OxfordAQA A-Level Chemistry: Optical isomerism (9620) – 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.

This set covers section 3.3.7 (Optical isomerism, subsection 3.3.7.1) of OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. The whole section is International A2 content, so it is International A-level only. Unit 4 assesses sections 3.3.7 to 3.3.16, and the synoptic Unit 5 can draw on any content. Questions 6, 9 and 10 also use reactions from sections 3.3.4.2 and 3.3.8.1, and question 11 uses E–Z isomerism from 3.3.1.3.

Learn the ideas first in the optical isomerism study guide and the revision notes. The course hub, the printable checklist and the free diagnostics help you decide what to work on next.

Data for this set. Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0. 3D drawing key: plain line = bond flat on the page; ▼ = wedge, out of the page; ⋮ = hashed bond, into the page.

Questions

1. The four dichloropropanes are 1,1-dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane and 2,2-dichloropropane. Identify the one with an asymmetric carbon atom and give the four groups bonded to that carbon. [2]

2. (a) State what is meant by a racemic mixture. [1] (b) Explain why a racemic mixture has no overall effect on plane-polarised light. [2]

3. Draw the structural formulas of the two enantiomers of 2-methylbutanoic acid, CH₃CH₂CH(CH₃)COOH, showing their 3D arrangement. [2]

4. Draw the displayed formulas of the two enantiomers of 2-chloropropan-1-ol, CH₃CHClCH₂OH. Show the 3D arrangement around the chiral carbon. [3]

5. The two enantiomers of a compound cannot be told apart by their melting points. Explain why, and state how they can be told apart. [2]

6. 2-methylbut-2-ene, (CH₃)₂C=CHCH₃, reacts with hydrogen bromide by electrophilic addition. Two products form.

(a) Name the major product and explain why it does not show optical isomerism. [2] (b) The minor product is 2-bromo-3-methylbutane. Explain why it forms as a racemic mixture. [3]

7. Four carboxylic acids have the molecular formula C₅H₁₀O₂.

(a) Name the one that shows optical isomerism. [1] (b) Give the total number of carboxylic acids with this formula, counting each enantiomer separately. Explain your answer. [2]

8. Three solutions of 2-methylbutanoic acid, P, R and S, have the same concentration. Under the same conditions, P rotates the plane of plane-polarised light by 8° clockwise, R rotates it by 8° anticlockwise and S does not rotate it.

(a) State the relationship between the compound in P and the compound in R. [1] (b) Suggest what S contains and explain why it does not rotate the plane. [2] (c) Equal volumes of P and R are mixed. Predict the effect of the mixture on plane-polarised light. [1]

9. A student adds an excess of KCN, followed by dilute acid, to 3.44 g of pentanal, CH₃CH₂CH₂CH₂CHO.

(a) Give the name of the hydroxynitrile formed. [1] (b) Explain why a racemic mixture of this product forms. [3] (c) Calculate the maximum mass of product that could form. [2] (d) The yield is 80.0%. Calculate the mass of each enantiomer obtained. [1]

10. Hexan-2-one, CH₃COCH₂CH₂CH₂CH₃, reacts with HCN to form a hydroxynitrile.

(a) Write an equation for this reaction, using structural formulas. [1] (b) Name the hydroxynitrile formed. [1] (c) Give the four groups bonded to the chiral centre in the product. [1] (d) Explain how the reaction gives equal amounts of the two enantiomers. [3] (e) A sample of the product made this way has no effect on plane-polarised light. A student concludes that the product has no chiral centre. Evaluate this conclusion. [2]

11. 3-chlorobut-1-ene, CH₂=CHCHClCH₃, and 1-chlorobut-2-ene, ClCH₂CH=CHCH₃, are structural isomers.

(a) Explain why 3-chlorobut-1-ene shows optical isomerism but not E–Z isomerism. [2] (b) Explain why 1-chlorobut-2-ene shows E–Z isomerism but not optical isomerism. [2] (c) State the total number of stereoisomers of the two compounds together. [1]

Answers

1. 1,2-dichloropropane [1]. C2 carries H, Cl, CH₃ and CH₂Cl [1] [2] Examiner insight: 1,1-dichloropropane is the usual wrong pick; its C1 has two Cl atoms, so it cannot gain the first mark or the second.

2. (a) A mixture of equal amounts of the two enantiomers [1] (b) The two enantiomers turn the plane of polarisation through equal angles but in opposite senses [1]. The amounts are equal, so the rotations cancel [1] [3] Examiner insight: An answer to (b) that says the molecules “are not chiral” contradicts the definition in (a) and earns nothing.

3. C2 is the chiral centre (H, CH₃, C₂H₅, COOH). First enantiomer drawn in 3D, using two flat bonds plus a wedge and a hashed bond [1]; second enantiomer a true reflection of the first [1] [2]

             CH3          :          CH3
             ▼            :          ▼
    C2H5 --  C  --  COOH  : HOOC --  C  --  C2H5
             ⋮            :          ⋮
             H            :          H

Examiner insight: Any sensible 3D layout works, provided the second drawing is a true reflection; check that each group sits on the same type of bond in both.

4. C2 is the chiral centre (H, Cl, CH₃, CH₂OH) [1]. First enantiomer as a displayed formula with a wedge and a hashed bond at C2 [1]. Second enantiomer as its mirror image, also fully displayed [1] [3]

         H     Cl    H
         |     ▼     |
    H -- C --- C --- C -- O -- H
         |     ⋮     |
         H     H     H

              H     Cl    H
              |     ▼     |
    H -- O -- C --- C --- C -- H
              |     ⋮     |
              H     H     H

Examiner insight: Writing CH₃ or OH as a group anywhere in a displayed formula loses the drawing marks, even when the 3D arrangement is right.

5. The enantiomers have the same bonds and the same intermolecular forces, so they have identical melting points [1]. Plane-polarised light separates them: one turns its plane clockwise and the other anticlockwise, by equal angles [1] [2] Examiner insight: “They react differently” is not a valid way to tell them apart here; the answer needs plane-polarised light.

6. (a) 2-bromo-2-methylbutane [1]. Its C2 carries two CH₃ groups (with Br and C₂H₅), so it has no asymmetric carbon [1] (b) The secondary carbocation (CH₃)₂CHC⁺HCH₃ has a flat (planar) arrangement around C⁺ [1]. Br⁻ may bond on the top face or the bottom face, and the two are equally probable [1]. The two enantiomers are therefore made in a 1:1 ratio [1] [5] Examiner insight: In (b), state where the plane is (the positive carbon); a vague “the molecule is flat” does not show which carbon becomes chiral.

7. (a) 2-methylbutanoic acid [1] (b) Pentanoic acid, 3-methylbutanoic acid and 2,2-dimethylpropanoic acid have no chiral centre; 2-methylbutanoic acid exists as two enantiomers [1]. Total = 4 + 1 = 5 [1] [3] Examiner insight: The explanation earns its own mark here, so name the acids rather than writing a bare number.

8. (a) They are the two enantiomers of 2-methylbutanoic acid [1] (b) S is a racemic mixture (equal amounts of both enantiomers) [1]. The equal and opposite rotations cancel [1] (c) No overall rotation, because it contains equal amounts of the two enantiomers [1] [4] Examiner insight: In (b), “S has no chiral centre” is wrong: all three solutions contain the same chiral compound.

9. (a) 2-hydroxyhexanenitrile, CH₃CH₂CH₂CH₂CH(OH)CN [1] (b) In pentanal the C=O carbon and its three attached atoms lie in one plane [1]. The cyanide ion can add to this carbon from above the plane or from below it [1]. Neither route is favoured, so a 50:50 mixture of the enantiomers results [1] (c) Mr of pentanal = 86.0, so n = 3.44 / 86.0 = 0.0400 mol [1]. Mr of C₆H₁₁NO = 113.0, so maximum mass = 0.0400 × 113.0 = 4.52 g [1] (d) 0.800 × 4.52 = 3.616 g; each enantiomer = 3.616 / 2 = 1.81 g [1] [7] Examiner insight: Show both Mr values in (c) and the moles of pentanal, so every step of your method is visible; a bare final answer gives no evidence of correct working if it turns out to be wrong.

10. (a) CH₃COCH₂CH₂CH₂CH₃ + HCN → CH₃C(OH)(CN)CH₂CH₂CH₂CH₃ [1] (b) 2-hydroxy-2-methylhexanenitrile [1] (c) OH, CN, CH₃ and CH₂CH₂CH₂CH₃ [1] (d) The C=O carbon in hexan-2-one is trigonal planar [1]. CN⁻ approaches it from one face or the other [1]. Both faces are attacked at the same rate, giving equal quantities of each enantiomer [1] (e) The conclusion is wrong: the product does have a chiral centre (C2) [1]. It forms as a racemate, so the equal and opposite rotations of the two enantiomers cancel [1] [8] Examiner insight: For (e), an “evaluate” answer needs a judgement plus a reason; “the student is wrong” on its own earns nothing.

11. (a) C3 carries H, Cl, CH₃ and CH=CH₂, so it is a chiral centre [1]. The terminal CH₂= carbon has two identical H atoms, so E and Z forms are impossible [1] (b) Rotation about the C=C is restricted and both alkene carbons have two unlike substituents (C2: H and CH₂Cl; C3: H and CH₃) [1]. No carbon has four different groups: C1 is CH₂Cl and C4 is CH₃ [1] (c) 2 enantiomers + 2 E–Z isomers = 4 [1] [5] Examiner insight: Deal with each type of stereoisomerism separately; one sentence covering both usually misses a reason for one of them.

Where marks are usually lost

• Calling a carbon chiral when two of its groups are the same, such as C1 in 1,1-dichloropropane.
• Leaving “non-superimposable” out of the meaning of enantiomers.
• Saying a racemate “has no chiral molecules” instead of explaining that equal and opposite rotations cancel.
• Drawing structural formulas with every bond flat, so the two enantiomers look identical.
• Showing CH₃ or OH as groups in a “displayed formula”.
• Explaining racemate formation without naming the planar carbon, the attack from either side and the equal chance of each.
• Treating the product of a symmetrical ketone or of methanal with KCN as chiral.
• Giving a bare total when counting isomers, with no list or reason.
• In yield calculations, halving the maximum mass but forgetting the percentage yield, or the other way round.

Next steps

• Optical isomerism revision notes
• Optical isomerism study guide
• Introduction to organic chemistry
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.7 Optical isomerism (3.3.7.1), International A2 content.