A Level Chemistry: Optical Isomerism and Chirality — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Work through these before looking at the answers. Write full answers — the marks are for the reasoning, not the conclusion.

Related: Optical Isomerism and Chirality revision notes

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Section A — short answer

1. State the condition for a carbon atom to be a chiral centre. [1]

2. Define the term enantiomers. [2]

3. State two properties that are identical between a pair of enantiomers, and one property that differs. [3]

4. Define the term racemic mixture, and state its effect on plane polarised light. [2]

5. Explain why propan-2-ol, CH₃CH(OH)CH₃, does not have a chiral centre. [2]

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Section B — structured

6. 2-hydroxypropanoic acid (lactic acid), CH₃CH(OH)COOH, has a chiral centre at C2.

(a) Identify the four different groups attached to the chiral carbon. [1]

(b) Explain, in terms of light, how the two enantiomers of this molecule could be distinguished experimentally. [2]

(c) A sample of 2-hydroxypropanoic acid, made in a school laboratory by a reaction that creates this chiral centre from a non-chiral starting material, shows no optical activity — even though it definitely contains chiral molecules. Explain this observation. [3]

7. Ibuprofen is a chiral drug; only one of its two enantiomers is an effective anti-inflammatory, while the other is essentially inactive.

(a) Explain, in terms of biological receptors, why the two enantiomers of ibuprofen can have different biological activity despite being otherwise chemically identical. [2]

(b) Suggest one reason why separating a racemic mixture of ibuprofen’s two enantiomers is more difficult than separating two ordinary (non-mirror-image) structural isomers. [2]

(c) Describe one strategy, other than separating a racemic mixture after synthesis, that a manufacturer could use to obtain mainly the active enantiomer. [2]

8. But-1-ene reacts with HBr to form 2-bromobutane, CH₃CHBrCH₂CH₃, which contains a chiral centre.

(a) Explain why this reaction produces a racemic mixture of 2-bromobutane rather than a single pure enantiomer. [3]

(b) State what would be observed if plane polarised light were passed through this racemic product. [1]

9. State the terms used to describe a single pure enantiomer that rotates plane polarised light clockwise, and one that rotates it anticlockwise. [2]

10. A molecule contains two separate chiral centres. State whether this molecule can still be described using the terms “chiral centre” and “enantiomer” from this topic, and state one term relating to stereoisomerism that is not required at this level for such a molecule. [2]

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Answers

1. The carbon is bonded to four different groups [1].

2. A pair of non-superimposable [1] mirror images of each other [1].

3. Identical: any two of melting point, boiling point, solubility, reactivity with non-chiral reagents [2]. Differs: optical activity (or biological activity) [1].

4. An equal (50:50) mixture of both enantiomers [1], which shows no net optical activity because the equal and opposite rotations cancel [1].

5. The central carbon is bonded to -OH, -H, -CH₃ and -CH₃ — two of the four groups are identical, so it is not bonded to four different groups and is not a chiral centre [2].

6. (a) -OH, -H, -CH₃, -COOH [1].

(b) Pass plane polarised light through a solution of each pure enantiomer separately [1]; the two enantiomers rotate the plane of polarisation by an equal angle in opposite directions (one clockwise, one anticlockwise) [1].

(c) The reaction creates a new chiral centre from a non-chiral starting material, and the attacking species is equally likely to react from either face [1], producing equal amounts of both enantiomers — a racemic mixture [1]. Because the two enantiomers’ rotations are equal and opposite, they cancel exactly, giving no net optical activity even though every molecule present is chiral [1].

7. (a) Biological receptors are themselves chiral and fit only one specific three-dimensional shape [1], so only one enantiomer binds effectively and produces the intended effect, while the other may bind poorly or not at all [1].

(b) Enantiomers share identical physical properties (boiling point, solubility, etc.) [1], and conventional separation techniques such as distillation or recrystallisation rely on differences in exactly those properties, so they cannot separate enantiomers [1].

(c) Using a chiral catalyst during synthesis (asymmetric synthesis) [1], to favour formation of one enantiomer directly, avoiding the need to separate a racemic mixture afterwards [1]. Any valid alternative strategy that avoids post-synthesis separation of a racemic mixture should be credited.

8. (a) The reaction creates a new chiral centre at C2 [1]. The intermediate the bromide ion attacks is planar, so it can be attacked from either face with equal probability [1], since nothing in the non-chiral starting materials or reagents favours one face — this gives equal amounts of both enantiomers [1].

(b) No (net) optical activity is observed — the plane of polarisation is not rotated. [1]

9. An enantiomer that rotates plane polarised light clockwise is dextrorotatory [1]; one that rotates it anticlockwise is laevorotatory [1].

10. Yes — each of the two chiral centres is still identified and described using “chiral centre” and “enantiomer” in the normal way [1]. The term not required at this level is diastereoisomer (the relationship between stereoisomers that are not mirror images of each other), along with classifying compounds as meso [1].

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Where marks are usually lost

• Stating a carbon with an -OH group is automatically chiral — all four attached groups must be checked.
• Describing a racemic mixture as “not chiral” instead of “chiral components whose rotations cancel.”
• Claiming enantiomers differ in melting point or boiling point — they don’t; only optical and biological activity differ.
• Explaining ibuprofen’s different biological activity without mentioning that receptors are themselves chiral.
• Forgetting why HCN/HBr-type addition to a planar intermediate gives a racemic product by default — equal probability of attack from either face.