AQA A-Level Chemistry: Optical isomerism (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 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 whole section is A-level only. At A-level it is assessed on Paper 2 (organic chemistry), and Paper 3 can test any content. Questions 7 and 9 also draw on reactions from 3.3.8 and 3.3.4, and question 10 on 3.3.13.3 (A-level only).

Learn the ideas first in the optical isomerism study guide and the revision notes. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics help you decide what to revise first.

Data for this set. Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0. In 3D drawings, a plain line is a bond in the plane of the paper, ▼ marks a wedge (towards you) and ⋮ a hashed bond (away from you).

Questions

1. State what is meant by a chiral centre. [1]

2. State the meaning of the term enantiomers. [2]

3. The four alcohols below are isomers of C₅H₁₂O. Identify the one that has a chiral centre and give the four groups attached to that carbon.

pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃; 2-methylbutan-2-ol, CH₃CH₂C(OH)(CH₃)₂; 3-methylbutan-1-ol, (CH₃)₂CHCH₂CH₂OH; pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃ [2]

4. Draw 3D representations of the two enantiomers of butan-2-amine, CH₃CH(NH₂)CH₂CH₃. [2]

5. Describe how plane-polarised light could be used to show that a solution of a single enantiomer is optically active and that a racemic mixture of the same compound is not. [4]

6. Explain why a racemic mixture of 2-bromobutane is optically inactive. [2]

7. Butanone, CH₃COCH₂CH₃, reacts with KCN followed by dilute acid.

(a) Name the organic product. [1] (b) Explain why the product has optical isomers. [1] (c) Explain why the product is formed as a racemic mixture. [3] (d) A student starts with 1.44 g of butanone. Calculate the maximum mass of product that could form, and the mass of each enantiomer in it. [3]

8. There are eight structural isomers with the formula C₅H₁₁Cl.

(a) Three of them have a chiral centre. Name these three. [3] (b) State the total number of isomers of C₅H₁₁Cl, including stereoisomers. [1]

9. Hydrogen bromide reacts with propene and with but-1-ene by electrophilic addition.

(a) Name the major product from propene and explain why it has no optical isomers. [2] (b) The major product from but-1-ene is 2-bromobutane. Explain why it forms as a racemic mixture. [3]

10. (A-level only, 3.3.13.3) A drug is made as a racemic mixture, but only one enantiomer acts on the target enzyme. Explain why. [3]

11. Compound P is pent-3-en-2-ol, CH₃CH=CHCH(OH)CH₃.

(a) Identify the chiral centre in P and give the four groups attached to it. [2] (b) Explain why P also shows E–Z isomerism. [2] (c) Deduce the number of stereoisomers of P. Explain your answer. [2] (d) A student says that 3-methylpent-1-ene, CH₂=CHCH(CH₃)CH₂CH₃, also has four stereoisomers. Evaluate this statement. [2]

Answers

1. A carbon atom bonded to four different atoms or groups [1] [1] Examiner insight: “Four different groups” with no mention of carbon (or “asymmetric carbon”) is vague; name the carbon atom.

2. Stereoisomers: same structural formula, different arrangement in space [1]. They are non-superimposable mirror images of each other [1] [2] Examiner insight: “Mirror images” alone is not enough for the second mark; every molecule has a mirror image, so write “non-superimposable”.

3. Pentan-2-ol [1]. C2 carries H, OH, CH₃ and CH₂CH₂CH₃ [1] [2] Examiner insight: Pentan-3-ol looks similar; its C3 carries two identical ethyl groups, so choosing it scores nothing here.

4. C2 is chiral (H, NH₂, CH₃, C₂H₅). A correct 3D shape for one enantiomer, with two bonds in the plane, one wedge and one hashed bond [1]; the second drawn as its mirror image [1] [2]

            CH3           mirror           CH3
             |              :               |
             C              :               C
           / ⋮ ▼            :             ▼ ⋮ \
       C2H5  H  NH2         :         H2N  H  C2H5

Examiner insight: Two flat drawings with all bonds as plain lines show no 3D arrangement and cannot show the two forms are different.

5. Pass plane-polarised light (light through a polarising filter) through the solution [1]. Use a second polarising filter after the sample and compare its angle with the angle for the pure solvent [1]. The single enantiomer rotates the plane of the polarised light, so the angle changes [1]. The racemic mixture gives no rotation, the same as the solvent [1] [4] Examiner insight: Comparing with the pure solvent is the control; without it, “the light is rotated” has nothing to be measured against.

6. Each enantiomer rotates the plane of plane-polarised light by the same angle in opposite directions [1]. There are equal amounts of each, so the rotations cancel [1] [2] Examiner insight: Saying the mixture “has no chiral centre” is wrong; each molecule is chiral, and the answer must be about cancelling rotations.

7. (a) 2-hydroxy-2-methylbutanenitrile, CH₃CH₂C(OH)(CH₃)CN [1] (b) C2 carries four different groups, OH, CN, CH₃ and C₂H₅, so it is a chiral centre [1] (c) The C=O group is planar (trigonal planar) [1]. CN⁻ can attack the carbonyl carbon from either side of the plane [1]. Attack from each side is equally likely, so equal amounts of the two enantiomers form [1] (d) n(butanone) = 1.44 / 72.0 = 0.0200 mol [1]. Mr of product (C₅H₉NO) = 99.0, so maximum mass = 0.0200 × 99.0 = 1.98 g [1]. Each enantiomer = 1.98 / 2 = 0.990 g [1] [8] Examiner insight: In (d), show the Mr and the mole calculation; with clear working, a slip in the Mr is easy to spot and the later steps can still be followed.

8. (a) 2-chloropentane [1]; 1-chloro-2-methylbutane [1]; 2-chloro-3-methylbutane [1] (b) 8 structural isomers + 3 extra enantiomers = 11 [1] [4] Examiner insight: 3-chloropentane is the common wrong choice in (a): its C3 carries two ethyl groups.

9. (a) 2-bromopropane [1]. Its C2 carries two CH₃ groups, so no carbon has four different groups [1] (b) The secondary carbocation, CH₃CH₂C⁺HCH₃, is planar around the positive carbon [1]. Br⁻ can attack from either side of the plane with equal probability [1]. So equal amounts of the two enantiomers form [1] [5] Examiner insight: Both “planar” and “equal probability” must appear in (b); “Br⁻ attacks from both sides” alone does not explain why the amounts are equal.

10. The active site of the enzyme is stereospecific, with a particular 3D shape [1]. Only one enantiomer has its groups arranged so that it fits and bonds to the active site [1]. The mirror-image enantiomer cannot bind in the same way, so only half of the racemic mixture is active [1] [3] Examiner insight: “The other enantiomer has a different shape” is not enough; link the mirror-image arrangement to the fit with the active site.

11. (a) C2, the carbon bonded to OH [1]. Its groups are H, OH, CH₃ and CH=CHCH₃ [1] (b) Rotation about the C=C bond is restricted [1]. Each carbon of the C=C carries two different groups (C3: H and CH(OH)CH₃; C4: H and CH₃) [1] (c) There are two E–Z forms, and each has the chiral centre, so each exists as two enantiomers [1]. 2 × 2 = 4 stereoisomers [1] (d) 3-methylpent-1-ene has a chiral centre at C3 (H, CH₃, C₂H₅, CH=CH₂), so it has two enantiomers [1]. C1 of the C=C carries two H atoms, so there is no E–Z isomerism; it has 2 stereoisomers and the student is wrong [1] [8] Examiner insight: In (c) and (d) the reasoning carries its own mark here, so deal with E–Z and optical isomerism separately rather than writing a bare number.

Where marks are usually lost

• Writing “mirror images” without “non-superimposable”.
• Missing “equal amounts” in a racemate definition, or “equal and opposite” in explaining optical inactivity.
• Choosing a carbon with two identical groups (such as C3 in pentan-3-ol or 3-chloropentane).
• Drawing enantiomers flat, or drawing the “mirror image” with a swapped group as well, which gives the same molecule twice.
• Leaving out “planar” or “equal probability of attack from either side” when explaining racemate formation.
• Calling the product of propanone and KCN chiral; its central carbon carries two CH₃ groups.
• Not showing Mr and moles in a calculation, so a slip cannot be traced.
• Counting stereoisomers without checking each type separately (E–Z and optical).

Next steps

• Optical isomerism revision notes
• Optical isomerism study guide
• Alkenes study guide
• AQA A-Level Chemistry exam preparation
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

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).