Edexcel International A Level Chemistry: Chirality (YCH11) – 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 Topic 15A: Chirality of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 15.1 to 15.5. Chirality is International A2 (IA2) material; the specification places it under Unit 4: Rates, Equilibria and Further Organic Chemistry. Questions 7 and 8 also use Topic 11 rate equations.

Related pages: the chirality study guide, the revision notes, the printable checklist, the course hub and a free diagnostic.

Questions

1. Explain the term chiral centre, and describe how the two enantiomers of a compound are related. [2]

2. Butane-1,2,3-triol has the structure HOCH₂CH(OH)CH(OH)CH₃. Identify each chiral centre in the molecule and list the four groups attached to each one. [3]

3. State whether each compound has optical isomers. Give a reason in each case.

(a) 1-chloro-2-methylpentane, ClCH₂CH(CH₃)CH₂CH₂CH₃ [1] (b) 3-bromo-3-ethylhexane, CH₃CH₂CBr(C₂H₅)CH₂CH₂CH₃ [1] (c) 2-methylheptanenitrile, CH₃CH(CN)CH₂CH₂CH₂CH₂CH₃ [1] (d) heptan-4-ol, CH₃CH₂CH₂CH(OH)CH₂CH₂CH₃ [1]

4. 2-aminopentanoic acid is CH₃CH₂CH₂CH(NH₂)COOH.

(a) Identify the chiral centre. [1] (b) Draw 3D diagrams of the two optical isomers of 2-aminopentanoic acid. [3]

5. A solution of one enantiomer of compound Z is placed in a polarimeter. The plane of polarisation is rotated by +14.6°.

(a) Define optical activity. [2] (b) State the reading for a solution of the other enantiomer at the same concentration, under the same conditions. [1] (c) State what is meant by a racemic mixture, and give the reading for a racemic mixture of Z at the same total concentration. [2] (d) Explain why the light used must be monochromatic. [1]

6. A single enantiomer of 3-iodohexane, CH₃CH₂CHICH₂CH₂CH₃, is hydrolysed to hexan-3-ol under two different sets of conditions. Under conditions A the hexan-3-ol rotates plane-polarised light. Under conditions B it has no effect on plane-polarised light.

Deduce the mechanism operating under each set of conditions. Explain your reasoning. [4]

7. A single enantiomer of 4-bromo-4-methyloctane, CH₃CH₂CH₂CBr(CH₃)CH₂CH₂CH₂CH₃, is hydrolysed by aqueous sodium hydroxide at a fixed temperature. The product is 4-methyloctan-4-ol.

Run	[RBr] / mol dm⁻³	[OH⁻] / mol dm⁻³	Rate at start / mol dm⁻³ s⁻¹
1	0.015	0.060	2.1 × 10⁻⁵
2	0.045	0.060	6.3 × 10⁻⁵
3	0.045	0.120	6.3 × 10⁻⁵

(a) Show that the reactant is chiral. [1] (b) Deduce the order with respect to each reactant and write the rate equation. [3] (c) Calculate the rate constant, with units. [2] (d) Predict whether the 4-methyloctan-4-ol formed rotates plane-polarised light. Explain your answer in terms of the mechanism. [3]

8. A single enantiomer of 2-chloroheptane, CH₃CHClCH₂CH₂CH₂CH₂CH₃, reacts with aqueous hydroxide ions to form heptan-2-ol. The heptan-2-ol formed is optically active.

Run	[RCl] / mol dm⁻³	[OH⁻] / mol dm⁻³	Rate at start / mol dm⁻³ s⁻¹
1	0.025	0.030	4.5 × 10⁻⁶
2	0.050	0.030	9.0 × 10⁻⁶
3	0.050	0.090	2.7 × 10⁻⁵

(a) Deduce the order with respect to each reactant and write the rate equation. [3] (b) Calculate the rate constant, with units. [2] (c) Explain how the optical activity of the product supports the mechanism suggested by the rate equation. [3] (d) The 2-chloroheptane used was the (+) enantiomer. A student predicts that the heptan-2-ol must be the (−) enantiomer. Comment on this prediction. [1]

9. Octan-2-one, CH₃COCH₂CH₂CH₂CH₂CH₂CH₃, reacts with HCN in the presence of KCN.

(a) Give the structural formula of the organic product. [1] (b) Identify the chiral centre in the product. [1] (c) Explain whether the product sample is optically active. [3]

10. A student wants optical activity data as evidence for the mechanism of HCN addition to a ketone, using either heptan-4-one, CH₃CH₂CH₂COCH₂CH₂CH₃, or heptan-3-one, CH₃CH₂COCH₂CH₂CH₂CH₃. Explain which ketone to choose. [3]

11. A single enantiomer of 1-chloro-2-methylpentane is hydrolysed by aqueous hydroxide ions. The 2-methylpentan-1-ol formed is optically active. A student concludes that this proves an SN2 mechanism. Evaluate this conclusion. [3]

Answers

1. A carbon atom that has four different atoms or groups attached to it [1]. Enantiomers are object and non-superimposable mirror images [1]. [2] Examiner insight: Writing only that they are “mirror images” misses the second mark; say non-superimposable.

2. C2 is a chiral centre: H, OH, CH₂OH and CH(OH)CH₃ [1]. C3 is a chiral centre: H, OH, CH₃ and CH(OH)CH₂OH [1]. Both sets of four groups are all different, so the molecule has two chiral centres, and C1 (CH₂) and C4 (CH₃) are not chiral [1]. [3] Examiner insight: Write each group out in full; listing “C” or “the chain” for a group does not show that the four groups are different.

3. (a) Yes. C2 carries H, CH₃, CH₂Cl and C₃H₇, four different groups [1]. (b) No. C3 carries Br, C₃H₇ and two identical C₂H₅ groups [1]. (c) Yes. C2 carries H, CN, CH₃ and C₅H₁₁ [1]. (d) No. C4 carries H, OH and two identical C₃H₇ groups [1]. [4] Examiner insight: A yes or no with no reason usually earns nothing; the mark is for the groups on the relevant carbon.

4. (a) C2, which carries H, NH₂, COOH and C₃H₇ [1]. (b)

        COOH         :         HOOC
       /             :             \
 C3H7--C===> NH2     :     H2N <===C--C3H7
       |             :             |
       H (hashed)    :    (hashed) H

C2 drawn as a tetrahedral centre: two in-plane bonds, a solid wedge and a hashed wedge [1]. All four groups bonded through the correct atom (N of NH₂, C of COOH) [1]. The second drawing is a true mirror image of the first [1]. [4] Examiner insight: Two drawings of the same molecule turned round earn no mirror-image mark, however neat the wedges.

5. (a) The ability of a single optical isomer to rotate the plane of polarisation [1] of plane-polarised monochromatic light [1]. (b) −14.6° (equal angle, opposite direction) [1]. (c) Both enantiomers mixed in equal amounts [1]; the reading is 0° because the equal and opposite rotations cancel [1]. (d) The angle of rotation depends on the wavelength of the light, so one wavelength gives a fixed, comparable value [1]. [6] Examiner insight: In (a), “bends light” or “rotates light” does not earn credit; it is the plane of polarisation that is rotated.

6. Reactant and product are both chiral at C3 (H, C₂H₅, C₃H₇ and I or OH), so the product’s rotation is valid evidence [1]. Conditions A: SN2 [1], because OH⁻ attacks from the side opposite the iodine as the C–I bond breaks, giving a single enantiomer with its arrangement inverted [1]. Conditions B: SN1, because the C–I bond breaks first to give a planar carbocation that OH⁻ attacks from either face with equal probability, giving a racemic mixture [1]. [4] Examiner insight: Naming SN1 and SN2 without the planar intermediate or the backside attack gains little; the explanation carries the marks.

7. (a) C4 carries CH₃, Br, C₃H₇ and C₄H₉: four different groups [1]. (b) Runs 1 and 2: [RBr] triples and the rate triples, so first order in RBr [1]. Runs 2 and 3: [OH⁻] doubles and the rate is unchanged, so zero order in OH⁻ [1]. Rate = k[RBr] [1]. (c) k = 2.1 × 10⁻⁵ ÷ 0.015 [1] = 1.4 × 10⁻³ s⁻¹ [1]. (d) The product will not rotate plane-polarised light [1]. The rate equation fits SN1: C–Br breaks in the rate-determining step, giving a planar carbocation [1]. OH⁻ attacks either face with equal probability, giving a racemic mixture of the chiral 4-methyloctan-4-ol [1]. [9] Examiner insight: In (c), a rate constant with missing or wrong units loses the final mark.

8. (a) Runs 1 and 2: [RCl] doubles and the rate doubles, so first order in RCl [1]. Runs 2 and 3: [OH⁻] triples and the rate triples, so first order in OH⁻ [1]. Rate = k[RCl][OH⁻] [1]. (b) k = 4.5 × 10⁻⁶ ÷ (0.025 × 0.030) [1] = 6.0 × 10⁻³ dm³ mol⁻¹ s⁻¹ [1]. (c) The rate equation shows both species take part in a single step, which fits SN2 [1]. OH⁻ attacks C2 from the side opposite the chlorine as the C–Cl bond breaks, so the arrangement around C2 is inverted [1]. Only one enantiomer of heptan-2-ol forms, so the product is optically active, not racemic as SN1 would give [1]. (d) Not justified: inversion changes the arrangement in space, but the product’s sign of rotation must be measured [1]. [9] Examiner insight: In (c), link both pieces of evidence; an answer that discusses only the rate equation, or only the optical activity, misses a mark.

9. (a) CH₃C(OH)(CN)CH₂CH₂CH₂CH₂CH₂CH₃ (2-hydroxy-2-methyloctanenitrile) [1]. (b) The carbon carrying OH and CN, with four different groups: CH₃, OH, CN and C₆H₁₃ [1]. (c) Not optically active [1]. The C=O group is planar, and CN⁻ attacks the carbonyl carbon from above or below with equal probability [1]. So the two enantiomers are made in a 1:1 ratio, a racemic mixture, and their rotations cancel [1]. [5] Examiner insight: Saying only “a racemic mixture forms” does not earn the mark for the reason; the planar C=O and equal attack from both sides are needed.

10. Heptan-3-one should be chosen [1]. Its HCN product carries C₂H₅, C₄H₉, OH and CN on the new carbon, so it is chiral and a racemic product shows as zero rotation [1]. The heptan-4-one product has two identical propyl groups there, so it is achiral and shows zero rotation whatever the mechanism [1]. [3] Examiner insight: “Choose heptan-3-one” with no reasoning earns at most the first mark; both products have to be considered.

11. The conclusion is not valid [1]. The chiral centre is C2, but substitution takes place at C1, so none of the bonds to C2 are broken [1]. The product stays optically active whichever mechanism operates, so the optical data cannot tell SN1 from SN2; the rate equation would be needed instead [1]. [3] Examiner insight: For “evaluate”, give a clear judgement and then the reason; describing SN2 on its own is not an evaluation.

Where marks are usually lost

• Missing a chiral centre that carries no hydrogen, such as C4 in 4-bromo-4-methyloctane.
• Counting a carbon with two identical chains, such as two propyl or two ethyl groups, as chiral.
• Drawing the second “enantiomer” as the same molecule rotated rather than reflected.
• Naming the mechanism without the planar carbocation, the planar C=O or the backside attack that explains the optical result.
• Using optical data as evidence when the product is achiral, or when the chiral centre is not the carbon being attacked.
• Predicting the sign of rotation of an SN2 product from the sign of the reactant.

Next steps

• Chirality revision notes
• Chirality study guide
• Course hub
• Printable checklist
• Try one of our all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 15A: Chirality, outcomes 15.1 to 15.5.