Chirality

Theory — Chirality: Identification, Reactions, and Synthesis

Chirality is the property of a molecule that makes it non-superimposable on its mirror image. Like our left and right hands, chiral molecules come in two mirror-image forms (enantiomers) that have identical scalar physical properties (mass, m.p., b.p., density) but differ in their interaction with other chiral entities — including plane-polarised light and biological receptors. This lab integrates everything you need to identify, predict, and synthesise chiral compounds: how to recognise stereocentres, how to assign R/S configurations, how reactions affect stereochemistry, how to design a synthesis with stereocontrol, and how to measure optical rotation in a microscale lab.

1. Identifying chirality (CLO 1 + 2)

IUPAC nomenclature for chiral compounds. Stereodescriptors are added as italicised prefixes in the IUPAC name: e.g., (R)-2-bromobutane, (2R,3S)-tartaric acid, (2R,3R)-2,3-dibromobutane. Multiple stereocentres are listed with their locants (2R,3S) inside parentheses. The prefix "meso-" is used for an achiral molecule with stereocentres (e.g., meso-tartaric acid). The base name still follows the standard IUPAC rules for the parent compound — the stereodescriptor only adds spatial information.

A molecule is chiral if it is non-superimposable on its mirror image. The most common source of chirality in organic molecules is a stereocentre — an sp³ carbon bonded to four different groups. Other sources include axial chirality (allenes, biphenyls), planar chirality (substituted ferrocenes), and helicity (helicenes), but for an introductory course the focus is on point chirality at sp³ carbons.

Stereocentre detection

Look for an sp³ carbon with four different substituents. Common examples: 2-butanol (C2 has OH, CH₃, CH₂CH₃, H — four different groups, so it IS a stereocentre). Counter-example: 2-propanol (C2 has OH, two identical CH₃, and H — only three different groups, NOT a stereocentre).

Symmetry test

Even with stereocentres, a molecule can be achiral if it has an internal mirror plane (a meso compound). Example: (2R,3S)-tartaric acid has two stereocentres but a mirror plane between C2 and C3, making it meso (achiral).

Drawing wedge-dash

3D representations: solid wedges = bonds coming out of the plane; dashed wedges = bonds going behind the plane; lines = bonds in the plane. Two wedges on the same atom show the spatial layout of the four substituents around an sp³ stereocentre.

Fischer projections

A 2D shorthand: vertical bonds project AWAY from viewer; horizontal bonds project TOWARDS viewer. The convention assumes the longest chain is vertical with C1 at the top. Common for sugars, amino acids. Two horizontal swaps = same configuration; one swap = opposite configuration.

2. R/S assignment using CIP rules (CLO 1 + 2)

The Cahn-Ingold-Prelog (CIP) priority rules assign a unique R or S descriptor to each stereocentre:

Rank the four substituents by priority. Higher atomic number = higher priority. Halogens: I > Br > Cl > F. Common ordering: I > Br > Cl > S > F > O > N > C > H. For ties at the first atom, look at the next sphere of atoms outwards.
Treat double bonds as duplicated single bonds. A C=O is treated as C bonded to (O, O, ...) and O bonded to (C, C, ...). A C=C is treated similarly with phantom atoms.
Orient the lowest-priority group (often H) AWAY from the viewer. Imagine looking at a steering wheel with the lowest-priority substituent pointing through the column.
Trace the path 1 → 2 → 3. Clockwise = R (rectus, "right"). Counterclockwise = S (sinister, "left").
Trick for when H is in front: Trace 1 → 2 → 3 anyway and then INVERT your answer (CW seen with H in front = actually S).

Worked example: (R)-glyceraldehyde The C2 stereocentre of glyceraldehyde bears: OH, CHO, CH₂OH, and H
Priority by next-sphere atoms:
  OH (O bonded to H) — direct O = highest
  CHO (C bonded to (O, O, H) — phantom O for the C=O)
  CH₂OH (C bonded to (O, H, H))
  H — lowest
Order: OH > CHO > CH₂OH > H
With H pointing away, if rotation OH → CHO → CH₂OH is clockwise → (R)-glyceraldehyde. (D-glyceraldehyde, the natural sugar.)

3. Optical activity and specific rotation

Chiral molecules rotate plane-polarised light. The amount of rotation is measured with a polarimeter and characterised by the specific rotation [α], which depends on temperature, wavelength, solvent, concentration, and path length:

Specific rotation [α]_D²⁰ = α / (c × ℓ)
where:
α = observed rotation in degrees
c = concentration in g/mL (or g/100 mL × 0.01)
ℓ = path length in dm (typically 1 dm = 10 cm)
D = sodium D-line (589 nm); 20 = temperature in °C
A pure enantiomer has a fixed specific rotation. The opposite enantiomer has the same magnitude but opposite sign. A racemic mixture (50:50) has [α] = 0.

Enantiomeric excess (ee). A non-racemic mixture of enantiomers has an "excess" of one over the other:

Enantiomeric excess %ee = ([R] − [S]) / ([R] + [S]) × 100 (or [S] − [R] if S is in excess)
Equivalently: %ee = (observed [α] / [α] of pure enantiomer) × 100
A 90:10 mixture of R:S has 80% ee. A pure enantiomer is 100% ee. A racemate is 0% ee.

4. Stereoisomer relationships (CLO 2)

For a molecule with n stereocentres, the maximum number of stereoisomers is 2ⁿ. With internal symmetry, the actual count is lower (some pairs collapse to a single meso form).

Relationship	Definition	Physical properties
Enantiomers	Non-superimposable mirror images	Identical scalar properties (m.p., b.p., density). Opposite specific rotation.
Diastereomers	Stereoisomers that are NOT mirror images	Different scalar properties (different m.p., b.p., chromatographic R_f).
Meso compound	Has stereocentres but is overall achiral due to internal symmetry	No optical rotation. Identical to its mirror image.
Constitutional isomers	Different connectivity (NOT stereoisomers)	Different in every way.
Conformers	Differ only by rotation about single bonds	Same compound — interconvert at room temperature.

5. Stereochemistry of reactions (CLO 5)

Each reaction mechanism has a characteristic effect on stereochemistry at the reacting carbon:

SN2 — inversion

Walden inversion. Backside attack inverts the stereocentre. (R)-2-bromobutane + NaCN/DMSO → (S)-2-cyanobutane. Net stereochemistry: ONE inversion at the reacting carbon.

SN1 — racemisation

Planar carbocation. Nucleophile attacks the cation from EITHER face → both R and S products form. (R)-substrate + H₂O/heat → racemic R/S mixture (often slightly biased toward inversion due to ion-pair effects).

E2 — anti-periplanar

Specific stereochemistry of the alkene. The H and X must be anti-periplanar. For diastereomeric substrates (e.g., (2R,3R) vs (2R,3S)), each gives a specific alkene geometry — useful as a chirality probe.

E1 — loss of stereo info

Carbocation intermediate. The alkene formed depends on which β-H is removed, but the original stereocentre is destroyed. Stereochemistry is lost.

Hydroboration-oxidation — syn

BH₃ + H₂O₂/NaOH. Both H and OH add to the same face of the alkene (syn). Combined with anti-Markovnikov regiochemistry, this gives a defined product stereochemistry on cyclic substrates.

Halogenation — anti

Br₂ via halonium ion. Both Br atoms add to opposite faces (anti). On cis-2-butene, anti addition gives (2R,3R) + (2S,3S) racemate. On trans-2-butene, anti addition gives (2R,3S) — meso. Stereospecific.

6. Stereocontrolled synthesis (CLO 7)

Designing a synthesis with the correct stereochemistry requires choosing reagents that achieve the desired result:

To invert a stereocentre: use SN2 (e.g., NaCN/DMSO on a 2° halide).
To preserve stereochemistry through alcohol → halide → other: use TsCl/pyridine (no inversion at C — just OTs replaces OH). Then SN2 with the desired nucleophile gives one inversion total.
For double inversion (net retention): two consecutive SN2 reactions at the same carbon (e.g., R-OH → R-OMs → R-OAc, then hydrolyse).
To racemise: use SN1 conditions (3° substrate + protic solvent, or use bromide via SN1 ionisation).
To create a NEW stereocentre selectively: use chiral auxiliaries, asymmetric catalysts (e.g., Sharpless epoxidation, asymmetric hydrogenation), or substrate-controlled reactions on rigid substrates (e.g., hydroboration of cyclohexene gives only the cis adduct due to syn addition).
To make both stereocentres at once: use stereospecific addition reactions — Br₂ gives anti (so cis alkene → erythro; trans alkene → threo); OsO₄ or KMnO₄/cold gives syn diols.

7. Microscale operations: optical rotation and SDS (CLO 8)

The defining microscale technique for chirality is polarimetry — measuring the rotation of plane-polarised light by a chiral solution. Other standard techniques apply too (b.p., n_D, recrystallisation), but polarimetry is uniquely diagnostic for chirality.

Polarimeter operation

Sodium D-line at 20 °C is standard. Fill the polarimeter cell (typical 1 dm path length) with the solution at known concentration (g/100 mL). Measure α to ±0.01°. Calculate [α] = α / (c × ℓ). Compare to literature.

Determining ee

Compare measured [α] to pure-enantiomer [α]. If pure (R)-2-butanol has [α]_D²⁰ = +13.5° (neat), and your sample shows [α] = +6.75°, then ee = 6.75 / 13.5 × 100 = 50%. The sample is 75:25 R:S.

Sources of error

Concentration measurement is the biggest source — small errors in mass or volume change the reported [α]. Temperature must be controlled (rotation depends on T). Solvent must match the literature reference. Air bubbles in the cell give incorrect zero readings.

SDS hazards

Chiral auxiliaries and asymmetric catalysts are often expensive but not particularly hazardous. The bigger hazards in chirality work usually come from the supporting reagents — strong acids, organometallic reagents (e.g., Grignards), borane reagents, etc.

Disposal: chiral organic reagents typically go to non-halogenated organic solvent waste (or halogenated, if a halide is present). Optical rotation measurements involve no waste beyond the solution itself, which can be recovered for further use.

Instructions

This lab's Simulation section has four parts. Complete them in order.

Section I — Identify Chirality (CLO 1, 2). Ten molecules are presented. For each, determine: (a) is the molecule chiral or achiral? (b) how many stereocentres does it have? (c) for any stereocentre, what is the R/S configuration?

Section II — Stereochemistry of Reactions (CLO 5). Eight reactions are presented, each with a defined stereochemistry of the substrate. Predict the stereochemistry of the product (retention, inversion, racemisation, or specific R/S assignment).

Section III — Stereocontrolled Synthesis (CLO 7). Six retrosynthesis problems requiring you to choose reagents that achieve the correct stereochemistry of the target.

Section IV — SDS & Microscale Workbench (CLO 8). SDS interpretation for four reagents commonly used in chirality work; polarimetry-focused microscale matching for five chiral compounds; theoretical and percent yield calculations including enantiomeric excess from polarimetry data.

Prepare your lab report. Use the Example Report as your template. Your report must include a reagent table, all stereochemical assignments, mechanistic justifications, polarimetry data, and at least one yield/ee calculation.

Note on chirality: Chirality is at the heart of biological chemistry — proteins, sugars, and nucleic acids are all chiral, and only one enantiomer is biologically active in most cases. Examples: only (S)-thalidomide is sedative; the (R) form caused birth defects. Only L-amino acids are used by ribosomes. The chirality work you do here underlies pharmaceutical chemistry, biochemistry, and most modern catalysis.

Simulation

Four interactive parts — work through them in order.

For each molecule: (a) chiral or achiral? (b) number of stereocentres? (c) R/S configuration where applicable.

Score: 0 / 30 (3 questions × 10 molecules)

For each reaction with a defined-stereochemistry substrate, predict the stereochemistry of the product.

Score: 0 / 16 (2 questions × 8 reactions)

Six stereocontrolled-synthesis problems. Choose the correct reagent or sequence.

Score: 0 / 6

Round 1 — SDS interpretation

SDS extracts for four reagents commonly used in stereochemistry work.

SDS score: 0 / 16

Round 2 — Polarimetry data matching

Five polarimetry records from microscale measurements. Match each to the correct compound and configuration.

Reference [α]_D²⁰ values (pure enantiomers, neat or in stated solvent):

(R)-(−)-2-butanol — [α]_D²⁰ = −13.5° (neat); enantiomer (S)-(+) = +13.5°
(S)-(+)-2-bromobutane — [α]_D²⁰ = +35.7° (neat); enantiomer (R)-(−) = −35.7°
(R)-(+)-limonene — [α]_D²⁰ = +123° (neat); enantiomer (S)-(−) = −123°
(R)-(+)-glyceraldehyde — [α]_D²⁰ = +8.7° (water, c = 2)
meso-tartaric acid — [α]_D²⁰ = 0° (always)

Polarimetry score: 0 / 5

Round 3 — Theoretical yield, percent yield, and enantiomeric excess

Three calculation problems including a polarimetry-based ee determination. ±5% tolerance.

Yield/ee score: 0 / 3

Team Questions

Discuss with your team. Each question targets one of the CLOs covered in this lab.

Question 1 — CLO 1. Identify the stereocentre(s) in 3-bromo-2-butanol [CH₃CH(OH)CH(Br)CH₃]. How many stereoisomers does this compound have, considering all R/S combinations?

Question 2 — CLO 2. Among the four stereoisomers of 3-bromo-2-butanol, identify which (if any) are enantiomer pairs and which are diastereomeric pairs. Are any meso? Justify briefly.

Question 3 — CLO 5. (R)-2-bromobutane is treated with NaCN in DMSO. Predict the product configuration and the operative mechanism. Now repeat the question for (R)-2-bromobutane heated in 50:50 water/ethanol with no added base.

Question 4 — CLO 5 (stereospecific addition). trans-2-butene is treated with Br₂ in CCl₄. The product has two stereocentres. What is its stereochemistry — does it form (2R,3R)/(2S,3S) racemate, or the (2R,3S) meso? Why?

Question 5 — CLO 7. Design a synthesis that converts (R)-2-butanol to (S)-2-butanol (a "stereoinversion" of the original substrate). Use TsCl chemistry plus an SN2 nucleophile.

Question 6 — CLO 8 (polarimetry). A student measures α = +5.4° for a sample of 2-butanol in a 1 dm cell at concentration 0.50 g/mL. Pure (S)-(+)-2-butanol has [α]_D²⁰ = +13.5°. Calculate (a) the observed [α], (b) the % ee, and (c) the ratio of (S):(R).

Example Lab Report

Sample report demonstrating the expected ACS-style format.

Chirality: Identification, Reaction Stereochemistry, and Stereocontrolled Synthesis

Submitted by: [Student Name]

Course: Organic Chemistry · Section: 221-A · Date: April 24, 2026

Abstract

This comprehensive lab examined chirality through identification, reaction stereochemistry, and stereocontrolled synthesis. Ten organic molecules were classified as chiral or achiral; their stereocentres were located and assigned R or S using CIP priority rules. Eight reactions of stereodefined substrates were analysed to predict the stereochemistry of the product (inversion in SN2, racemisation in SN1, anti-periplanar selectivity in E2, syn vs anti addition to alkenes). Six stereocontrolled syntheses required choosing reagent sequences that achieved a target configuration. SDSs for four reagents were interpreted for hazards, PPE, and disposal. Polarimetry data for five chiral compounds were analysed to confirm identity and configuration. Theoretical yield, percent yield, and enantiomeric excess (ee) were calculated for the SN2 conversion of (R)-2-bromobutane to (S)-2-cyanobutane (achieved 88% yield, 96% ee).

Introduction

Chirality is the central organising principle of three-dimensional organic chemistry. Two enantiomers have identical scalar properties (m.p., b.p., density, and most spectroscopic properties) but differ in their interaction with chiral environments — including biological receptors and plane-polarised light. The mechanism of every reaction has a definite consequence for stereochemistry: SN2 inverts a stereocentre; SN1 racemises; E2 requires anti-periplanar geometry; addition reactions to alkenes are either syn or anti depending on mechanism. Predicting the stereochemistry of a product is therefore as important as predicting its connectivity. This lab tested all five CLOs through this lens.

Reagents and Hazards (drawn from SDS)

Reagent	Major hazard	PPE	Key §9 data	Disposal
NaCN	Acutely toxic (H300, H310, H330); evolves HCN if acidified	Goggles, double gloves, lab coat, fume hood; never acidify	White hygroscopic solid; M.W. 49.01	Strict cyanide waste container; never mix with acid
TsCl	Lachrymator; H315, H318, H335	Goggles, gloves, lab coat, fume hood mandatory	White solid; m.p. 65–69 °C	Hydrolyse in aqueous NaOH, then aqueous waste
BH₃·THF	Flammable (H225); reacts violently with water (H260)	Goggles, gloves, lab coat, fume hood; dry and inert	Colourless solution; pungent	Quench cautiously with isopropanol then water; aqueous waste
Br₂ in CCl₄	Toxic by inhalation (H330); CCl₄ suspected carcinogen (H351)	Goggles, double gloves, fume hood mandatory	Reddish-brown solution; volatile	Reduce excess Br₂ with Na₂S₂O₃, halogenated waste
OsO₄	Acutely toxic (H300, H330); volatile	Goggles, double gloves, fume hood; small quantities only	Pale yellow crystals; volatile; toxic vapour	Reduce excess with Na₂SO₃, then heavy-metal waste
Sharpless reagent (Ti(OiPr)₄ + DET)	Mild irritant; flammable solvent	Goggles, gloves, lab coat, fume hood; dry and cold	Pale yellow solution in CH₂Cl₂	Quench with water; halogenated organic waste

Results — Section I: Chirality Identification (10 molecules)

#	Molecule	Chiral?	Stereocentres	Configuration
1	2-butanol	Yes	1 (C2)	R or S — depends on which enantiomer drawn
2	2-propanol	No	0	—
3	(R)-glyceraldehyde	Yes	1 (C2)	R
4	(2R,3R)-tartaric acid	Yes	2 (C2, C3)	R, R
5	(2R,3S)-tartaric acid (meso)	No (overall achiral)	2 (C2, C3)	R, S — meso
6	cis-1,2-dimethylcyclohexane	No (mirror plane)	2	R, S — meso
7	trans-1,2-dimethylcyclohexane	Yes	2	R, R or S, S (one pair of enantiomers)
8	(R)-1-phenylethanol	Yes	1	R
9	2-methyl-2-butanol	No (not 4 different groups)	0	—
10	(2S,3R)-2-bromo-3-chlorobutane	Yes	2	S, R (NOT meso — different halogens; different priorities at the two centres)

Results — Section II: Stereochemistry of Reactions

#	Substrate	Reagent	Stereo outcome	Mechanism
1	(R)-2-bromobutane	NaCN, DMSO	(S)-2-cyanobutane (inversion)	SN2 — Walden inversion
2	(R)-2-iodobutane	aqueous EtOH, heat	Racemic 2-butanol	SN1 — planar cation
3	(2R,3R)-2-bromo-3-methylbutane	NaOEt, EtOH, heat	(E)-2-methyl-2-butene (anti-periplanar removal)	E2
4	(2R,3S)-2-bromo-3-methylbutane	NaOEt, EtOH, heat	(Z)-2-methyl-2-butene (anti-periplanar from the other diastereomer)	E2 — diastereomers give different alkene geometry
5	cis-2-butene	Br₂ in CCl₄	(2R,3R) + (2S,3S) racemic mixture	Halonium opening: anti addition
6	trans-2-butene	Br₂ in CCl₄	(2R,3S) — meso product	Anti addition to trans alkene
7	cyclohexene	BH₃; then H₂O₂/NaOH	cis-2-methylcyclohexanol — but only racemic from cyclohexene (no pre-existing chirality)	Hydroboration: syn addition; oxidation: retention
8	1-methylcyclohexene	BH₃; then H₂O₂/NaOH	trans-2-methylcyclohexanol (anti-Markovnikov, syn addition)	Syn addition + anti-Markov regiochem

Polarimetry Verification

The product of the (R)-2-bromobutane → (S)-2-cyanobutane SN2 reaction was characterised by polarimetry to confirm both identity and stereochemical purity:

Measurement	Conditions	Observed	Interpretation
Observed rotation α	1 dm cell, c = 0.50 g/mL (neat sample), 20 °C, Na D-line	+11.4°	—
Calculated [α]_D²⁰	α / (c × ℓ)	+22.8°	—
Literature [α]_D²⁰ for (S)-2-cyanobutane	neat	+23.7°	—
Enantiomeric excess	22.8 / 23.7 × 100	96.2% ee	98:2 (S):(R)

The high ee (96%) confirms that the SN2 mechanism is highly stereospecific. The small loss (4%) is consistent with a minor SN1 contribution or trace racemisation during workup.

Theoretical and Percent Yield

Experiment — (R)-2-bromobutane → (S)-2-cyanobutane (SN2 with NaCN):

Quantity	Value
Starting material: (R)-2-bromobutane (M.W. 137.02)	0.685 g (5.00 mmol)
Stoichiometry	1:1 with NaCN
Theoretical mass: (S)-2-cyanobutane (M.W. 83.13)	5.00 × 10⁻³ × 83.13 = 0.4157 g
Mass isolated	0.366 g
Percent yield	0.366 / 0.4157 × 100 = 88.0%
Enantiomeric excess (from polarimetry)	96.2%
Mass of (S)-(+)-2-cyanobutane (the desired enantiomer)	0.366 × 0.981 = 0.359 g (98.1% of isolated mass is S)

Discussion

SN2 stereochemistry confirmed by polarimetry. The (R)-bromide gave 96% ee of the (S)-cyanide — a high but not perfect Walden inversion. The 4% loss of ee can be attributed to a small SN1 contribution (the 2° substrate has some SN1 character in DMSO at higher temperatures), or to trace HCN formation that protonates and racemises the cyanide. The high ee confirms the SN2 mechanism and demonstrates the value of polarimetry as a stereochemical probe.

Stereospecificity of Br₂ addition. Entries 5 and 6 of Section II demonstrate that anti addition to cis-2-butene gives the racemic (2R,3R)/(2S,3S) pair, while anti addition to trans-2-butene gives only the meso (2R,3S) form. This is a classic illustration of stereospecificity: different starting geometries give different stereochemical outcomes by the same mechanism. The result is also a powerful diagnostic — running Br₂ on an unknown alkene and observing whether the product is meso or chiral identifies the alkene geometry.

Hydroboration: syn + anti-Markovnikov. Entry 8 (1-methylcyclohexene + BH₃; then H₂O₂/NaOH) gives trans-2-methylcyclohexanol — the OH ends up on the LESS substituted carbon (anti-Markovnikov), and the H and OH are added to the SAME face (syn). On a cyclic substrate, "same face" forces the cis arrangement of B and H; oxidation with retention gives trans-OH and CH₃ on opposite faces of the ring.

The role of polarimetry. Without polarimetry, the SN2 inversion (R → S) would be invisible — the chemical reaction "looks" the same as any other halide-to-cyanide conversion. The optical rotation measurement is what tells us the mechanism is happening with high fidelity. Combined with a chemical-ee determination (e.g., chiral HPLC or NMR with a chiral shift reagent), polarimetry is the standard tool for assessing stereochemical purity.

Conclusions

All five CLOs covered in this lab were addressed. CLO 1 was demonstrated through structural classification of the 10 molecules. CLO 2 was demonstrated through R/S assignments and the recognition of meso compounds (entries 5, 6). CLO 5 was demonstrated through correct stereochemistry predictions for the 8 reactions. CLO 7 was demonstrated through six stereocontrolled syntheses. CLO 8 was demonstrated through correct interpretation of four SDSs, accurate matching of polarimetry data to compound identity, and successful yield + ee calculations.

References

1. Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed., Oxford University Press, 2012, Ch. 14, 16, 41.
2. Eliel, E. L.; Wilen, S. H. Stereochemistry of Organic Compounds, Wiley, 1994.
3. Sigma-Aldrich Safety Data Sheets, accessed online March 2026.
4. Polarimeter operating manual, Rudolph Research Autopol IV, 2024.

Practice Questions

Test your understanding. Try each one before peeking at the hint.

Practice 1 — CLO 1+2

2,3-dichlorobutane has how many stereoisomers? List them. Which is meso? Are the chiral pair enantiomers or diastereomers of the meso form?

Hint: Three stereoisomers total: (2R,3R), (2S,3S), (2R,3S) — the (2S,3R) is the same compound as (2R,3S) due to the internal mirror plane. (2R,3R) and (2S,3S) are an enantiomeric pair. The meso form (2R,3S) is a diastereomer of both members of the chiral pair.

Practice 2 — CLO 1 (R/S assignment)

Assign R or S to the stereocentre in: (a) (CH₃)₂CHCH(OH)CH₃ — drawn with OH on a wedge (forward); (b) ethyl 2-bromopropanoate, drawn with Br on a dash (back).

Hint: (a) priority OH > CH(CH₃)₂ > CH₃ > H. With OH coming forward (wedge), H is behind (dash). Trace 1→2→3: clockwise → R. (b) Ethyl 2-bromopropanoate: priority Br > C(=O)OEt > CH₃ > H. With Br dashed back, H is forward (wedge). Trace 1→2→3 and INVERT (because H is in front): the visible CW becomes actually S; CCW becomes R.

Practice 3 — CLO 2 (specific rotation)

Pure (R)-(+)-glyceraldehyde has [α]_D²⁰ = +8.7° (water, c = 2). A student measures α = +1.5° in a 1 dm cell at c = 0.5 g/mL. What is the % ee, and which enantiomer is in excess?

Hint: [α] = α / (c × ℓ) = 1.5 / (0.5 × 1) = +3.0°. Compare to pure: 3.0 / 8.7 × 100 = 34.5% ee. Sample is in excess (R)-(+) (positive sign). Composition: 67.2% (R), 32.8% (S).

Practice 4 — CLO 5 (SN2 vs SN1)

(S)-2-iodobutane is dissolved in: (a) DMSO + NaCN; (b) 50% aqueous ethanol with no added base. Predict the stereochemistry of the major product in each case.

Hint: (a) SN2 with strong nucleophile in polar aprotic solvent → backside attack with inversion → (R)-2-cyanobutane. (b) SN1/E1 in protic solvent through 2° cation → racemic 2-butanol (plus some 2-butene). The stereochemistry is lost in the cation intermediate.

Practice 5 — CLO 5 (stereospecific addition)

Predict the product of: (a) trans-2-pentene + Br₂; (b) cis-2-pentene + Br₂; (c) cyclohexene + BH₃, then H₂O₂/NaOH. State the relationship between the products of (a) and (b) — same compound, enantiomers, or diastereomers?

Hint: Br₂ adds anti via halonium. (a) trans-2-pentene + Br₂ → (2R,3S) and (2S,3R) — but here C2 and C3 have different substituents (CH₃ vs CH₂CH₃), so the two are NOT meso; they are an enantiomeric pair of (2R,3S)/(2S,3R). (b) cis-2-pentene + Br₂ → (2R,3R) and (2S,3S) — also an enantiomeric pair. The products of (a) and (b) are diastereomers (different relative configuration at the two stereocentres). (c) cyclohexene + BH₃ then H₂O₂/NaOH → racemic cyclohexanol (no stereocentre, since both ring carbons after addition are equivalent).

Practice 6 — CLO 7 (synthesis)

Design a synthesis of (S)-2-aminobutane from (R)-2-butanol. Use SN2 chemistry to invert the stereocentre. Note: amines are typically introduced via the azide intermediate.

Hint: Three steps. (1) (R)-2-butanol + TsCl/pyridine → (R)-2-butyl tosylate (no inversion at C — only OH is replaced). (2) + NaN₃ in DMSO → (S)-2-azidobutane (SN2 with one inversion: R → S). (3) Reduction with H₂/Pd-C or LiAlH₄ → (S)-2-aminobutane (reduction at N does not affect C stereocentre).

Practice 7 — CLO 7 (resolution)

A racemic mixture of 2-bromobutane is obtained from a synthesis. Outline a method to resolve the two enantiomers (separate them). What property of the (R)/(S) pair is exploited?

Hint: Direct separation is impossible because R and S have identical scalar properties. The classic method: convert each enantiomer into a diastereomeric salt or ester using a CHIRAL resolving agent (e.g., react with (R)-mandelic acid via the carboxylate of the corresponding chiral acid). The two diastereomeric esters have different scalar properties (m.p., solubility) and can be separated by recrystallisation. After separation, hydrolyse to recover the resolved 2-bromobutanes.

Practice 8 — CLO 8 (SDS)

Sodium azide (NaN₃) is commonly used in stereochemistry work for the alkyl halide → azide → amine sequence (Practice 6). Before opening the bottle, the student reads the SDS. State (a) the most serious hazard, (b) the special incompatibility (something it should NEVER contact), (c) the disposal route, (d) the first-aid response if NaN₃ contacts skin.

Hint: (a) Acutely toxic by all routes (H300, H310 — inhibits cytochrome oxidase like cyanide). (b) NEVER contact heavy metals (especially Cu pipework!) — forms shock-sensitive heavy-metal azides that are violently explosive. Lab plumbing must NEVER carry NaN₃ waste. (c) Reduce with sodium nitrite + dilute acid (NaN₃ + HNO₂ → N₂ + N₂O + H₂O), then aqueous waste. (d) Brush off any solid, then water rinse 15 min, immediate medical attention.

Practice 9 — CLO 8 (polarimetry)

A student isolates a sample of synthetic (S)-2-butanol after reductive amination. Polarimetry: 0.250 g of sample dissolved in 10.0 mL of ethanol gives α = +0.27° in a 1 dm cell at 20 °C. Pure (S)-(+)-2-butanol has [α]_D²⁰ = +13.5° (in ethanol). Calculate (a) the observed [α], (b) the % ee, (c) the (S):(R) ratio.

Hint: c = 0.250 g / 10.0 mL = 0.025 g/mL. [α] = 0.27 / (0.025 × 1) = +10.8°. % ee = 10.8 / 13.5 × 100 = 80%. Composition: (S):(R) = 90:10 (because in 80% ee, the major is (100+80)/2 = 90% and the minor is (100-80)/2 = 10%).

Practice 10 — CLO 8 (yield and ee)

A two-step asymmetric synthesis: (1) (R)-2-bromobutane + NaCN/DMSO → (S)-2-cyanobutane (Step 1: 88% chemical yield, 96% ee). (2) Hydrolysis with H₃O⁺/H₂O → (S)-2-methylbutanoic acid (Step 2: 95% chemical yield, ee preserved). Starting from 1.50 g of (R)-2-bromobutane (M.W. 137.02), calculate the mass of (S)-2-methylbutanoic acid (M.W. 102.13) and the final ee.

Hint: moles 2-bromobutane = 1.50 / 137.02 = 0.01095 mol. After Step 1 (88%): 0.01095 × 0.88 = 0.00964 mol of nitrile. After Step 2 (95%): 0.00964 × 0.95 = 0.00915 mol of acid. Mass = 0.00915 × 102.13 = 0.935 g. Overall chemical yield = 0.88 × 0.95 = 83.6%. ee preserved through hydrolysis (no stereocentre is touched), so final ee = 96%.