Explain when a reaction creates equal amounts of enantiomers and why a racemate has no net optical rotation.
Cancellation does not make each molecule achiral
A racemic mixture contains equal amounts of the two enantiomers. Their equal and opposite rotations cancel, so the mixture has no net effect on plane-polarised light. Each molecule remains chiral. A zero rotation could also arise from an achiral substance, so optical inactivity alone does not identify a racemate.
A mixture containing unequal amounts can be optically active. Extension example: if pure enantiomers at a fixed total concentration would give +10.0° and −10.0°, a 70:30 mixture gives 0.70(10.0) + 0.30(−10.0) = +4.0°, assuming rotations are additive under unchanged conditions. Do not treat this optional calculation as a required new AQA formula.
Two equally accessible faces can make a racemate
A trigonal-planar carbonyl group can be attacked from either face. When an achiral reactant reacts in an achiral environment and neither face is favoured, formation of a new chiral carbon gives equal amounts of its two configurations. The explanation needs both planarity and equal probability of attack, followed by checking that the product really has four different groups.
For ethanal plus HCN, the product CH₃CH(OH)CN has CH₃, H, OH and CN at its new tetrahedral centre and can form a racemate. Methanal gives HOCH₂CN, which still has two H atoms at that carbon, so it does not give this enantiomer pair. Propanone gives (CH₃)₂C(OH)CN, also achiral because the two methyl groups are identical.
An unsymmetrical ketone such as butan-2-one gives CH₃C(OH)(CN)CH₂CH₃, whose four groups differ. The attacking nucleophile is CN⁻, using its carbon lone pair; HCN describes the overall addition stoichiometry, not the attacking species in the mechanism.
Diagram placeholder
Two-face attack on ethanal to add
Labels to include:
- Trigonal-planar CH₃–C(=O)–H
- Cδ⁺ and Oδ⁻
- CN⁻ carbon lone pair approaching from either face
- Curly arrow from carbon lone pair to carbonyl C
- Curly arrow from C=O π bond to O
- Tetrahedral O⁻ intermediate, then O protonation
- Mirror products CH₃CH(OH)CN with equal proportions under achiral conditions
Show approach above and below the original carbonyl plane, then two wedge/dash products with the same connectivity. The carbonyl carbon becomes a chiral tetrahedral carbon; the nitrile carbon is an additional carbon atom.
The same logic applies to a planar carbocation
A planar carbocation intermediate may be attacked from either face by a nucleophile. In addition of HBr to but-1-ene, the pathway giving 2-bromobutane can therefore yield its two enantiomers in equal amounts under achiral conditions. This does not mean every structural product is produced in equal amounts: the ratio of 1-bromobutane to 2-bromobutane is a separate issue.
June 2023 Paper 2 Q04.2 assessed a different alkene and required a linked explanation from the planar carbocation to attack from either side and optical isomers. The lesson is to identify the actual intermediate and the four product groups, rather than claiming every addition reaction gives optical isomers.
A chiral binding site can distinguish the pair
Enzymes and receptors have three-dimensional, chiral binding sites. One enantiomer may align the necessary groups for binding while its mirror image cannot make the same interactions. The two can therefore differ in biological activity even though their properties in an achiral environment are alike.
Do not assume the second enantiomer is always inactive or harmful, or that purifying one automatically removes every risk. Its behaviour must be established for the particular molecule; some enantiomers can interconvert in biological conditions. The specified enzyme and drug applications are developed in Amino Acids, Proteins and DNA.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Why is a racemate optically inactive?Show answer
It contains equal amounts of enantiomers whose rotations are equal and opposite under the same conditions. Their effects cancel; the individual molecules are still chiral.
Q2. Why does adding HCN to propanone not produce optical isomers at the former carbonyl carbon?Show answer
The product has OH, CN and two identical CH₃ groups at that carbon. It is not bonded to four different groups.
Q3. Give the full reasoning for a racemate from an achiral unsymmetrical ketone and CN⁻.Show answer
The carbonyl carbon is trigonal planar. In an achiral environment the two faces are equally accessible, so attack occurs equally from either side. If the product has four different groups, equal amounts of its enantiomers form.
Q4. Does a racemic 2-bromobutane product imply a 50:50 mixture of 1-bromobutane and 2-bromobutane?Show answer
No. Racemic describes the ratio between the two enantiomers of 2-bromobutane. The relative amounts of constitutional products are a different ratio.
Q5. Explain why enantiomers may have different effects on an enzyme.Show answer
The enzyme active site is chiral. Only one arrangement may position all the required groups for the relevant binding interactions, so the mirror image can bind differently or fail to bind effectively.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- AQA 7405 organic chemistry specification — 3.3.7 coverage and required skills.
- Chemrevise: Optical isomerism — Coverage checklist; explanations, data exercises and quick checks on this page are original Finesse material.
- AQA June 2023 Paper 2 mark scheme — Q04.2 pp17–18 and examiner report Q04.2 p3: planar carbocation, attack from either face and product chirality. Single-centre scope follows the specification.
- AQA June 2023 Paper 2 examiner report — Read alongside the question-specific marking guidance; not a universal wording checklist.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.