Haloalkanes and Haloarenes · Lesson 9 of 16
Chirality and Substitution Stereochemistry
“Connect three-dimensional molecular structure with optical activity, inversion, retention and racemisation.”
• Distinguish chiral molecules, achiral molecules and enantiomeric pairs. • Interpret wedge–dash drawings without confusing rotation of a drawing with bond inversion. • Separate optical rotation from absolute configuration and chemical identity. • Explain the stereochemical consequences and limitations of SN2 and SN1 models.
A mirror image may have the same connections but a different identity
Your two hands have the same general parts but cannot be superimposed in three dimensions while keeping all parts aligned. A molecule is chiral when its mirror image cannot be superimposed by rotation and translation. Mirror-related, non-superimposable stereoisomers are enantiomers. An achiral molecule is superimposable on its mirror image. Molecular rotation does not break bonds; reflection can change handedness in a way that rotation cannot undo.
For the simple molecules considered here, one tetrahedral carbon bearing four different substituents is a useful sufficient signal of chirality. In propan-2-ol the central carbon has OH,H,CH₃,CH₃, so two groups are identical and the molecule is achiral. In butan-2-ol it has OH,H,CH₃,C₂H₅, four distinct groups, producing a pair of enantiomers. The pair of mirror drawings of propan-2-ol are not enantiomers because they represent the same superimposable molecule. More complicated molecules require whole-molecule symmetry analysis rather than simply counting marked carbons.
Problem
Classify CH₃CH(OH)Br, CH₃CHBr₂, pentan-2-ol, pentan-3-ol, 2-bromobutane and 1-bromobutane.
- 1.CH₃CH(OH)Br has H,CH₃,OH,Br at one carbon and is chiral; CH₃CHBr₂ has two identical Br groups and is achiral.
- 2.Pentan-2-ol has methyl and propyl groups at its OH-bearing carbon and is chiral. Pentan-3-ol has two identical ethyl groups and is achiral.
- 3.2-Bromobutane has H,Br,methyl,ethyl and is chiral. 1-Bromobutane has no tetrahedral carbon with four different substituents and is achiral.
Other single-centre examples include 2-chlorobutane, CHO–CH(OH)–CH₂OH, bromochloroiodomethane HC(Br)(Cl)I and 2-bromopropanoic acid CH₃CHBrCOOH. Check the complete substituent, not only its first atom: CH₃ and CH₂OH both begin with carbon but differ further along the branch, so they are not identical groups.
Wedges show depth, not stronger chemical bonds
A plain line lies in the drawing plane. A solid wedge projects toward the viewer; a hashed wedge projects away. These conventions show a tetrahedral arrangement without drawing a physically flat carbon. Rotating the whole model or the viewing direction can change the appearance while preserving configuration. Exchanging two groups at a single stereocentre reverses handedness; two successive exchanges restore the original arrangement. A useful check is whether all four attachments can be matched by a genuine spatial rotation.
Absolute configurations may be labelled R or S after assigning substituent priorities and viewing the lowest-priority group away. Br outranks O, O outranks C, and C outranks H by directly attached atomic number; when two attached atoms are the same, compare outward atoms at the first difference. For 2-bromobutane, the priority order is Br>ethyl>methyl>H. With H directed away, clockwise 1→2→3 is R and counterclockwise is S. The label must be recalculated after a chemical transformation if substituent priorities change; geometric inversion is not defined merely as exchanging the written letter R with S.
What a polarimeter actually measures
Ordinary light contains oscillations in many planes; a polariser selects plane-polarised light. Some samples rotate its plane as it passes through. A polarimeter measures the rotation under specified wavelength, temperature, path length, concentration and solvent conditions. A positive sign denotes the conventional dextrorotatory direction and a negative sign the opposite direction. Equal amounts of a pair of enantiomers have equal and opposite rotations under the same measurement conditions.
The optical sign (+)/(−) is experimental information, not something deduced from the name R/S or from a wedge leaning right. R does not mean positive rotation. A racemic mixture contains equal amounts of the two enantiomers and has zero net rotation from that pair. Thus zero rotation can arise either from achiral molecules or from cancellation in a racemate; it does not by itself prove every molecule in the sample is achiral.
Problem
In a fixed measurement setup, pure A would rotate +20° and its enantiomer B −20° at the same total concentration. Predict a 50:50 and a 75:25 A:B mixture.
- 1.For 50:50, the two contributions cancel: 0.50×20+0.50×(−20)=0°. This is a racemate.
- 2.For 75:25, rotation is 0.75×20−0.25×20=+10°, assuming ideal additive rotation at the stated total concentration.
- 3.The mixture is enriched in A and still consists of individually chiral molecules. The numerical signs are supplied measurements, not predictions from an R/S drawing.
SN2 inversion and retention at an untouched centre
Backside SN2 attack replaces the leaving group from the opposite side and inverts the arrangement at the carbon actually undergoing substitution. In a chiral alkyl halide that carbon can therefore yield a single inverted substitution configuration under an ideal clean SN2 pathway. The product’s optical sign is not guaranteed to be opposite to the reactant’s, because they are different chemical compounds. Likewise a changed R/S descriptor must be checked against the new priority list rather than used alone as proof of the geometry.
Problem
Consider replacing the terminal OH of chiral HOCH₂CH(CH₃)CH₂CH₃ by Cl.
- 1.The stereocentre is C2, not the terminal CH₂OH carbon. The replacement occurs at C1.
- 2.If the transformation occurs without rearrangement or reaction at C2, its four bonds retain their spatial arrangement. This is retention at the untouched stereocentre.
- 3.The product is 1-chloro-2-methylbutane. Its rotation can have a different sign from that of 2-methylbutan-1-ol because a new chemical substance has been made. Optical sign alone cannot establish geometric inversion.
SN1 opens two faces, but does not guarantee exactly equal attack
An ideal freely solvated carbocation at the reacting centre is approximately planar. A nucleophile can approach either face, forming the two configurations of a substituted tetrahedral product. This explains the tendency toward racemisation when an optically enriched halide follows a stepwise pathway. A perfectly equal two-face probability would give a 50:50 product mixture.
Real SN1 reactions often give incomplete racemisation because the leaving anion and solvent environment can remain associated with the cation, shielding one face. A result enriched in inversion does not automatically prove pure SN2, and a planar intermediate does not guarantee an experimentally exact 50:50 mixture. Other stereocentres or chiral surroundings can also remove equivalence of the two faces. These qualifications preserve the useful ideal model without confusing it with a universal product ratio.
| Outcome | What is preserved or changed | What cannot be inferred automatically |
|---|---|---|
| Retention | Same spatial arrangement at the specified centre | Same optical sign for different compounds |
| Inversion | Opposite local tetrahedral arrangement | Opposite optical sign without measurement |
| Racemisation | Formation of an equal enantiomer pair in the ideal full limit | That all stepwise substitutions are exactly 50:50 |
| Rotation of a model | Same molecule and configuration | That a turned drawing is a new stereoisomer |
Quick Quiz
Quiz
Which molecule has one simple tetrahedral stereocentre?
Explanation: Its OH-bearing carbon has four different groups.
What does a racemic mixture contain?
Explanation: Equal opposite rotations cancel under identical conditions.
What does a solid wedge indicate?
Explanation: It is a depth convention for a three-dimensional drawing.
What can be concluded from a product being (+) while its chemically different reactant was (−)?
Explanation: Rotation is a property of each measured substance, not a universal configuration code.
Why can an SN1 product fail to be exactly racemic?
Explanation: Associated leaving ions and solvent can preserve a facial bias.
Practice Problems
- Explain why butan-2-ol is chiral but propan-2-ol is not.
- Can a zero-rotation sample contain chiral molecules?
- Explain why an SN2 transition at a stereocentre gives inversion.
- A mixture gives +6° when the pure positive enantiomer would give +30° at the same total concentration. Find the ideal A:B proportions.
- Why can terminal substitution in 2-methylbutan-1-ol retain its stereocentre?
- Explain why “SN1 always gives exactly 50:50” is too strong.
Butan-2-ol has OH,H,methyl,ethyl at its tetrahedral C2, four different groups. Propan-2-ol has two methyl groups, so its mirror image can be superimposed. Identical molecular formula connectivity between two mirror drawings does not alone decide chirality; the three-dimensional superposition test does.
Key Takeaways
• Chirality means non-superimposability on a mirror image. • One tetrahedral carbon with four different groups provides a useful simple chirality test. • Enantiomers share connectivity but differ in handedness. • Wedge–dash notation shows depth; rotating a drawing does not invert a stereocentre. • Optical rotation sign is measured and is not the same as R/S configuration. • A racemate has cancelling rotations even though its molecules can be chiral. • SN2 gives geometric inversion, while SN1 can give partial or complete racemisation. • Retention at an untouched centre does not require unchanged optical sign after a different compound is formed.