tetrahedral stereocenter’s configuration. Flipping two groups at one tetrahedral stereocenter is different: it changes handedness. These operations can look deceptively similar on a flat page, so identify which physical or drawing operation is being performed.

An ordinary line represents a bond approximately in the page, a solid wedge points toward the viewer, and a hashed wedge points away. Wedge width should be interpreted with the stated convention and the stereocenter identified. A skeletal drawing with no stereochemical information may represent an unspecified configuration rather than an achiral molecule or an explicitly racemic sample.

A useful comparison order is formula → connectivity → permitted rotations → stereochemical relationships. Skipping the connectivity step can lead to calling an alcohol and ether enantiomers merely because the drawings look reflected. Skipping the rotation step can lead to counting the same molecule twice. Configuration labels help, but only after consistent atom mapping and substituent priorities are established.

02 / UNDERSTAND & APPLY

2. Structural isomerism changes the bonding map

Chain isomerism changes the carbon skeleton. For C₅H₁₂, the distinct skeletons are pentane, 2-methylbutane and 2,2-dimethylpropane. A bent drawing of pentane remains pentane. Renumbering a chain from the opposite end does not create a fourth skeleton. Position isomerism keeps the main type of skeleton and functional group while moving its position, as in pentan-1-ol and pentan-2-ol.

Functional-group isomerism changes the group class while preserving the formula. Propanal and propanone share C₃H₆O but place the carbonyl in an aldehyde or ketone environment. For C₃H₈O, propan-1-ol, propan-2-ol and methoxyethane illustrate both position and alcohol–ether functional relationships. A formula does not uniquely identify a functional group.

Metamerism is a traditional description for different carbon-group distributions around a polyvalent connecting functional unit. For example, ethoxyethane and 1-methoxypropane share C₄H₁₀O and the ether functional group but distribute the four carbons as 2+2 or 1+3 around oxygen. They are constitutional isomers, not a separate fundamental category outside structural isomerism.

Ring–chain relationships occur when the same formula permits a ring or an unsaturated open chain, as in cyclobutane and but-1-ene, both C₄H₈. Degree of unsaturation can help organize possibilities: for ordinary neutral compounds containing C, H, N and halogens X, DBE=(2C+2+N−H−X)/2; O and divalent S do not change that simple count. One ring or double bond contributes one unit; a triple bond contributes two.

Tautomers are interconverting constitutional isomers with a changed proton position and associated bonding pattern, such as a keto–enol pair. They are not resonance contributors: resonance moves electron bookkeeping without changing atom positions, whereas tautomerization transfers a nucleus and changes connectivity. Equilibrium abundances can be highly unequal, so drawing two tautomers does not imply a 1:1 mixture.

ORIGINAL WORKED EXAMPLE

Enumerate a restricted formula without double counting

List all neutral acyclic saturated alcohol and ether constitutional isomers of C₃H₈O.

  1. For an alcohol, the three-carbon skeleton is propane; attaching OH at an end gives propan-1-ol.
  2. Attaching OH to the middle gives propan-2-ol. The two ends are symmetry-equivalent.
  3. For an ether, split three carbons across oxygen into positive-sized groups: only 1+2 is possible.
  4. This gives CH₃OCH₂CH₃, methoxyethane. No stereocenter occurs in these three structures.

Answer Three: propan-1-ol, propan-2-ol and methoxyethane.

03 / UNDERSTAND & APPLY

3. Conformation changes through rotation, with an energy cost

In a Newman projection, look along a specified C–C bond. The front carbon is represented at the centre and its three outward bonds start there; the rear carbon is represented by a circle and its bonds start at the circle’s edge. The six visible substituent bonds belong to two different carbons, not a single six-coordinate carbon.

Ethane’s staggered arrangements place rear C–H directions between front C–H directions, while eclipsed arrangements align them in the end-on view. Rotation is possible but not perfectly barrier-free. Torsional effects make eclipsed ethane higher in energy than staggered ethane. A diagram of energy against dihedral angle is not a reaction profile for breaking the C–C bond.

For butane viewed along C2–C3, anti describes a 180° dihedral angle between the methyl groups and gauche describes a 60° relationship. Both are staggered, but anti is ordinarily lower in energy for isolated butane because the bulky groups are farther apart. Eclipsed conformations are higher-energy comparisons. The label gauche is not synonymous with eclipsed.

Cyclohexane chair flipping exchanges axial and equatorial positions while preserving whether each substituent is up or down. Consequently, a cis pair remains cis after a chair flip. Cis/trans describes the relative ring faces; axial/equatorial describes a particular conformation. For flexible molecules, a single snapshot can hide symmetry or make an achiral species appear deceptively handed.

Conformers commonly interconvert too rapidly to isolate under ordinary conditions, but sufficiently hindered rotation can create isolable atropisomers. Thus “rotation around a single bond” does not automatically mean an experimentally irrelevant distinction. Always attend to the molecule, barrier, temperature and observation timescale.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Anti and gauche conformations in a specified Newman viewLooking from carbon2 toward carbon3 of butane, the front carbon dot carries a methyl at twelve oclock and two hydrogens. The rear carbon circle carries one methyl and two hydrogens. In the first panel the rear methyl is at six oclock, giving an anti180degree methyl dihedral. In the second it is at two oclock, giving gauche60degrees. Both are staggered.Butane viewed along C2 → C3Front carbon = dot; rear carbon = circleCH₃HHCH₃HHAnti: methyl–methyl dihedral 180°Staggered, ordinarily lower energy for butaneCH₃HHCH₃HHGauche: methyl–methyl dihedral 60°Also staggered; it is not an eclipsed conformer.Rotate about C2–C3; connectivity stays the same.
Front bonds are rust coloured; rear bonds are green. A conformational change by C2–C3 rotation does not change connectivity or turn this into a configurational E/Z pair.

04 / UNDERSTAND & APPLY

4. Chirality belongs to the whole molecular arrangement

A chiral molecule is not superimposable on its mirror image by proper rotations and translations. A common source is a tetrahedral carbon attached to four different groups. Butan-2-ol has OH, H, CH₃ and CH₂CH₃ at C2 and therefore gives an enantiomeric pair. Propan-2-ol has two identical methyl groups at its OH-bearing carbon and lacks that stereocenter.

“Different groups” means different complete substituent paths, not merely different first atoms. In butan-2-ol the methyl and ethyl substituents both begin with carbon but are different after that. In a substituted ring, two bonds leaving a candidate stereocenter may lead along different ring paths; compare them until a difference or full equivalence is established.

An internal mirror plane in the relevant molecular arrangement is sufficient to establish achirality. Its absence in one arbitrary sketch is not by itself proof of chirality; other improper symmetries and conformational equivalence must be considered. For elementary flexible molecules, compare a suitable conformation with its mirror rather than treating one twisted drawing as the entire identity.

Chirality can also arise without a tetrahedral carbon stereocenter. An allene R¹R²C=C=CR³R⁴ can be axially chiral when the substituents at each terminal carbon are different, because the two terminal substituent planes are perpendicular. Appropriately hindered substituted biaryls can also show axial chirality. Conversely, multiple stereocenters do not guarantee a chiral whole molecule: meso compounds are the central counterexample.

The usual enantiomer comparison assumes the same chemical and isotopic composition. Isotopic substitution can break a previously present equivalence. A tetrahedral carbon attached to H, D, OH and CH₃ has four different groups even though H and D have the same atomic number; their isotope masses distinguish them.

Stereogenic atoms need not be carbon. An ordinary pyramidal amine with three different groups can have a chiral instantaneous arrangement, but rapid nitrogen inversion usually prevents isolation of its two nitrogen configurations at room temperature. A tetrahedral quaternary ammonium ion with four different substituents does not undergo that ordinary lone-pair umbrella inversion. Appropriate phosphines, sulfonium ions and sulfoxides can also be configurationally stable; stability must be assessed for the actual structure rather than inferred from a lone pair alone.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

End-on mirror comparison of a substituted alleneThe stated allene connectivity is Br(H)C=C=C(F)Cl. The C=C=C carbon axis is linear. Viewed down that axis, the front terminal carbon has Br above and H below, while the rear terminal carbon has F left and Cl right. In the second end-on projection the rear left-right positions are reflected while the front top-bottom positions remain. Both ends have distinct substituents; the arrangements are nonsuperimposable mirror partners. The dot and surrounding circle are depth cues for different terminal carbons, not a six-coordinate atom.Allene: perpendicular terminal planesBr(H)C=C=C(F)ClLook end-on along the linear C=C=C axis.BrHFClOne handed arrangementBrHClFEnd-on mirror partnerRust: front terminal substituents Br and H.Green: rear terminal substituents F and Cl.The terminal planes are perpendicular in 3D.Different groups at each end allow axial chirality.
End-on allene projection with perpendicular terminal substituent planes. It resembles a cross but is not a Fischer projection. No R/S axial descriptor or optical-rotation sign is assigned.

ORIGINAL WORKED EXAMPLE

Identify a stereocenter by complete paths

Does C3 of 3-methylhexane have four different substituents?

  1. C3 is bonded to H and a methyl substituent.
  2. The path toward C1 is ethyl, CH₂CH₃.
  3. The path toward C6 is propyl, CH₂CH₂CH₃.
  4. H, methyl, ethyl and propyl are four different substituents. Both longer paths begin with CH₂, but differ farther outward.

Answer Yes. C3 is a tetrahedral stereocenter, giving an R/S enantiomeric pair.

05 / UNDERSTAND & APPLY

5. Assign R/S by priority, orientation and a final check

The Cahn–Ingold–Prelog procedure first ranks the atoms directly attached to the stereocenter by atomic number: a directly attached O outranks N, which outranks C, which outranks H. Do not use the total mass or apparent size of an entire substituent. If two directly attached atoms are isotopes of the same element, the larger mass number has higher priority, so D outranks H.

When first atoms tie, compare the atoms attached to each tied atom, excluding the stereocenter, ordered from highest to lowest atomic number. Compare the ordered lists at the first difference. For CH₂OH, CHO and CH₂CH₃ attached through carbon, the lists are (O,H,H), (O,O,H) and (C,H,H), respectively, using the multiple-bond duplicate convention. Thus CHO outranks CH₂OH, which outranks ethyl.

A double bond is handled with duplicate-atom bookkeeping for ranking, not by pretending a physically separate extra oxygen exists in the molecule. Continue outward only as far as needed. A farther heavy atom cannot override an earlier difference: the first point of difference decides the ordering.

Once priorities 1 to 4 are set, view the centre with group 4 pointing away. A clockwise path 1→2→3 is R; a counterclockwise path is S. If group 4 points toward you in the standard projection, reverse the apparent clockwise/counterclockwise result. If it lies in the drawing plane, reorient the structure or use a controlled swap, keeping track of the change.

Swapping any two substituents once inverts a tetrahedral configuration; two swaps restore it. A cyclic permutation of three groups while the fourth remains fixed preserves the handedness. These parity rules provide a useful check on a redrawing, but they must be applied to the same centre and the same actual substituents. R/S labels can change after a chemical reaction merely because priorities changed even if the spatial arrangement was retained.

ORIGINAL WORKED EXAMPLE

Assign priorities before following the circle

A tetrahedral carbon is bonded to OH, CHO, CH₂OH and H. H is directed away; the projected sequence OH→CHO→CH₂OH is clockwise. Assign configuration.

  1. O in OH outranks the carbon atoms of the two carbon-containing groups. H is priority 4.
  2. For CHO, the next-atom list is O,O,H; for CH₂OH it is O,H,H.
  3. Therefore priorities are OH 1, CHO 2, CH₂OH 3, H 4.
  4. With 4 away, the clockwise 1→2→3 path gives R.

Answer R. Ranking by total substituent size would be an unreliable shortcut.

06 / UNDERSTAND & APPLY

6. Fischer projections encode toward and away bonds

A Fischer cross is a stereochemical convention: horizontal bonds point toward the viewer and vertical bonds point away. The intersection is a tetrahedral centre even though the drawing is a flat cross. A long vertical chain with several crosses is a bookkeeping projection, not a claim that every atom lies in one physical plane.

Rotating an entire Fischer projection by 180° in the plane preserves its configuration. Rotating it by 90° and then interpreting it again as a Fischer projection generally inverts each represented stereocenter. This does not contradict the fact that a real molecule can be rotated harmlessly in space: the projection convention reassigns which bonds point toward and away after that paper rotation.

For a direct R/S assignment, rank substituents first. If priority 4 is on a vertical arm, the ordinary clockwise R/counterclockwise S rule applies. If priority 4 is on a horizontal arm, reverse the apparent result because that group points toward the viewer. This avoids unnecessary swaps and the risk of forgetting to reverse their effects.

The biochemical D/L system specifies a relative configuration through a reference convention. It is not interchangeable with R/S, and neither system predicts the experimentally measured sign of optical rotation. Uppercase D and L should also not be confused with informal lowercase d/l rotation notation. A D sugar is not automatically dextrorotatory.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Fischer projection handedness and the viewing-direction correctionAn upper Fischer projection has methyl top, ethyl bottom, bromine left and hydrogen right. Horizontal bonds project toward the viewer and vertical bonds away. Priorities are Br1,ethyl2,methyl3,H4. The projected 1-2-3 sequence is counterclockwise, but H points toward the viewer, so the centre is R. The lower drawing reflects left and right to give S.Fischer: horizontal bonds point toward youVertical bonds point awayCH₃CH₂CH₃BrH1 = Br; 2 = ethyl; 3 = methyl; 4 = H1 → 2 → 3 looks counterclockwise.H is horizontal and toward you: reverse.Configuration: RIts reflected Fischer projectionCH₃CH₂CH₃HBrMirror partner: SA 180° paper rotation preserves a Fischer structure.A 90° redraw under the same convention inverts it.
The crosses encode tetrahedral stereochemistry rather than planar carbon. R/S does not predict optical rotation sign.

ORIGINAL WORKED EXAMPLE

A Fischer centre with the lowest priority toward the viewer

For a Fischer projection of CH₃–CH(Br)–CH₂CH₃, place CH₃ at the top, CH₂CH₃ at the bottom, Br at the left and H at the right. Assign C2.

  1. Br is priority 1 and H is 4.
  2. Ethyl outranks methyl, so the bottom group is 2 and top group is 3.
  3. The apparent left→bottom→top path is counterclockwise.
  4. H is horizontal and points toward the viewer, so reverse that result.

Answer C2 is R. Its left/right-reflected Fischer drawing is S.

07 / UNDERSTAND & APPLY

7. Mirror pairs, nonmirror pairs and internal compensation

Enantiomers are nonsuperimposable mirror images with the same connectivity. In an achiral environment, a pure enantiomer pair has the same ordinary melting point, boiling point and many spectral properties, while their optical rotations are equal and opposite under the same conditions. Interactions with a chiral environment, such as an enzyme or chiral chromatography phase, can differ.

Diastereomers are stereoisomers that are not mirror images. They can have different physical properties and reactivity even in achiral surroundings. Cis/trans and E/Z pairs are familiar diastereomers. For a molecule with two nonsymmetry-related tetrahedral stereocenters, changing one configuration while retaining the other normally gives a diastereomer; inverting both gives the enantiomer.

A meso compound is achiral despite having stereogenic centres, because the whole arrangement is superimposable on its mirror image through internal equivalence. For 2,3-dibromobutane, (2R,3R) and (2S,3S) form an enantiomeric pair, while the (2R,3S) and (2S,3R) descriptions represent one meso form after appropriate end-for-end mapping. The carbon chain has identical ends, which is essential to that symmetry reduction.

A meso sample and a racemate can both show zero net optical rotation, but for different reasons. Each meso molecule is achiral; a racemate contains equal amounts of two chiral enantiomers whose rotations cancel. A meso compound cannot be resolved into a pair of enantiomers merely by separating the molecules already present, because that pair is not what the sample contains.

Do not decide chirality from “one R and one S” alone. A molecule with inequivalent ends may have an R,S form that is chiral. Similarly, do not decide that two structures are identical simply because their labels look alike unless the atom numbering and connectivity correspond. Symmetry is a property of the full molecular graph and spatial arrangement.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Enantiomeric pair and meso symmetry in 2,3-dibromobutaneThe upper Fischer projections with opposite-side bromines are 2R3R and 2S3S mirror partners. The lower projection has both bromines left and both hydrogens right, giving 2R3S with identical terminal methyl groups. Reflection across the horizontal midpoint exchanges equivalent halves of that Fischer representation. Reversing the numbering yields the same meso species described as 2S3R. There are three distinct configurations in total.2,3-Dibromobutane: three unique formsFischer convention at every crossCH₃CH₃BrHHBr(2R,3R)CH₃CH₃HBrBrH(2S,3S)Non-superimposable mirror pairMeso form: (2R,3S) = (2S,3R)CH₃CH₃BrHBrHThe dashed line marks a reflection symmetryof this Fischer representation with identical ends.Meso: one achiral species, not a 50:50 mixture.
Fischer symmetry bookkeeping, not a literal all-planar conformation. Horizontal bonds point toward and vertical bonds away. The meso molecule is achiral; the upper enantiomers are individually chiral.

ORIGINAL WORKED EXAMPLE

Use symmetry before multiplying configurations

How many distinct configurational stereoisomers does 2,3-dibromobutane have when conformers are not counted?

  1. Two stereocenters give a preliminary list RR, SS, RS and SR.
  2. RR and SS are different mirror partners.
  3. Because both chain ends are CH₃ and the substitutions match, RS and SR describe the same meso compound after suitable atom mapping.
  4. The duplicate pair reduces four labels to three different molecules.

Answer Three: one enantiomeric pair plus one meso form.

08 / UNDERSTAND & APPLY

8. E/Z and ring cis/trans describe restricted relative arrangements

An ordinary alkene can have E/Z stereoisomerism only when each double-bond carbon has two different substituents. CH₂=CHCH₂CH₃ fails this test at its CH₂ end; but-2-ene passes. A double bond prevents free interconversion of the two configurations by ordinary low-energy rotation because twisting disrupts the pi interaction.

Assign the higher-priority substituent independently at each end using CIP rules. If those two selected groups lie on the same side of the C=C reference line, the configuration is Z; if they lie on opposite sides, it is E. Compare two priorities at each carbon, not all four substituents in one global ranking.

Cis/trans compares an explicitly identifiable pair of like or reference groups. It is especially convenient for a disubstituted alkene such as but-2-ene, where cis corresponds to Z and trans to E. That correspondence is not universal. With different substituent patterns, a pair of identical groups may be cis while the two highest-priority groups are opposite and the alkene is E.

In rings, cis means substituents are on the same ring face and trans means opposite faces. A ring drawing may use wedges and dashes to communicate this. These labels do not depend on whether substituents are axial or equatorial in one chair. Ring constraints can also restrict which formal geometries are physically accessible, so multiplying possible double-bond labels without checking ring size can overcount.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

E and Z are determined by local CIP prioritiesTwo structures of CH3-C(Cl)=C(H)-CH3 are compared. In the first, both methyl groups are above the double-bond axis. Chlorine below at the left carbon outranks methyl, while methyl above at the right outranks H, so this drawing is E. In the second, methyl groups are opposite but the two high-priority groups are both below, giving Z.Choose the high-priority group at each endCCCH₃ClCH₃HLeft: Cl > CH₃. Right: CH₃ > H.High priorities opposite: EThe two methyl groups are on the same side.CCCH₃ClHCH₃High priorities same side: ZThe two methyl groups are on opposite sides.E/Z compares CIP priorities, not just like groups.Each alkene carbon must bear two distinct groups.
Each carbon is trigonal planar. Double-bond drawings specify connectivity and relative sides; precise bond lengths and angles are schematic.

ORIGINAL WORKED EXAMPLE

Why cis does not always mean Z

For CH₃–C(Cl)=C(H)–CH₃, draw both methyl groups above the double-bond line, with Cl and H below their respective carbons. Assign E/Z.

  1. At the left alkene carbon, Cl outranks the methyl carbon. Thus the higher-priority group is below.
  2. At the right alkene carbon, methyl carbon outranks H. Thus the higher-priority group is above.
  3. The higher-priority groups are on opposite sides.
  4. The two methyl groups are cis in the stated drawing, but the CIP descriptor is E.

Answer E, despite the same-side methyl groups.

09 / UNDERSTAND & APPLY

9. Count unique structures and keep rotation separate from configuration

For n independent ordinary tetrahedral stereocenters with no symmetry reduction, 2ⁿ is the maximum number of configurations. An independent E/Z double bond can introduce another factor of two. These are upper bounds unless all combinations are possible and distinct. Meso forms, equivalent centres, ring restrictions and other structural constraints can reduce the count.

For 2-bromo-3-chloropentane, two distinct stereocenters and unequal chain ends give four ordinary configurations: RR, SS, RS and SR. There is no meso reduction based on interchanging its different substituents. RR/SS and RS/SR form two enantiomeric pairs; other pairings are diastereomeric. By contrast, the symmetric dibromobutane example has only three.

Observed optical rotation depends on the sample and measurement conditions. In a solution convention with path length l in decimetres and concentration c in g mL⁻¹, specific rotation is [α]=α_obs/(lc), at a stated temperature and wavelength. A concentration stated in g per 100 mL must first be converted. Rotation sign, solvent, temperature and wavelength are not optional details when comparing numerical values.

For a mixture containing only two enantiomers under conditions where their rotations add linearly, enantiomeric excess is ee=|x_major−x_minor|×100%. If the pure major enantiomer’s specific rotation is known under exactly matching conditions, the magnitude ratio of mixture to pure specific rotation equals ee as a fraction. This inference fails if other optically active species contribute or conditions differ.

A zero rotation can mean a racemate, an achiral compound, a meso species or a cancellation involving a more complicated mixture. Conversely, R does not mean positive rotation and S does not mean negative rotation. Absolute configuration must come from stereochemical evidence; optical rotation is a separate measured property. Resolution exploits a chiral distinction, for example converting enantiomers to diastereomeric derivatives or using a chiral stationary phase.

x_major=(1+ee_fraction)/2; x_minor=(1−ee_fraction)/2

For a two-enantiomer mixture only, with mole fractions summing to one.

ORIGINAL WORKED EXAMPLE

Recover a composition from matched optical data

A solution containing only an enantiomer pair has c=0.0200 g mL⁻¹ and l=2.00 dm. Its observed rotation is +1.20°. The pure positive enantiomer has [α]=+50.0° mL g⁻¹ dm⁻¹ under the same conditions. Find the composition.

  1. The mixture specific rotation is 1.20/(2.00×0.0200)=+30.0 in the stated units.
  2. The excess of the positive enantiomer is 30.0/50.0=0.600, or 60.0% ee.
  3. The positive fraction is (1+0.600)/2=0.800.
  4. The negative fraction is 0.200. Neither fraction supplies an R/S assignment without separate information.

Answer 80.0% positive enantiomer and 20.0% negative enantiomer; 60.0% ee.

10 / UNDERSTAND & APPLY

10. Solved practice with an identity check

For every pair, first confirm molecular formula and connectivity. Then track the viewing convention, compare configurations under consistent numbering, and inspect symmetry. The following exercises are original and do not represent a past-paper question bank.

ORIGINAL PRACTICE · 1

Are pentan-1-ol and pentan-2-ol enantiomers?

Reveal answer & reasoning

No. They are positional constitutional isomers.

Both have C₅H₁₂O, but the OH group is attached to a different carbon. Connectivity differs before stereochemistry is considered.

ORIGINAL PRACTICE · 2

Are CH₃COCH₃ and CH₂=C(OH)CH₃ resonance contributors?

Reveal answer & reasoning

No. They are keto–enol tautomers.

A hydrogen nucleus changes its bonded atom and the bonding pattern changes. Resonance contributors keep the nuclear framework fixed.

ORIGINAL PRACTICE · 3

Rank OH, CH₂CH₃, CH₃ and H at a tetrahedral carbon.

Reveal answer & reasoning

OH > CH₂CH₃ > CH₃ > H.

O outranks C and H. The tied carbon groups compare (C,H,H) against (H,H,H), making ethyl higher than methyl.

ORIGINAL PRACTICE · 4

Priority 4 points away and the 1→2→3 sequence is counterclockwise. What is the configuration?

Reveal answer & reasoning

S.

The standard away-view rule is counterclockwise S. If priority 4 pointed toward the viewer in the corresponding projection, that apparent result would be reversed.

ORIGINAL PRACTICE · 5

A single-centre Fischer projection is rotated 180° in the page. Does its configuration change?

Reveal answer & reasoning

No.

The horizontal/toward and vertical/away convention is preserved. The equivalent permutation is even. A 90° paper rotation interpreted as a new Fischer projection would not preserve the configuration.

ORIGINAL PRACTICE · 6

For 2-bromo-3-chloropentane, compare (2R,3S) with (2S,3R), then with (2R,3R).

Reveal answer & reasoning

The first comparison gives enantiomers; the second gives diastereomers.

Both inequivalent centres invert in the first pair, giving the mirror image. Only C3 changes in the second pair; the molecules are not mirror images.

ORIGINAL PRACTICE · 7

Can 2-methylbut-2-ene have an E/Z pair?

Reveal answer & reasoning

No.

One double-bond carbon bears two identical methyl substituents. E/Z requires two distinct groups at each end.

ORIGINAL PRACTICE · 8

A sample contains 65.0% of one enantiomer and 35.0% of the other. Find ee.

Reveal answer & reasoning

30.0% ee in favour of the major enantiomer.

Subtract the fractions: 0.650−0.350=0.300. The major fraction itself is not the enantiomeric excess.

ORIGINAL PRACTICE · 9

A compound has three independent tetrahedral stereocenters with no symmetry reduction and no additional stereogenic features. What is the maximum stereoisomer count and number of enantiomeric pairs?

Reveal answer & reasoning

Eight stereoisomers and four enantiomeric pairs.

There are 2³ configurations. Under the stated absence of meso/symmetry reduction, each configuration has one distinct fully inverted mirror partner.

ORIGINAL PRACTICE · 10

Does a chair flip convert cis-1,2-disubstituted cyclohexane to trans?

Reveal answer & reasoning

No.

The flip exchanges axial and equatorial placement while retaining the up/down identity of each substituent. Relative ring-face configuration remains cis.

KEEP THE BIG IDEAS

Quick revision

  • Connectivity separates constitutional isomers from stereoisomers.
  • Priority rules and viewing direction jointly determine R/S.
  • Fischer horizontal bonds point toward the viewer.
  • Meso and racemic are different reasons for zero rotation.
  • E/Z uses local priorities and is not a universal synonym for trans/cis.
  • Count symmetry-distinct configurations and state the assumptions behind optical calculations.
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