Skip to notes
wise otter.Chemistry
Contents
9476 / 2027
Organic Chemistry

Lesson 5 of 34

Chirality is non-superimposability on a mirror image

A tetrahedral centre with four different groups is a common cause, but molecular symmetry still matters.

A-Level 9476 (2026-2027)

  • I can identify a chiral centre.Syllabus

    Syllabus 9476, 11.2(c). Identify a chiral centre.

  • I can assess molecular chirality using centres and symmetry.Syllabus

    Syllabus 9476, 11.2(d). Assess molecular chirality using centres and symmetry.

  • I can connect optical activity with chiral molecules.Syllabus

    Syllabus 9476, 11.2(e). Connect optical activity with chiral molecules.

  • I can compare enantiomer physical properties.Syllabus

    Syllabus 9476, 11.2(f). Compare enantiomer physical properties.

  • I can compare enantiomer chemical properties.Syllabus

    Syllabus 9476, 11.2(g). Compare enantiomer chemical properties.

  • I can relate stereoisomerism to biological properties.Syllabus

    Syllabus 9476, 11.2(h). Relate stereoisomerism to biological properties.

  • I can identify stereochemical features from structures.Syllabus

    Syllabus 9476, 11.2(j). Identify stereochemical features from structures.

Chirality is non-superimposability on a mirror image

A tetrahedral centre with four different groups is a common cause, but molecular symmetry still matters.

A chiral centre is usually a carbon bonded to four different atoms or groups. In butan-2-ol, the carbon with the OH is joined to OH, H, CH3 and CH2CH3. So there are two arrangements that are mirror images and cannot be placed on top of each other. These are enantiomers. Compare whole groups, not just the first atom. Methyl and ethyl both start with carbon, but they are different groups.

A pair of enantiomers shown with wedges

Each tetrahedral carbon has CH3 and H attached by two ordinary bonds in the plane of the page, with a solid wedge to OH towards the viewer and a hashed wedge to CO2H away. Reflection in the vertical mirror exchanges every left-right position, giving non-superimposable arrangements.

The solid wedge points towards you; the separated hashed segments point away. Reflect every attached group to draw the partner. A planar cross without wedge information would not establish the stereochemistry.

When a molecule has several chiral centres, look at the whole molecule. A plane of symmetry inside it can make it achiral, even though it has chiral centres. A molecule can also be chiral for other reasons. So use the real shape and symmetry of the structure you are given. Do not just count centres.

Two chiral centres, but an achiral whole molecule

One stereoisomer of butane-2,3-diol has a plane perpendicular to the page through the midpoint of the central carbon-carbon bond. Reflection exchanges the two central carbons, the two methyl groups, the two OH groups projecting towards the viewer and the two H atoms pointing away. The complete structure maps onto itself.

The dashed line is the edge of an internal mirror plane. Both OH bonds are solid wedges; both C-H bonds are hashed wedges. The symmetry conclusion concerns this specific stereoisomer, not every possible arrangement of butane-2,3-diol.

Worked exampleCheck centres first, then the whole structure

Is the illustrated stereoisomer of CH3CH(OH)CH(OH)CH3 chiral?

Try it first. Then show the worked stepsHide the worked steps
  1. At either starred carbon the four groups are H, OH, CH3 and CH(OH)CH3. Each is therefore a chiral centre.
  2. Reflect the entire structure in the marked plane. A front-facing OH maps to the other front-facing OH, and a rear H to the other rear H; the carbon skeleton and methyl groups also match.
  3. The molecule has an internal plane of symmetry, so its mirror image is superimposable. Counting two centres alone would give the wrong conclusion.
  4. If the OH and H positions at only one centre are exchanged, this internal plane is lost and a different stereoisomer results. It is essential to inspect the wedges, not only the condensed formula.
Answer

This illustrated stereoisomer is achiral despite its two chiral centres. It is often called a meso form; the symmetry reasoning is what establishes the answer.

Properties of an enantiomer pair in an achiral environment
PropertyComparison
Melting point, boiling point and ordinary solubilityIdentical under the same achiral conditions.
Plane-polarised lightEqual concentrations of pure enantiomers rotate it by equal magnitudes in opposite directions under matching conditions.
Reactions with achiral reagentsThe same ordinary chemical properties and rates under matching conditions.
Interactions with chiral moleculesCan differ, for example in binding to an enzyme or receptor.

An optically active sample rotates plane-polarised light. It contains chiral molecules whose rotations do not cancel. A racemic mixture has equal amounts of two enantiomers, so their rotations cancel. So a sample that does not rotate light may still contain chiral molecules. The direction of rotation must be measured. You cannot read it from a wedge drawing.

Receptors and enzymes in the body are chiral. Two stereoisomers can bind to them differently, so they can differ in activity, side effects or how the body breaks them down. That is why the molecular formula and functional groups alone cannot tell you how a drug behaves. The three-dimensional shape matters too. This does not mean one particular form is always the better one.

Check your understanding

Is 2-methylpropan-2-ol chiral, since its OH-bearing carbon is tetrahedral?

  1. Yes. Any sp3 carbon bonded to an OH group has a mirror image that cannot be superimposed.
  2. No. Chiral centres must carry a hydrogen atom, and this carbon carries none.
  3. No. That carbon carries three identical methyl groups, so it is not a chiral centre.
Show the answer

No. That carbon carries three identical methyl groups, so it is not a chiral centre.

A tetrahedral carbon is chiral only if its four groups are all different. Here three of the four groups are CH3, so the molecule matches its mirror image.