AQA A-Level Chemistry Paper 3, 2017: Question 13

Optical Isomerism & Formation of Racemic Mixtures

📋 What this question tests

This multiple-choice question assesses your ability to predict reaction products from electrophilic and nucleophilic additions, identify chiral centres (asymmetric carbon atoms), and determine whether an attack on a planar intermediate/group produces an optically inactive racemic mixture (racemate).

Electrophilic addition to unsymmetrical and symmetrical alkenes via planar carbocation intermediates.
Nucleophilic addition of cyanide (:CN⁻) to aldehydes and ketones via a planar carbonyl group (C=O).
Recognition of a chiral centre (a carbon atom bonded to four different groups).

Question 13 • Multiple Choice [1 Mark]

Identifying Reactions That Do Not Form a Racemate

Question: Which pair of compounds does not form a racemic mixture when the compounds react?

✅ Correct Answer: D

D: Propanone + HCN

When propanone reacts with HCN, the product formed is 2-hydroxy-2-methylpropanenitrile:

CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN

The central carbon is bonded to two identical methyl groups (-CH₃), along with -OH and -CN. Because it does not have four different groups attached, it is achiral. An achiral molecule cannot exhibit optical isomerism, so no racemic mixture can be formed.

💡 Key Knowledge: Why A, B, and C Form Racemates

A (But-2-ene + HCl): Forms a planar carbocation intermediate [CH₃CH(+)CH₂CH₃]. The Cl⁻ ion attacks equally from above or below the plane, yielding an equimolar (racemic) mixture of (R)- and (S)-2-chlorobutane.
B (Propanal + HCN): The trigonal planar carbonyl group (C=O) of propanal is attacked with equal probability from above or below by :CN⁻, yielding a racemic mixture of 2-hydroxybutanenitrile (chiral centre has: -H, -OH, -CN, -CH₂CH₃).
C (But-1-ene + HCl): The major product follows Markovnikov's rule via the secondary carbocation to form 2-chlorobutane (chiral), which is produced as a racemic mixture.

🧠 Exam Technique: Quick Chiral Check

Notice the negative: The question asks which does NOT form a racemic mixture. Highlight words like not immediately.
Inspect the starting carbonyl/alkene for symmetry:
A symmetrical ketone like propanone (CH₃COCH₃) already has two identical groups attached to the carbonyl carbon.
Shortcut: Adding any single group to a symmetrical ketone with identical alkyl groups will leave those two alkyl groups attached to the same carbon, guaranteeing that the product is achiral. You can answer this in under 15 seconds!

❌ Common Errors & Pitfalls

Assuming all carbonyl additions give racemates: Students often remember that nucleophilic addition of HCN to carbonyls produces a racemic mixture, forgetting that this only applies when the starting carbonyl is an aldehyde (except methanal) or an unsymmetrical ketone.
Overlooking the minor product in Option C: But-1-ene gives 1-chlorobutane as a minor product, but the reaction forms 2-chlorobutane as the major product, which does form a racemate. Therefore, C cannot be the answer.
Missing the requirement for a chiral centre: Attack on a planar intermediate from both sides only gives a racemate if the resulting carbon is attached to four different groups.

Mark Scheme Breakdown:
• D = 1 Mark (any other response scores 0).

AQA A-Level Chemistry Paper 3, 2017: Question 13

Question

Mark scheme

How to answer it