AQA A-Level Chemistry: Optical isomerism (7405)

This guide teaches section 3.3.7 Optical isomerism of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards. The specification marks the whole section A-level only, so it is not part of the AS course. At A-level, Paper 2 assesses organic chemistry (section 3.3), and Paper 3 can test any content and any practical skill.

For quick recall, use the revision notes. To test yourself, use the practice questions. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are.

What this unit covers

Spec point (3.3.7)	What you must be able to do	Status
Optical isomerism as stereoisomerism	Explain that it arises from chirality, limited to molecules with a single chiral centre	A-level only
Asymmetric carbon atoms	Recognise a chiral centre in a 2D or 3D structure	A-level only
Enantiomers	Describe them as non-superimposable mirror images that differ in their effect on plane-polarised light	A-level only
Drawing enantiomers	Draw the structural formulas and displayed formulas of both enantiomers, showing the 3D arrangement	A-level only
Racemic mixtures	Define a racemate; explain how racemates form and why they are optically inactive	A-level only
Practical skills	Pass polarised light through a sucrose solution (a skills opportunity, not a required practical)	A-level only

Two later sections build on this one. Section 3.3.8 asks you to explain why KCN followed by dilute acid gives a mixture of enantiomers, and 3.3.13.3 asks why a stereospecific active site bonds to only one enantiomer. Both are A-level only and are previewed below. Optical isomerism in octahedral complexes with bidentate ligands is part of 3.2.5.3 and is covered in the transition metals guide.

Stereoisomers: a quick recap

Section 3.3.1.3 defines stereoisomers as compounds with the same structural formula but a different arrangement of atoms in space. You have met one kind already: E–Z isomerism, caused by restricted rotation about a C=C bond (see the introduction to organic chemistry). Optical isomerism is the second kind. It has a different cause: a carbon atom bonded to four different atoms or groups.

Chiral centres

An asymmetric carbon atom is a carbon bonded to four different atoms or groups. It is a chiral centre. A molecule with one chiral centre is chiral: it cannot be superimposed on its mirror image.

The four bonds around a saturated carbon point to the corners of a tetrahedron (bond angle 109.5°, from the bonding guide). When all four groups differ, the molecule and its mirror image cannot be turned to match, in the same way your left and right hands cannot.

Method: finding a chiral centre

1. Ignore every carbon with a double or triple bond. It has fewer than four groups.
2. Ignore any carbon with two or more hydrogen atoms (CH₃, CH₂). Two of its groups are identical.
3. For each remaining carbon, write its four groups in full. Compare whole groups, not just the first atom: –CH₂CH₃ and –CH₂CH₂CH₃ both start with CH₂, but they are different groups.
4. If all four are different, mark that carbon with an asterisk (*).

Worked example 1: spot the chiral centre

Which of these has a chiral centre? Mark it.

(a) 3-methylhexane, CH₃CH₂CH(CH₃)CH₂CH₂CH₃ (b) 3-methylpentane, CH₃CH₂CH(CH₃)CH₂CH₃ (c) 2-hydroxypropanoic acid, CH₃CH(OH)COOH

(a) The only carbon with one H is C3. Its groups are H, CH₃, CH₂CH₃ and CH₂CH₂CH₃. All four differ, so C3 is chiral: CH₃CH₂C*H(CH₃)CH₂CH₂CH₃.

(b) C3 carries H, CH₃, CH₂CH₃ and CH₂CH₃. Two ethyl groups are the same, so there is no chiral centre.

(c) C2 carries H, CH₃, OH and COOH. All four differ: CH₃C*H(OH)COOH. The COOH carbon has a C=O bond, so it cannot be chiral.

Enantiomers

A chiral molecule exists as two optical isomers, called enantiomers. They are non-superimposable mirror images of each other. Both have the same structural formula, so they are stereoisomers.

Enantiomers have the same physical properties, such as melting point and boiling point, and they react in the same way with most reagents. They differ in two ways that matter at A-level:

• Effect on plane-polarised light. One enantiomer rotates the plane of plane-polarised light clockwise. The other rotates it by the same angle anticlockwise. A substance that rotates the plane is optically active.
• Interaction with other chiral molecules, such as enzymes (section 3.3.13.3, below).

Drawing enantiomers in 3D

A flat structural formula such as CH₃CH(NH₂)COOH looks the same for both enantiomers, so you must show the 3D arrangement around the chiral carbon. Use these conventions:

• a plain line is a bond in the plane of the paper
• a wedge (drawn solid, written ▼ below) is a bond coming towards you
• a hashed bond (written ⋮ below) is a bond going away from you

Draw two bonds in the plane, one wedge and one hashed bond. Then draw the second molecule as a reflection, with a mirror line between them.

Worked example 2: the enantiomers of 2-aminopropanoic acid

Draw the two enantiomers of 2-aminopropanoic acid, CH₃CH(NH₂)COOH.

Step 1. Find the chiral centre. C2 carries H, CH₃, NH₂ and COOH, so it is chiral.

Step 2. Draw the first enantiomer: COOH up in the plane, CH₃ lower left in the plane, NH₂ lower right on a wedge, and H behind on a hashed bond.

Step 3. Reflect it. Groups that were on the left move to the right. The bond types do not change.

            COOH          mirror          COOH
             |              :              |
             C              :              C
           / ⋮ ▼            :            ▼ ⋮ \
        H3C  H  NH2         :         H2N  H  CH3

This is a structural formula in 3D, because CH₃, NH₂ and COOH are written as groups. For a displayed formula, keep the same arrangement but draw every bond inside each group, for example C=O and O–H in the COOH group and all three C–H bonds in the methyl group. The 3D arrangement around the chiral carbon is the same in both kinds of drawing.

A useful check. Swapping any two groups on a chiral carbon turns one enantiomer into the other. If you cannot tell whether two drawings are the same molecule, count the swaps needed to turn one into the other: one swap means they are enantiomers, two swaps means they are the same molecule.

Plane-polarised light

Ordinary light vibrates in all planes. A polarising filter lets through light vibrating in one plane only: plane-polarised light.

If plane-polarised light passes through a solution of one enantiomer, the plane is rotated. To measure this, a second polarising filter (the analyser) is placed after the sample. You turn it until the light is darkest and compare the angle with the angle needed for the pure solvent. A difference means the sample is optically active.

The specification suggests passing polarised light through a solution of sucrose to see this. Set up two polarising filters with water between them and turn one until the light is darkest. Replace the water with sucrose solution: the light comes back, and the filter must be turned through an angle to darken it again. Sucrose has several chiral centres, so it is a demonstration only; drawing questions are limited to molecules with a single chiral centre.

Racemic mixtures

A racemic mixture, or racemate, is a mixture of equal amounts of the two enantiomers.

Why a racemate is optically inactive. Each enantiomer rotates the plane of polarised light by the same angle, in opposite directions. In a racemate there are equal amounts of each, so the rotations cancel and there is no overall rotation.

How racemates form

A racemate forms when a reaction makes a new chiral centre from a planar carbon. The reagent can attack either face of the planar group, and each face is equally likely to be attacked, so equal amounts of the two enantiomers form.

Nucleophilic addition to a carbonyl group (section 3.3.8). The C=O carbon in an aldehyde or ketone is trigonal planar. In the reaction with KCN followed by dilute acid, the cyanide ion attacks the carbonyl carbon from above or below the plane with equal probability. If the product’s carbon carries four different groups, the product is a racemate. The full mechanism belongs to section 3.3.8, which explains why aldehydes and unsymmetrical ketones give mixtures of enantiomers.

Electrophilic addition to an alkene. Adding HBr to but-1-ene gives 2-bromobutane as the major product (see the alkenes guide). The secondary carbocation is planar around the positive carbon, so the bromide ion attacks either face equally. The 2-bromobutane formed is racemic.

Worked example 3: will the product be a racemate?

Each carbonyl compound reacts with KCN followed by dilute acid. Say whether each product is a racemate.

Carbonyl compound	Product	Groups on the new C–OH carbon	Racemate?
Propanal, CH₃CH₂CHO	2-hydroxybutanenitrile, CH₃CH₂CH(OH)CN	H, OH, CN, CH₂CH₃	Yes
Propanone, CH₃COCH₃	2-hydroxy-2-methylpropanenitrile, (CH₃)₂C(OH)CN	OH, CN, CH₃, CH₃	No: two CH₃ groups, so no chiral centre
Pentan-3-one, CH₃CH₂COCH₂CH₃	2-ethyl-2-hydroxybutanenitrile, (C₂H₅)₂C(OH)CN	OH, CN, C₂H₅, C₂H₅	No: two ethyl groups

An aldehyde (other than methanal) or an unsymmetrical ketone gives a racemate. A symmetrical ketone gives a product with two identical groups, so there is no chiral centre.

Stereospecific enzymes (preview of 3.3.13.3)

An enzyme’s active site is stereospecific: its shape fits only one enantiomer of a substrate or drug, because only that enantiomer holds its groups in the right positions to bond to the site. So if a drug is made as a racemate, only half of it may fit the target active site.

Worked example 4: counting isomers

How many isomers, including stereoisomers, have the formula C₄H₉Br?

Step 1. List the structural isomers: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. That is 4.

Step 2. Check each for a chiral centre. Only 2-bromobutane has one: C2 carries H, Br, CH₃ and CH₂CH₃. 2-bromo-2-methylpropane has three CH₃ groups on C2, so it is not chiral.

Step 3. 2-bromobutane exists as two enantiomers, so it counts twice. None of the four has a C=C bond, so there is no E–Z isomerism.

Total: 4 structural isomers + 1 extra stereoisomer = 5 isomers.

Common errors

• Comparing only the first atom of each group, so two different chains (such as ethyl and propyl) are called identical, or two ethyl groups are missed.
• Marking a CH₂ or C=O carbon as a chiral centre.
• Writing “mirror images” without “non-superimposable”. Every molecule has a mirror image; enantiomers are mirror images that cannot be superimposed.
• Drawing the second enantiomer with a wedge changed to a hashed bond as well as reflecting it. A reflection plus an extra swap gives the same molecule back.
• Explaining racemate inactivity without saying the rotations are equal and opposite, or without saying the amounts are equal.
• Explaining racemate formation without naming the planar group and attack from both sides with equal probability.

Where to go next

Work through the practice questions, then use the revision notes for final checks. The halogenoalkanes and alcohols guides cover reactions of compounds such as 2-bromobutane and butan-2-ol, and the exam preparation guide explains how the papers are set out.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.7 Optical isomerism (A-level only).